OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:
OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:
5th Edition
ISBN: 9781285460369
Author: STANITSKI
Publisher: Cengage Learning
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Chapter 1, Problem 21QRT

Perform these calculations and express the result with the proper number of significant figures.

  1. (a) 4.850 g 2.34 g 1.3 mL
  2. (b) V = 4 3 π r 3 where r = 4.112 cm
  3. (c) (4.66 × 10−3) × 4.666
  4. (d) 0.003400 65.2

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The following calculations has to be performed and the result with the proper number of significant figures has to be expressed.

  4.850g-2.34g1.3mL

Concept Introduction:

Significant digits:

In a number, the digits which contribute to the precision of the number are said to be significant digits.

Rules for determination of significant digits in a number:

  • All non-zero digits are significant.
  • The zero’s appearing between two non-zero digits are significant.
  • The zero’s before the non-zero numbers are not significant.
  • Zero’s after non-zero number without decimal are non-significant.
  • Zero’s after non-zero number with decimal are significant.
  • Zero’s after the decimal point are significant.
  • Any numbers with scientific notation are significant.

The sum of the significant digits has to be given in the lowest decimal present in the values.

The significant digit in multiplication and division has to be given as the quantity with the fewest significant figures.

Explanation of Solution

The given calculation is

  4.850g-2.34g1.3mL

The nominator part is subtracted.  The significant figures in the subtraction gives two decimal points as the first value (4.850) has three decimals and the second value (2.34) has two decimals and the lowest decimals are given in significant figure.

  2.51g1.3mL

The resulted value is divided by 1.3mL.  The nominator part is with three significant figures whereas the denominator consists of two significant figures as all non-zero digits are significant.  The resulting value contains two significant figures.

  2.51g1.3mL=1.9g/mL

The result contains two significant figures (1and9).

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The following calculations has to be performed and the result with the proper number of significant figures has to be expressed.

  V=43πr3wherer = 4.112cm

Concept Introduction:

Refer part (a).

Explanation of Solution

Given, V=43πr3wherer = 4.112cm.

The value of r is given and the value of π = 3.1415926....  The values present in division are single digits.  The value of r contains four significant figures and the value of π contains eight significant figures as all non-zero digits are significant.  The resulting value has to be given with four significant figures.

Substitute the values in the formula.

  V=43πr3V=43(3.1415926)(4.112cm)3V=43(3.1415926)(69.52cm)V=43(218.4)V=291.2

The result contains four significant figures (2,9,1and2).

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The following calculations has to be performed and the result with the proper number of significant figures has to be expressed.

  (4.66 × 10-3) × 4.666

Concept Introduction:

Refer part (a).

Explanation of Solution

Given (4.66 × 10-3) × 4.666.

The calculation is of multiplication.  The first value (4.66 × 103) consists of three significant figures and the second value (4.666) consists of four significant figures as all non-zero digits are significant.  The resulting value has to be given with three significant figures.

  (4.66 × 10-3) × 4.666= 0.00466 × 4.666= 0.0217

As the zero’s present before the non-zero digits are non-significant, the resulted value contains three significant figures (2,1and7).

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The following calculations has to be performed and the result with the proper number of significant figures has to be expressed.

  0.00340065.2

Concept Introduction:

Refer part (a).

Explanation of Solution

Given, 0.00340065.2.

The calculation is division.  The numerator consists of four significant figures as zero’s present before non-zero digits are not significant.  The denominator consists of three significant figures as all non-zero digits are significant.  The result has to be with three significant figures.

  0.00340065.2=5.21×10-5

The result consists of three significant figures (5,2and1).

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Chapter 1 Solutions

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:

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