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# MP Board Class 11th Chemistry Important Questions Chapter 8 Redox Reactions

## MP Board Class 11th Chemistry Important Questions Chapter 8 Redox Reactions

### Redox Reactions Important Questions

Redox Reactions Objective Type Questions

Question 1.

Question 1.
Metals used in Daniel cell are:
(a) N and Cu
(b) Zn and Ag
(c) Ag and Cu
(d) Zn and Cu
(d) Zn and Cu

Question 2.
Which of the following is strongly reducing:
(a) r
(b) cr
(c) Bf
(d) r
(d) r

Question 3.
Oxidation in electrolyte occurs:
(a) At anode
(b) At cathode
(c) At both the electrodes
(d) None of these
(a) At anode

Question 4.
Which of the following statement is correct? Galvanic cell converts:
(a) Chemical energy to electrical energy
(b) Electrical energy to chemical energy
(c) Electrical state of metal to combined state
(d) Electrolytes to ions
(b) Electrical energy to chemical energy

Question 5.
Oxidation number of chlorine in HOCl is:
(a) – 1
(b) 0
(c) +I
(d) +2
(c) +I

Question 6.
Oxidation number of Mn in K2MnO4 is:
(a) +2
(b) +6
(c) +7
(d) 0
(b) +6

Question 7.
Oxidation number of Cr in K2Cr2O7 is:
(a) -6
(b) +6
(c) +2
(d) -2.
(b) +6

Question 8.
Oxidation number of Ni in Ni(CO)4 is:
(a) 0
(b) +2
(c) +1
(d) – 1.
(a) 0

Question 9.
Unit of cell constant is:
(a) ohm-1cm-1
(b) ohm cm
(c) cm
(d) cm-1
(d) cm-1

Question 10.
Oxidation state of nitrogen is maximum in:
(a) N3H
(b) NH2OH
(c) N2H4
(d) NH3
(a) N3H

Question 11.
Oxidation state of oxygen is zero in:
(a) CO
(b) O3
(c) SO2
(d) H2O2
(b) O3

Question 12.
Oxidation state of Fe in K4[Fe(CN)6] is:
(a) +2
(b) +6
(c) +3
(d) +4
(a) +2

Question 13.
Oxidation state of S In H2S2O8 is:
(a) +2
(b) +4
(c) +6
(d) +7
(c) +6

Question 14.
Oxidation state of Mn in KMnO4 is:
(a) +4
(b) +6
(c) +7
(d) +5
(c) +7

Question 15.
oxidation state of chlorine is in:
(a) HCl
(b) HClO4
(c) ICI
(d) Cl2O
(d) Cl2O

Question 16.
Oxidation state of Cl in ClO3 ion is:
(a) +4
(b) +5
(c) +3
(d) +2
(b) +5

Question 17.
Halogen which is reduced most easily is:
(a) I2
(b) Br2
(c) Cl2
(d) F2
(a) I2

Question 18.
Oxidation state of C in CCl4 is:
(a) +4
(b) -4
(c) +6
(d) – 6
(a) +4

Question 19.
Oxidation state of S in S04-2 is:
(a) +6
(b) -6
(c) +5
(d) – 5
(a) +6

Question 20.
Oxidizing agent are:
(a) Electron acceptor
(b) Electron donor
(c) Proton acceptor
(d) Neutron acceptor
(a) Electron acceptor

Question 2.
Fill in the blanks:

1. Process of loss of electrons is called ……………………..
2. Process of gain of electrons is called …………………….
3. The deterioration of metals in presence of atmospheric gases and moisture is called …………………………
4. A device in which electric energy gets converted to chemical energy is called …………………….. cell
5. In electrochemical series the ability to reduce ………………………… on moving down
6. Most strong reducing element is ………………………….
7. Most strong oxidizing element is ………………………..
8. Theory of dissociation of electrolyte was proposed by ……………………….

1. Oxidation
2. Reduction
3. Corrosion
4. Electrochemical cell
5. Decreases
6. Lithium
7. Fluorine
8. Arrhenius

Question 3.

1. What is balanced redox reaction?
2. Why does iron displaced copper from its salt solution?
3. What is the oxidation state of chlorine in Cl2O?
4. Why does an electrolyte dissociate into ions when dissolved in water?
5. A saturated solution of KNO3 is used for the formation of salt bridge. Why?
6. In SnCl2 + 2 FeCl3 → SnCl4 + 2 FeCl2, reaction which element is oxidizing agent?
7. What is oxidation state of Xe in XeO3?

1. Redox reaction in which amount of oxidation and reduction is to the same extent.
2. Because standard eletrode potential value of iron is less then the standard electrode potential of copper.
3. +1
4. Because the electrostate attractive force between the ions break due to water.
5. Because the speed of K+ and NO3 and is almost same.
6. FeCl3
7. +6

Question 4.
Match the following:

1. (c)
2. (d)
3. (a)
4. (b)
5. (e)

Redox Reactions Very Short Answer Type Questions

Question 1.
Why iron displaces copper from its salt solution?
Because the standard electrode potential of iron is less than copper.

Question 2.
What is the oxidation number of Cl in Cl2O?
+1.

Question 3.
When an electrolyte is dissolved in water, it dissociates into ions? Why?
Because due to water the electrostatic attraction force between the ions breaks.

Question 4.
Why the saturated solution of KNO3 is used to prepare salt bridge?
Because the speed of K+ and NO3 ions is similar.

Question 5.
In SnCl2 + 2FeCl3 → SnCl4 + 2FeCl2, which is oxidizing agent?
FeCl3.

Question 6.
What is the oxidation number of Xe in XeO3?
+6.

Question 7.
Which metal is used in Daniel cell?
Lithium.

Question 8.
What is the oxidation state of Mn in K2MnO4?
+6.

Question 9.
Which is the most reducing element?
Lithium.

Question 10.
Who maintains the neutrality among the two half cells?
Salt bridge.

Question 11.
What is the standard electrode potential of SHE?
0.00V (Zero).

Question 12.
Which is the best conducting metal?
Silver (Ag).

Question 13.
The reaction 3ClOaq → Cl3 + 2Cl is an example of?
Disproportionation reaction.

Question 14.
Which halogen reduces very easily?
F2.

Question 15.
Which reaction is called electron acceptance reaction?
Reduction.

Question 16.
Where oxidation occurs in electrolysis?
At anode.

Redox Reactions Short Answer Type Questions – I

Counting atoms calculator in my website step-by-step explanation.

Question 1.
What is oxidation process? Explain redox reaction with example?
Oxidation state of an element is defined as “The residual charge left on its atom”. When all the other atoms are removed from the molecule as ion.

2Mg + O2 → 2MgO [Addition of oxygen]
2FeCl2 → 2FeCl3 [Addition of electronegativity element]
H2S + Cl2 → S + 2HCl [Removal of hydrogen]
2Ki + H2O2 → 2KOH + I2 [Removal of electropositive element]

Question 2.
What is reduction process? Explain with example?
The ability of any substance to accept hydrogen atom or +ve electrolytic element or to release oxygen or – ve element is called reduction process.
CuO + H2 → Cu + H2O [Remove of oxygen]
2FeCl3 + H2 S → 2FeCl2 + 2HCl + S [Removal of electromagnetive element]
Cl2 + H2S → 2 HCl + S [Addition of H]
S + Fe → Fes [Addition of electropositive element]

Question 3.
AgF2 is unstable compound, but if it is formed then it will acts as strong oxidising agent Why?
In AgF2, Ag has oxidation state +2, it is highly unstable, it very easily accept electron and forms oxidation state +1. Because Ag+ has stable configuration so it is stable.
Ag2+ + e → Ag+
This is the reason why AgF2 behaves as a strong oxidizing agent.

Question 4.
By the structure of S4O62-clarify that the oxidation state of S is +5?
The oxidation state of two central sulphur atom is 0. As the electron pair remains in the centre cf the bond. So the remaining sulphur atom has oxidation state 4 – 5.

⇒ 6(-2) + 2x = -2
⇒ -12 + 2x = -2
⇒ 2x = -2 + 12 = 10
⇒ x = $$\frac{10}{2}$$ = 5.

Question 5.
What is electrochemical series?
Electrochemical series:
When different elements and ions are arranged in increasing order of their standard electrode potential, a series is obtained which is called electrochemical series as activity series.

Question 6.
What is salt bridge? Write its two uses?
1. It connects the solution in two half ceil and completes the cell circuit

2. The salt bridge which contains the solution of an electrolyte say K+NO3 maintains the electrical neutrality by diffusion of ion through it. As positive ion begins to be formed in oxidation half cell negative ion from the salt bridge diffuse out into this half cell. Similarly, as the positive Ion begin to be consumed in the reduction.

Hair cell positive:
On from the salt bridge diffuse out into this half cell. As a result the solution in the two half cell remain electrically neutral and the current continous to flow.

Question 7.
On the basis of electron transfer explain oxidation and reduction reactions?
Reactions involving exchange of electrons are called redox reaction. Reaction given below
2FeCl3 + SnCl2 → 2FeCl2 + SnCl4
It is an example of redox reaction in which FeCl3 is oxidizing agent while SnCl2 is a reducing agent.

According to electronic concept during oxidation substance loses one or more electron and its positive oxidation number increases while negative oxidation number decreases. (MPBoardSolutions.com) On the other hand, during reduction substance gains one or more electron and is positive oxidation number decreases while negative oxidation number increases.

Question 8.
What is electrochemical equivalent?
The mass of any substance deposited at any electrode is directly proportional to the quantity of electricity passed.

Thus, if W gm substance is deposited on passing Q coulomb electricity, then W ∝ Q or W = ZQ
Where, Z is a constant of proportionality and is called electrochemical equivalent of the substance deposited. If a current of I ampere is passed for t second, then Q = I × t. (MPBoardSolutions.com) So that, W = Z × Q = Z × I × r

Thus, if, Q = 1 coulomb, 1=1 ampere and t = 1 second, then W = Z. Hence, electrochemical equivalent of a substance may be defined as, “The mass of the substance deposited when a current of one ampere is passed for one second”.

As one Faraday (96500 C) deposits one gram equivalent of the substance, hence, electrochemical equivalent can be calculated from the equivalent mass.
$$\frac { Equivalentmassofthesubstance\quad }{ 96500 }$$

Question 9.
What happens when Zn is kept in CuS04 solution? Explain with equation?
The reaction occurring in the beaker may be written as

Cancelling the common S042+ ions
Zn(s) + Cu2+(eq) → Zn2+(eq) + Cu(s)
In this case Zinc metal loses electron and gets oxidised to Zn++ ions which goes into the solution. Due to this, weight of zinc plate gradually decreases. On the other hand, electrons lost by Zinc atom are taken up by Cu2+ ions of CuS04 solution.

Question 10.
What do you mean by standard electrode potential?
Standard electrode potential: Standard electrode potential (E°) of a half cell (electrode) is the potential difference which is produced when one electrode is dipped in the solution of molar concentration of its ion at 298 K. (MPBoardSolutions.com) If electrode is gaseous, the pressure of gas must be one atmosphere.
In IUPAC system reduction potentials are known as standard electrode potential.

Question 11.
The reaction Cl2(g) + 2OH(aq) → ClO(aq) + Cl(aq) + H2O(l) represents the process of bleaching. Identify and name species that bleaches the substances due to its oxidizing action?

In this reaction, O. N. of Cl increases from 0 in Cl2 to +1 in ClO and decreases to -1 in Cl. Therefore, Cl2 is both oxidized to ClO and reduced to Cl (Disproportionation reaction). Since, Cl ion cannot act as oxidizing agent (Because it cannot decrease its O.N. lower than -1) therefore, Cl2 bleaches substance due to oxidizing action of ClO (Hypochlorite) ion.

Question 12.
Define oxidizing agent and reducing agent?
Oxidizing agent:
A substance which brings oxidation is called oxidising agent. On the basis of electronic theory, an oxidizing agent may be defined as an electron acceptor. i.e. (MPBoardSolutions.com) in the reaction between Zn atom and Cu2+ ion in aqueous solution of Cu2+ salt, Cu2+ ion is the oxidizing agent as it brings about the oxidation of Zn atom by gaining the electron lost by it.
Zn + Cu2+ → Zn2 + Cu
(Gains e)
(∴ Oxidising agent)
Thus, oxidizing agent or oxidant undergoes gain of electrons during the reaction.
Reducing agent:
A substance which brings about reduction is called a reducing agent. On the basis of electronic theory, a reducing agent may be defined as an electron donor, e.g., in reaction between Zn atom (MPBoardSolutions.com) Cu2+ ion in an aqueous salt solution of Cu2+, Zn atom is the reducing agent as it brings about reduction of Cu2+ ion by losing the electrons.
Zn + Cu2+ → Zn2+ + Cu
(Loses e)
(∴Reducing agent)

Question 13.
If iron rod is dipped in CuS04 solution then copper is displaced from its solution by iron but when copper rod is dipped in FeS04 solution. Why iron is not displaced by copper? Explain?
Fe occupies higher position in electrochemical series than Cu. So, Fe is more active than Cu. That is why Fe displaces Cu from CuS04 solution, but Cu is not able to displace Fe from FeS04 solution.

Question 14.
What is oxidation number?
Oxidation number of an element is defined as, “The residual charge left on its atom when all the other atoms are removed from the molecule as ions.” Atoms can have positive, negative or zero oxidation state depending upon their state of combination. (MPBoardSolutions.com) In fact, oxidation number is the charge assigned to the atom in a species according to some rules. It is also known as oxidation state.

Question 15.
In electrochemical series, the activity of metals increases or decreases in which sequence?
Species present in the right of electrochemical series are reducing agent. Smaller the value of E°, stronger is the reducing agent. The activity of metals decrease on going down the electrochemical series.

Question 16.
What is E° value be + ve shows in Galvanic cell?
The positive value of E° shows:

1. Oxidation occurs at anode and
2. The electrode where reduction occurs can be taken as cathode.

Question 17.
What is electrode potential? Its value depends on what factors?
The electrical potential difference setup between the metal and its solution is known as electrode potential.
The electrode potential depends upon:

1. Concentration of ions in the solution
2. Temperature.

Question 18.
On the basis of standard electrode potential given below, arrange the metals in their increasing reducing power?
K+/ K = – 2.93V, Ag+/ Ag = 0.80V, Hg2+/ Hg = 0.79V, Mg2+/ Mg = -2.37V, Cr3+/ Cr = -0.74V?
Less the value of E°, the reducing power increases of the metal. So the sequence is Ag < Hg < Cr < Mg < K.

Question 19.
Among MgO, ZnO, CuO and CaO which oxide will reduce by hydrogen?
Mg, Zn and Ca are more reactive than H in electrochemical series. So they will not reduced by H. Cu is present below H, so the reactivity of Cu is less than H. So CuO will easily get reduced by H.

Question 20.
The electrode potential (E°) of Ag, Ba, Mg and Au are +0.80, -2.90, -2.37 and +1.42 volt. Among these metals which will displace H from acids and which will not?
Those metals which have E° value negative they are more reactive than hydrogen and can easily displace H from weak acids. So, Ba and Mg can displace H from acid solution.

Question 21.
Can be put CuSO4 solution in silver container? Why?
In electrochemical series Ag comes below Cu, so reactivity of Ag is less than Cu. There Ag will not displace Cu from CuSO4 solution. CuSO4 can be kept in Ag container.

Redox Reactions Short Answer Type Questions – II

Question 1.
Fluorine reacts with ice as follows:
H2O(s) → HF(g) + HOF(g)
Justify that, this reaction is a redox reaction.

Since, fluorine can undergo oxidation as well as reduction, it is an example of redox reaction.

Question 2.
In acidic medium MnO42- shows disproportionation reaction but MnO4 does not show why?
In oxidation state of Mn in MnO4-2 is +6. It can increase oxidation state (+7) and decrease (+4, +3, +2, 0 etc.). So in acidic medium it shows disproportionation reaction.

In MnO4, Mn has maximum oxidation state (+7). It can only decrease its oxidation state. Due to which it cannot show disproportionation reaction.

Question 3.
In the following reaction whose oxidation and whose reduction occur:
PbS + 4H2O2 → PbSO4 + 4H2O
In this reaction PbS is oxidized as PbSO4 and H2O2 is reduced to form H2O.

Question 4.
Differentiate between oxidation number and valency?
Differences between Oxidation number and Valency:
Oxidation:

1. It is the residual charge left on the atom of an element when all other atoms are removed from molecule or ion.
2. It refers to charge and can be positive, negative or zero, e.g., oxidation number of C in CO2 is +4 and that of oxygen is -2.
3. It may have fractional value.
4. Elements like C, N, O have variable oxidation number.

Valency:

1. It is the combining capacity of the element expressed as number of H – atom or double the number of oxygen atoms which combine with one atom of the element.
2. It refers to number only and does not carry any sign, e.g., in CO2 valency of carbon is 4 and that of oxygen is 2.
3. It is always a whole number.
4. Elements like C, N, O exhibit constant valency.

Question 5.
Why fluorine doesn’t show disproportionation reaction?
Like chlorine, bromine and iodine also undergo similar disproportionation reactions but fluorine does not. The reason for this anomalous behaviour is that fluorine being the strongest oxidizing agent does not show positive oxidation states. It, therefore, reacts in a different way forming oxygen difluoride (OF2).
2F2(g) + 2OH(g) → 2F(aq) + OF2(g) + H2O(l)

Question 6.
Show that galvanic cell where following reaction occurs:
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2 Ag(s)
Now explain :

1. Which electrode is -ve?
2. What is the motive of electricity in cell?
3. What is the reaction occurring on each electrode?

Galvanic cell Zn(s) ∥ Zn2+(aq)Ag(aq) ∥ Ag+(aq) ∥ Ag(s)

1. Zn electrode is negatively charged. Zn is oxidized into Zn+2.
2. Electricity flows from Ag electrode to Zn electrode and electrons flow from Zn to Ag electrode.
3. Reactions on electrode:

Cathode — Zn(s) → Zn+2(s) + 2e
Anode — 2Ag+(aq) + 2e → 2Ag(s)

Question 7.
Write the definition of electrochemical cell? Write the chemical reactions of Daniel cell?
An electrochemical cell (Galvanic cell) is a device in which the redox reaction takes place indirectly and the decrease in free energy appears in the form of electrical work i.e. chemical energy is converted into electrical energy.

Question 8.
Write the factors affecting electrode potential?
The factors affecting electrode potential are:

1. Tendency of metal to donate electrons: More the tendency to release electron more will be electrode potential.
2. Temperature: With increasing temperature electrode potential increases.
3. Concentration of solution: More the concentration of solution more will be electrode potential.

Question 9.
Determine the oxidation number of following:

1. Cr in K2Cr2O7
2. Mn in KMnO4
3. S in Na2S4O6.

Solution:
1. Cr in K2Cr2O7
2 (+1) + 2x + 7 (-2) = 0
or 2 + 2x – 14 = 0
or 2x = 12
or x = +6.

2. Mn in KMnO4
1 (+1) + 1x + 4(-2) = 0
or 1 + x – 8 = 0
or x = +7.

3. S in Na2S4O6
2 (+1) + 4x + 6 (-2) = 0
or 2 + 4x – 12 = 0
or 4x = 10
or x = 2.5.

Question 10.
Why does the following reactions occur:
XeO6(aq)4- + 2F(aq) + 6H+(aq) → XeO3(g) + F2(g) + 3H2O(l)?
What conclusion about the compound Na4XeO6 (of which XeO64- is a part) K2MnF6 can be drawn from the reaction?

O.N. of Xe decreases from +8 to +6. This shows that XeO64- is an oxidizing agent. It oxidises F to F2. This reaction shows that Na2XeO6 (or XeO6) is a stronger oxidizing agent than F2.

Question 11.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water?

Multiplying Cl by 2 (As 2 atoms are in Cl2)
Cl2(aq) + SO2(aq) + H2O(l) → 2Cl(aq) + SO42-(aq)
On adding 4H+ in left side and multiplying in H2O by 2,
Cl2(aq) + SO2(aq) + 2H2O(l) → 2Cl(aq) + SO42- + 4H+
It is a balanced disproportionation reaction.

Question 12.
Differentiate between electrochemical cell and electrolytic cell?
Differences between electrochemical cell and electrolytic cell:
Electrochemical (Galvanic) cell:

1. In Galvanic cell, chemical energy is changed into electrical energy.
2. Reaction in Galvanic cell is spontaneous.
3. Oxidation and reduction occurs in different half cell.
4. Anode is negative while cathode is positive terminal.
5. The electrons move from anode to cathode in the external circuit.

Electrolytic cell:

1. In electrolytic cell, electrical energy is changed into chemical energy.
2. Reaction in electrolytic cell is non – spontaneous.
3. Oxidation and reduction occurs in same container.
4. Anode is positive while cathode is negative terminal.
5. The electrons are supplied by the external circuit which enter through the cathode and came out through anode.

Question 13.
What is Nernst equation? Derive relation between E and E°?
In Standard cell, the concentration of the solution used is 1M and the temperature is maintained at 25°C (For 298K). But the Galvanic cells are often constructed under non standard condition. In such cases the cell voltage (or EMF) is calculated by Nernst equation as follows:
E = E° – $$\frac{RT}{nF}$$ loge $$\frac { M_{ (s) } }{ M^{ n+ }_{ (aq) } }$$
or E = E° + $$\frac{RT}{nF}$$ loge $$\frac { M^{ n+ }_{ (aq) } }{ M_{ (s) } }$$
In changing loge in log10
or E = E° + $$\frac{2.303}{nF}$$ log10 $$\frac { M^{ n+ }_{ (aq) } }{ M_{ (s) } }$$ [M(s) = 1]
or E = E° + $$\frac{2.303RT}{nF}$$ log10[Mn+(aq)]
or E = E° + $$\frac{0.0591}{nF}$$ log10 [Mn+(aq)].

Question 14.
Consider the reactions:
2S2O32-(aq) + I2(s) → S4O62-(aq) + 2I(aq)
S2O32-(aq) → 2SO42-(aq) + 4Br(aq) + 10H+(aq)
Why does the same reductant, thiosulphate reacts differently with iodine and bromine?
Br2 is stronger oxidizing agent than I2, it oxidises SO32- to SO42- i. e. from +2 state to +6 state of sulphur. However, I2 being weaker oxidizing agent, oxidises S2O32- to S4O62- ion i.e. from + 2 to + 2.5 state of sulphur.

Redox Reactions Long Answer Type Questions – I

Question 1.
What is standard hydrogen electrode? How it is prepared?
Standard hydrogen electrode: This consists of gas at 1 atmospheric pressure bubbling over a platinum electrode immersed in 1 M HCl at 25°C (298 K) as shown in figure. The platinum electrode is coated with platinum black to atmospheric increase its surface. (MPBoardSolutions.com) The hydrogen electrode thus Pressure constructed forms a half cell which on coupling with any other half cell begins to work on the principle of oxidation or reduction. Electrode depending upon the 1 M H+solution circumstances works both as anode or cathode.

Cell reaction of standard hydrogen electrode platinum electrode (SHE) when it acts as anode is
H2(g) → 2H+ + 2e
It is represented as
H2(g)(1 atm) Pt | H3O+(aq)(1.0 M)
When it acts as cathode, the cell reaction is
2H+ + 2e → H2(aq)
and it is represented as
H3O+(aq) (1.0 M) | H2(g) (1 atm) Pt
Standard hydrogen electrode (SHE) is arbitarily assigned a potential of zero.

Question 2.
On the basis of electrode potential explain which reaction is feasible:
Cu2+/Cu = 0.34 V, E°Zn2+/Zn = -0.76 V, E°Mg2+/Mg = 2.37 V, E°Fe2+/Fe = -0.74V

1. Cu + Zn2+ → Cu2+ + Zn
2. Mg + Fe2+ → Mg2+ + Fe.

Solution:
1. Cu + Zn2+ → Cu2+ + Zn
Cu2+/Cu = 0.34 V and E°Zn2+/Zn = – 0.76 V
In given cell, Cu oxidised to Cu2+. So Cu2+/Cu will act as anode and Zn2+/Zn acts as cathode.
cell = E°cathode – E°anode
cell = – 0.74 – (-2.37) = +1.63V
cell = +Ve, shows that the reaction is feasible.

2. Mg + Fe2+ → Mg2+ + Fe
Mg2+/Mg = 2.37 V and E°Cu2+/Cu = – 0.74 V
Mg oxidised to Mg2+ and acts as anode Mg2+/Mg.
Fe2+ reduced to Fe and acts as cathode Fe2+/Fe.
cell = E°cathode – E°anode
cell = – 0.74 – (-2.37) = +1.63 V

Question 3.
While sulphur dioxide and hydrogen peroxide can act as oxidizing as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants, why?
In sulphur dioxide (SO2) and hydrogen peroxide (H2O2), the oxidation states of sulphur and oxygen are +4 and -1 respectively. Since, they can increase as well as decrease when there compounds take part in chemical reaction, they can act as oxidizing as well as reducing agents. For example,
Increase in O.N.

In ozone (O3), the oxidation state of oxygen is zero while in nitric acid (HNO3), the oxidation state of nitrogen is +5. Since, both of them can undergo decrease in oxidation state and not an increase in its value, they can act only as oxidizing agents and not as reducing agents.

In ozone (O3), the oxidation state of oxygen is zero while in nitric acid (HNO3), the oxidation state of nitrogen is + 5. Since both them can undergo decrease in oxidation state and not an increase in its value, they can act only as oxidizing agents and not as reducing agents.

Question 4.
How do you count for the following observation though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant, why? Write a balanced redox equation for the reaction?
Toluene can be oxidized to benzoic acid in acidic, alkaline and neutral medium by using potassium permanganate.

1. In acidic medium:
[MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(l)] × 6

In the manufacture of benzoic acid alcoholic KMnO4 is more useful.
Alcohol if used as solvent will help in the formation of a homogeneous mixture between toluene (non – polar) and KMnO4 (ionic). Actually alcohol has non – polar alkyl group as well as polar OH group.

Question 5.
How the redox reactions are balanced by ion – electron method?
The redox reactions are balanced by ion – electron method are as follows:
1. Indicate oxidation number of each element and identify the elements which undergoes change in oxidation number.

2. Indicate increase and decrease in oxidation number per atom. Multiply the increase or decrease in oxidation number with number of atoms undergoing the change.

3. Multiply the formula with suitable integer to equalise the increase and decrease in oxidation number.

4. Balance O atoms:
K2Cr2O7 + 6HCl + xHCl → 2KCl + 2CrCl3 + yCl2 + H2O + 6H2O

5. Balance H atoms:
K2Cr2O7 + 6HCl + 8 HCl → 2KCl + 2CrCl3 + 3Cl2 + 7H2O
Here x = 8 and y = 3 to balance Cl atoms
K2Cr22-O7 + 14 HCl → 2KCl + 2rCl3 + 3Cl2 + 7H2O.
Example 2. Cr2O72- + Fe2+ + H+ → Cr3+ + Fe3+ + H2O.
Solution:
Step 1. (Cr2O7)2- + Fe2+ + H+ → Cr3+ + Fe3+ + H2O.

Step 2.

Step 3.
Multiply Fe by 6 to equalise increase or decrease in O.N.
Cr2O72- + 6Fe2+ + H+ → Cr3+ + Fe3+ + H2O

Step 4.
Balance atoms other than O and H
Cr2O72- + 6Fe2+ + H+ → 2Cr3+ + 6Fe3+ + H2O

Step 5.
Balance O atoms
Cr2O72- + 6Fe2+ + H+ → 2Cr3+ + 6Fe3+ + H2O + 6H2O

Step 6.
Balance H atoms
Cr2O72- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O.

Question 6.
On the basis of electrode potential values given below? Determine reaction between reactants will occur or not?

For feasibility of any reaction E° = +ve.
cell = E°cathode – E°anode
(a) Fe3+(aq) and I(aq):
2Fe3+ + 2I(aq) → 2Fe2+(aq) + 2I2g
Half cell reaction (oxidation)
2I(aq) → I2(g) + 2e E° = – 0.54 V
Reducion half cell,
2Fe3+(aq) + 2e → 2Fe2+(aq) E° = +0.77 V
Complete reaction,
2Fe3+(aq) + 2I(aq) → 2Fe2+(aq) + I2(g)cell = +0.23 V
E° = +ve, reaction is feasible.

(b) Ag+(aq) and Cu(s)2+:
2Ag(aq) + Cu(s) → Cu2+(aq) + 2Ag(aq)
Oxidation half reaction,
Cu(s) → Cu2+(aq) + 2e E° = – 0.34 V
Reduction half reaction,
2Ag+(aq) + 2e → 2Ag(s) E° = – 0.80 V
Complete reaction,
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)cell = +0.46 V

Question 7.
Give the main characteristics of electrochemical series?
The main characteristics of electrochemical series are:

1. Tendency to loose electron becomes greater and the corresponding element or ion behaves as strong reducing agent if it has greater negative value of E°.
2. The substances above hydrogen are weak oxidising agents and substances below it are strong oxidising agents.
3. If the reducing agent of electrode pair is strong reducing agent, its oxidising agent is weak and when reducing agent of electrode pair is weak, its oxidising agent is strong.
4. In series the reducing efficiency decreases from above to downwards.
5. The more the positive value of E°(MPBoardSolutions.com) the stronger is its oxidising properties. In the series the oxidising power of cation decreases as we move upwards from downwards.
6. In series the electropositive character of metals decreases from upwards to downwards.
7. Along the series the electronegative character of non – metal increases from upward to downwards.
8. The metals which come prior to hydrogen displaces hydrogen from acids,
9. The metals which occupy higher position in the series displaces metals which come later in series from their salt solution.

Question 8.
Balance the equation by oxidation number method and identify the oxidising agent and reducing agent?
Cl2O7(g) → ClO2(aq) + O2(g).
Oxidation number method:

Balance the increase and decrease in O.N. multiply H2O2 by 4,
Cl2O7 + 4H2O2 → ClO2 + O2
Balance atom other,
Cl2O7(g) + 4H2O2(aq) → 2ClO2(aq) + 4O2(g)
Balance ‘H’,
Cl2O7(g) + 4H2O2(aq) + 2OH(aq) → 2ClO2(aq) + 4O2(g) + 5H2O.

Question 9.
Whenever a reaction between an oxidizing agent and a reducing agent is carried out compound of lower oxidation state is formed if the reducing agent is ¡n excess and compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement by giving three illustrations?
1. Let us consider the reaction between carbon and oxygen.
Reducing agent + Oxidising agent

2. Let us consider the reaction between white phosphorous (P4) and Cl2(g)

3. Let us consider the reaction between sulphur and oxygen.
Reducing agent + Oxidising agent

Redox Reactions Long Answer Type Questions – II

Question 1.
Balance the following reaction by ion – electron method:
Cr2O72- + Fe2+ + H+ → Cr+3 + Fe+3.
Solution:
Cr2O72- + Fe2+ → Cr+3 + Fe+3 + H2O

1. Write the oxidation number of each atom in skeleton equation.

Species undergo change is Cr and Fe.

2. Dividing the equation into two half reaction.
Cr2O72- → Cr+3 (Reduction half)
Fe+2 → Fe+3 (Oxidation half)

3. Balancing reduction half:
(a) Balancing Cr atom,
Cr2O72- → 2Cr+3

(b) Adding e to make up the difference in O.N.

(Since each Cr atom gain 3e ∴ for 2 Cr atom 6e)

(iv) Balancing oxidation half:
(a) Balancing Fe atom (Balanced)
(b) Adding e to make up the difference in O.N.
Fe+2 → Fe+3 + e

(v) Balancing ‘O’ atom,
Cr2O72- + 6e → 2Cr+3 + 7H2O
Fe+2 → Fe+3 + e

(vi) Balancing H by adding H+ ions to the side which is deficient on H atoms.
Cr2O72- + 6e + 14H+ → 2Cr+3 + 7H2O
Fe2+ → Fe+3 + e

(vii) Equalising e,

Question 2.
To determine the oxidising power/reducing power in a solution which method is used? Explain with example?
Determination of standard electrode potential ; To determine electrode potential of an electrode, a cell is setup using this electrode and standard hydrogen electrode. The EMF of the cell is measured with the help of a voltameter (or more accurately by potentiometer). (MPBoardSolutions.com) The electrode potential of hydrogen electrode is taken as zero. Therefore, the E.M.F. of such a cell will directly give the value for electrode potential for the given electrode at standard condition (298 K, 1M, 1 atm pressure).

1. If the deflection in galvanometer is towards metal electrode (experimental electrode) then it is anode (-ve terminal) of the cell. It is allotted negative value of standard reduction electrode potential.
For example, in the determination of E° value of zinc, zinc acts as anode and has got negative value of E°.

cell = E°cathode – E°anode
∴ 0.76 V = 0.00V – E°zn+2/Zn = -0.76V

Question 3.
What is oxidation number? Write its main rules of determination?
Oxidation number of an element is defined as, “The residual charge left on its atom when all the other atoms are removed from the molecule as ions”. Atoms can have positive, negative or zero oxidation state depending upon their state of combination. (MPBoardSolutions.com) In fact, oxidation number is the charge assigned to the atom in a species according to some rules. It is also known as oxidation state.

Rules for assigning oxidation number:
The assignment of oxidation numbers is arbitrary and is usually governed by the following conclusions:

1. The oxidation number of an element in free state i. e., elementary state is regarded as zero. For example,
Oxidation state of hydrogen in H2 = 0
Oxidation state of helium in He = 0
Oxidation state of sulphur in S8 = 0
Oxidation state of sodium in Na = 0
Oxidation state of magnesium in Mg = 0

2. In a compound, the more electronegative elements are assigned negative (-) oxidation state and less electronegative elements are assigned positive (+) values. For example, in HCl as chlorine is more electronegative than hydrogen its oxidation state will have negative value while that of hydrogen will be positive.

3. In the formula of a compound, the sum of the negative oxidation states is equal to the sum of the positive oxidation states.

4. The sum of oxidation number of all the atoms in a neutral molecule is taken as zero.

5. Hydrogen has an oxidation state +1 in compounds like H2S, H2O, HCl, etc. Exceptionally it has the oxidation number -1 in metallic hydrides, such as NaH, CaH2, etc.

6. Oxygen is usually assigned oxidation number -2, except in H2O2 and in oxide of fluorine [F2O], in which the oxidation number -1 and +2 respectively. In all, oxidation number of oxygen is -1.

7. Fluorine being the most electronegative element is assigned oxidation number -1 in all its compounds. Other halogens also show -1 oxidation number.

8. The number of monoatomic ion in an ionic compound is equal to its electric charge. Thus, the elements of group IA of the periodic table (Li, Na, K, Rb, Cs) all have oxidation number + 1, while the alkaline earth metals of group IIA (Ca, Sr, Ba) have oxidation number + 2.

Question 4.
Balance the following reaction by ion – electron method?
Ion – electron method:
Oxidation half equation is:

The oxidation number is balanced by adding 2 electrons as:
H2O2(aq) → O2(g) + 2e
The charge is balanced by adding 2OH ions as:
H2O2(aq) + 2OH(aq) → O2(g) + 2e
The oxygen atoms are balanced by adding 2H2O as:
H2O2(aq) + 20H(aq) → O2(g) + 2H2O(s) + 2e ……………… (1)
The reduction half equation is:

The Cl atoms are balanced as:
Cl2O7(g) → 2ClO2(aq)
The oxidation number is balanced by adding 8 electrons as:
Cl2O7(g) + 8e → 2ClO2(aq)
The charge is balanced by adding 6OH as:
Cl2O7(g) + 8e → 2ClO2(aq) + 6OH(aq) ………………………… (2)
The balancing equation can be obtained by multiplying equation (1) with 4 and adding equation (2) to it as:
Cl2O7(g) + 4H2O2(aq) + 2OH(aq) → 2ClO2(aq) + 4O2(g) + 5H2O(l)

Question 5.
Write the main uses of electrochemical series?
Electrochemical series:
When different elements and ions are arranged in increasing order of their standard electrode potentials, a series is obtained which is called electrochemical series or activity series.
The uses of electrochemical series are:
1. Electropositive character of metal:
Metal which loses electron easily from its outermost shell shows high electropositive character while the atom which loses electron with difficulty shows least electropositive character. In electrochemical series, Li shows maximum electropositive character while fluorine shows least electropositive character.

2. Comparison of reactivity of metals:
Metal having negative value of E° loses electron easily. Hence, greater the negative value of E° more is the reactivity and reducing character of a metal.

3. Knowledge of oxidizing agent and reducing agent:
Substance which gains electron and shows maximum value of standard electrode potential E° is strongest oxidizing agent. On the basis of this F is strongest oxidizing agent while Li is weakest oxidizing agent.

4. Displacement of elements:
Metal having greater tendency to form ion displaces others. Therefore, elements getting priority in the series displace ions of the elements following them. For example. When Cu turnings are added to AgNO3 solution, Ag gets precipitated.
2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s)

5. Electroplating:
Process of depositing layer of gold or silver on copper, brass, iron etc. is called electroplating. It makes articles lustrous and attractive.

6. Metallurgy:
More reactive metal displaces less reactive metal from their aqueous salt solution.

7. Corrosion of metals:
Destruction of metals due to action of chemicals, air and moisture is called corrosion.
e.g., iron rust in moist air. To prevent this a layer of more reactive metal like Sn or Zn is deposited on iron. Coating iron with zinc is called galvanization.

Question 6.
What information did you get from the following reaction:
(CN)2(g) + 20H(aq) → CN(aq) + CNO(aq) + H2O(l)
(CN)2(g) + 2OH(aq) → CNO(aq) + H2O(l).
1. Let the O.N. of C in (CN)2 = x
2x + 2 (-3) = 0
or x = +3.

2. Let the O.N. of C in CN = x
x + (-3) = -1
or x = +2.

3. Let the O.N. of C in CNO = x
x + (-3) + (-2) = -1 or
x = +4.

Following informations obtained from above equation:

1. It is a disproportionation reaction.
2. The reaction occurs in basic medium.
3. O.N. of N in (CN)2 is -3 and that in CN is -2 and in CNO it is -5.
4. Cyanogen (CN)2 gets simultaneously reduced to CN ion as well as oxidized to cyanate, CNO ion.

Question 7.
The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H+ ion. Write a balanced ionic equation for the reaction?


text:
Home > Geometry, Problem Solving > Regular polygons and lattice points.

## Regular polygons and lattice points.

Se say that a point in the plane is a lattice point if both its coordinates are integers. Prove that for $n\neq 4$ we can’t find a regular polygon with $n$ egdes, its vertices being lattice points.

Solution:
Looking at Pick’s Formula, we see that two times the area of a polygon with vertices in lattice points is an integer. But if we take an equilateral triangle with side of length $a$ with $a^2 \in \mathbb{Z}$ (distance formula) then its area is $\frac{a^2 \sqrt{3}}{4}$ which is not a rational number. Contradiction. As suggested in the comments, we need to take care of the case $n=6$ separately, but this is a simple consequence of $n=3$.
Suppose now $n \geq 5,\ n\neq 6$ and denote our polygon $P_1...P_n$. Then the angles of such a polygon are greater than $90^\circ$. Therefore, if we construct $Q_i$ such that $\overrightarrow{P_iQ_i}=\overrightarrow{P_{i+1}P_{i+2}}$. Then $Q_1...Q_n$ is a regular polygon with vertices in lattice points, which is strictly inside the initial polygon. If $P_1...P_n$ exists, then we can repeat this construction as many times we want, but this is a contradiction, since there are only a finite number of lattice points inside $P_1..P_n$.

Yes, you are right. The case $n=6$ should be treated separately, but as you point out, it is a consequence of the case $n=3$.

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text:
# Updates on the experiments X17

• I
kodama
TL;DR Summary
. Assuming the ATOMKI evaluation of the electron-pair production rate from X17
your thoughts on this -- Status of the X17 search in Montreal

they'll verify or refute a clear signal after about two weeks of data taking with a 2 μA proton beam

if Montreal Tandem accelerator confirm X17 a clear signal after about two weeks of data taking with a 2 μA proton beam would you accept it ? when it is ready - data 2023

ARIEL experiments and theory and The Positron Annihilation to Dark Matter Experiment also seeking X17 not sure when it is ready

[Submitted on 21 Nov 2022]

### Status of the X17 search in Montreal

G. Azuelos, B. Broerman, D. Bryman, W.C. Chen, H.N. da Luz, L. Doria, A. Gupta, L-A. Hamel, M. Laurin, K. Leach, G. Lefebvre, J-P. Martin, A. Robinson, N. Starinski, R. Sykora, D. Tiwari, U. Wichoski, V. Zacek

At the Montreal Tandem accelerator, an experiment is being set up to measure internal pair creation from the decay of nuclear excited states using a multiwire proportional chamber and scintillator bars surrounding it from the DAPHNE experiment. The acceptance covers a solid angle of nearly 4π. Preamplifiers and the data acquisition hardware have been designed and tested. The water-cooled 7LiF target, mounted on an Al foil is in a thin carbon fiber section of the beamline. The experiment will focus at first on a measurement of the internal pair creation from the 18.15 MeV state of 8Be. Assuming the ATOMKI evaluation of the electron-pair production rate from X17, a Geant4 simulation predicts observation of a clear signal after about two weeks of data taking with a 2 μA proton beam. The IPC measurement could eventually be extended to the giant dipole resonance of 8Be, as well as to other nuclei, in particular to 10B.

 Comments: 5 pages, 4 figures, Proceedings contribution, TRIUMF Ariel Workshop, May 25-27 2022 Subjects: Instrumentation and Detectors (physics.ins-det); Nuclear Experiment (nucl-ex) Cite as: arXiv:2211.11900 [physics.ins-det] (or arXiv:2211.11900v1 [physics.ins-det] for this version) https://doi.org/10.48550/arXiv.2211.11900

[Submitted on 16 Oct 2022]
ARIEL experiments and theory
Petr Navratil

I present an overview of experiments at TRIUMF ARIEL and ISAC facilities covering both the current and the future envisioned programs. I also briefly review theory program at TRIUMF that relates to the ARIEL experimental program. I highlight several recent experimental results from the nuclear astrophysics, nuclear structure, fundamental symmetries, and the sterile neutrino search. Finally, I mention ongoing theoretical ab initio calculations of the proton capture on 7Li related to the X17 boson observation.

Comments: Contribution to proceedings of the workshop New Scientific Opportunities with the TRIUMF ARIEL e-linac, 10 pages, 5 figures
Subjects: Nuclear Theory (nucl-th); Nuclear Experiment (nucl-ex)
Cite as: arXiv:2210.08438 [nucl-th]

[Submitted on 29 Sep 2022]

### Dark sector studies with the PADME experiment

A.P. Caricato, M. Martino, I. Oceano, S. Spagnolo, G. Chiodini, F. Bossi, R. De Sangro, C. Di Giulio, D. Domenici, G. Finocchiaro, L.G. Foggetta, M. Garattini, A. Ghigo, P. Gianotti, T. Spadaro, E. Spiriti, C. Taruggi, E. Vilucchi, V. Kozhuharov, S. Ivanov, Sv. Ivanov, R. Simeonov, G. Georgiev, F. Ferrarotto, E. Leonardi, P. Valente, E. Long, G.C. Organtini, G. Piperno, M. Raggi, S. Fiore, P. Branchini, D. Tagnani, V. Capirossi, F. Pinna, A. Frankenthal

The Positron Annihilation to Dark Matter Experiment (PADME) uses the positron beam of the DAΦNE Beam-Test Facility, at the Laboratori Nazionali di Frascati (LNF) to search for a Dark Photon A′. The search technique studies the missing mass spectrum of single-photon final states in e+e−→A′γ annihilation in a positron-on-thin-target experiment. This approach facilitates searches for new particles such as long lived Axion-Like-Particles, protophobic X bosons and Dark Higgs. This talk illustrated the scientific program of the experiment and its first physics results. In particular, the measurement of the cross-section of the SM process e+e−→γγ at s√=21 MeV was shown.

 Subjects: High Energy Physics - Experiment (hep-ex); Instrumentation and Detectors (physics.ins-det) Cite as: arXiv:2209.14755 [hep-ex]

[Submitted on 22 Sep 2022 (v1), last revised 1 Nov 2022 (this version, v2)]
New anomaly observed in 12C supports the existence and the vector character of the hypothetical X17 boson
A.J. Krasznahorkay, A. Krasznahorkay, M. Begala, M. Csatlós, L. Csige, J. Gulyás, A. Krakó, J. Timár, I. Rajta, I. Vajda, N.J. Sas

Employing the 11B(p,γ)12C nuclear reaction, the angular correlation of e+e− pairs was investigated in the angular range of 40∘Θ≤175∘ for five different proton energies between Ep = 1.50 - 2.50 MeV. At small angles (Θ≤120∘), the results can be well interpreted by the internal pair creation process of electromagnetic radiations with E1 and M1 multipolarities and by the external pair creation in the target backing. However, at angles greater than 120∘, additional count excess and anomalies were observed, which could be well accounted for by the existence of the previously suggested hypotetical X17 particle. Our results show that the X17 particle was generated mainly in E1 radiation. The derived mass of the particle is mXc2=17.03±0.11(stat)±0.20(syst) MeV. According to the mass, and to the derived branching ratio (Bx=3.6(3)×10−6), this is likely the same X17 particle, which we recently suggested for describing the anomaly observed in the decay of 8Be and 4He.

Comments: 5 pages, 4 figures. arXiv admin note: text overlap with arXiv:2104.10075, arXiv:2205.07744, arXiv:1910.10459

[Submitted on 19 Sep 2022]
Resonant search for the X17 boson at PADME
Luc Darmé, Marco Mancini, Enrico Nardi, Mauro Raggi

We discuss the experimental reach of the Frascati PADME experiment in searching for new light bosons via their resonant production in positron annihilation on fixed target atomic electrons. A scan in the mass range around 17 MeV will thoroughly probe the particle physics interpretation of the anomaly observed by the ATOMKI nuclear physics experiment. In particular, for the case of a spin-1 boson, the viable parameter space can be fully covered in a few months of data taking.

Comments: 8 pages, 5 figures and 1 table
Subjects: High Energy Physics - Phenomenology (hep-ph); High Energy Physics - Experiment (hep-ex)
Cite as: arXiv:2209.09261 [hep-ph]

Nuclear Experiment
[Submitted on 8 Aug 2022 (v1), last revised 15 Aug 2022 (this version, v2)]
Searching for New Physics with DarkLight at the ARIEL Electron-Linac
The DarkLight Collaboration, E. Cline, R. Corliss, J. C. Bernauer, R. Alarcon, R. Baartman, S. Benson, J. Bessuille, D. Ciarniello, A. Christopher, A. Colon, W. Deconinck, K. Dehmelt, A. Deshpande, J. Dilling, D. H. Dongwi, P. Fisher, T. Gautam, M. Gericke, D. Hasell, M. Hasinoff, E. Ihloff, R. Johnston, R. Kanungo, J. Kelsey, O. Kester, M. Kohl, I. Korover, R. Laxdal, S. Lee, X. Li, C. Ma, A. Mahon, J. W. Martin, R. Milner, M. Moore, P. Moran, J. Nazeer, K. Pachal, T. Patel, T. Planche, M. Rathnayake, M. Suresh, C. Vidal, Y. Wang, S. Yen

The search for a dark photon holds considerable interest in the physics community. Such a force carrier would begin to illuminate the dark sector. Many experiments have searched for such a particle, but so far it has proven elusive. In recent years the concept of a low mass dark photon has gained popularity in the physics community. Of particular recent interest is the 8Be and 4He anomaly, which could be explained by a new fifth force carrier with a mass of 17 MeV/c2. The proposed DarkLight experiment would search for this potential low mass force carrier at ARIEL in the 10-20 MeV e+e− invariant mass range. This proceeding will focus on the experimental design and physics case of the DarkLight experiment.

Comments: 7 pages, 4 figures, to be submitted as part of the proceedings on "New Scientific Opportunities with the TRIUMF ARIEL e-linac"
Subjects: Nuclear Experiment (nucl-ex); Instrumentation and Detectors (physics.ins-det)
Cite as: arXiv:2208.04120 [nucl-ex]
(or arXiv:2208.04120v2 [nucl-ex] for this version)

[Submitted on 18 Nov 2021 (v1), last revised 10 Feb 2022 (this version, v2)]
Neutrino Physics Opportunities with the IsoDAR Source at Yemilab
J. Alonso, C.A. Argüelles, A. Bungau, J.M. Conrad, B. Dutta, Y.D. Kim, E. Marzec, D. Mishins, S.H. Seo, M. Shaevitz, J. Spitz, A. Thompson, L. Waites, D. Winklehner

IsoDAR seeks to place a high-power-cyclotron and target combination, as an intense source of ν¯e at the level of ∼1023/year, close to a kiloton-scale neutrino detector in order to gain sensitivity to very short-baseline neutrino oscillations (ν¯e→ν¯e) and perform precision tests of the weak interaction, among other physics opportunities. Recently, IsoDAR has received preliminary approval to be paired with the 2.26~kton target volume liquid scintillator detector at the Yemi Underground Laboratory (Yemilab) in Korea, at a 17~m center-to-center baseline, and cavern excavation for IsoDAR is now complete. In this paper, we present the physics capabilities of IsoDAR@Yemilab in terms of sensitivity to oscillations (via inverse beta decay, IBD; ν¯e+p→e++n), including initial-state wavepacket effects, and the weak mixing angle (via elastic scattering off atomic electrons, ν¯e+e−→ν¯e+e−). We also introduce a study of IsoDAR sensitivity to new particles, such as a light X boson, produced in the target that decays to νeν¯e.

Comments: 17 pages, 16 figures; this version presents a number of new physics topics and studies
Subjects: High Energy Physics - Experiment (hep-ex); High Energy Physics - Phenomenology (hep-ph)
Cite as: arXiv:2111.09480 [hep-ex]

High Energy Physics - Phenomenology
[Submitted on 12 Oct 2021]
X17 discovery potential in the γN→e+e−N process at electron scattering facilities
Johannes Backens, Marc Vanderhaeghen

We propose a direct search for the X17 particle, which was conjectured to explain the ATOMKI 8Be and 4He anomalies, through the dilepton photoproduction process on a nucleon in the photon energy range below or around the pion production threshold. For the scenarios of either pseudoscalar, vector, or axial-vector quantum numbers of the conjectured X17, we use existing constraints to estimate the X17 signal process. For dilepton resolutions which have been achieved in previous experiments, a signal-to-background ratio of up to an order of magnitude is found for a neutron target, and in particular for the pseudoscalar and vector X17 scenarios.

Subjects: High Energy Physics - Phenomenology (hep-ph); Nuclear Theory (nucl-th)
Cite as: arXiv:2110.06055 [hep-ph]

Last edited:

Staff Emeritus
Don't understand why there needs to be so many threads on this rather than having the messages all in one place. And don't see what the advantage is to guessing what the results might be rather than waiting to see what they are.

kodama
i find it exciting that 2023 we will have answers

if x17 is confirmed what exactly does it do ?

as a force what new interaction with fermions?

is it work with susy?

Last edited:
kodama
could x17 explain why Neutrino have mass

Staff Emeritus
No.

topsquark and ohwilleke
kodama
No.
if in 2023 Montreal Tandem accelerator or other like Positron Annihilation to Dark Matter Experiment and ARIEL and ISAC facilities and DarkLight experiment. rules out x17 i would accept it

if in 2023 Montreal Tandem accelerator or other experiments like PADME and ARIEL and ISAC facilities and DarkLight experiment. confirm x17 would you accept it ?

Staff Emeritus
You asked a question. "could x17 explain why Neutrino have mass". The answer is "no". That has nothing to do with what other experiments do or do not see. If these experiments see positive results, it still wouldn't explain neutrino masses, the results of the last election in [pick your favorite country]. why Nicholas Cage finds himself making stinker after stinker, or why red licorice is yummy and black licorice is nasty.

Further, the thing to do is not to speculate on what they might see, but wait and see what they do see. You know, actual science.

kodama
You asked a question. "could x17 explain why Neutrino have mass". The answer is "no". That has nothing to do with what other experiments do or do not see. If these experiments see positive results, it still wouldn't explain neutrino masses, the results of the last election in [pick your favorite country]. why Nicholas Cage finds himself making stinker after stinker, or why red licorice is yummy and black licorice is nasty.

Further, the thing to do is not to speculate on what they might see, but wait and see what they do see. You know, actual science.
do you think Montreal Tandem accelerator or other like Positron Annihilation to Dark Matter Experiment and ARIEL and ISAC facilities and DarkLight experiment could prove that x17 existed if successful and positive for a signal ?

Staff Emeritus
Just as posting multiple threads on the same topic doesn't yield different answers, asking the same question multiple times doesn't either.

topsquark, phinds and Tom.G
kodama
Volume 414 - 41st International Conference on High Energy physics (ICHEP2022) - Beyond the Standard Model
Latest results and future prospects of the NA64 experiment at CERN SPS
L. Marsicano
Full text: pdf
Pre-published on: November 21, 2022
Published on: —
Abstract
The search for Dark Matter (DM) is one of the hottest topics of modern physics. Despite the various astrophysical and cosmological observations proving its existence, its elementary properties remain to date unknown. In addition to gravity, DM could interact with ordinary matter through a new force, mediated by a new vector boson (Dark Photon, Heavy Photon or A′
), kinetically mixed with the Standard Model photon. The NA64-e experiment at CERN fits in this scenario, aiming to produce DM particles using the 100 GeV SPS electron beam impinging on a thick active target (electromagnetic calorimeter). In this setup the DM production signature consists in a large observed missing energy, defined as the difference between the energy of the incoming electron and the energy measured in the calorimeter, coupled with null activity in the downstream veto systems. Recently, following the growing interest in positron annihilation mechanisms for DM production, the NA64 collaboration has performed preliminary studies with the aim to run the experiment with a positron beam, as planned within the POKER (POsitron resonant annihilation into darK mattER) project.
This work presents the latest NA64-e
results and its future prospects, reporting on the progresses in the positron beam run and discussing the sensitivity of the experiment to alternative variations of to the dark photon paradigm.
DOI: https://doi.org/10.22323/1.414.0174

Staff Emeritus
There is very little value in just dumping an abstract here.

Did you read the paper? It spends only a few sentences on the X17, and what there is negative. No observation.

kodama
There is very little value in just dumping an abstract here.

Did you read the paper? It spends only a few sentences on the X17, and what there is negative. No observation.
Did you read the paper? Did you read section 4.2 page 6

Last edited:
Staff Emeritus
Yes. The key words are "has been proposed". That means "does not exist".

malawi_glenn
kodama
Yes. The key words are "has been proposed". That means "does not exist".

### Updates on the experiments X17

any and all present or future experiments search for x17

this paper

Latest results and future prospects of the NA64 experiment at CERN SPS

experiments search for x17

it also explains why NA64 experiment at CERN SPS hadn't seen x17 and what NA64 experiment at CERN SPS will do future experiments search for x17

Last edited:
Staff Emeritus
it also explains why NA64 experiment at CERN SPS hadn't seen x17
Yeah, it seems not to be there.

kodama
Yeah, it seems not to be there.
:)

the exciting thing i think is that we should get results as early as 2023. When I first heard about it in the news in 2016 and again in 2019 I wondered how soon will we know. These papers seem to imply 2023

kodama
Volume 406 - Corfu Summer Institute 2021 "School and Workshops on Elementary Particle Physics and Gravity" (CORFU2021) - Workshop on Connecting Insights in Fundamental Physics: Standard Model and Beyond

Searching for light dark matter with the PADME experiment

I. Oceano*, A.P. Caricato, M. Martino, S. Spagnolo, G. Chiodini, F. Bossi, et al. (click to show)

Full text: pdf

Published on: November 23, 2022

Abstract
The search for dark matter at accelerators has gained a lot of attention in recent years. Among the theoretical scenarios that can be studied, an attractive one is postulating a new UD(1)
gauge symmetry mediated by a massive boson often named dark photon A′. The A′ could be the bridge between the Standard Model (SM) and a hypothetical Dark Sector (DM), having a small coupling ϵ with SM particles.
PADME (Positron Annihilation to Dark Matter Experiment) is the first fixed target experiment searching for the A′ produced in association with a photon in e+e− annihilations. It exploits the missing mass technique and does not make any assumption on the decay mode of the A′. PADME is located at the Laboratori Nazionali di Frascati (LNF) of INFN and it is designed to be sensitive to the production of a dark photon with mass MA′≤23.7 MeV. Its setup is also ideal to investigate other scenarios of low mass dark matter: axion-like particles (ALPs), dark Higgs, and the X17
boson claimed to explain anomalies observed recently in a nuclear physics experiment.

DOI: https://doi.org/10.22323/1.406.0040

kodama

### Nuclear Instruments and Methods in Physics Research Section A: Accelerators, Spectrometers, Detectors and Associated Equipment

Volume 1047, February 2023, 167858

### A spectrometer for the measurement of anomalies in the angular correlation of electron and positron internally produced in excited 8Be and 4He

https://doi.org/10.1016/j.nima.2022.167858Get rights and content

### Abstract

In this work we describe the ongoing construction of a Time Projection Chamber-based (TPC) spectrometer for light charged particles utilising magnetic field as a means for energy measurement, combining Multi Wire Proportional Chambers (MWPC) with Timepix3 pixel detectors for improved spatial and angular resolution. The spectrometer will be operated at the Institute of Experimental and Applied Physics (IEAP) Van-de-Graaff facility with the goal of verifying the so-called ATOMKI anomaly (deviation in the distribution of the opening angle between electrons and positrons originating in internal pair creations – IPC – in the decay of 8Be and 4He excited nuclei). Description of the detectors is provided along with a few results.

## Introduction

Recent studies reported a 6.8-

anomaly in the distribution of the opening angle of ee pairs produced in a Be transition [1], and identified a similar behaviour in the decay of
He [2]. These results have prompted the interest of the scientific community. The reported effect can be related with unidentified nuclear reactions, experimental effects, or even the production of a new boson, which has been put forward by several theoreticians [3] but that no other experiment has observed so far.

In order to provide an independent assessment of the anomaly, the Institute of Experimental and Applied Physics (IEAP) at the Czech Technical University in Prague is developing a dedicated experiment. To check the anomaly we plan to use our Van de Graaff accelerator, where a proton beam will induce the needed

Be and He excited states (using Li or
H, respectively, as fixed targets).

To detect the decay products of

Be and He, a spectrometer will be mounted surrounding the target. It will consist of 6 independent modules, each consisting of a Timepix3 (TPX3) detector [4] and a Multi-Wire Proportional Chamber (MWPC) [5] to track and measure the angular correlation of the particles, followed by a Time Projection Chamber (TPC) [6]. The 15-mm gaps between the TPC modules contain permanent magnets that provide a magnetic field to identify the particles and measure their momenta [7] (see Fig. 1). The potential sources of background are the scattered protons and the decay of Be – easily shielded; and gamma production, which is suppressed by combining the information of the different detection layers.

Gold Member
On the plus side, it doesn't look like a terribly expensive experiment - probably well under $1 million Canadian. kodama kodama On the plus side, it doesn't look like a terribly expensive experiment - probably well under$1 million Canadian.
by LHC cost standards sure
why's it taking so long time to get answers to x17

Gold Member
by LHC cost standards sure
why's it taking so long time to get answers to x17
It took many decades to figure out the particle basis of the Periodic Table of Elements. It took 40 years to find the Higgs boson. It took about sixty years to figure out that neutrinos have a non-zero mass and we still have only the most vague idea about the extent to which there is CP violation by neutrinos. It took decades to figure out that the nuclear binding force was mediated by pions and other light mesons. It took several centuries to figure out that Newtonian gravity had to be modified with General Relativity.

Real life isn't a movie. Not everything gets resolved in an hour and a half.

kodama
kodama
It took many decades to figure out the particle basis of the Periodic Table of Elements. It took 40 years to find the Higgs boson. It took about sixty years to figure out that neutrinos have a non-zero mass and we still have only the most vague idea about the extent to which there is CP violation by neutrinos. It took decades to figure out that the nuclear binding force was mediated by pions and other light mesons. It took several centuries to figure out that Newtonian gravity had to be modified with General Relativity.

Real life isn't a movie. Not everything gets resolved in an hour and a half.

the original results were reported in 2015
apparently its 2025 we get some analysts say for x17 fire turned research

kodama
High Energy Physics - Experiment
arXiv:2009.02756 (hep-ex)
[Submitted on 6 Sep 2020 (v1), last revised 8 Sep 2020 (this version, v2)]
Hunting down the X17 boson at the CERN SPS
E. Depero, Yu. M. Andreev, D. Banerjee, J. Bernhard, V. Burtsev, A . Chumakov, D. Cooke, A. Dermenev, S. Donskov, R. Dusaev, T. Enik, N. Charitonidis, A. Feshchenko, V. Frolov, A. Gardikiotis, S. Gerassimov, S. Girod, S. Gninenko, M. Hosgen, V. Kachanov, A. Karneyeu, G. Kekelidze, B. Ketzer, D. Kirpichnikov, M. Kirsanov, V. Kolosov, I. Konorov, S. Kovalenko, V. Kramarenko, L. Kravchuk, N. Krasnikov, S. Kuleshov, V. Lyubovitskij, V. Lysan, V. Matveev, Yu. Mikhailov, L. Molina Bueno, D. Peshekhonov, V. Polyakov, B. Radics, R. Rojas, A. Rubbia, V. Samoylenko, D. Shchukin, H. Sieber, V. Tikhomirov, I. Tlisova, D. Tlisov, A. Toropin, A. Trifonov, B. Vasilishin, G. Vasquez, P. Volkov, V. Volkov, P. Ulloa, P. Crivelli

Recently, the ATOMKI experiment has reported new evidence for the excess of e+e− events with a mass ∼17 MeV in the nuclear transitions of 4He, that they previously observed in measurements with 8Be. These observations could be explained by the existence of a new vector X17 boson. So far, the search for the decay X17→e+e− with the NA64 experiment at the CERN SPS gave negative results. Here, we present a new technique that could be implemented in NA64 aiming to improve the sensitivity and to cover the remaining X17 parameter space. If a signal-like event is detected, an unambiguous observation is achieved by reconstructing the invariant mass of the X17 decay with the proposed method. To reach this goal an optimization of the X17 production target, as well as an efficient and accurate reconstruction of two close decay tracks, is required. A dedicated analysis of the available experimental data making use of the trackers information is presented. This method provides independent confirmation of the NA64 published results [Phys. Rev. D101, 071101 (2020)], validating the tracking procedure. The detailed Monte Carlo study of the proposed setup and the background estimate shows that the goal of the proposed search is feasible.

Subjects: High Energy Physics - Experiment (hep-ex); Instrumentation and Detectors (physics.ins-det)
Cite as: arXiv:2009.02756 [hep-ex]

ohwilleke
kodama
if x17 is real what does it do ?

does it mediate a force between dark matter particles ?

kodama
Nuclear Experiment
[Submitted on 20 Jan 2023]
A new direct detection electron scattering experiment to search for the X17 particle
D. Dutta (1), H. Gao (2), A. Gasparian (3), T. J. Hague (3 and 4), N. Liyanage (5), R. Paremuzyan (6), C. Peng (7), W. Xiong (8), P. Achenbach (6), A. Ahmidouch (3), S. Ali (5), H. Avakian (6), C. Ayerbe-Gayoso (1), X. Bai (5), M. Battaglieri (9), H. Bhatt (1), A. Bianconi (10 and 11), J. Boyd (5), D. Byer (2), P. L. Cole (12), G. Costantini (10 and 11), S. Davis (3), M. De Napoli (13), R. De Vita (9), B. Devkota (1), B. Dharmasena (5), J. Dunne (1), L. El Fassi (1), V. Gamage (5), L. Gan (14), K. Gnanvo (6), G. Gosta (10 and 11), D. Higinbotham (6), C. Howell (2), S. Jeffas (5), S. Jian (5), A. Karki (1), B. Karki (2), V. Khachatryan (2), M. Khandaker (15), V. Kubarovsky (6), I. Larin (16), M. Leali (10 and 11), V. Mascagna (17 and 11), G. Matousek (2), S. Migliorati (10 and 11), R. Miskimen (16), P. Mohanmurthy (1), H. Nguyen (5), E. Pasyuk (6), A. Rathnayake (5), J. Rittenhouse West (4), A. Shahinyan (18), A. Smith (2), S. Stepanyan (6), E. van Nieuwenhuizen (2), L. Venturelli (10 and 11), B. Yu (2), Z. Zhao (2), J. Zhou (2 and 4) ((1) Mississippi State University, (2) Duke University, (3) North Carolina A&T State University, (4) Lawrence Berkeley National Laboratory, (5) University of Virginia, (6) Thomas Jefferson National Accelerator Facility, (7) Argonne National Laboratory, (8) Shandong University, (9) Isituto Nazionale di Fisica Nucleare, Sezione di Genova, (10) Dipartimento di Ingegneria dell'Informazione, Università di Brescia, (11) Isituto Nazionale di Fisica Nucleare, sezione di Pavia, (12) Lamar University, (13) Isituto Nazionale di Fisica Nucleare, Sezione di Catania, (14) University of North Carolina, Wilmington, (15) Energy Systems, (16) University of Massachusetts, (17) DiSAT, Università dell'Insubria, (18) Yerevan Physics Institute)

A new electron scattering experiment (E12-21-003) to verify and understand the nature of hidden sector particles, with particular emphasis on the so-called X17 particle, has been approved at Jefferson Lab. The search for these particles is motivated by new hidden sector models introduced to account for a variety of experimental and observational puzzles: excess in e+e− pairs observed in multiple nuclear transitions, the 4.2σ disagreement between experiments and the standard model prediction for the muon anomalous magnetic moment, and the small-scale structure puzzle in cosmological simulations. The aforementioned X17 particle has been hypothesized to account for the excess in e+e− pairs observed from the 8Be M1, 4He M0, and, most recently, 12C E1 nuclear transitions to their ground states observed by the ATOMKI group. This experiment will use a high resolution electromagnetic calorimeter to search for or set new limits on the production rate of the X17 and other hidden sector particles in the 3−60 MeV mass range via their e+e− decay (or γγ decay with limited tracking). In these models, the 1−100 MeV mass range is particularly well-motivated and the lower part of this range still remains unexplored. Our proposed direct detection experiment will use a magnetic-spectrometer-free setup (the PRad apparatus) to detect all three final state particles in the visible decay of a hidden sector particle for an effective control of the background and will cover the proposed mass range in a single setting. The use of the well-demonstrated PRad setup allows for an essentially ready-to-run and uniquely cost-effective search for hidden sector particles in the 3−60 MeV mass range with a sensitivity of 8.9×10−8 - 5.8×10−9 to ϵ2, the square of the kinetic mixing interaction constant between hidden and visible sectors.

Comments: 6 pages, 7 figures. arXiv admin note: substantial text overlap with arXiv:2108.13276
Subjects: Nuclear Experiment (nucl-ex)
Cite as: arXiv:2301.08768 [nucl-ex]

kodama
[Submitted on 24 Jan 2023]

### Status and prospects of the NA64 experiment at the CERN SPS

P. Crivelli

NA64 is a fixed target experiment at the CERN SPS designed as a hermetic general purpose detector to search for Dark Sector physics in missing energy events from electron/positron, hadrons, and muon scattering off nuclei. In this contribution to the FIPs 2022 workshop, we review the current status and prospects of NA64.

 Comments: 5 pages, 4 figures, prepared for the proceedings of the FIPs 2022 workshop. arXiv admin note: text overlap with arXiv:2102.01885 Subjects: High Energy Physics - Experiment (hep-ex) Cite as: arXiv:2301.09905 [hep-ex] (or arXiv:2301.09905v1 [hep-ex] for this version) https://doi.org/10.48550/arXiv.2301.09905

kodama

is this initial data taking consistent with x17 or rule it out?

#### Attachments

• Screenshot 2023-01-25 at 00-42-13 ICHEP2022_Benmansour - ICHEP2022_Benmansour_0707.pdf.png
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Staff Emeritus
Generalissimo Francisco Franco is still dead.

ohwilleke
kodama
Generalissimo Francisco Franco is still dead.
could you elaborate?

what does this data mean?

2022 Award
ohwilleke
kodama
Franco still dead. I take it V50 means 'nothing new here'.
okay but what does this means

what is actually happened here ?
it looks like e+e- pairs are emissions at an angle off the target
is the SM or x17?

Last edited:
Staff Emeritus
If the experiment doesn't say they see something, having an outsider say that the do ois crackpotty. It puts the potty in cracpotty.
Seeing an e+e- pair is a very different think from seeing an X17.

kodama
If the experiment doesn't say they see something, having an outsider say that the do ois crackpotty. It puts the potty in cracpotty.
Seeing an e+e- pair is a very different think from seeing an X17.

a small fraction of the decay seems to be a wide angle e+e- pair the green and dotted red from brown dot target is what you would expect x17


text:
Geometry: Common Core (15th Edition)

$C$
The definition of bisecting is that something is cut in two equal portions. If $\overline {BD}$ bisects $\angle ABC$, then $\overline {BD}$ cuts $\angle ABC$ into exactly two congruent angles. In the answer options, only option $C$ shows that $\angle ABC$ is cut into exactly two congruent angles: $\angle ABD$ and $\angle CBD$. These angles are marked as being congruent in the diagram.


text:
# Mechanics

calculus-based physics for scientists and engineers

## Newton's Third Law in Two Dimensions

### Newton's Third Law in Two Dimensions

Since we have some experience with Newton's Third Law, should we attempt a definition? Here is my offering:

For every force that thing A (as agent) exerts on thing B (as object) there is another force that thing B (as agent) exerts on thing A (as object). These forces are the same type of force, have the same magnitude, but point in opposite directions.

How we find them has not changed. Create a sentence in the following form: "The (type) force on the (object) by the (agent)" Switch the objects of the prepositions -- change the names of the object and agent -- and you have found the Newton's Third Law pair.

Newton's Third Law force pairs always act on different objects. They can never act on the same object nor be on the same free body diagram. Also, never forget, even though these forces may have the same magnitude, the effects of the forces may be very different because the objects have significantly different mass or made out of different materials.

The video does the following example: A block $$m_A$$ is sliding frictionlessly along the ground. A second block, $$m_B$$ falls on it from above. Assume $$m_B$$ doesn't bounce, but slides along $$m_A$$ until it comes to rest on top of $$m_A$$ -- it is moving the same speed as $$m_A$$. The combined set of blocks is still moving though. What is that speed?

### Solution

1. Two blocks with masses $$m_A$$ and $$m_B$$ are on an incline set at angle $$\theta$$ from the horizontal and have coefficients of kinetic friction $$\mu_a$$ and $$\mu_b$$, where block $$a$$ is in front and $$\mu_a > \mu_b$$. When in contact, what is their acceleration down the incline?

### Solution

1. A block is pulled along the ground by a horizontal force F. A rope is attached to the other side of the block, goes round a massless pulley attached to a wall, and attaches to another block on top of the first block. The coefficients of kinetic friction between the blocks is $$\mu_k$$ and between the block and ground is $$\mu_g$$. What is the acceleration of the block?

### Solution

2. (Advanced) The sled slides frictionlessly along the ground. The coefficients of kinetic and static friction between the block and sled are $$\mu_k$$ and $$\mu_s$$. Assuming the system starts at rest, what is the maximum force you can pull on the (massless) rope such that the block will not slide with respect to the sled? What is the acceleration of the system? Again assuming the system starts at rest, what force will result in no acceleration for the block?

### Solution

3. A car is pulling a trailer. The car exerts a pulling force on the trailer and the trailer exerts a pulling force on the car. Which force has greater magnitude? They are the same! That is a key result of N3. But, wait... If they pull on each other with the same magnitude, how does the car pull the trailer at all?


text:
15/03/2020 997

# Mark the letter A, B, C, or D on your answer sheet to indicate the word whose underlined part differs from the other three in pronunciation in each of the following questions.

Đáp án chính xác

Quảng cáo

### Trả lời:

Giải bởi Vietjack

Chọn B

### Câu 1:

#### Mark the letter A, B, C, or D on your answer sheet to indicate the sentence that best completes each of the following exchanges.Don’t forget to send your parents my regards.” - “_________”

15/03/2020 15,029

15/03/2020 9,675

15/03/2020 9,104

15/03/2020 8,724

15/03/2020 8,708

15/03/2020 7,950

15/03/2020 7,914


text:
# Understanding the Anova (car package) output in r

I am running a two-way anova test using Anova from car package. My data looks like this:

> head(x)
Type Bin Score
1    0  SI 2.120
2    0   R 2.246
3    0   R 2.246
4    0   R 2.511
5    0   R 2.420
6    0   R 2.270

> summary(x)
Type    Bin           Score
0:906   I :  68   Min.   :1.202
1:258   R :1570   1st Qu.:2.000
2:346   SI: 328   Median :2.280
3:436             Mean   :2.299
4: 20             3rd Qu.:2.622
Max.   :3.233

> lapply(x,class)
$Type [1] "factor"$Bin
[1] "factor"

$Score [1] "numeric"  With an unbalanced design as follows: > table(x$Type,x\$Bin)

I   R  SI
0  42 702 162
1   6 190  62
2   2 296  48
3  18 362  56
4   0  20   0

So I after much research, I ended up using Anova from car package with Type III SS method. Here is how I am conducting the test:

> Anova(lm(Score ~ Bin * Type, data=x, contrasts=list(Bin=contr.sum, Type=contr.sum)),singular.ok = TRUE, type=3)

With results as follows:

Note: model has aliased coefficients
sums of squares computed by model comparison

Anova Table (Type III tests)

Response: Score
Sum Sq   Df F values   Pr(>F)
Bin         0.000    0
Type        2.081    3   4.3875 0.004381 **
Bin:Type    5.494    6   5.7927 5.42e-06 ***
Residuals 308.700 1953
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

1) This particular line from the results? Why might the SS and DF be zero for Bin?

           Sum Sq   Df F values   Pr(>F)
Bin         0.000    0

2) How this particular model still has aliased coefficients if Bin is not even being included in the final result?

Note: model has aliased coefficients
sums of squares computed by model comparison

3) Is there a better way to conduct this test in R?


text:
Numpy Testing For NaN Values

## Description

We can use the isnan command to test a value to see if it is not a number (i.e. $\frac{1}{0}$).

## Sage Cell

#### Code

import numpy as np
a = np.nan
b = 5
print np.isnan(a)
print np.isnan(b)

none

## Tags

Primary Tags:

Secondary Tags:

A list of possible tags can be found at The WeBWorK Open Problem Library. For linear algebra tags see the Curated Courses Project.

## Related Cells

Any related cells go here. Provide a link to the page containing the information about the cell.


text:
# Calculate the covariance matrix of $\hat{\beta}$

$y=X\beta+u$ where $u \sim N(0,\Sigma)$ and $\Sigma$ is symmetric & idempotent.
$X: n*k$, $y:n*1$, $\beta=k*1$, $u:n*1$ vector.
Suppose you apply LS(least square) to the model.
Calculate the covariance matrix of $\hat{\beta}$.

Since $\hat{\beta}=(X'X)^{-1}X'y$,
$E(\hat{\beta})=\beta$ so $\hat{\beta}$ is an unbiased estimator,
$Cov(\hat{\beta})=E(\hat{\beta}-E(\hat{\beta}))(\hat{\beta}-E(\hat{\beta}))'=E[(X'X)^{-1}X'uu'X(X'X)^{-1}]=\Sigma (X'X)^{-1}$
Is it right?

• You might want to look at this, since $\hat{\beta}$ is just a linear transformation of $y$. – Stefan Hansen Oct 15 '13 at 11:59
• @StefanHansen I think I skipped too many parts so now I added them. – dont Oct 15 '13 at 12:00

you may use this property: $Cov\left( Ay \right)=ACov\left( y \right)A'$ where $A={{\left( {X}'X \right)}^{-1}}{X}',Cov\left( y \right)=\text{ }\!\!\Sigma\!\!\text{ }$
so ${{\left( {X}'X \right)}^{-1}}{X}'\text{ }\!\!\Sigma\!\!\text{ }X{{\left( {X}'X \right)}^{-1}}$.
• How does it simplify to $\Sigma (X'X)^{-1}$? – Stefan Hansen Oct 15 '13 at 12:33
• But $AB$ does not equal $BA$ in general for matrices $A$ and $B$, so you can't move $\Sigma$ to the front. – Stefan Hansen Oct 15 '13 at 12:41
• thanks it escaped my notice. So we get the conclusion $Cov={{\left( {X}'X \right)}^{-1}}{X}'\text{ }\Sigma \text{ }X{{\left( {X}'X \right)}^{-1}}$ and it cant be written as $(X'X)^{-1}$ – mert Oct 15 '13 at 12:53
• Without further assumptions on $\Sigma$, I don't think we can reduce it any more than $(X'X)^{-1} X'\Sigma X(X'X)^{-1}$. – Stefan Hansen Oct 15 '13 at 12:56
• The only assumption i know is $\text{ }\!\!\Sigma\!\!\text{ }={{\sigma }^{2}}I$ when we get ${{\sigma }^{2}}{{\left( {X}'X \right)}^{-1}}$ – mert Oct 15 '13 at 12:59


text:
# Comparing Cox Proportional Hazards Models (variable selection)

I am using a cox proportional hazards model to run a survival analysis in r on a number of non-nested, distinct covariates such as Age, Blood Type, Cancer, etc:

 A, B, C, D, E

When I run the model on the omnibus null hypothesis:

surv ~ A + B + C + D

The effects of all of the covariates are insignificant because the number of subjects that have measurements for every covariate is relatively small. However, when I isolate single or other combinations of covariates in different cox models:

surv ~ A
surv ~ A + C
surv ~ B + D

I'm showing significant effects because the sample set is larger (i.e. the number of observations discarded by the model shrinks).

What I'm having difficulty understanding is how to do the following:

• Comparing the different cox models for the best fit, i.e. is surv ~ A + B + D a better model than surv ~ A + C ? Should I be comparing the likelihood, wald or logrank scores?
• Is it possible to run every possible combination of covariates to determine the best model? I have about 15 covariates.


text:
kidzsearch.com > wiki   Explore:web images videos games

# Norwegian language

Norwegian
norsk
Pronunciation[nɔʂk] (East and North)
[nɔʁsk] (West)
Native toNorway
EthnicityNorwegians
Native speakers4.3 million  (2012)[1]
Language family
Early forms:
Standard forms
written Bokmål (official)
• written Riksmål (unofficial)
written Nynorsk (official)
• written Høgnorsk (unofficial)
Writing systemLatin (Norwegian alphabet)
Norwegian Braille
Official status
Official language in Norway
Nordic Council
Regulated byLanguage Council of Norway (Bokmål and Nynorsk)
Ivar Aasen-sambandet (Høgnorsk)
Language codes
ISO 639-1no – inclusive code

Individual codes:
nbBokmål

nnNynorsk
ISO 639-2nor
ISO 639-3norinclusive code
Individual codes:
nob – Bokmål
nno – Nynorsk
Linguasphere52-AAA-ba to -be;
52-AAA-cf to -cg
Areas where Norwegian is spoken, including North Dakota (where 0.4% of the population speaks Norwegian) and Minnesota (0.1% of the population) (Data: U.S. Census 2000).
 This page contains IPA phonetic symbols in Unicode. Without proper rendering support, you may see question marks, boxes, or other symbols instead of Unicode characters.

The Norwegian language is the official language of Norway. It is spoken by over four and a half million people, and it belongs to the group of North Germanic languages which are spoken in Scandinavia. These include Swedish, Danish, Icelandic and Faeroese.

Two forms of the language exist: bokmål (which means "book language") and nynorsk (which means "new Norwegian").

## History of the Norwegian language

### Old Norse

Old Norse is the language that was spoken hundreds of years ago in Scandinavia at the time of the Vikings. It is very similar to today’s Icelandic language. This is because many Vikings sailed from Norway to Iceland in order to escape from the rule of the Norwegian kings who were making people pay lots of tax.

### Bokmål

During the 13th century the Black Death killed two thirds of the population of Norway. The Danish kings and queens noticed that Norway was weak and defenceless, so they annexed Norway (made it part of Denmark). For hundreds of years Norway was ruled by the Danes. All the rulers, priests, estate owners and noblemen were Danish. Many of them settled in Norway. This is why today’s standard Norwegian (Bokmål) is similar to Danish. Norwegians were not allowed to print books in Norwegian. Anyone wanting to study had to go to Denmark or Germany.

In 1814 Denmark lost a war and had to give Norway to Sweden. Then the Norwegians were allowed to have their own university. Gradually the Danish language was mixed up with the Norwegian dialects and became today’s Norwegian language. Norwegian and Danish look very similar when they are written, but when they are spoken they sound different. In Danish a lot of the sounds are swallowed.

### Nynorsk

During the 19th century a slightly different form of Norwegian was made up by several people. This eventually became known as “Nynorsk.” It was based on old forms of Norwegian and dialects. During the 20th century some attempts were made to join Bokmål and Nynorsk into one language, but they did not succeed. Today about one person in nine or ten in Norway writes Nynorsk. Children in school have to learn both forms.

## The Norwegian alphabet

The Norwegian alphabet has 29 letters. These are the same letters as the English alphabet plus three extra vowels: æ ø å

The letters c, q, w, x and z are only used for words that have been borrowed from other languages.

## References

1. De Smedt, Koenraad; Lyse, Gunn Inger; Gjesdal, Anje Müller; Losnegaard, Gyri S. (2012). The Norwegian Language in the Digital Age. White Paper Series. Berlin, Heidelberg: Springer Berlin Heidelberg. pp. 45. . . "Norwegian is the common spoken and written language in Norway and is the native language of the vast majority of the Norwegian population (more than 90%) and has about 4,320,000 speakers at present.".


text:
# 颜色区间统计

historgram 函数的使用

## code

# Image Histogram Info Example
#
# This script computes the histogram of the image and prints it out.

import sensor, image, time

sensor.reset()
sensor.set_pixformat(sensor.GRAYSCALE) # or RGB565.
sensor.set_framesize(sensor.QVGA)
sensor.skip_frames(time = 2000)
sensor.set_auto_gain(False) # must be turned off for color tracking
sensor.set_auto_whitebal(False) # must be turned off for color tracking
clock = time.clock()

while(True):
clock.tick()
img = sensor.snapshot()
# Gets the grayscale histogram for the image into 8 bins.
# Bins defaults to 256 and may be between 2 and 256.
print(img.get_histogram(bins=8))
print(clock.fps())

# You can also pass get_histogram() an "roi=" to get just the histogram of that area.
# get_histogram() allows you to quickly determine the color channel information of
# any any area in the image.


text:
# tril

Lower triangular part of a matrix

### tril(X)

• X should be a matrix, i.e., ndims(X) equal to 2.
• It returns the lower triangular part of X, i.e., elements below and on the main diagonal of X.
• If X is a character array, elements above the main diagonal are filled with the white-space character.
• It returns an empty matrix if X is empty.

### tril(X, k)

• k should be an integer scalar, not equal to NaN or Inf.
• If k < 0, it returns elements below and on the lower k-th diagonal of X.
• If k > 0, it returns elements below and on the upper k-th diagonal of X.
• If k == 0, it returns the same result as tril(X).

Example 1: In the following, tril() extracts the lower triangular part of a character array. The upper part is filled with spaces.

a=['abcd';'efgh';'ijkl';'mnop']
tril(a)

a =
abcd
efgh
ijkl
mnop

ans =
a
ef
ijk
mnop


text:
# Special Relativity Exercise (A.P. French)

Thread moved from the technical forums, so no Homework Template is shown
I just solved an exercise in the special relativity book of A.P. French. I would like you to tell me if the answer is correct. Yes I suposse is a very easy exersice.
I hope you can help me, because I really like learn relativity and cosmology, but I have not had teachers of this, and I am completely self-taught in the subject.

Exercise.

In a certain reference system it is observed that a particle has a total energy of 5GeV, and an P of 3GeV/c, (that is to say cP, which has units of energy, is 3GeV).

a) What is the energy of this particle in a system where P is 4GeV/c ?
b) What will be your mass at rest in uma?
c) What is the relative speed of the two reference systems?

My atempt of solution.
Question a)
Ok, there are two systems of reference
System 1: E_1, P_1
System 2: E_2 (unknow), P_2
I apply the relativistic invariant for the four moment enegy ; $E_0^2=E^2-c^2p^2$, then: $$E_1^2-c^2p_1^2=E_2^2-c^2p_2^2$$
I make sustitution of terms: E_1=5GeV; P_1=3GeV/c; E_2=?; P_2=4GeV/c$$5^2-c^2(\frac 3 c)^2=E_2^2-c^2(\frac 4 c)^2$$ I solve and the result is: $E_2=4\sqrt2 \text { GeV}$

Question b)
I use the reference system 2 for compute te invariant energy-momentum $$E_0^2=E_2^2 - c^2p^2$$ I put the values into the above equation: $$E_0^2=(4\sqrt 2)^2-c^2(\frac 4 c)^2$$ Solution is $E_0=4\text{GeV}$
We can use the conversión factors and acheive a rest mass $m_0=4.28\text{u}$, (atomic units of mass)

Question c)
For easyest calculations I use masses in GeV in this question.
In the system 2 particle have a rest mass m_0 of 4GeV, but
In ths system 1 particle have a mass of 5GeV. Then $$\frac{ m_1} {m_0}=\gamma$$$$\frac{ 5} {4}=\gamma=1.25$$
Then $\gamma=\frac{1}{\sqrt{1-\beta^2}}$.
$\beta=0.6$, and then $v=0.6c$
******************************************************************************************************************
Ok indications are welcome, thanks a lot.

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Orodruin
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(a) and (b) are correct. In (c) you are not asked for the velocity relative to the rest frame of the particle, you are asked for the relative velocity between the different inertial frames. This cannot be answered without additional assumptions (such as the particle moving in the x-direction in both frames).

Also
In the system 2 particle have a rest mass m_0 of 4GeV
the rest mass (or just mass, we really do not talk about "relativistic mass" anymore) is invariant, i.e., it is the same in all frames.

Ok, then how can I solve question c)
I'm a teacher in a High School. And I work in relativity only because I like to learn about this subject. Then you can say me the answer without problem, I'm not doing my homework for tomorrow!!! (the post was moved here)

berkeman
Mentor
Then you can say me the answer without problem, I'm not doing my homework for tomorrow!!!
We treat all schoolwork questions the same here at the PF. Whether it is for self study or for an assignment for class. We require that you show your work and try to figure out the problems -- we are not allowed to give answers to schoolwork-type questions here.

PeroK
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Ok, then how can I solve question c)
Have you heard of the "energy-momentum transformation"?

First of all, I'm sorry if I did not follow any rules. Rules are for all people, of course!! You can be sure that I have my honesty : -)

Have you heard of the "energy-momentum transformation"?
I suposse transform with a Lorentz transformation, like time- space. Is it?

PeroK
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First of all, I'm sorry if I did not follow any rules. Rules are for all people, of course!! You can be sure that I have my honesty : -)

I suposse transform with a Lorentz transformation, like time- space. Is it?
Yes, it should be in AP French's book somewhere.

Ok, I need to make Tle lorentz transformation between E and E'!! How I did not realize that!
Thank you very much.

Orodruin
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A (much) more convenient way (if you are familiar with it) than applying the Lorentz transformation is to use the Minkowski inner product between the 4-momenta.

Orodruin said:
A (much) more convenient way (if you are familiar with it) than applying the Lorentz transformation is to use the Minkowski inner product between the 4-momenta.
Ok, thanks, this for tomorrow, for today is adequate for me.

I have made this transformation: $$E'=\frac{E-p_x·v}{\sqrt {1-\beta^2}}$$E=5 GeV; Px=3/c GeV; E'=4·sqr(2) GeV; $$4\sqrt {2}=\frac{5-\frac 3 c·v}{\sqrt {1-\beta^2}}$$$$4\sqrt {2}=\frac{5-3\beta}{\sqrt {1-\beta^2}}$$
Solving: $$41\beta^2-30\beta-7=0$$$$\beta^2-0.731\beta-0.17=0$$
$$\beta=\frac {0.731\pm\sqrt{0.731^2+4·0.17}}{2}=\frac{0.731\pm1.10}{2}=0.9155$$and $v=0.9155c$

It's ok?

Orodruin
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No. You cannot just throw away one of the roots. (This is part of the indeterminacy I already mentioned.)

Ok Orodruin, I will think in these questios tomorow in the afternoon, today is too later.
I also think in calculate whit the rest mass the two velocities of the two reference frames respect to the rest frame, and then whit the formula or relatives velocities calculate it betwen the two, but I don't now if this procedure are ok.
Ok is too late I go to sleep... zzzzzz

A (much) more convenient way (if you are familiar with it) than applying the Lorentz transformation is to use the Minkowski inner product between the 4-momenta.
Ok, I do not know very well what you mean, because my knowledge is very limited, but I have tried this, looking for the angle between the 4-momenta, in the hope that this angle are related with beta. Units are in GeV.
$\vec {p_1}=\begin{pmatrix}E_1\\p_1·c\end{pmatrix}=\begin{pmatrix}5\\3\end{pmatrix};\vec{p_2}=\begin{pmatrix}E_2\\p_2·c\end{pmatrix}=\begin{pmatrix}4·\sqrt2\\4\end{pmatrix}$
Minkosky metric and inner product: $$\eta_{ab}=\begin{pmatrix}-1&0\\0&1\end{pmatrix}; \vec X·\vec Y=\eta_{ab}·X^a·Y^b$$
We calculate, only terms in that a=b do not vanishing, then is not too hard:$$\vec {p_1}·\vec {p_1}=p_1^2=-1·5^2+1·3^2; p_1=\sqrt{-16}$$$$\vec {p_2}·\vec {p_2}=p_2^2=-1·(4·\sqrt2)^2+1·4^2; p_2=\sqrt{-16}$$
Ok this seems to be invariant. Now we calculate P_1·P_2.$$\vec {p_1}·\vec {p_2}=-1·5·4\sqrt2+1·3·4=-16,28$$
But now I have a problem, because if I try to calculate the angle:$$cos(\theta)=\frac{\vec p_1·\vec p_2}{p_1·p_2}=\frac{-16,28}{-16}>1$$
Then a this point I don't understand anything. Then I need your help.
And Orodruin do you refers to this calculations when you says the QUOTE above?

Orodruin
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"Angles" work differently in Minkowski space. What you are looking for is the rapidity between the frames. Due to the hyperbolic nature, the rapidity between two 4-vectors $P_1$ and $P_2$ is given by
$$\cosh(\theta) = \pm \frac{P_1 \cdot P_2}{\sqrt{P_1^2 P_2^2}}$$
rather than $\cos(\theta)$. The $\pm$ depends on the sign convention you are using for the inner product, most particle physicists would use $+$ in the time component.

Edit: Also, the relation between the rapidity and the relative speed is
$$\beta = \tanh(\theta).$$

Wow!!!!
Ok. Today I have an institutional act, and I will not be able to continue this work until tomorrow.

I've used -1 for temporal component, but from now one I will use the +1 sign.
Ok, another question, if I use the sign +1 in the minkowsky metric then also is + in the above equation?
I'm not to much accustomed to the hyperbolics trigonometrics functions, and will need to buy a handle calculator with them. Those expressions of exponentials and logarithms are tedious for handle.

Thanks a lot!!! I will post the answer tomorrow, when I have finished the calculations.

Orodruin
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I'm not to much accustomed to the hyperbolics trigonometrics functions, and will need to buy a handle calculator with them. Those expressions of exponentials and logarithms are tedious for handle.
They do not differ that much from the usual trigonometric counterparts apart from the appearance of some relative minus signs. In particular, a minus sign in the hyperbolic one as compared to the trigonometric one:
$$\cosh^2(\theta) - \sinh^2(\theta) = 1.$$
In particular, this means that
$$\tanh(\theta) = \frac{\sinh(\theta)}{\cosh(\theta)} = \frac{\sqrt{\cosh^2(\theta) - 1}}{\cosh(\theta)},$$
which is all you need here.

Ok, then for calculate betha, with the above results, we do: $$cosh(\theta)=\frac{\vec p_1·\vec p_2}{p_1·p_2}=\frac {-16.28}{ -16}=1.0175$$ and:
$$\beta=tanh(\theta)=\frac{\sqrt{[cosh(\theta)]^2-1}}{cosh(\theta)}=\frac{0.1879}{1.0175}=0,1847$$$$v=0.1847·c$$
I suposse I'ts ok : -)

Orodruin
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So this reproduces the (absolute value of the) result that you got from using the $-$ root of your previous second order equation. How would you reproduce the $+$ root?

Ok, the bull is not dead yet...

Orodruin
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Well, since you found two solutions before, you should ask yourself why you found two solutions.

$$sinh(\theta)=\pm\sqrt{[cosh(\theta)]^2-1}$$
is a square equation, then there are two solutions. Is it?

Orodruin
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No, that is not it. That is just a redefinition of the positive direction of $\theta$. The difference comes (much) earlier and has to do with your assumptions when you derived your original second order equation in relation to your assumptions when you wrote down the 4-momenta of the particles.

Ok, I've been thinking about that question for an hour, but I'm sorry, I think my current knowledge is not enough to answer it

Orodruin
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Ok, I've been thinking about that question for an hour, but I'm sorry, I think my current knowledge is not enough to answer it
I believe they should be, so go back to look at your assumptions.

Ok, I'm going to check everything, from the beginning.
In the Lorentz transforms, I understand, E, E ', + Px · V, -Px · v, and the labels in the vectors .... All that can lead to different results.

All this will take me time, maybe one day. Be patient, when I have a result I will post it.
And lots of thanks for your help!!!


text:
# What is the product of a reaction between an acyl chloride and a secondary alcohol?

For the following two reactants, I want to predict the major product.

I would have thought the second reactant $\ce{CH3CH(OH)CH3}$ splits into $\ce{CH3CHCH+}$ and $\ce{OH-}$ , and the $\ce{OH-}$ would replace the $\ce{Cl-}$ in the first reactant $\ce{Phenyl-COCl}$, giving B. My main reason for thinking this would occur is that I always see e.g. water form $\ce{H+ and OH-}$.

However, the answer is C. The $\ce{CH3CH(OH)CH3}$ gives up only its $\ce{H+}$, and the rest of it joins $\ce{Phenyl-COCl}$ to form C.

How do I predict which atom(s) will leave/join? And for this particular question, how do I know the bond in the 2nd reactant cleaves at O-H not at C-OH?

• I don't understand what you're asking. The answer is C. Why or what are you asking about D? – Zhe Feb 15 '17 at 3:18
• Sorry! Typo. I am asking why the H+ leaves the CH3CH(OH)CH3, leaving the CH3CH(O)CH3- to swap with the Cl- (giving C). Rather than CH3CH(OH)CH splitting into CH3CHCH+ and OH-, and the OH- replacing the Cl- in the first reactant. – K-Feldspar Feb 15 '17 at 3:20
• In other words, why does the 2nd reactant split at the O-H bond and not the C-OH bond. Thank you. Let me know if my question still doesn't make sense and I will try and re-word it. – K-Feldspar Feb 15 '17 at 3:23