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College Physics

11th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781305952300
Chapter 30, Problem 33P
Textbook Problem
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Find the number of electrons, and of each species of quark, in 1.00 L of water.

To determine
Number of electrons and quark particle in 1.00L .

Explanation of Solution

Given info:

Mass of water is 1.00L .

Avogadro number is 6.02×1023molecules/mol .

The number of gram per mole of water is 18.0g/mol .

Formula to the total number of molecules in water is,

N=mnNA

  •   N is the number of molecules
  • m is the mass of water
  • n is the number of gram per mole of water
  • NA is the Avogadro number

Substitute 1.00L for m , 6.02×1023molecules/mol for NA and 18.0g/mol for n to find N .

N=(1.00L)(1000g1L)(18.0g/mol)(6.02×1023molecules/mol)=3.34×1025molecules

Each molecule contains 10 protons, 10 electrons, and 8 neutrons.

Formula to the total number of electron in 1.00L of water is,

Ne=neN

  •   N is the number of molecules
  • ne is the number of electron in one molecule of water
  • Ne is the total number of electron in 1.00L of water

Substitute 3.34×1025molecules for N and 10electrons/molecule for ne to find Ne .

Ne=(10electrons/molecule)(3.34×1025molecules)=3.34×1026electrons

Thus, the total number of electrons in 1.00L of water is 3.34×1026electrons .

Formula to the total number of proton in 1.00L of water is,

Np=npN

  • np is the number of proton in one molecule of water
  • Np is the total number of proton in 1.00L of water

Substitute 3.34×1025molecules for N and 10protons/molecule for np to find Np .

Np=(10protons/molecule)(3.34×1025molecules)=3.34×1026protons

Thus, the total number of protons in 1

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Chapter 30 Solutions

College Physics
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