Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
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Chapter 4, Problem 4.7QAP
Interpretation Introduction

(a)

Interpretation:

The given operation should be performed by using binary numbers and after that the final binary number obtained should be converted back to decimal.

Concept introduction:

Rules of Binary addition:

Binary sum of 0 and 0 results 0 and yields 0 carry.

Binary sum of 0 and 1 results 1 and yields 0 carry.

Binary sum of 1 and 1 results 0 and yields 1 carry.

Expert Solution
Check Mark

Answer to Problem 4.7QAP

810+510=(1000)2+(101)2=(1101)2=1310

Explanation of Solution

We will convert the decimal to binary as follows:

Divide 8 by 2, we will get 4, remainder will be 0(LSB).

Now divide 4 by 2, we will get 2 remainder will be 0.

Now divide 2 by 2, we will get 1(MSB) remainder will be 0.

Hence the binary equivalent of 8 will be 1000, it can be written as follows:

810=(1000)2

Similarly, we can find the binary equivalent of 5 as,

510=(101)2

The sum is,

1000

+101

----------

1101

-----------

The decimal equivalent of sum is, (1101)2=1310

Interpretation Introduction

(b)

Interpretation:

The given operation should be performed by using binary numbers and after that the final binary number obtained should be converted back to decimal.

Concept introduction:

Rules of Binary addition:

Binary sum of 0 and 0 results in 0 and yields 0 carry

Binary sum of 0 and 1 results 1 and yields 0 carry

Binary sum of 1 and 1 results 0 and yields 1 carry

Expert Solution
Check Mark

Answer to Problem 4.7QAP

33910+2510=(101010011)2+(11001)2=(101101100)2=36410

Explanation of Solution

we will convert the decimal to binary as follows:

Divide 339 by 2, we will get 169 ,remainder will be 1(LSB).

Now divide 169 by 2, we will get 84 remainder will be 1.

Now divide 84 by 2, we will get 42 remainder will be 0.

....................................…………

....................................…………

Now divide 2 by 2, we will get 1(MSB) remainder will be 0.

Hence the binary equivalent of 339 will be 101010011, it can be written as follows:

33910=(101010011)2

Similarly we can find the binary equivalent of 25.

2510=(11001)2

101010011

+11001

----------------

101101100

-----------------

(101101100)2=36410

Interpretation Introduction

(c)

Interpretation:

The given operation should be performed by using binary numbers and after that the final binary number obtained should be converted back to decimal.

Concept introduction:

Rules of Binary addition:

Binary sum of 0 and 0 results 0 and yields 0 carry

Binary sum of 0 and 1 results 1 and yields 0 carry

Binary sum of 1 and 1 results 0 and yields 1 carry

Expert Solution
Check Mark

Answer to Problem 4.7QAP

4710+1610=(101111)2+(10000)2=(111111)2=6310

Explanation of Solution

we will convert the decimal to binary as follows:

Divide 47 by 2, we will get 23 ,remainder will be 1(LSB).

Now divide 23 by 2, we will get 11 remainder will be 1.

....................................…………

....................................…………

Now divide 2 by 2, we will get 1(MSB) remainder will be 0.

Hence the binary equivalent of 47 will be 101111, it can be written as follows:

4710=(101111)2

Similarly, we can find the binary equivalent of 16.

1610=(10000)2

101111

+10000

------------

111111

-------------

(111111)2=6310

Interpretation Introduction

(d)

Interpretation:

The given operation should be performed by using binary numbers and after that the final binary number obtained should be converted back to decimal.

Concept introduction:

Rules of Binary multiplication:

Binary product of 0 and 0 results 0

Binary product of 0 and 1 results 0

Binary product of 1 and 1 results 1

Expert Solution
Check Mark

Answer to Problem 4.7QAP

210×910=(1001)2×(10)2=(10010)2=1810

Explanation of Solution

we will convert the decimal to binary as follows:

Divide 9 by 2, we will get 4 ,remainder will be 1(LSB).

Now divide 4 by 2, we will get 2 remainder will be 0.

....................................…………

....................................…………

Now divide 2 by 2, we will get 1(MSB) remainder will be 0.

Hence the binary equivalent of 9 will be 1001, it can be written as follows:

910=(1001)2

Similarly, we can find the binary equivalent of 2.

210=(10)2

1001

×10

-----------

0000

1001

--------------

10010

--------------

(10010)2=1810

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