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Anit [1.1K]
3 years ago
12

Be sure to answer all parts. the average atomic mass of nitrogen is 14.0067. the atomic masses of the two stable isotopes of nit

rogen, 14n and 15n, are 14.003074002 and 15.00010897 amu, respectively. use this information to determine the percent abundance of 14n.
Chemistry
1 answer:
Sergio039 [100]3 years ago
8 0
To answe this question, we will assume that the percentage abundance of 14n is y. Since the percentage abundance of 14n + percentage abundance of 15n = 100% = 1
Therefore, percentage abundance of 15n = 1 - y.

Now we know that the average atomic mass of nitrogen is 14.0067
Therefore:
average atomic mass of nitrogen = atomic mass of 14n x its percentage abundance + atomic mass of 14n x its percentage abundance
14.0067 = 14.003074002 y + <span>15.00010897 (1 - y)
</span>14.0067 = 14.003074002 y + 15.00010897 - 15.00010897 y
14.0067 - 15.00010897 = 14.003074002 y  - 15.00010897 y 
<span>0.99703497 y =  0.99340897
</span>
Therefore y = <span> 0.99636

Based on this, the percentage abundance of 14n is </span> 0.99636x100= <span>99.636%
while percentage abundance of 15n = 100% - </span>99.636% = <span>0.364%</span>
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Explanation :

The values of standard reduction electrode potential of the cell are:

E^0_{[Ni^{2+}/Ni]}=-0.23V

E^0_{[Zn^{2+}/Zn]}=-0.76V

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : Zn\rightarrow Zn^{2+}+2e^-     E^0_{[Zn^{2+}/Zn]}=-0.76V

Reaction at cathode (reduction) : Ni^{2+}+2e^-\rightarrow Ni     E^0_{[Ni^{2+}/Ni]}=-0.23V

The balanced cell reaction will be,  

Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}

E^o=(-0.23V)-(-0.76V)=0.53V

(a) Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}

E_{cell}=0.49V

(b) Now we have to calculate the cell potential when the concentration of Ni^{2+} has fallen to 0.500 M.

New concentration of Ni^{2+} = 1.50 - x = 0.500

x = 1 M

New concentration of Zn^{2+} = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}

E_{cell}=0.52V

(c) Now we have to calculate the concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}

Now put all the given values in the above equation, we get:

0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}

x=1.49M

The concentration of Ni^{2+} = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of Zn^{2+} = 0.100 + x = 0.100 + 1.49 = 1.59 M

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3 years ago
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