Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 3.1, Problem 17LC

a)

Interpretation Introduction

Interpretation : The given problem is to be solved and to be expressed in scientific notation and to correct the number of significant figures.

Concept Introduction : A given number is stated as the product of two numbers in scientific notation: a coefficient and 10 raised to a power.

a)

Expert Solution
Check Mark

Answer to Problem 17LC

The answer to the problem in scientific notation is 6.6×104 .

Explanation of Solution

Every number bigger than one and less than 10 is a coefficient.

There is an integer exponent.

In the case of an addition, the exponents must be the same otherwise they should be aligned and then the addition is carried out.

The given problem is 5.3×104+1.3×104 .

Since the exponents are the same, the coefficients are added.

  5.3×104+1.3×104=5.3+1.3×104=6.6×104

b)

Interpretation Introduction

Interpretation : The given problem is to be solved and to be expressed in scientific notation and to correct the number of significant figures.

Concept Introduction : A given number is stated as the product of two numbers in scientific notation: a coefficient and 10 raised to a power.

b)

Expert Solution
Check Mark

Answer to Problem 17LC

The answer to the problem in scientific notation is 4.0×107 .

Explanation of Solution

Every number bigger than one and less than 10 is a coefficient.

There is an integer exponent.

The division sign is used between the numerator and denominator, so the coefficients are divided while the exponents are subtracted.

The given problem is 7.2×104÷1.8×103 .

The coefficients are divided and the exponents are subtracted.

  7.2×104÷1.8×103=7.21.8×1043=4.0×107

c)

Interpretation Introduction

Interpretation : The given problem is to be solved and to be expressed in scientific notation and to correct the number of significant figures.

Concept Introduction : A given number is stated as the product of two numbers in scientific notation: a coefficient and 10 raised to a power.

c)

Expert Solution
Check Mark

Answer to Problem 17LC

The answer to the problem in scientific notation is 107 .

Explanation of Solution

A given number is stated as the product of two numbers in scientific notation: a coefficient and 10 raised to a power.

Every number bigger than one and less than 10 is a coefficient.

There is an integer exponent.

The multiplication sign is used in the denominator so the coefficients are multiplied while the exponents are added up.

The given problem is 104×103×106 .

Since there are only exponents in multiplication, these exponents are added up as it has the same base value.

  104×103×106=1043+6=107

d)

Interpretation Introduction

Interpretation : The given problem is to be solved and to be expressed in scientific notation and to correct the number of significant figures.

Concept Introduction : A given number is stated as the product of two numbers in scientific notation: a coefficient and 10 raised to a power.

d)

Expert Solution
Check Mark

Answer to Problem 17LC

The answer to the problem in scientific notation is 8.70×101 .

Explanation of Solution

A given number is stated as the product of two numbers in scientific notation: a coefficient and 10 raised to a power.

Every number bigger than one and less than 10 is a coefficient.

There is an integer exponent.

In the case of subtraction, the exponents must be the same otherwise they should be aligned, and then the subtraction is carried out.

The given problem is 9.12×1014.7×102 .

The exponents are first aligned to the same power.

  9.12×1014.7×102=9.12×1010.47×101

Subtract the coefficients.

  9.12×1010.47×101=9.120.47×101=8.65×101=8.70×101

e)

Interpretation Introduction

Interpretation: The given problem is to be solved and to be expressed in scientific notation and to correct the number of significant figures.

Concept Introduction : A given number is stated as the product of two numbers in scientific notation: a coefficient and 10 raised to a power.

e)

Expert Solution
Check Mark

Answer to Problem 17LC

The answer to the problem in scientific notation is 1.9×1014 .

Explanation of Solution

A given number is stated as the product of two numbers in scientific notation: a coefficient and 10 raised to a power.

Every number bigger than one and less than 10 is a coefficient.

There is an integer exponent.

The multiplication sign is used in the denominator so the coefficients are multiplied while the exponents are added up.

The given problem is 5.4×104×3.5×109 .

The coefficients are multiplied and the exponents are added up.

  5.4×104×3.5×109=5.4×3.5×104+9=18.9×1013=1.89×1014=1.9×1014

Chapter 3 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 3.1 - Prob. 11SPCh. 3.1 - Prob. 12LCCh. 3.1 - Prob. 13LCCh. 3.1 - Prob. 14LCCh. 3.1 - Prob. 15LCCh. 3.1 - Prob. 16LCCh. 3.1 - Prob. 17LCCh. 3.1 - Prob. 18LCCh. 3.2 - Prob. 19SPCh. 3.2 - Prob. 20SPCh. 3.2 - Prob. 21SPCh. 3.2 - Prob. 22SPCh. 3.2 - Prob. 23LCCh. 3.2 - Prob. 24LCCh. 3.2 - Prob. 25LCCh. 3.2 - Prob. 26LCCh. 3.2 - Prob. 27LCCh. 3.2 - Prob. 28LCCh. 3.2 - Prob. 29LCCh. 3.2 - Prob. 30LCCh. 3.2 - Prob. 31LCCh. 3.2 - Prob. 32LCCh. 3.2 - Prob. 33LCCh. 3.2 - Prob. 34LCCh. 3.2 - Prob. 35LCCh. 3.3 - Prob. 36SPCh. 3.3 - Prob. 37SPCh. 3.3 - Prob. 38SPCh. 3.3 - Prob. 39SPCh. 3.3 - Prob. 40SPCh. 3.3 - Prob. 41SPCh. 3.3 - Prob. 42SPCh. 3.3 - Prob. 43SPCh. 3.3 - Prob. 44SPCh. 3.3 - Prob. 45SPCh. 3.3 - Prob. 46SPCh. 3.3 - Prob. 47SPCh. 3.3 - Prob. 48SPCh. 3.3 - Prob. 49SPCh. 3.3 - Prob. 50LCCh. 3.3 - Prob. 51LCCh. 3.3 - Prob. 52LCCh. 3.3 - Prob. 53LCCh. 3.3 - Prob. 54LCCh. 3.3 - Prob. 55LCCh. 3 - Prob. 56ACh. 3 - Prob. 57ACh. 3 - Prob. 58ACh. 3 - Prob. 59ACh. 3 - Prob. 60ACh. 3 - Prob. 61ACh. 3 - Prob. 62ACh. 3 - Prob. 63ACh. 3 - Prob. 65ACh. 3 - Prob. 66ACh. 3 - Prob. 67ACh. 3 - Prob. 68ACh. 3 - Prob. 69ACh. 3 - Prob. 70ACh. 3 - Prob. 71ACh. 3 - Prob. 72ACh. 3 - Prob. 73ACh. 3 - Prob. 74ACh. 3 - Prob. 75ACh. 3 - Prob. 76ACh. 3 - Prob. 77ACh. 3 - Prob. 78ACh. 3 - Prob. 79ACh. 3 - Prob. 80ACh. 3 - Prob. 81ACh. 3 - Prob. 82ACh. 3 - Prob. 83ACh. 3 - Prob. 84ACh. 3 - Prob. 85ACh. 3 - Prob. 86ACh. 3 - Prob. 87ACh. 3 - Prob. 88ACh. 3 - Prob. 89ACh. 3 - Prob. 90ACh. 3 - Prob. 91ACh. 3 - Prob. 92ACh. 3 - Prob. 93ACh. 3 - Prob. 94ACh. 3 - Prob. 95ACh. 3 - Prob. 96ACh. 3 - Prob. 97ACh. 3 - Prob. 98ACh. 3 - Prob. 99ACh. 3 - Prob. 100ACh. 3 - Prob. 101ACh. 3 - Prob. 102ACh. 3 - Prob. 103ACh. 3 - Prob. 104ACh. 3 - Prob. 105ACh. 3 - Prob. 106ACh. 3 - Prob. 107ACh. 3 - Prob. 108ACh. 3 - Prob. 1STPCh. 3 - Prob. 2STPCh. 3 - Prob. 3STPCh. 3 - Prob. 4STPCh. 3 - Prob. 5STPCh. 3 - Prob. 6STPCh. 3 - Prob. 7STPCh. 3 - Prob. 8STPCh. 3 - Prob. 9STPCh. 3 - Prob. 10STPCh. 3 - Prob. 11STPCh. 3 - Prob. 12STP
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