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# NCERT Solutions for Class 10 Maths Exercise 1.4 Chapter 1 Real Numbers

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Ex 1.4) Exercise 1.4

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## NCERT Solutions for Class 10 Maths Chapter 1

### Exercise 1.4

1. Without actually performing the long division, state whether the following  rational numbers will have a terminating decimal expansion or a non terminating repeating decimal expansion:

(i) $\mathbf{\frac{13}{3125}}$

Ans: Given a rational number $\frac{13}{3125}$.

If the denominator of a rational number has prime factors of the form 2n  5m, where, m and n are positive integers. Then the rational number has a terminating decimal  expansion. If the denominator has factors other than 2and 5, then it has non terminating decimal expansion.

The denominator of the given number is 3125.

Then, factors of 3125 are

⇒312= 5$\times$ 5$\times$ 5$\times$ 5$\times$ 5$\times$ 5

⇒ 3125 =  55

Here, the factors of  denominator are of the form 5m

Therefore, $\frac{13}{3125}$  has terminating decimal expansion.

(ii) $\mathbf{\frac{17}{8}}$

Ans: Given a rational number $\frac{13}{3125}$.

If the denominator of a rational number has prime factors of the form 2n 5m, where, m and n are positive integers. Then the rational number has terminating decimal  expansion. If the denominator has factors other than 2 and 5, then it has non terminating decimal expansion.

The denominator of the given number is 8 .

Then, factors of 8 are

⇒ 8= 28 $\times$ 28 $\times$ 2

⇒ 8 =23

Here, the factors of denominator are of the form 2 n

Therefore,$\frac{13}{3125}$ has terminating decimal expansion.

(iii) $\mathbf{\frac{64}{455}}$

Ans: Given a rational number $\frac{64}{455}$.

If the denominator of a rational number has prime factors of the form 2n 5m , where, m and n are positive integers. Then the rational number has a terminating decimal  expansion. If the denominator has factors other than 2and 5, then it has non terminating decimal expansion.

The denominator of the given number is 455.

Then, factors of 455 are

⇒ 455 =  5 $\times$ 7 $\times$ 13

Here, the factors of denominator are not in the form  2n 5m. The denominator has factors  other than 2and 5.

Therefore, $\frac{64}{455}$ has non-terminating repeating decimal expansion.

(iv) $\mathbf{\frac{15}{1600}}$

Ans: Given a rational number $\frac{15}{1600}$

If the denominator of a rational number has prime factors of the form 2n 5m, where, m and n are positive integers. Then the rational number has terminating decimal  expansion. If the denominator has factors other than 2and 5, then it has non terminating decimal expansion.

The denominator of the given number is 1600.

Then, factors of 1600 are

⇒  1600 =  2$\times$ 2$\times$ 2$\times$ 2$\times$ 2$\times$ 2$\times$ 5$\times$ 5

⇒ 1600 =  26 $\times$ 55

Here, the factors of denominator are of the form 2n 5m .

Therefore,$\frac{15}{1600}$ has terminating decimal expansion.

(v)$\mathbf{\frac{29}{343}}$

Ans: Given a rational number $\frac{29}{343}$

If the denominator of a rational number has prime factors of the form 2n  5m, where, m and n are positive integers. Then the rational number has a terminating decimal  expansion. If the denominator has factors other than 2and 5, then it has non terminating decimal expansion.

The denominator of the given number is 343.

Then, factors of 343 are

⇒343=  7$\times$ 7$\times$4 7

⇒343 =  73

Here, the factors of denominator are not in the form 2n 5m. The denominator has factors  other than 2and 5.

Therefore, $\frac{29}{343}$  has non-terminating repeating decimal expansion.

(vi) $\mathbf{\frac{23}{2^3 5^2}}$

Ans: Given a rational number $\frac{23}{2^3 5^2}$

If the denominator of a rational number has prime factors of the form 2n 5m, where, m and n are positive integers. Then the rational number has a terminating decimal  expansion. If the denominator has factors other than 2 and 5, then it has non terminating decimal expansion.

The denominator of the given number is 2352

Here, the denominator is of the form 2n 5m

Therefore,$\frac{23}{2^3 5^2}$ has terminating decimal expansion.

(vii)  $\mathbf{\frac{129}{2^2 5^7 7^5}}$

Ans: Given a rational number $\frac{129}{2^2 5^7 7^5}$

If the denominator of a rational number has prime factors of the form  2n  5m, where, m and n are positive integers. Then the rational number has terminating decimal  expansion. If the denominator has factors other than 2 and 5, then it has non terminating decimal expansion.

The denominator of the given number is  225775

Here, the denominator is of the form 2n 5m but also has factors other than 2 and 5.

Therefore, $\frac{129}{2^2 5^7 7^5}$has non-terminating repeating decimal expansion.

(viii)  $\mathbf{\frac{6}{15}}$

Ans: Given a rational number $\frac{6}{15}$

If the denominator of a rational number has prime factors of the form 2n 5m, where, m and n are positive integers. Then the rational number has a terminating decimal  expansion. If the denominator has factors other than 2and 5, then it has non terminating decimal expansion.

The denominator of the given number is 15.

Then, factors of 3125 are

⇒ 15 =  3 $\times$  5

But we can write the numerator of the given number as

$\frac{6}{15}$ = $\frac{2 \times 3}{3 \times 5}$ = $\frac{2}{5}$

Here, the factors of denominator are of the form  5m

Therefore, $\frac{6}{15}$ has terminating decimal expansion.

(ix)  $\mathbf{\frac{35}{50}}$

Ans: Given a rational number $\frac{35}{50}$

If the denominator of a rational number has prime factors of the form 2n 5m, where, m and n are positive integers. Then the rational number has a terminating decimal expansion. If the denominator has factors other than 2and 5, then it has non terminating decimal expansion.

The denominator of the given number is 50.

Then, factors of 50 are

⇒ 50  = 10  \$times$ 5

But we can write the numerator of the given number as

$\frac{35}{50}$ =  $\frac{7 \times5}{5 \times 10}$

⇒ 10 =  2 $\times$ 5

Here, the factors of denominator are of the form 2n  5m  .

Therefore, $\frac{35}{50}$ has terminating decimal expansion.

(x)  $\mathbf{\frac{77}{210}}$

Ans: Given a rational number  $\frac{77}{210}$

If the denominator of a rational number has prime factors of the form 2n  5m , where, m and n are positive integers. Then the rational number has a terminating decimal  expansion. If the denominator has factors other than 2 and 5, then it has non terminating decimal expansion.

The denominator of the given number is 210 .

Then, factors of 210 are

⇒ 210 =  2 $\times$ 3 $\times$5 $\times$7

Here, the denominator has factors other than 2and 5.

Therefore,$\frac{77}{210}$  has non-terminating repeating decimal expansion.

2. Write down the decimal expansions of those rational numbers in Question 1  above which have terminating decimal expansions.

(i) $\mathbf{\frac{13}{3125}}$

Ans: To find the decimal expansion of $\frac{13}{3125}$, we will divide the numerator of the

number by the denominator using a long division method. We get

$\;\;\;\;\;\;\;\;\;\;0.00416\\3125\;\;)\overline{13.00000}\\\;\;\;\;\;\;\;\;\;\; 0 \\\;\;\;\;\;\;\;\;\;\;\overline{13000}\\\;\;\;\;\;\;\;\;\;\;12500\\\;\;\;\;\;\;\;\;\;\;\overline{5000}\\\;\;\;\;\;\;\;\;\;\;3125\\\;\;\;\;\;\;\;\;\;\;\overline{18750}\\\;\;\;\;\;\;\;\;\;\;18750\\\;\;\;\;\;\;\;\;\;\;\overline{0}$

Therefore, the decimal expansion of  $\frac{13}{3125}$ is 0.00416.

(ii) $\mathbf{\frac{17}{8}}$

Ans: To find the decimal expansion of $\frac{17}{8}$,  we will divide the numerator of the number  by denominator using long division method. We get

$\;\;\;\;\;\;\;\;\;\;2.125\\8\;\;)\overline{17}\\\;\;\;\;\;\;16\\\;\;\;\;\;\;\;\overline{10}\\\;\;\;\;\;\;\;8\\\;\;\;\;\;\;\;\overline{20}\\\;\;\;\;\;\;\;16\\\;\;\;\;\;\;\;\overline{40}\\\;\;\;\;\;\;\;40\\\;\;\;\;\;\;\;\overline{0}$

Therefore, the decimal expansion of  $\frac{17}{8}$  is 2.125.

(iii) $\mathbf{\frac{15}{1600}}$

Ans: To find the decimal expansion of $\frac{15}{1600}$, we will divide the numerator of the

number by the denominator using a long division method. We get

$\;\;\;\;\;\;\;\;\;\;0.005375\\1600\;\;)\overline{15.000000}\\\;\;\;\;\;\;\;\;\;0\\\;\;\;\;\;\;\;\;\;\;\overline{150}\\\;\;\;\;\;\;\;\;\;\;0\\\;\;\;\;\;\;\;\;\;\;\overline{15000}\\\;\;\;\;\;\;\;\;\;\;14400\\\;\;\;\;\;\;\;\;\;\;\overline{6000}\\\;\;\;\;\;\;\;\;\;\;4800\\\;\;\;\;\;\;\;\;\;\;\overline{12000}\\\;\;\;\;\;\;\;\;\;\;11200\\\;\;\;\;\;\;\;\;\;\;\overline{8000}\\\;\;\;\;\;\;\;\;\;\;8000\;\;\;\;\;\;\;\;\;\;\\ \overline{0}$

Therefore, the decimal expansion of $\frac{15}{1600}$  is 0.009375.

(iv) $\mathbf{\frac{23}{2^3 5^2}}$

Ans: To find the decimal expansion of    $\frac{23}{2^3 5^2}$

we will divide the numerator of the number

by the denominator using a long division method. We get

$\frac{23}{2^3 5^2}$ = $\frac{23}{200}$

$\;\;00.115\\\;\;\;\;200) \overline{23.000}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0\\\;\;\;\;\;\;\;\;\;\; \overline{23}\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\\ \;\;\;\;\;\;\;\;\;\;\;\;\;0 \\\;\;\;\;\;\;\;\;\;\;\; \overline {230} \\\;\;\;\;\;\;\;\;\;\;\; 200\\\;\;\;\;\;\;\;\;\;\;\; \overline{1000}\\ \;\;\;\;\;\;\;\;\;\;\; 1000 \\\;\;\;\;\;\;\;\;\;\;\; \overline{0}$

Therefore, the decimal expansion of  $\frac{23}{200}$ is 00.115.

(v) $\mathbf{\frac{6}{15}}$

Ans: To find the decimal expansion of $\frac{6}{15}$, we will divide the numerator of the number  by the denominator using a long division method. We get

$\frac{6}{15}$ = $\frac{2 \times 3}{3 \times 5}$= $\frac{2}{5}$

$\;\;\;\;\;\;\;\;0.4\\\;\;\;\;\;5)\overline{2.0}\\ \;\;\;\;\;\; \;\;\;0 \\ \;\;\;\;\;\;\;\;\overline{20}\\\;\;\;\;\;\;\;\;20\\ \;\;\;\;\;\;\;\; \overline{0}$

Therefore, the decimal expansion of $\frac{6}{15}$ is 0.4.

(vi) $\mathbf{\frac{35}{50}}$

Ans: To find the decimal expansion of $\frac{35}{50}$,  we will divide the numerator of the number  by the denominator using a long division method. We get

$\;\;\;\;\;\;\;\;0.7\\\;\;\;\;\;50)\overline{35.0}\\ \;\;\;\;\;\; \;\;\;0 \\ \;\;\;\;\;\;\;\;\overline{350}\\\;\;\;\;\;\;\;\;350\\ \;\;\;\;\;\;\;\; \overline{0}$

Therefore, the decimal expansion of $\frac{6}{15}$  is 0.7.

3. The following real numbers have decimal expansions as given below. In each  case, decide whether they are rational or not. If they are rational, and of the form $\mathbf{\frac{p}{q}}$ what can you say about the prime factors of q?

(i) 43.123456789

Ans: Given a decimal expansion 43.123456789.

The given number has terminating expansion, we can write the number as $\frac{43124356789}{1000000000}$, , which is of the form pq.

Therefore, the number 43.123456789is a rational number.

Since the number has terminating decimal expansion, the factors of q must be of the  form 2n 5m

(ii) 0.120120012000120000......

Ans: Given a decimal expansion 0.120120012000120000.......

When we observe the given expansion we can say that the number has non-terminating  and non-repeating decimal expansion. Hence we cannot express it in the form of $\frac{p}{q}$ Therefore, the number is irrational.

(iii) 43.123456789

Ans: Given the decimal expansion 43.123456789.

The given number has non-terminating but repeating decimal expansion. So the  number will be of the form$\frac{p}{q}$ .

Therefore, the number 43.123456789is a rational number.

But the factors of denominator are not of the form 2n 5m. Denominator also has factors  other than 2and 5.

### NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.4

Opting for the NCERT solutions for Ex 1.4 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 1.4 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 1 Exercise 1.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 1 Exercise 1.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 1 Exercise 1.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

### Decimal Expansions of Rational Numbers

Rational Numbers: Rational numbers are numbers that are written in the form p/q, (p in the numerator and q in the denominator) where p and q are integers and q≠0. We can say all fractions are rational numbers.

Examples of rational numbers: 1/3, 3/10, -7/10,−3

### Terminating and nonterminating decimals

1. Terminating decimals: The decimal values whose value is definite and terminate at a certain digit are said to be Terminating decimals.

Example: 0.4, 5.8, and so on.

2. Non-terminating: The decimal values in which the digits after the decimal point do not terminate and the decimal value is not exactly known are called Non-terminating decimals.

Example: 0.333333….., 0.13135235343…

Non-terminating decimals can be classified again as:

• Non-terminating recurring Decimals: In this type of decimal number, a part of the decimal indefinitely repeats itself. Eg: 0.142869142869142869….(where 142869 is repeating after the decimal point).

• Non-terminating non-recurring Decimals: In this type of decimal number, no part of the decimal repeats indefinitely. Example: π=3.1415926535…

### How to check whether a given rational number is terminating or not?

If a rational number a/b is given, then its decimal expansion of the rational number will terminate if it satisfies the below two conditions:

a) The H.C.F of a (numerator) and b (denominator) of the given rational number should be 1.

b) The denominator of the given rational number(b) can be expressed as a prime factorization of 2 and 5 i.e b=2m×5n where m or n, or both can be 0. If the prime factorization of b contains any number other than 2 or 5, then the decimal expansion of that number will be a non-terminating decimal.

### NCERT Solutions for Class 10 Maths Chapter 1 Exercises

 Chapter 1 - Real Numbers Exercises in PDF Format Exercise 1.1 5 Questions & Solutions (4 Long Answers, 1 Short Answer) Exercise 1.2 7 Questions & Solutions (4 Long Answers, 3 Short Answers) 3 Questions & Solutions (3 Short Answers)


text:
Natalie Evans

King's College

## Time:

Thursday, October 21, 2021 - 10:00am to 10:50am

## Location:

https://uci.zoom.us/j/95642648816

The Hardy-Littlewood generalised twin prime conjecture states an asymptotic formula for the number of primes $p\le X$ such that $p+h$ is prime for any non-zero even integer $h$. While this conjecture remains wide open, Matom\"{a}ki, Radziwi{\l}{\l} and Tao proved that it holds on average over $h$, improving on a previous result of Mikawa. In this talk we will discuss an almost prime analogue of the Hardy-Littlewood conjecture for which we can go beyond what is known for primes. We will describe some recent work in which we prove an asymptotic formula for the number of almost primes $n=p_1p_2 \le X$ such that $n+h$ has exactly two prime factors which holds for a very short average over $h$.

format: plain


text:
See on GitHub

Onefetch is a command-line Git information tool written in Rust that displays project information and code statistics for a local Git repository directly on your terminal. The tool is completely offline - no network access is required.

By default, the repo’s information is displayed alongside the dominant language’s logo, but you can further configure onefetch to instead use an image - on supported terminals -, a text input or nothing at all.

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As of now, onefetch supports more than 50 different programming languages; if your language of choice isn’t supported: Open up an issue and support will be added.


text:
# well-founded relation

A binary relation $R$ on a class (http://planetmath.org/Class) $X$ is well-founded if and only if

• each nonempty subclass of $X$ contains an $R$-minimal element and,

• for each $x\in X$, $\{y\mid y\,R\,x\}$ is a set.

The notion of a well-founded relation is a generalization of that of a well-ordering relation: proof by induction and definition by recursion may be carried out over well-founded relations.

Title well-founded relation WellfoundedRelation 2013-03-22 17:24:31 2013-03-22 17:24:31 ratboy (4018) ratboy (4018) 9 ratboy (4018) Definition msc 03E20 Relation RMinimalElement well-founded


text:
0
Research Papers

# A Haptic Simulator for Training the Application of Range of Motion Exercise to Premature Infants

[+] Author and Article Information

Department of Biomedical Engineering, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975kareem.adnan@gmail.com

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440iahmad@uci.edu

Maria Coussens

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440macousse@uci.edu

Alon Eliakim

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440aeliaki2@uci.edu

Susan Gallitto

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440sagallit@uci.edu

Donna Grochow

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440dmgrocho@uci.edu

Robin Koeppel

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440rkoeppel@uci.edu

Dan Nemet

Child Health and Sports Center, Pediatrics Meir Medical Center, 44281, Israeldnemet@uci.edu

Julia Rich

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440jkrich@uci.edu

Feizal Waffarn

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440fwaffarn@uci.edu

Dan M. Cooper

Department of Pediatrics, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975 91440dcooper@uci.edu

David J. Reinkensmeyer

Department of Biomedical Engineering, Department of Mechanical and Aerospace Engineering, University of California at Irvine, 4200 Engineering Gateway, Irvine, CA 92697-3975;dreinken@uci.edu

J. Med. Devices 3(4), 041008 (Dec 09, 2009) (7 pages) doi:10.1115/1.4000430 History: Received September 02, 2008; Revised October 06, 2009; Published December 09, 2009; Online December 09, 2009

## Abstract

The range of motion exercise is an experimental therapy for improving bone and muscle growth in premature infants but little is known about the magnitude of pressures that must be applied to the limbs during this exercise to elicit a physiological benefit and novice caregivers currently must rely on subjective instruction to learn to apply appropriate pressures. The goal of this study was to quantify the pressures applied by experienced caregivers during application of this exercise and to create a haptic simulator that could be used to train novice caregivers such as parents to apply the same pressures. We quantified the pressure applied by two neonatal intensive care nurses (“experts”) to the wrists of nine newborn, premature infants of varying gestational ages using an infant blood pressure cuff modified to act as a finger pressure sensor. The experts applied statistically significant different pressures depending on gestational age but did not differ significantly between themselves in the pressure they applied. We then created a robotic simulator of the premature infant wrist and programmed it to respond with the measured pressure-angle properties of the actual infants’ wrists. The novice adult participants $(n=19)$ used the simulator to learn to apply target pressures for simulated wrists that corresponded to three different gestational ages. Training with the simulator for 30 min allowed the participants to learn to apply pressures significantly more like those of the experts. The performance improvement persisted at a retention test several days later. These results quantify for the first time the pressures applied during assisted exercise, include novel observations about joint flexibility and maturation early in life and suggest a strategy for teaching exercise intervention teams to provide assisted exercise within a more reproducible range using haptic simulation technology.

<>

## Figures

Figure 1

Left: Infant blood pressure cuffs worn on the thumb and index finger of the clinician. Right: Experiment set-up—the nurse ranged the infant’s wrist as we recorded pressure readings and wrist movement with a video camera.

Figure 2

Example pressure-wrist angle curve for one infant. Individual data points from each video capture frame are represented as dots, for five repetitions of the exercise. Polynomial curve fit is shown by the thick line.

Figure 3

Robotic baby wrist simulator. From left to right: 3D CAD model of mechanical components of robot, CAD model of assembled components, picture of actual robot showing silicone covering hand and arm to create a more realistic feel.

Figure 4

Feedback screen for the haptic training program. Images of the babies were shown to participants to indicate which baby the robot was behaving like. During the training phase participants were given knowledge of results in the form of a light indicating if the previous trial was: too soft, too hard, or good with a yellow, red, or green circle shown, respectively. The large number is a counter that indicated which trial number the participant had just completed.

Figure 5

Pressure-wrist angle curves. The first three graphs show each group’s pressure versus angle curves for each nurse, infant, and wrist within that group, where group 1 has the youngest infants (gestational ages for group 1: 30–33 weeks, group 2: 34–37 weeks, and group 3: 38–41 weeks). The dashed line is the mean curve for all the curves in that group. The fourth graph shows only the mean curves.

Figure 6

Average maximum (flexion) and minimum (extension) stretching pressure applied to the infant wrist, as a function of infant group (i.e., infant age). Nurse 1 is denoted with an x and nurse 2 with an o. Pressure depended significantly on infant age (ANOVA, linear contrast, p<0.02). The bars show 1 standard deviation across infants (three in each group) and wrists (left, right).

Figure 7

Left: mean pressure matching error. The mean pressure error across participants during the baseline, short-term retention test, and long-term retention test with bars used to denote the standard deviation of the force error across participants. Right: mean of the absolute pressure matching error. Averages of absolute value of error and with error bars to indicate one standard deviation. Participants showed significant improvement as a result of training.

Figure 8

Variability of pressure matching error. Standard deviation of error averaged across all participants and all babies. Error bars indicate one standard deviation of the individual standard deviations. Significant reduction (p=0.01) of the average standard deviation from baseline to retention test indicates that participants became more consistent at applying the desired pressure with training.

Figure 9

Actual pressure versus target pressure. These plots show each participant’s average pressure applied to each infant (open circles) and the mean of all participants’ average pressure for each infant (filled circles near regression lines) versus the target pressure for each infant.

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text:
## MTALHAHASSAN2 one year ago can how we find the derivative of the following function 3sqrt x -1+x/x^4

1. anonymous

ill help but you also have to try.

2. MTALHAHASSAN2

ok sure

3. anonymous

4. anonymous

Now I need a medal

5. MTALHAHASSAN2

@Redneckk__Lifee but how you get that

6. anonymous

online duhh. and why did you ask my name you already know it.

7. anonymous

I'm pretty sure I'm waiting for a medal!!!!!!

8. marigirl

Okay so we are differentiating $\sqrt[3]{(x-1)}+ \frac{ x }{ x^4 }$ Is that the equation? I want to make sure its correct before we begin

9. marigirl

@MTALHAHASSAN2

10. MTALHAHASSAN2

@marigirl nope its wrong

11. marigirl

I meant- did i type your QUESTION correctly

12. MTALHAHASSAN2

|dw:1443775238584:dw|

13. FireKat97

so its $f(x) = 3\sqrt{x} - \frac{ 1 + x }{ x^4 }$ ???

14. MTALHAHASSAN2

nope

15. FireKat97

no, I mean is that the question?

16. FireKat97

@MTALHAHASSAN2

17. MTALHAHASSAN2

yes it is

18. MTALHAHASSAN2

ok wait a sec

19. FireKat97

okay, so what you want to start off doing, is bringing the $x^4$ to the top, can you show me how you may do this?

20. MTALHAHASSAN2

@FireKat97 wait a sec

21. FireKat97

okay just tag me back whenever you're ready :)

22. MTALHAHASSAN2

|dw:1443777396971:dw|

23. MTALHAHASSAN2

here is the question

24. MTALHAHASSAN2

ok so we need to find the derivative of this

25. FireKat97

so its just like the one I'd typed up above?

26. MTALHAHASSAN2

oh yeah know i see

27. FireKat97

so firstly you want to start off by bringing the x^4 up to the top, can you do this and show me what you get?

28. MTALHAHASSAN2

29. FireKat97

nah, no problem :)

30. MTALHAHASSAN2

|dw:1443777684653:dw|

31. anonymous

haha

32. MTALHAHASSAN2

wait but why do we move 4

33. FireKat97

close, so I see you have brought up and distributed the x^4 and differentiated the 3x^1/2 and - x^-4, but you forgot about the - x^-3, do you see what i mean by this?

34. FireKat97

I find it easier to move it to the top, distribute it and then differentiate each of the terms

35. FireKat97

I'll show you what I mean

36. MTALHAHASSAN2

yeah plz i am confuse

37. FireKat97

$f(x) = 3\sqrt{x} - \frac{ 1 + x }{ x^4 }$ = $3\sqrt{x} - x ^{-4} (1 + x)$ do you see what I did here?

38. MTALHAHASSAN2

yeah know that's make a lot of scence

39. FireKat97

I simply brought the x^-4 into the numerator position and am now going to distribute it

40. FireKat97

yup so now we can distribute it like so $3\sqrt{x} - x^{-4} - x^{-3}$ and now we can differentiate each of the terms, have a go

41. FireKat97

do you see how I got that?

42. MTALHAHASSAN2

wait how you get -x^3

43. MTALHAHASSAN2

oh i got u

44. FireKat97

okay so when we open up $-x^{-4} (1 + x)$ we do $-x^{-4} (1) -x^{-4} (x^1)$ and according to index laws, when we multiply something with the same base, we add th powers, so for e.g. $(a^2).(a^3) = a^5$ so in the same way, $(-x^{-4}).(x^1) = -x^{-4 + 1} = -x^{-3}$

45. FireKat97

and now we have three separate terms, which we can differentiate

46. FireKat97

@MTALHAHASSAN2 did you try differentiating?

47. MTALHAHASSAN2

oh yeah i am trying it

48. FireKat97

oh okay :)

49. MTALHAHASSAN2

so are we left up with 3sqrt -x^-4 -x^-3

50. FireKat97

yup

51. MTALHAHASSAN2

and know we have to differentiating?

52. MTALHAHASSAN2

right

53. FireKat97

yeah we differentiate each of the terms

54. MTALHAHASSAN2

so for the sqrt is it be 1/2sqrt x

55. FireKat97

close

56. MTALHAHASSAN2

|dw:1443779376906:dw|

57. FireKat97

yup thats right!

58. MTALHAHASSAN2

59. FireKat97

the $-x^{-4} - x^{-3}$??

60. MTALHAHASSAN2

is it goona go in the denominater

61. FireKat97

we haven't differentiated those two terms yet..

62. FireKat97

not quite...

63. FireKat97

do you want me to show you?

64. MTALHAHASSAN2

yes plz

65. FireKat97

okay so you know how we have $f(x) = 3x^{1/2} -x^{-4} -x^{-3}$ you did well in correctly being able to differentiate the first term, but we still have two terms left to differentiate so we should get $f'(x) = \frac{ 3 }{ 2\sqrt{x} } -(-4)x^{-4-1} -(-3)x^{-3-1}$ which can further simplify down to $f'(x) = \frac{ 3 }{ 2\sqrt{x} } + 4x^{-5} + 3x^{-4}$ does that make sense?


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# 2789. Hanoi Towers game

In this problem we consider a modification of the classical Hanoi Towers game. The rules of our modification are as follows:

There are three pegs: peg A, peg B and peg C.

Each peg initially contains zero or more discs.

All discs have the same size and there are three types of discs: type A, type B and type C.

In one move you can move a disc from the top of one peg onto the top of another peg.

Your goal is to have all discs of type A on peg A, all discs of type B on peg B and all discs of type C on peg C.

You must achieve your goal in the minimum possible number of moves.

The initial locations of the discs are given in the strings pegA, pegB and pegC. Each character of pegA represents a disc located on peg A, where ‘A’ represents a disc of type A, ‘B’ represents a disc of type B and ‘C’ represents a disc of type C. The discs are given in order, from bottom to top. pegB and pegC describe the discs on pegs B and C, respectively, in the same format.

Print the minimum number of moves required to achieve the goal of the game.

note:

We guarantee it’s always possible to achieve the goal of the game.

pegA, pegB and pegC will contain only the characters ‘A’, ‘B’ and ‘C’.

The total number of characters in pegA, pegB and pegC will be between 1 and 10, inclusive.

### 输入格式

Only one test case.

It consists of three integers a ,b, c which represent number of characters in pegA, pegB and pegC, and 3 lines follow.

### 输出格式

Print the minimum number of moves required to achieve the goal of the game.

### 样例

Input
1 1 1
B
C
A

Output
5

### 提示

There is exactly one disc of each type in this example, so we will refer to each disc by its type. The following sequence of moves is the shortest:
1. Move disc A to peg A.
2. Move disc C to peg C.
3. Move disc A to peg C.
4. Move disc B to peg B.
5. Move disc A to peg A.

7 人解决，11 人已尝试。

7 份提交通过，共有 12 份提交。

6.4 EMB 奖励。


text:
# compare quasi-Poisson models with different response variables

My response variable is a combining of two variables which I would like to rescale with different weights. ratio:0.5/0.5; 0.25/0.75; 0.1/0.9. Now I would like to test which is the best fit. The three models have overdispersion so I used quasipoisson regression.

Following former instructions in this matter, I used the F test, but it didn't work because of the different response variable. I got this massage:

    anova(glm0.5_0.5,glm0.75_0.25,glm0.9_0.1, test = "F")

Response: data$dep0.5_0.5 Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev F Pr(>F) NULL 575 11386.0 academic95 1 3720.1 574 7666.0 231.505 < 2.2e-16 *** israeli_ac95 1 1239.1 573 6426.8 77.113 < 2.2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Warning message: In anova.glmlist(c(list(object), dotargs), dispersion = dispersion, : models with response ‘c("data$dep0.75_0.25", "data\$dep0.9_0.1")’ removed because response differs from model 1


text:
# Calculating binding energies for excitated levels in potassium?

Suppose a laser is used to excite electronic transitions in potassium from the $3p$ orbitals to ones higher than $4s$. When I attempt to use the Bohr model to calculate binding energies, what I obtain for $n\le4$ disagrees significantly with this source, calling into doubt my results for $n>4$. I think it might be possible to use the Rydberg formula, but I'm having trouble finding a working example that computes the right energies; in addition, there is another form, and the two seem inconsistent. How does one calculate the binding and transition energies in potassium for levels $n>4$? Are the quantum defects for arbitrary elements available online?

What changes here if we're talking not about pure potassium but potassium chloride?

• Perhaps relevant: chemistry.stackexchange.com/questions/32350/… Oct 25, 2015 at 11:41
• What do you expect by using the Victorian technology (Bohr model)? :) More than a century ago Bohr himself attempted to use his model to calculate energy levels in many-electron atoms and quickly found that it gives unacceptable numbers already for the lightest of then, $\ce{He}$ atom. For the next couple of decades after this discovery a scientific thought in that area had developed revolutionary and a good physical theory that gives way better predictions for atomic energy levels (amongst many other things) was formulated. Oct 26, 2015 at 21:41
• Fair enough. :) But my problem is a practical one, and it would be nice to have realistic transition energies, so that I will know what photon wavelengths are needed to excite the transitions. Any help is appreciated. Oct 26, 2015 at 22:26


text:
The open source CFD toolbox
Boundary conditions
Collaboration diagram for Boundary conditions:

Wall functions

## Classes

class  fixedShearStressFvPatchVectorField
This boundary condition sets a user-defined shear stress constant and uniform across a given patch by using the expression: More...

## Detailed Description

This group contains turbulence model boundary conditions


text:
# recserar¶

## Purpose¶

Computes a vector of autoregressive recursive series.

## Format¶

y = recserar(x, y0, rho)
Parameters
• x (NxK matrix) – If simulating an AR process, this would contain the error term and constant if included in the model.

• y0 (PxK matrix) – The starting values for the series.

• rho (PxK matrix) – The AR parameters.

Returns

y (NxK matrix) – Contains the series.

## Examples¶

### AR(1) without constant¶

// Starting value for the time series
y0 = 0;

// AR(1) parameter
rho = 0.6;

// Innovations
eps = rndn(10, 1);

// Simulate AR(1) model
y = recserar(eps, y0, rho);

### AR(2) with constant¶

// Starting value for the time series
y0 = { 0, 0 };

// AR(2) parameters
rho = { 0.6, -0.3 };

// Constant term
const = 1.3;

// Innovations
eps = rndn(10, 1);

// Simulate AR(2) model with constant
y = recserar(eps + const, y0, rho);

### Example 3¶

n = 10;

sig = { 1 -.3, -.3 1 };
mu = { 0, 0 };
e = rndMVn(n, mu, sig);

x = ones(n, 1)~rndn(n, 3);

b = 1|2|3|4;

rho = { 0.5, 0.3 };
y0 = zeros(1, 2);
y = recserar(x*b+e, y0, rho);

In this example, two autoregressive series are formed using simulated data. The general form of the series can be written:

y[1,t] = rho[1,1]*y[1,t-1] + x[t,.]*b + e[1,t];
y[2,t] = rho[2,1]*y[2,t-1] + x[t,.]*b + e[2,t];

The error terms ($$e[1,t]$$ and $$e[2,t]$$) are not individually serially correlated, but they are contemporaneously correlated with each other. The variance-covariance matrix is $$\sigma$$.

## Remarks¶

recserar() is particularly useful in dealing with time series.

Typically, the result would be thought of as $$K$$ vectors of length $$N$$.

y0 contains the first $$P$$ values of each of these vectors (thus, these are prespecified). The remaining elements are constructed by computing a $$P^{th}$$ order “autoregressive” recursion, with weights given by a, and then by adding the result to the corresponding elements of x. That is, the $$t^{th}$$ row of y is given by:

y[t,.] = x[t,.] + a[1,.] * y[t-1,.] +...+ a[P,.] * y[t-p,.], t = P + 1,...N

and

y[t,.] = y0[t,.], t = 1,...,P

Note that the first $$P$$ rows of x are not used.


text:
# SPIM Registration

TODO component: {"content"=>"sc.fiji:SPIM\\_Registration"}

## Important Note

 ”””"”Please Note: This version of the software is outdated. It will be part of Fiji for the time being, but I highly recommend using the new Multiview Reconstruction Plugin. It is much more powerful, flexible and completely integrated with the BigDataViewer.”””””

## Citation

Please note that the SPIM registration plugin available through Fiji, is based on a publication. If you use it successfully for your research please be so kind to cite our work:

• S. Preibisch, S. Saalfeld, J. Schindelin and P. Tomancak (2010) “Software for bead-based registration of selective plane illumination microscopy data”, Nature Methods, 7(6):418-419. Webpage PDF Supplement
• S. Preibisch, F. Amat, E. Stamataki, M. Sarov, R.H. Singer, E. Myers and P. Tomancak (2014) “Efficient Bayesian-based Multiview Deconvolution”, Nature Methods, 11(6):645-648. Webpage

## Introduction & Overview

Selective Plane Illumination Microscopy (SPIM, Science, 305(5686):1007-9) allows in toto imaging of large specimens by acquiring image stacks from multiple angles. However, to realize the full potential of these acquisitions the data needs to be reconstructed:

We developed several algorithms for the registration and fusion of multi-angle SPIM acquisitions. This plugin collection allows you to reconstruct an isotropic output image from several input images, called views. This process can be applied to timelapse acquisitions as well. The complete reconstruction process is split into two parts, where you have several choices for each step:

• Multi-view registration: Aligns all views of the dataset

* Bead-based registration: Uses flourescent beads to achieve the alignment. This method is very fast and solves timeseries registration, but relies on the incorporation of fluorescent beads. The plugin can be found in Plugins->SPIM Registration->Bead-based registration.

• Segmentation-based registration: Uses structures within the sample like nuclei to achieve registration. It is slower and timelapse registration might be a sample dependent problem, but does not require any special sample prepration.
• Multi-view fusion: Computes the output image(s) based on the registration results

* Multi-view fusion: Performs a weighted average fusion of the dataset, which includes blending and/or content-based weightening. The plugin can be found in Plugins->SPIM Registration->Multi-view fusion.

• Multi-view deconvolution: Performs a multi-view deconvolution of the dataset which requires an estimation of the point spread function.

Please note: the SPIM registration has been rewritten and now replaces the “old” plugin collection (Registration, Advanced Registration, MultiChannel Registration) which has been moved to the deprecated folder. For information regarding these outdated plugins please refer to this page.

## Detailed tutorials

SPIM registration is particularly useful for processing of data acquired on OpenSPIM. Detailed step-by-step tutorials are available on the OpenSPIM wiki.

The pipeline is relatively linear as described above, we highlight the steps where alternative routes exist.

## How do I view fused, saved output image(s)

The fused images will be saved in the output directory of the dataset as a collection of two-dimensional planes. The created output images can be viewed for example with Plugins->Image5D->Virtual Image 5D Opener or via the Plugins->LOCI->Bio-Formats Importer (check the option Group files with similar names).


text:
19

# Questions

9
26
4
1k
views

### Is there a simple proof that a group of linear growth is quasi-isometric to Z?

may 5 at 19:58 balan-antoine 1

2
12
3
725
views

### How to shuffle a deck by parts?

aug 19 10 at 10:34 Benoît Kloeckner 5,4891829

3
11
1
530
views

### Reference for a particular Radon transform on non-positively curved spaces

jun 30 11 at 20:22 MathOverflow 123

4
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1
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views

sep 11 10 at 5:51 Agol 23.9k13384

7
1
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views

### Smoothing of piecewise Euclidean Riemannian metrics

feb 14 at 0:22 MathOverflow 123

1
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3
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views

### What do singular, atomless invariant measures of $\times d$ look like?

nov 22 10 at 23:12 Pablo Shmerkin 1,163512

3
5
3
390
views

### Quantitative measurement of infinite dimensionality

oct 15 11 at 17:41 Gerald Edgar 15.1k12867

5
5
2
259
views

### Solution uniqueness for ODE

mar 14 12 at 11:39 Pietro Majer 19.7k12469

1
5
0
185
views

### Spectrum of an operator arising in a dynamical problem

oct 21 10 at 16:11 Benoît Kloeckner 5,4891829

1
4


text:
# svn diff without checkout

Is it possible to find out a recursive diff between two tags of a project in remote repository?

In other words if i know the location of 2 tags are https://myrepo.com/project/tags/v1 and https://myrepo.com/project/tags/v2 respectively, can i find the recursive diff between the 2 without having the code checked out on my machine?

-

svn diff works with URLs, too. If you have the SVN command line tools already installed, all you have to do is to run:

svn diff https://myrepo.com/project/tags/v1 https://myrepo.com/project/tags/v2 > v1tov2.diff

After this v1tov2.diff will contain the unified diff between the two tags.

-
Ok I did this but the output has many details that I want to avoid. Can I just have file names from the output of this command? –  Mustafa Apr 11 '12 at 4:20
When i do a svn diff --diff-cmd -x "-q" myrepo.com/project/tags/v1 myrepo.com/project/tags/v2 I get svn: Can't set process 'diff' child outfile: The parameter is incorrect. Any idea what's wrong here? –  Mustafa Apr 11 '12 at 4:53
You may pass '--summarize' as an additional argument to svn diff. Then you get only the filenames and a code indication what happened with the file (added, modified, deleted). '--diff-cmd' is used for an entirely different purpose: this allows you to specify a external program which SVN will use to calculate the difference between two changed files. So don't pass the flag. –  sebasgo Apr 11 '12 at 16:23
I tried svn diff --summarize and the output asks me to look at svn help which has no summarize option :( –  Mustafa Apr 11 '12 at 18:31
This works only for URLs, not paths. –  sebasgo Apr 11 '12 at 18:33

With the command line tool:

svn diff https://myrepo.com/project/tags/v1 https://myrepo.com/project/tags/v2

-


text:
# SEARCH CONTENT

## You are looking at 1 - 10 of 10,440 items :

• General Linguistics
Clear All
Essays on Semantic and Pragmatic Topics
OPEN ACCESS

## Abstract

Sound change in the form of plosive mergers has been reported for a variety of languages and is the result of a reduction of phonetic distance between two (or more) sounds. The present study is concerned with the opposite development of phonetic differentiation in plosives (akin to a phonetic split), a less commonly reported phenomenon that is taking place in Austrian German at the moment. A previously small (or null) phonetic distinction between fortis and lenis plosives – a presumed near-merger – is gradually developing into a clear phonetic contrast in younger speakers. In the present study, voice onset time of word-initial plosives was measured in two generations of Austrian speakers (born in the middle and at the end of the 20th century), yielding an ongoing phonetic differentiation where the voice onset time of lenis consonants is shortened while, at the same time, that of fortis consonants is lengthened. These results present an insight into the recent diachronic development of Austrian German and the changes in plosive production that are currently taking place.

FREE ACCESS
OPEN ACCESS

## Abstract

In the literature on relative clauses (e. g. : 4), it is occasionally observed that the German complex definite determiner d-jenige (roughly ‘the one’) must share company with a restrictive relative clause, in contrast to bare determiners der/die/das ( : 213–215; ; ). Previous works such as (: 378–379) and () treat the relative clause as a complement of D to account for its mandatory occurrence. While such syntactic analyses have intuitive appeal, they pose problems for a compositional semantic analysis.

The goal of this paper is twofold. First, we report on two rating studies providing empirical evidence for the obligatoriness of relative clauses in German DPs introduced by the complex determiner d-jenige. Secondly, following (, ), we provide an analysis of the phenomenon at the syntax-semantics interface that captures familiar ( ) as well as novel related observations. Particularly, the analysis accounts for the facts that postnominal modifiers can figure in d-jenige-DPs and that the element can have anaphoric demonstrative pronominal uses.

## Abstract

There is a use of the German third person neuter pronoun $es$ in the prefield, known as prefield-$es$, which is characterized by being neither referential, nor an argument of the verb. According to Speyer’s (, ) optimality theoretic prefield ranking, this should only occur if a sentence contains no alternative element eligible to be moved to the prefield. This paper investigates a so far unnoticed use of $es$ in the prefield in combination with a demonstrative pronoun dies and a copula verb ist, which will be referred to as Es ist dies-sentence. This construction is an instance of prefield-$es$, but contravenes the expectations about the use of prefield-$es$ postulated by Speyer, since Es ist dies-sentences do contain a suitable candidate to fill the prefield, the demonstrative pronoun dies. In a corpus study, Es ist dies-sentences are compared to a sample of Dies ist-sentences. According to the corpus data, Es ist dies occurs predominantly in southern dialects. Significant differences between the two samples concern 1) the distance to the antecedent of dies and 2) the type of content of the sentence. An online rating study, that compared acceptability judgments of Es ist dies-sentences between speakers from different regions, confirmed that Es ist dies-sentences are a phenomenon of southern dialects. In the light of these results, a modification of Speyer’s (, ) Stochastic OT model is proposed.


text:
## SI unit prefixes

Even with scientific notation, very large numbers and very small numbers can be awkward to work with. That is why scientists have developed a set of standard prefixes that are attached to a unit to indicate multiples of that unit. Each of the prefixes represents a particular power of ten. There are currently 20 prefixes in use:

Prefixes greater than 1

 Factor Name symbol 1024 yotta Y 1021 zetta Z 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 101 deka da

Prefixes less than 1

 Factor Name symbol 10-1 deci d 10-2 centi c 10-3 milli m 10-6 micro μ 10-9 nano n 10-12 pico p 10-15 femto f 10-18 atto a 10-21 zepto z 10-24 yocto y

Do you have to memorize all of these? No. If you are a high school chemistry or physics student, you should know the prefixes in blue above. If you are an AP chemistry or AP physics student, you should also learn the prefixes in red. If you plan to be a scientist or an engineer, it wouldn’t hurt you to learn the rest, but the extreme prefixes are not so common.

Why do the prefixes skip three powers of 10? This allows us to use numbers that are always between 1 and 999. This is easiest to explain with an example. Let’s say we are measuring very small electrical currents. The following table shows how we would state the values without scientific notation, with scientific notation, and with prefixes. You can see how convenient this prefix method is:

 Current Scientific notation Prefix notation 0.50 A 5.0 x 10-1 A 500 mA 0.050 A 5.0 x 10-2 A 50 mA 0.0050 A 5.0 x 10-3 A 5 mA 0.00050 A 5.0 x 10-4 A 500 μA 0.000050 A 5.0 x 10-5 A 50 μA 0.0000050 A 5.0 x 10-6 A 5 μA 0.00000050 A 5.0 x 10-7 A 500 nA 0.000000050 A 5.0 x 10-8 A 50 nA 0.0000000050 A 5.0 x 10-9 A 5 nA 0.00000000050 A 5.0 x 10-10 A 500 pA

## How to use ICE charts

When you are given the initial concentrations of the components in a reaction and you want to know the final concentrations at equilibrium, an ICE chart can be very useful. The ICE chart (some teachers call it an OCE chart or an IΔE chart) is a tool that helps you set up the proper algebraic expression in order to calculate the concentration change of each compound. It’s based on a very simple concept: the initial concentration of a compound plus the change in concentration of that compound must equal its final (equilibrium) concentration. To set up an ICE chart, follow these easy rules:

1. Write out the balanced equilibrium equation.
2. On the first line (I), put down the initial concentrations of each compound. If a compound is not in the initial mix, write 0. Be sure that you calculate the molarity, and not just the number of moles.
3. Pick a direction for the reaction to proceed. It’s usually pretty obvious, but if you aren’t sure which way the reaction will go, just choose one direction. It doesn’t matter if you guess wrong.
4. On the second line (C), use a variable to write the change in concentration for each compound using the balanced equation as a guide. For each compound that is consumed, write –Ax, where A is the coefficient of the compound in the balanced reaction. For each compound that is being formed, write +Ax. See the example below.
5. On the third line (E), write the sum of the first two lines. This must be the final concentration of each compound.
6. Use the equilibrium law and the value of Kc to write an expression with the equilibrium concentrations. Then solve this equation for x.
7. Use this value to determine the equilibrium concentrations of each compound. [Note: if the value of x is negative, this means you guessed wrong in step 3, and the reaction actually proceeds in the opposite direction. You will still get the correct final values of each compound if you add or subtract this negative value in this step.]

Example 1: 1.0 mole of H2 and 1.0 mole of I2 are placed in a 2.0 L flask. The following reaction takes place at 400°C: H2 (g) + I2 (g) → 2 HI (g). What are the concentrations of the three compounds at equilibrium? The Kc for the formation of hydrogen iodide (HI) from hydrogen and iodine gases is 64.1 at 400°C.

Solution: Start by writing the equation and the initial concentrations. Don’t forget to divide by the volume to find the molarities of all the compounds.

Next, put in the changes in concentration in terms of x. This reaction must go to the right, and we get two moles of HI for each mole of the reactants:

Then add the two columns to calculate to equilibrium concentrations:

Finally, write the equilibrium expression and solve:

$K_C=64.1= \dfrac{(2x)^2}{(0.5-x)^2}$

$\sqrt{64.1} = \sqrt{\dfrac{(2x)^2}{(0.5-x)^2}}$

$8.0= \dfrac{2x}{0.5-x}$

$4-8x=2x$

$4=10x$

$x=0.4$

$\therefore [ \text{H}_2]=[ \text{I}_2]=0.1 \, M; \, [ \text{HI}]=0.80 \, M$

Note that if you are given the change in concentration or the initial and final concentrations of a compound, you do not need to use an ICE chart. You can calculate the equilibrium values with a little bit of arithmetic. See the following example:

Example 2: The reaction in Example 1 is repeated at a different temperature with the same initial concentrations, and the final concentration of HI is determined to be 0.60 M. Determine the value of Kc at this new temperature.

Solution: To solve for Kc, we need to know the final concentrations of all three components. We are given the final value of HI, and knowing that we produced 0.60 M HI means that we must have lost 0.30 M of each of the reactants (use the balanced reaction to determine this). Since we started with 0.50 M of each, the final concentrations of each reactant must be 0.20 M (just subtract the numbers). So the equilibrium expression gives us:

$K_C= \dfrac{(0.6)^2}{(0.2)(0.2)}=9.0$

## Balancing redox reactions in basic solutions

When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in a basic solution.

Example : Iodide ions are oxidized in a basic solution by permanganate ions to produce iodine and Mn2+ ions. Show the balanced reaction.

Solution: I’ve made this example a little harder than in my other redox reaction posts because we first need to figure out from the example what the half reactions look like. This is not as complicated as it sounds. Just follow the English and turn it into chemical formulas. Our problem should therefore look like this:

I + MnO4 → I2 + Mn2+

Now comes the tricky part. It turns out it isn’t very easy to balance a redox reaction in basic solutions by adding OH ions, because when we balance the hydrogen atoms, we throw off the balance of the oxygen atoms. So we employ a clever technique. We first balance the reaction as though it were an acidic solution, then we change the solution from acidic to basic. This is a five-step process, and you’ll notice the first four steps are identical to the process for balancing a redox reaction in an acidic solution:

1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
2. Balance the oxygen atoms by adding water molecules.
3. Balance the hydrogen atoms by adding H+ ions.
4. Check you answer by verifying that the charges are balanced.
5. Cancel out the H+ ions by adding an equal number of OH ions to both sides, which changes the H+ ions into water.

Let’s see how this works on our example. The iodine atom goes from an oxidation number of 1- on the left to 0 on the right, so each iodine atom loses one electron. Since there are two iodine atoms in I2, we think of this half-reaction as losing two electrons. The manganese goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. To balance the electrons, we need a total transfer of ten electrons. Multiply the iodine compounds by five and the manganese compounds by two. (Note that the iodide ion was multiplied by 10 rather than 5 to balance the number of iodine atoms):

10 I + 2 MnO4 → 5 I2 + 2 Mn2+

Now we balance the oxygen atoms. There are 8 oxygen atoms on the left and none on the right, so we add 8 water molecules to the right side:

10 I + 2 MnO4 → 5 I2 + 2 Mn2+ + 8 H2O

Next we balance the hydrogen atoms by adding 16 H+ ions to the left (we pretend we have an acidic solution):

10 I + 2 MnO4 + 16 H+ → 5 I2 + 2 Mn2+ + 8 H2O

The equation should be fully balanced, and we check our work by verifying the charges balance. 10- (the iodide ions), 2- (the permanganate ions) and 16+ (the hydrogen ions) gives a net charge on the left of 4+. On the right side, we also have a total charge of 4+.

Finally, we are ready to change the reaction from acidic to basic. We cancel out the 16 H+ ions by adding 16 OH ions to both sides:

10 I + 2 MnO4 + 16 H+ + 16 OH → 5 I2 + 2 Mn2+ + 8 H2O + 16 OH

Because H+ and OH ions react to form water molecules, we simplify the left side:

10 I + 2 MnO4 + 16 H2→ 5 I2 + 2 Mn2+ + 8 H2O + 16 OH

Our last step is to simplify the equation by subtracting the 8 extra water molecules from the right side. We have to subtract the same number from the left side to keep the reaction balanced:

10 I + 2 MnO4 + 8 H2O → 5 I2 + 2 Mn2+ + 16 OH

(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in acidic solutions.)

## Balancing redox reactions in acidic solutions

When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in an acidic solution.

Example 1: Balance the following reaction:

MnO4 + NO + H+ → Mn2+ + H2O + NO3

Solution: To balance reactions like this, use the following four-step procedure:

1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
2. Balance the oxygen atoms by adding water molecules.
3. Balance the hydrogen atoms by adding H+ ions.
4. Check you answer by verifying that the charges are balanced.

In this reaction, the Mn atom goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. The nitrogen atom goes from an oxidation number of 2+ in the nitric oxide to 5+ in the nitrate ion. It has lost three electrons. To balance the electrons, we need to make the total number of electrons transferred equal 15 (because that’s the least common multiple of three and five!). We multiply the Mn ions by 3 and the N compounds by 5:

3 MnO4 + 5 NO + H+ → 3 Mn2+ + H2O + 5 NO3

The next step is to balance the oxygen atoms. There are 17 oxygen atoms on the left side of the reaction and 15 on the right side (before you include the water). Therefore we need two water molecules on the right:

3 MnO4 + 5 NO + H+ → 3 Mn2+ + 2 H2O + 5 NO3

Finally, we balance the hydrogen atoms (there are four of them in the water molecule on the right) by adding 4 hydrogen ions on the left:

3 MnO4 + 5 NO + 4 H+ → 3 Mn2+ + 2 H2O + 5 NO3

We check our work by verifying the charges balance. 3- (the permanganate ions) and 4+ (the hydrogen ions) gives a net charge on the left of 1+. 6+ (the manganese ions) and 5- (the nitrate ions) on the right also gives a net charge of 1+. We have succeeded!

Example 2: Complete and balance the following reaction. Assume it takes place in an acidic solution:

Cr2O72 + HSO3 → SO42 + Cr3+

Solution: In this reaction, the Cr atoms go from an oxidation number of 6+ in the dichromate ion to an oxidation number of 3+ in the chromate ion. Therefore, each chromium atom gains three electrons. But since there are two chromium atoms in the dichromate ion, we must think of this half-reaction as gaining six electrons overall. The sulfur atom goes from an oxidation number of 4+ to 6+, so it loses two electrons.

In order to balance the electrons, we need a net transfer of six electrons in each half-reaction. We multiply the ions with the sulfur atoms by three. Note that we also multiply the chromium ion on the right by two to balance the number of Cr atoms:

Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+

The next step is to balance the oxygen atoms.  There are 16 oxygen atoms on the left and 12 on the right, so we add 4 water molecules to the right side:

Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+ + 4 H2O

Finally, we balance the hydrogen atoms by adding H+ ions. There are three H atoms on the left and 8 H atoms on the right, so we need five H+ ions on the left:

5 H+ + Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+ + 4 H2O

We check our answer by checking the charge on each side. The left side has charges of 5+ (the H ions), 2- (the dichromate ion) and 3- (the hydrogen sulfite ions) for a net of zero. The right side has charges of 6- (the sulfate ions) and 6+ (the chromate ions) also for a net of zero.

(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in basic solutions.)

## Balancing redox reactions– the basics

When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how the process works when both the oxidizing agent and the reducing agent are single elements.

Example (a very simple example): Balance the following reaction:

… Mg(s) + … Al3+(aq) → … Mg2+(aq) + … Al(s)

Solution: At first glance, it may appear that the reaction is already balanced because there is one magnesium atom and one aluminum atom on each side. But notice that the charges do not balance. Because aluminum gains three electrons in this reaction, but magnesium atoms lose only two electrons, you will need more magnesiums than aluminums. To find the right ratio, determine the least common multiple of 2 and 3, which is 6. Then multiply the magnesiums by 3 and the aluminums by 2 so that there is a total transfer of 6 electrons in this reaction:

3 Mg(s) + 2 Al3+(aq) → 3 Mg2+(aq) + 2 Al(s)

The charges balance, as required.

(Look for other tips on this website if you need help with balancing redox reactions in acidic or basic solutions.)

## Balancing simple chemical reactions

When balancing equations, start with the atoms that show up in only one compound on each side. Figure out those coefficients first, and then work back and forth from the reactants side to the products side to find the coefficients of each of the other compounds.

Example 1: Balance the following equation:

… C5H12 + … O2 → … CO2 + … H2O

Solution: This reaction showing the combustion of pentane is not too complicated. Notice that carbon appears in only one compound on each side, so it’s a good place to start. The five carbon atoms on the left need to be balanced by five carbon atoms on the right:

1 C5H12 + … O2 → 5 CO2 + … H2O

There are 12 hydrogen atoms on the left, so balance these on the right:

1 C5H12 + … O2 5 CO2 + 6 H2O

Finally, count up all the oxygen atoms on the right (there are 16 of them!) and balance with the O2 on the left:

1 C5H128 O2 5 CO2 + 6 H2O

Example 2: Balance the following equation:

… C6H14 + … O2 → … CO2 + … H2O

Solution: Changing the pentane to hexane has made this equation more complicated. Let’s see why. First, let’s balance the carbon and hydrogen as before:

1 C6H14 + … O2 6 CO27 H2O

There are now 19 oxygen atoms on the right side. How do you balance this on the left side? Some teachers will let you use fractions for your coefficients, and 19/2 is the correct fraction in this case. But most teachers want you to use only whole numbers for the coefficients. If you encounter a fraction, you need to multiply the entire equation by the denominator to finish the problem. Here, we multiply the equation above by 2:

2 C6H14 + … O2 12 CO2 + 14 H2O

Notice there are now 38 oxygen atoms on the right side, and the last step is simple:

2 C6H14 + 19 O2 12 CO2 + 14 H2O

Here’s a trick you can apply when the equation is a double replacement reaction:

Example 3: Balance the following reaction:

… Na2SO4 + … Al(NO3)3 → … NaNO3 + … Al2(SO4)3

Solution: Let’s start by balancing the aluminum:

… Na2SO4 + 2 Al(NO3)3 → … NaNO3 + 1 Al2(SO4)3

Now instead of balancing the sulfur atoms, we can save time by balancing the sulfate ions instead. We treat them as a group rather than as separate atoms. There are three sulfates on the right side, so we need three on the left:

3 Na2SO4 + 2 Al(NO3)3 → … NaNO3 + 1 Al2(SO4)3

Note that you can only balance polyatomic ions like this when the ions do not change from reactant side to product side. Finally, you can see there are six sodium atoms on the left (or, six nitrate ions), so the final answer is:

3 Na2SO4 + 2 Al(NO3)3 6 NaNO3 + 1 Al2(SO4)3

## Figuring out conversion factors

A conversion factor expresses the relationship between the unit in the numerator and the unit in the denominator. The quantities are actually the same, but the numbers are different because the units are different. Any time you can express a fixed relationship between two quantities, you can make two conversion factors out of that relationship. (Flipping the original conversion factor upside down gives you the second one!)

Examples

•  2.54 cm = 1 in. so the conversion factor is:

$\dfrac {2.54 \, cm}{1 \, in} \text{ or } \dfrac{1 \, in}{2.54 \, cm}$

•   1.00 cal = 4.184 J, so the conversion factor is:

$\dfrac {1.00 \, cal}{4.184 \, J} \text{ or } \dfrac{4.184 \, J}{1.00 \, cal}$

•  1.00 mole O2 has a mass of 15.999 g, so the conversion factor is:

$\dfrac {15.999 \, g \text{ O}_2}{1.00 \, mol \text{ O}_2} \text{ or } \dfrac{1.00 \, mol \text{ O}_2}{15.999 \, g \text{ O}_2}$

• In the production of ammonia, N2 + 3 H2 → 2 NH3, two moles of ammonia are produced for every 3 mol of H2 consumed, so the conversion factor is:

$\dfrac {2 \, mol \text{ NH}_3}{3 \, mol \text{ H}_2} \text{ or } \dfrac{3 \, mol \text{ H}_2}{2 \, mol \text{ NH}_3}$

## Ideal Gas Law vs. Combined Gas Law

When you are doing gas law problems, your first decision should be whether to use the Ideal gas law or the combined gas law. How do you know which to use?

The Ideal Gas Law is used when you have one set of conditions for your system. Then if you know any three of the values for pressure, volume, moles and temperature, you can solve for the unknown quantity. The Combined Gas Law is used when you have two sets of conditions. In other words, you know the initial pressure, volume and temperature, and two of those quantities change to a new set of conditions. Then you can calculate the new value of the third quantity.

Example 1: What is the volume of a container that holds 0.561 mol of a gas at 1.15 atm at 27°C?

Answer: This problem describes a single set of conditions, so use the Ideal Gas Law:

$PV=nRT$

$(1.15 \, atm)(V)=(0.561 \, mol) \! \left ( 0.0821 \, \frac{\ell \cdot atm}{mol \cdot K} \right ) \! (300 \, K)$

$V=12.0 \, L$

Example 2: A 3.00 L container holds an ideal gas at STP. What will the new pressure be if the volume is increased to 3.75 L and the temperature is increased to 500 K?

Answer: This problem describes an initial condition and a new set of conditions, so use the combined gas law:

$\dfrac{P_1V_1}{T_1}= \dfrac{P_2V_2}{T_2}$

$\dfrac{(1.00 \, atm)(3.00 \, L)}{273 \, K}= \dfrac{(P)(3.75 \, L)}{500 \, K}$

$P=1.47 \, atm$

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