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statuscvo [17]
3 years ago
10

Can someone please help me ? Click the picture

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
8 0

Explanation:

Keq =[NO2] ^2 / [N2 O4]

4.64 x 10^-3

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Calcium has a cubic closest packed structure (fcc) as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate t
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Answer:

1.536 g/cm^3 is the density of solid calcium.

Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

We have:

r = 197 pm = 197\times 10^{-12} m = 197\times 10^{-12}\times 100 cm

r=197\times 10^{-10} m

a=r\times 2\sqrt{2}

a=197\times 10^{-10} m\times 2\sqrt{2}=557.200\times 10^{-10} cm

M = 40 g/mol

Z = 4

On substituting all the given values , we will get the value of 'a'.

\rho =\frac{4\times 40 g/mol}{6.022\times 10^{23} mol^{-1}\times (557.200\times 10^{-10} cm)^{3}}

\rho =1.536 g/cm^3

1.536 g/cm^3 is the density of solid calcium.

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Answer:

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Answer:

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