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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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. Aluminum exists in several isotopic forms, including A 13 27 l , A 13 28 l , and A 13 29 l AI. Indicate the number of protons and the number of neutrons in each of these isotopes.

Interpretation Introduction

Interpretation:

Interpret number of proton or number of neutron for 1 3 Al27, 1 3 Al28, 1 3 Al29.

Concept Introduction:

The symbol of element can be shown as Z XA

Where, Z is atomic number or number of proton

A is mass number

Number of neutron = A - Z.

Explanation

For 1 3 Al27

Number of proton = 13

Number of neutron = 27 − 13 = 14

For 1 3 Al28

Nu...

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