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3 years ago

How many moles of N are in 0.173 g of N2O?

Chemistry

1 answer:

krok68 [10]3 years ago

3 0

Find the RMM of both then cross multiply keeping 0.173 under N2O and you will get the answer

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The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

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3 years ago

How much heat is absorbed in the complete reaction of 3.00 grams of SiO2 with excess carbon in the reaction SiO2(g) + 3C(s) → Si

defon

Answer:

31.24 kJ

Explanation:

SiO₂(g) + 3C(s) → SiC(s) + 2CO(g) ΔH° = 624.7 kJ/mol

First we convert 3.00 grams of SiO₂ to moles, using its molar mass:

3.00 g SiO₂ ÷ 60.08 g/mol = 0.05 mol

Now we calculate the heat absorbed, using the given ΔH°:

If the complete reaction of 1 mol of SiO₂ absorbs 624.7 kJ, then with 0.05 mol:

0.05 mol * 624.7 kJ/mol = 31.24 kJ of heat would be absorbed.

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True... there are better ways using less materials and printing is that

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Vladimir79 [104]

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Explanation:

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