EBK INTRODUCTORY CHEMISTRY
EBK INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 9780100480483
Author: DECOSTE
Publisher: YUZU
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Chapter 3, Problem 15CR
Interpretation Introduction

(a)

Interpretation:

The value of 493.2 g should be converted to kilograms.

Concept Introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Expert Solution
Check Mark

Answer to Problem 15CR

Value of 493.2 g in kilograms is 0.4932 kg.

Explanation of Solution

Since,1 g = 0.001 kg.

So,

  493.2 g = 493.2 g × 0.001 kg1 g  = 0.4932 kg

Interpretation Introduction

(b)

Interpretation:

The value of 493.2 g should be converted to pounds.

Concept Introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Expert Solution
Check Mark

Answer to Problem 15CR

Value of 493.2 g in pounds is 1.085 pound.

Explanation of Solution

Since, 1 g = 0.0022pound.

So,

  493.2 g = 493.2 g × 0.0022 pound 1 g = 1.085 pound

Interpretation Introduction

(c)

Interpretation:

The value of 9.312 mi should be converted to kilometers.

Concept Introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Expert Solution
Check Mark

Answer to Problem 15CR

Value of 9.312mi in kilometers is 14.986211 km.

Explanation of Solution

Since, 1 mi = 1.06934 km.

So,

  9.312 mi = 9.312 mi × 1.06934 km1 mi  = 14.986211 km

Interpretation Introduction

(d)

Interpretation:

The value of 9.312 mi should be converted to feet.

Concept Introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Expert Solution
Check Mark

Answer to Problem 15CR

Value of 9.312 mi in feet is 49167.36 feet.

Explanation of Solution

Since, 1 mi = 5280 feet.

So,

  9.312 mi = 9.312 mi × 5280feet1 mi = 49167.36 feet

Interpretation Introduction

(e)

Interpretation:

The value of 4.219 m should be converted to feet.

Concept Introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Expert Solution
Check Mark

Answer to Problem 15CR

Value of 4.219 m in feet is 13.841864 feet.

Explanation of Solution

Since, 1 m = 3.28084 feet.

So,

  4.219 m = 4.219 m × 3.28084 feet 1 m = 13.841864 feet

Interpretation Introduction

(f)

Interpretation:

The value of 4.219 m should be converted to centimeters.

Concept Introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Expert Solution
Check Mark

Answer to Problem 15CR

Value of 4.219 m in centimeters is 421.9 cm.

Explanation of Solution

Since, 1 m = 100 cm.

So,

  4.219 m = 4.219 m ×100 cm1 m  = 421.9 cm

Interpretation Introduction

(g)

Interpretation:

The value of 429.2 mL should be converted to liters.

Concept Introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Expert Solution
Check Mark

Answer to Problem 15CR

Value of 429.2 mL in liters is 0.4292 L.

Explanation of Solution

Since, 1 mL = 0.001 L.

So,

  429.2 mL = 429.2 mL × 0.001 L1 mL  = 0.4292L

Interpretation Introduction

(h)

Interpretation:

The value of 2.934 L should be converted to quarts.

Concept Introduction:

An arithmetical multiplier which is used for converting a quantity expressed in one unit into another equivalent set of units is said to be conversion factor.

Expert Solution
Check Mark

Answer to Problem 15CR

Value of 3.934 L in quarts is 3.1003232 quarts.

Explanation of Solution

Since, 1 L = 1.05669 quarts.

So,

  3.934 L = 3.934 L × 1.05669 quarts1 L = 3.1003232 quarts

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Chapter 3 Solutions

EBK INTRODUCTORY CHEMISTRY

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