Number of electrons and quark particle in 1.00 L .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 30, Problem 33P

To determine

Number of electrons and quark particle in $1.00 L$ .

Expert Solution & Answer

Answer to Problem 33P

There are

$3.34×1026$ electrons,

$9.36×1026$ up quarks and

$8.70×1026$ down quarks.

Explanation of Solution

Given info:

Mass of water is $1.00 L$ .

Avogadro number is $6.02×1023 molecules/mol$ .

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

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

$N=(1.00 L)(1000 g1 L)(18.0 g/mol)(6.02×1023 molecules/mol)=3.34×1025 molecules$

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

Formula to the total number of electron in $1.00 L$ 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 $1.00 L$ of water

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

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

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

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

$Np=npN$

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

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

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

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

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

$Nn=nnN$

$nn$ is the number of neutron in one molecule of water
$Nn$ is the total number of neutron in $1.00 L$ of water

Substitute $3.34×1025 molecules$ for $N$ and $8 neutron/molecule$ for $nn$ to find $Nn$ .

$Nn=(8 neutron/molecule)(3.34×1025 molecules)=2.68×1026 protons$

Each proton contains $2$ up quarks and $1$ down quark. Each neutron contains $1$ up quark and $2$ down quarks.

Formula to the total number of up quarks in $1.00 L$ of water is,

$Nu=2Np+Nn$

$Nu$ is the total number of up quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nu$ .

$Nu=2(3.34×1026 protons)+(2.26×1026 neutrons)=9.36×1026 up quarks$

Formula to the total number of down quarks in $1.00 L$ of water is,

$Nd=Np+2Nn$

$Nd$ is the total number of down quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nd$ .

$Nd=(3.34×1026 protons)+2(2.26×1026 neutrons)=8.70×1026 up quarks$

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

Conclusion:

There are $3.34×1026$ electrons, $9.36×1026$ up quarks and $8.70×1026$ down quarks.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Two conductors having net charges of +14.0 µC and -14.0 µC have a potential difference of 14.0 V between them. (a) Determine the capacitance of the system. F (b) What is the potential difference between the two conductors if the charges on each are increased to +196.0 µC and -196.0 µC? V

Please see the attached image and answer the set of questions with proof.

How, Please type the whole transcript correctly using comma and periods as needed. I have uploaded the picture of a video on YouTube. Thanks,

Answer 2

Question

Chapter 30, Problem 33P

To determine

Number of electrons and quark particle in $1.00 L$ .

Expert Solution & Answer

Answer to Problem 33P

There are

$3.34×1026$ electrons,

$9.36×1026$ up quarks and

$8.70×1026$ down quarks.

Explanation of Solution

Given info:

Mass of water is $1.00 L$ .

Avogadro number is $6.02×1023 molecules/mol$ .

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

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

$N=(1.00 L)(1000 g1 L)(18.0 g/mol)(6.02×1023 molecules/mol)=3.34×1025 molecules$

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

Formula to the total number of electron in $1.00 L$ 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 $1.00 L$ of water

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

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

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

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

$Np=npN$

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

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

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

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

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

$Nn=nnN$

$nn$ is the number of neutron in one molecule of water
$Nn$ is the total number of neutron in $1.00 L$ of water

Substitute $3.34×1025 molecules$ for $N$ and $8 neutron/molecule$ for $nn$ to find $Nn$ .

$Nn=(8 neutron/molecule)(3.34×1025 molecules)=2.68×1026 protons$

Each proton contains $2$ up quarks and $1$ down quark. Each neutron contains $1$ up quark and $2$ down quarks.

Formula to the total number of up quarks in $1.00 L$ of water is,

$Nu=2Np+Nn$

$Nu$ is the total number of up quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nu$ .

$Nu=2(3.34×1026 protons)+(2.26×1026 neutrons)=9.36×1026 up quarks$

Formula to the total number of down quarks in $1.00 L$ of water is,

$Nd=Np+2Nn$

$Nd$ is the total number of down quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nd$ .

$Nd=(3.34×1026 protons)+2(2.26×1026 neutrons)=8.70×1026 up quarks$

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

Conclusion:

There are $3.34×1026$ electrons, $9.36×1026$ up quarks and $8.70×1026$ down quarks.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 30, Problem 33P

To determine

Number of electrons and quark particle in $1.00 L$ .

Expert Solution & Answer

Answer to Problem 33P

There are

$3.34×1026$ electrons,

$9.36×1026$ up quarks and

$8.70×1026$ down quarks.

Explanation of Solution

Given info:

Mass of water is $1.00 L$ .

Avogadro number is $6.02×1023 molecules/mol$ .

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

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

$N=(1.00 L)(1000 g1 L)(18.0 g/mol)(6.02×1023 molecules/mol)=3.34×1025 molecules$

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

Formula to the total number of electron in $1.00 L$ 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 $1.00 L$ of water

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

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

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

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

$Np=npN$

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

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

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

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

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

$Nn=nnN$

$nn$ is the number of neutron in one molecule of water
$Nn$ is the total number of neutron in $1.00 L$ of water

Substitute $3.34×1025 molecules$ for $N$ and $8 neutron/molecule$ for $nn$ to find $Nn$ .

$Nn=(8 neutron/molecule)(3.34×1025 molecules)=2.68×1026 protons$

Each proton contains $2$ up quarks and $1$ down quark. Each neutron contains $1$ up quark and $2$ down quarks.

Formula to the total number of up quarks in $1.00 L$ of water is,

$Nu=2Np+Nn$

$Nu$ is the total number of up quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nu$ .

$Nu=2(3.34×1026 protons)+(2.26×1026 neutrons)=9.36×1026 up quarks$

Formula to the total number of down quarks in $1.00 L$ of water is,

$Nd=Np+2Nn$

$Nd$ is the total number of down quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nd$ .

$Nd=(3.34×1026 protons)+2(2.26×1026 neutrons)=8.70×1026 up quarks$

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

Conclusion:

There are $3.34×1026$ electrons, $9.36×1026$ up quarks and $8.70×1026$ down quarks.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 30, Problem 33P

To determine

Number of electrons and quark particle in $1.00 L$ .

Expert Solution & Answer

Answer to Problem 33P

There are

$3.34×1026$ electrons,

$9.36×1026$ up quarks and

$8.70×1026$ down quarks.

Explanation of Solution

Given info:

Mass of water is $1.00 L$ .

Avogadro number is $6.02×1023 molecules/mol$ .

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

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

$N=(1.00 L)(1000 g1 L)(18.0 g/mol)(6.02×1023 molecules/mol)=3.34×1025 molecules$

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

Formula to the total number of electron in $1.00 L$ 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 $1.00 L$ of water

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

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

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

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

$Np=npN$

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

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

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

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

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

$Nn=nnN$

$nn$ is the number of neutron in one molecule of water
$Nn$ is the total number of neutron in $1.00 L$ of water

Substitute $3.34×1025 molecules$ for $N$ and $8 neutron/molecule$ for $nn$ to find $Nn$ .

$Nn=(8 neutron/molecule)(3.34×1025 molecules)=2.68×1026 protons$

Each proton contains $2$ up quarks and $1$ down quark. Each neutron contains $1$ up quark and $2$ down quarks.

Formula to the total number of up quarks in $1.00 L$ of water is,

$Nu=2Np+Nn$

$Nu$ is the total number of up quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nu$ .

$Nu=2(3.34×1026 protons)+(2.26×1026 neutrons)=9.36×1026 up quarks$

Formula to the total number of down quarks in $1.00 L$ of water is,

$Nd=Np+2Nn$

$Nd$ is the total number of down quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nd$ .

$Nd=(3.34×1026 protons)+2(2.26×1026 neutrons)=8.70×1026 up quarks$

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

Conclusion:

There are $3.34×1026$ electrons, $9.36×1026$ up quarks and $8.70×1026$ down quarks.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 30, Problem 33P

To determine

Number of electrons and quark particle in $1.00 L$ .

Expert Solution & Answer

Answer to Problem 33P

There are

$3.34×1026$ electrons,

$9.36×1026$ up quarks and

$8.70×1026$ down quarks.

Explanation of Solution

Given info:

Mass of water is $1.00 L$ .

Avogadro number is $6.02×1023 molecules/mol$ .

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

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

$N=(1.00 L)(1000 g1 L)(18.0 g/mol)(6.02×1023 molecules/mol)=3.34×1025 molecules$

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

Formula to the total number of electron in $1.00 L$ 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 $1.00 L$ of water

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

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

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

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

$Np=npN$

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

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

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

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

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

$Nn=nnN$

$nn$ is the number of neutron in one molecule of water
$Nn$ is the total number of neutron in $1.00 L$ of water

Substitute $3.34×1025 molecules$ for $N$ and $8 neutron/molecule$ for $nn$ to find $Nn$ .

$Nn=(8 neutron/molecule)(3.34×1025 molecules)=2.68×1026 protons$

Each proton contains $2$ up quarks and $1$ down quark. Each neutron contains $1$ up quark and $2$ down quarks.

Formula to the total number of up quarks in $1.00 L$ of water is,

$Nu=2Np+Nn$

$Nu$ is the total number of up quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nu$ .

$Nu=2(3.34×1026 protons)+(2.26×1026 neutrons)=9.36×1026 up quarks$

Formula to the total number of down quarks in $1.00 L$ of water is,

$Nd=Np+2Nn$

$Nd$ is the total number of down quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nd$ .

$Nd=(3.34×1026 protons)+2(2.26×1026 neutrons)=8.70×1026 up quarks$

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

Conclusion:

There are $3.34×1026$ electrons, $9.36×1026$ up quarks and $8.70×1026$ down quarks.

Answer 6

Chapter 30, Problem 33P

To determine

Number of electrons and quark particle in $1.00 L$ .

Expert Solution & Answer

Answer to Problem 33P

There are

$3.34×1026$ electrons,

$9.36×1026$ up quarks and

$8.70×1026$ down quarks.

Explanation of Solution

Given info:

Mass of water is $1.00 L$ .

Avogadro number is $6.02×1023 molecules/mol$ .

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

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

$N=(1.00 L)(1000 g1 L)(18.0 g/mol)(6.02×1023 molecules/mol)=3.34×1025 molecules$

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

Formula to the total number of electron in $1.00 L$ 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 $1.00 L$ of water

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

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

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

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

$Np=npN$

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

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

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

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

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

$Nn=nnN$

$nn$ is the number of neutron in one molecule of water
$Nn$ is the total number of neutron in $1.00 L$ of water

Substitute $3.34×1025 molecules$ for $N$ and $8 neutron/molecule$ for $nn$ to find $Nn$ .

$Nn=(8 neutron/molecule)(3.34×1025 molecules)=2.68×1026 protons$

Each proton contains $2$ up quarks and $1$ down quark. Each neutron contains $1$ up quark and $2$ down quarks.

Formula to the total number of up quarks in $1.00 L$ of water is,

$Nu=2Np+Nn$

$Nu$ is the total number of up quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nu$ .

$Nu=2(3.34×1026 protons)+(2.26×1026 neutrons)=9.36×1026 up quarks$

Formula to the total number of down quarks in $1.00 L$ of water is,

$Nd=Np+2Nn$

$Nd$ is the total number of down quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nd$ .

$Nd=(3.34×1026 protons)+2(2.26×1026 neutrons)=8.70×1026 up quarks$

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

Conclusion:

There are $3.34×1026$ electrons, $9.36×1026$ up quarks and $8.70×1026$ down quarks.

Answer 7

Chapter 30, Problem 33P

To determine

Number of electrons and quark particle in $1.00 L$ .

Answer 8

Expert Solution & Answer

Answer to Problem 33P

There are

$3.34×1026$ electrons,

$9.36×1026$ up quarks and

$8.70×1026$ down quarks.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given info:

Mass of water is $1.00 L$ .

Avogadro number is $6.02×1023 molecules/mol$ .

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

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

$N=(1.00 L)(1000 g1 L)(18.0 g/mol)(6.02×1023 molecules/mol)=3.34×1025 molecules$

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

Formula to the total number of electron in $1.00 L$ 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 $1.00 L$ of water

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

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

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

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

$Np=npN$

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

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

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

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

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

$Nn=nnN$

$nn$ is the number of neutron in one molecule of water
$Nn$ is the total number of neutron in $1.00 L$ of water

Substitute $3.34×1025 molecules$ for $N$ and $8 neutron/molecule$ for $nn$ to find $Nn$ .

$Nn=(8 neutron/molecule)(3.34×1025 molecules)=2.68×1026 protons$

Each proton contains $2$ up quarks and $1$ down quark. Each neutron contains $1$ up quark and $2$ down quarks.

Formula to the total number of up quarks in $1.00 L$ of water is,

$Nu=2Np+Nn$

$Nu$ is the total number of up quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nu$ .

$Nu=2(3.34×1026 protons)+(2.26×1026 neutrons)=9.36×1026 up quarks$

Formula to the total number of down quarks in $1.00 L$ of water is,

$Nd=Np+2Nn$

$Nd$ is the total number of down quarks

Substitute $3.34×1026 protons$ for $Np$ and $2.26×1026 neutrons$ for $Nn$ to find $Nd$ .

$Nd=(3.34×1026 protons)+2(2.26×1026 neutrons)=8.70×1026 up quarks$

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

Conclusion:

There are $3.34×1026$ electrons, $9.36×1026$ up quarks and $8.70×1026$ down quarks.

Answer 12

Ch. 30.6 - Prob. 30.1QQ Ch. 30.6 - Prob. 30.2QQ Ch. 30 - Prob. 1CQ Ch. 30 - Prob. 2CQ Ch. 30 - Prob. 3CQ Ch. 30 - Prob. 4CQ Ch. 30 - Prob. 5CQ Ch. 30 - Prob. 6CQ Ch. 30 - Prob. 7CQ Ch. 30 - Prob. 8CQ

Number of electrons and quark particle in 1.00 L .

Concept explainers

Number of electrons and quark particle in $1.00 L$ .

Answer to Problem 33P

Explanation of Solution

Want to see more full solutions like this?

Chapter 30 Solutions

Number of electrons and quark particle in 1.00 L .

Concept explainers

Number of electrons and quark particle in 1.00 L<m:math><m:mrow><m:mn>1.00</m:mn><m:mtext> </m:mtext><m:mtext>L</m:mtext></m:mrow> </m:math>.

Answer to Problem 33P

Explanation of Solution

Want to see more full solutions like this?

Chapter 30 Solutions

Number of electrons and quark particle in $1.00 L$ .