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Chapter 19, Problem 1P

(a)

To determine

The charge and the mass of an ionized hydrogen atom to three significant digits.

(a)

Expert Solution
Check Mark

Answer to Problem 1P

The charge of ionized hydrogen atom is +1.60×1019C_, and the mass of ionized hydrogen atom is 1.67×1027kg_.

Explanation of Solution

The ionized hydrogen atom H+ is the loss of on electron from the hydrogen atom.

Write the expression for charge on ionized hydrogen atom H+.

    q=ne        (I)

Here, q is the charge on ionized hydrogen atom, n is the number of electron lost, and e is the charge of single electron.

Write the expression for mass of ionized hydrogen atom.

    m=(1.0079u)mpme        (II)

Here, m is the mass of the ionized hydrogen atom, mp is the mass of proton, and me is the mass of electron.

Conclusion:

Substitute 1 for n, and 1.60×1019C for e in equation (I), to find q.

    q=(1)(1.60×1019C)=1.60×1019C

Substitute 1.660×1027kg for mp, and 9.11×1031kg for me in equation (II), to find m

    m=(1.0079u)(1.660×1027kg)(9.11×1031kg)=1.67×1027kg

Therefore, the charge of ionized hydrogen atom is +1.60×1019C_, and the mass of ionized hydrogen atom is 1.67×1027kg_.

(b)

To determine

The charge and the mass of an ionized sodium atom to three significant digits.

(b)

Expert Solution
Check Mark

Answer to Problem 1P

The charge of ionized sodium atom is +1.60×1019C_, and the mass of ionized hydrogen atom is 3.82×1026kg_.

Explanation of Solution

Write the expression for mass of ionized sodium atom.

    m=(22.99u)mpme        (III)

Here, m is the mass of the ionized sodium atom, mp is the mass of proton, and me is the mass of electron.

Conclusion:

Substitute 1 for n, and 1.60×1019C for e in equation (I), to find q.

    q=(1)(1.60×1019C)=1.60×1019C

Substitute 1.660×1027kg for mp, and 9.11×1031kg for me in equation (III), to find m

    m=(22.99u)(1.660×1027kg)(9.11×1031kg)=3.82×1026kg

Therefore, the charge of ionized sodium atom is +1.60×1019C_, and the mass of ionized hydrogen atom is 3.82×1026kg_.

(c)

To determine

The charge and the mass of a chloride ion cl to three significant digits.

(c)

Expert Solution
Check Mark

Answer to Problem 1P

The charge of a chloride ion cl is 1.60×1019C_, the mass of a chloride ion cl is 5.89×1026kg_.

Explanation of Solution

Write the expression for mass of a chloride ion cl.

    m=(35.453u)mpme        (IV)

Here, m is the mass of a chloride ion cl, mp is the mass of proton, and me is the mass of electron.

Conclusion:

Substitute 1 for n, and 1.60×1019C for e in equation (I), to find q.

    q=(1)(1.60×1019C)=1.60×1019C

Substitute 1.660×1027kg for mp, and 9.11×1031kg for me in equation (IV), to find m

    m=(35.453u)(1.660×1027kg)(9.11×1031kg)=5.89×1026kg

Therefore, The charge of a chloride ion cl is 1.60×1019C_, the mass of a chloride ion cl is 5.89×1026kg_.

(d)

To determine

The charge and the mass of a doubly ionized calcium atom Ca++ to three significant digits.

(d)

Expert Solution
Check Mark

Answer to Problem 1P

The charge of a doubly ionized calcium atom Ca++ is 3.20×1019C_, and the mass of a doubly ionized calcium atom Ca++ is 6.65×1026kg_.

Explanation of Solution

Write the expression for a doubly ionized calcium atom Ca++.

    m=(40.078u)mpnme        (V)

Here, m is the mass of a doubly ionized calcium atom Ca++, mp is the mass of proton, n is the number of electron lost, and me is the mass of electron.

Conclusion:

Substitute 2 for n, and 1.60×1019C for e in equation (I), to find q.

    q=(2)(1.60×1019C)=+3.20×1019C

Substitute 1.660×1027kg for mp, 2 for n, and 9.11×1031kg for me in equation (V), to find m

    m=(40.078u)(1.660×1027kg)2(9.11×1031kg)=6.65×1026kg

Therefore, the charge of a doubly ionized calcium atom Ca++ is 3.20×1019C_, and the mass of a doubly ionized calcium atom Ca++ is 6.65×1026kg_.

(e)

To determine

The charge and the mass of a N3 ion to three significant digits.

(e)

Expert Solution
Check Mark

Answer to Problem 1P

The charge of a N3 ion is 4.80×1019C_, and the mass of a N3 ion is 2.33×1026kg_.

Explanation of Solution

Write the expression for a N3 ion

    m=(14.007u)mp+nme        (VI)

Here, m is the mass of a N3 ion, mp is the mass of proton, n is the number of electron gained, and me is the mass of electron.

Conclusion:

Substitute 3 for n, and 1.60×1019C for e in equation (I), to find q.

    q=(3)(1.60×1019C)=4.80×1019C

Substitute 1.660×1027kg for mp, 3 for n, and 9.11×1031kg for me in equation (VI), to find m

    m=(14.007u)(1.660×1027kg)+3(9.11×1031kg)=2.33×1026kg

Therefore, the charge of a N3 ion is 4.80×1019C_, and the mass of a N3 ion is 2.33×1026kg_.

(f)

To determine

The charge and the mass of a N4+ ion to three significant digits.

(f)

Expert Solution
Check Mark

Answer to Problem 1P

The charge of a N4+ ion is 6.40×1019C_, and the mass of a N4+ ion is 2.32×1026kg_.

Explanation of Solution

Write the expression for a N4+ ion

    m=(14.007u)mpnme        (VII)

Here, m is the mass of a N4+ ion, mp is the mass of proton, n is the number of electron lost, and me is the mass of electron.

Conclusion:

Substitute 4 for n, and 1.60×1019C for e in equation (I), to find q.

    q=(4)(1.60×1019C)=+6.40×1019C

Substitute 1.660×1027kg for mp, 4 for n, and 9.11×1031kg for me in equation (VII), to find m

    m=(14.007u)(1.660×1027kg)4(9.11×1031kg)=2.32×1026kg

Therefore, The charge of a N4+ ion is 6.40×1019C_, and the mass of a N4+ ion is 2.32×1026kg_.

(g)

To determine

The charge and the mass of the nucleus of nitrogen atom to three significant digits.

(g)

Expert Solution
Check Mark

Answer to Problem 1P

The charge of the nucleus of nitrogen atom is 1.12×1018C_, and the mass of the nucleus of nitrogen atom is 2.32×1026kg_.

Explanation of Solution

Consider nitrogen atom as seven times ionized nitrogen atom N7+.

Write the expression for a N7+ ion

    m=(14.007u)mpnme        (VIII)

Here, m is the mass of a ionized nitrogen atom N7+, mp is the mass of proton, n is the number of electron lost, and me is the mass of electron.

Conclusion:

Substitute 7 for n, and 1.60×1019C for e in equation (I), to find q.

    q=(7)(1.60×1019C)=+1.12×1018C

Substitute 1.660×1027kg for mp, 7 for n, and 9.11×1031kg for me in equation (VIII), to find m

    m=(14.007u)(1.660×1027kg)7(9.11×1031kg)=2.32×1026kg

Therefore, the charge of the nucleus of nitrogen atom is 1.12×1018C_, and the mass of the nucleus of nitrogen atom is 2.32×1026kg_.

(h)

To determine

The charge and the mass of molecular ion H2O to three significant digits.

(h)

Expert Solution
Check Mark

Answer to Problem 1P

The charge of molecular ion H2O is 1.60×1019C_, and the mass of molecular ion H2O is 2.99×1026kg_.

Explanation of Solution

Write the expression for charge of H2O molecular ion

    m=[2(1.0079u+15.999)]mp+nme        (IX)

Here, m is the mass of molecular ion H2O, mp is the mass of proton, n is the number of electron lost, and me is the mass of electron.

Conclusion:

Substitute 1 for n, and 1.60×1019C for e in equation (I), to find q.

    q=(1)(1.60×1019C)=1.60×1019C

Substitute 1.660×1027kg for mp, 1 for n, and 9.11×1031kg for me in equation ((IX), to find m

    m=[2(1.0079u+15.999)](1.660×1027kg)+1(9.11×1031kg)=2.99×1026kg

Therefore, The charge of molecular ion H2O is 1.60×1019C_, and the mass of molecular ion H2O is 2.99×1026kg_.

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Chapter 19 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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