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inna [77]
4 years ago
11

I need help solving this for chemistry. Don’t know where to start:/

Chemistry
1 answer:
Juli2301 [7.4K]4 years ago
8 0
1 electron has charge =1.602* 10⁻¹⁹ C
1 mole of electrons have 1.602* 10⁻¹⁹*6.02*10²³C = 9.64*10⁴ C/1mol

One ion Co²⁺   takes 2e⁻ to become Co⁰.
1 mol of Co²⁺  ions take 2 mole of e⁻ to become Co⁰, so
 0.30 mol Co²⁺  ions take mole of 0.60 mol e⁻ to become Co⁰

9.64*10⁴(C/1mol) *0.60 (mol)≈ 5.8 *10⁴ Coulombs.
Correct answer is C
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Nickel has an FCC structure and an atomic radius of 0.124 nm. (12 points) a. What is the coordination number? b. Calculate the A
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Answer:

a)CN=12

b)APF=74 %

c)a=0.35 nm

d)ρ=9090.9 Kg/m^3

Explanation:

Given that

Nickel have FCC structure

We know that in FCC structure ,in FCC 8 atoms at corner with 1/8 th part in one unit cell and 6 atoms at faces with 1/2 part in one unit cell .

Z=8 x 1/8 + 1/2 x 6 =4

Z=4

Coordination number (CN)

 The number of atoms which touch the second atoms is known as coordination number.In other word the number of nearest atoms.

CN=12

Coordination number of FCC structure is 12.These 12 atoms are 4 atoms at the at corner ,4 atoms at 4 faces and 4 atoms of next unit cell.

APF

APF=\dfrac{Z\times \dfrac{4}{3}\pi R^3}{a^3}

We know that for FCC

4R=\sqrt{2}\ a

Now by putting the values

APF=\dfrac{4\times \dfrac{4}{3}\pi R^3}{\left(\dfrac{4R}{\sqrt2}\right)^3}

APF=0.74

APF=74 %

4R=\sqrt{2}\ a

4\times 0.124=\sqrt{2}\ a

a=0.35 nm

Density

\rho=\dfrac{Z\times M}{N_A\times a^3}

We know that M for Ni

M=58.69 g/mol

a=0.35 nm

N_A=6.023\times 10^{23}\ atom/mol

\rho=\dfrac{Z\times M}{N_A\times a^3}

\rho=\dfrac{4\times 58.69}{6.023\times 10^{23}\times( 0.35\times 10^{-9})^3}\ g/m^3

ρ=9090.9 Kg/m^3

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Question 4 (1 point)
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Answer:

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Determine the number of unpaired electrons expected for [Fe(NO2)6]3−and for [FeF6]3− in terms of crystal field theory.
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Answer:

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Explanation:

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