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Rzqust [24]
2 years ago
10

How many moles of ammonia can be formed from 4.0 mol H2

Chemistry
1 answer:
Bess [88]2 years ago
3 0

Answer: 1. 2.7 moles  of ammonia are formed

2. 12.0 moles  of hydrogen are required

3. 2.0 moles of nitrogen are required

Explanation:

The balanced chemical equation is:

N_2+3H_2\rightarrow 2NH_3

According to stoichiometry:

3 moles of hydrogen form = 2 moles of ammonia

Thus 4.0 moles of hydrogen form =\frac{2}{3}\times 4.0=2.7moles of ammonia

According to stoichiometry:

2 moles of ammonia are formed by = 3 moles of hydrogen

Thus 8.0 moles of ammonia are formed by  =\frac{3}{2}\times 8.0=12.0moles of hydrogen

According to stoichiometry:

3 moles of hydrogen react with = 1 mole of nitrogen

Thus 6.0 moles of hydrogen react with =\frac{1}{3}\times 6.0=2.0moles of nitrogen

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Which of the following is NOT true regarding Rutherford's Gold Foil experiment?
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Answer:

The area around the nucleus must be of low mass.

Explanation:

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3 years ago
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
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<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

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

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

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