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3 years ago

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hydrogen is found primarily as two isotopes in nature: (1.0078u) and (2.0140u). Caluclate the percentage abundance of each isoto

pe based on hydrogens average atomic mass

1 answer:

kotegsom [21]3 years ago

8 0

<span>hydrogens average atomic mass = 1,01u

x = isotope about mass </span><span>(1.0078u)
y = isotope about mass </span><span>(2.0140u)
</span><span>
x + y = 100%
so:
x = 100% - y

</span> $hydrogens ~average~atomic~mass=\frac{(1.0078\cdot x)+(2.0140\cdot y)}{100}\\
1.01=\frac{(1.0078\cdot (100-y))+(2.0140y)}{100}||\cdot 100\\
101=100.78-1.0078y+2.0140\\
101-100.78=1.0062y\\
0,22=1.0062y||:1.0062\\
0.219\%=y\\\\
x=100\%-y\\
x=100\%-0.219\%\\
x=99.781\%\\\\
$
<span>
answer:
</span> $x=99.781\%\\
y=0.219\%$ <span>

</span>

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(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

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

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$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

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Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

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

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}$

where,

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

$E_{cell}$ = 0.507 V

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

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

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3 years ago