Chapter 1, Problem 26RPS

(a)

Interpretation Introduction

Interpretation: The correct number of significant figure has to be given..

Concept introduction:

Significant of zeroes:

• Zeroes between two other significant digits are significant.
• Zeroes to the right of a nonzero number, and also to the right of decimal place, are significant.
• Zeroes that are placeholders are not significant.

Rules used for significant figures in calculations:

Rule 1: When adding or subtracting numbers, the number of decimal places in the answer is equal to the number of decimal places in the number with fewest digits after the decimal.

Rule 2: In multiplication or division, the number of significant figures in the answer is determined by the quantity with the fewest significant figures.

Rule 3: When a number is rounded off, the last digit to be retained is increased by one only if the following digit is 5 or greater.

Answer to Problem 26RPS

$\text{2}\text{.44}\times {10}^8$; The number of significant figures is three.

Explanation of Solution

In $6.25\times {10}^2$, can be written as $625$ has no decimal numbers with no zeroes and has only the significant numbers.  So, the significant number is 3.

The multiplication of ${\left(625\right)}^3$:

  $\underset{\_}{\begin{array}{l}\text{×}\underset{\_}{\begin{array}{l}\text{ 625}\text{ significant}\text{figure}\text{3}\\ \text{625 }\text{significant}\text{figure}\text{3}\\ \text{625 }\text{significant}\text{figure}\text{3}\end{array}}\\ \text{=}\text{244140625 }\end{array}}$

The number of significant figures should the answer contain is the fewest number of significant figures of the given quantity.  All the three quantities possess same significant figures.  Hence, the answer should be written as $\text{2}\text{.44}\times {10}^8$ in scientific figure in order to achieve the significant figure three.

(b)

Interpretation Introduction

Interpretation: The correct number of significant figure has to be given..

Concept introduction:

Significant of zeroes:

• Zeroes between two other significant digits are significant.
• Zeroes to the right of a nonzero number, and also to the right of decimal place, are significant.
• Zeroes that are placeholders are not significant.

Rules used for significant figures in calculations:

Rule 1: When adding or subtracting numbers, the number of decimal places in the answer is equal to the number of decimal places in the number with fewest digits after the decimal.

Rule 2: In multiplication or division, the number of significant figures in the answer is determined by the quantity with the fewest significant figures.

Rule 3: When a number is rounded off, the last digit to be retained is increased by one only if the following digit is 5 or greater.

Answer to Problem 26RPS

$0.00485$; The number of significant figures is three.

Explanation of Solution

In $\sqrt{2.35\times {10}^{-3}}$, contains decimal numbers and only the significant numbers.  So, the significant number is 3.

The value of $\sqrt{2.35\times {10}^{-3}}$ is $0.004847$, in order to achieve significant figure of 3, the value can be written as $0.00485$ where only $\text{4,8,and}\text{5}$ are significant.  Hence, the significant figure is three.

(c)

Interpretation Introduction

Interpretation: The correct number of significant figure has to be given..

Concept introduction:

Significant of zeroes:

• Zeroes between two other significant digits are significant.
• Zeroes to the right of a nonzero number, and also to the right of decimal place, are significant.
• Zeroes that are placeholders are not significant.

Rules used for significant figures in calculations:

Rule 1: When adding or subtracting numbers, the number of decimal places in the answer is equal to the number of decimal places in the number with fewest digits after the decimal.

Rule 2: In multiplication or division, the number of significant figures in the answer is determined by the quantity with the fewest significant figures.

Rule 3: When a number is rounded off, the last digit to be retained is increased by one only if the following digit is 5 or greater.

Answer to Problem 26RPS

$0.133$; The number of significant figures is three.

Explanation of Solution

In ${\left(2.35\times {10}^{-3}\right)}^{\frac{1}{3}}$, contains decimal numbers and only the significant numbers.  So, the significant number is 3.

The value of ${\left(2.35\times {10}^{-3}\right)}^{\frac{1}{3}}$ is $0.132950$, in order to achieve significant figure of 3, the value can be written as $0.133$ where only $\text{1,3,and}\text{3}$ are significant.  Hence, the significant figure is three.

(d)

Interpretation Introduction

Interpretation: The correct number of significant figure has to be given..

Concept introduction:

Significant of zeroes:

• Zeroes between two other significant digits are significant.
• Zeroes to the right of a nonzero number, and also to the right of decimal place, are significant.
• Zeroes that are placeholders are not significant.

Rules used for significant figures in calculations:

Rule 1: When adding or subtracting numbers, the number of decimal places in the answer is equal to the number of decimal places in the number with fewest digits after the decimal.

Rule 2: In multiplication or division, the number of significant figures in the answer is determined by the quantity with the fewest significant figures.

Rule 3: When a number is rounded off, the last digit to be retained is increased by one only if the following digit is 5 or greater.

Answer to Problem 26RPS

$\text{0}\text{.0286 (rounding off)}$; The number of significant figures is three.

Explanation of Solution

The division and multiplication of:

  $\begin{array}{l}\text{(1}\text{.68)}\left[\frac{\text{23}\text{.56-2}\text{.3}}{\text{1}\text{.248}\times {\text{10}}^3}\right]\\ \underset{\_}{\text{Subtraction of 23}\text{.56 and 2}\text{.3:}}\\ \underset{\_}{\begin{array}{l}-\underset{\_}{\begin{array}{l}\text{ 23}\text{.56}\text{ decimal place}\text{2}\\ \text{ 2}\text{.3 decimal place}\text{1}\end{array}}\\ \text{=}\text{21}\text{.26 should be written as 21}\text{.2 (decimal place}\text{1})\end{array}}\\ \\ \underset{\_}{\text{Division of 21}\text{.2 and 1}\text{.248}\times {\text{10}}^3\text{:}}\\ \underset{\_}{\begin{array}{l}\text{÷}\underset{\_}{\begin{array}{l}\text{ 21}\text{.2 significant}\text{figure}3\\ \text{1248}\text{significant}\text{figure}\text{4}\end{array}}\\ \text{=}\text{0}\text{.0170 significant}\text{figure}\text{3 }\end{array}}\\ \\ \text{Finally,}\\ \underset{\_}{\text{Multiplication of 1}\text{.68 and 0}\text{.0170:}}\\ \underset{\_}{\begin{array}{l}\text{×}\underset{\_}{\begin{array}{l}\text{1}\text{.68}\text{ significant}\text{figure}\text{3}\\ \text{0}\text{.0170 significant}\text{figure}\text{3}\end{array}}\\ \text{=}\text{0}\text{.02856 significant}\text{figure}\text{3 }\end{array}}\\ \end{array}$

Finally, on the multiplication of $\text{1}\text{.68 and 0}\text{.0170}$ gives the value $\text{0}\text{.02856}$.  Out of the two obtained quantity $\text{1}\text{.68}$ possesses 3 significant figures.  Hence, the answer should be written as $\text{0}\text{.0286 (rounding off)}$.  The significant figure is three.