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marin [14]
3 years ago
14

Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 125 pm . Calculate the density of s

olid crystalline chromium in grams per cubic centimeter.
Chemistry
1 answer:
mafiozo [28]3 years ago
8 0

Answer:

\rho=7.15\ g/cm^3

Explanation:

The expression for density is:

\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}

N_a=6.023\times 10^{23}\ {mol}^{-1}

M is molar mass of Chromium = 51.9961 g/mol

For body-centered cubic unit cell , Z= 2

\rho is the density  

Radius = 125 pm = 1.25\times 10^{-8}\ cm

Also, for BCC, Edge\ length=\frac{4}{\sqrt{3}}\times radius=\frac{4}{\sqrt{3}}\times 1.25\times 10^{-8}\ cm=2.89\times 10^{-8}\ cm

Thus,  

\rho=\frac{2\times \:51.9961}{6.023\times \:10^{23}\times \left(2.89\times 10^{-8}\right)^3}\ g/cm^3

\rho=7.15\ g/cm^3

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What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.
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From the given cell representation, we conclude that

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The oxidation-reduction half cell reaction will be,

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

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o{cell} of the reaction, we use the equation:

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Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

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

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E_{cell} = ?

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

E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}

E_{cell}=1.14V

Therefore, the voltage of the voltaic cell is 1.14 V

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