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Ipatiy [6.2K]
3 years ago
6

Choose all of the following mole ratios that are correct for the reaction given below. Be sure to choose all that apply.

Chemistry
2 answers:
BigorU [14]3 years ago
7 0

Answer:

3 h2=2nh3

2nh3=1n2

3h2=1n2

Explanation:

Dennis_Churaev [7]3 years ago
5 0

Answer:

2 moles NH3 = 1 mole of N2

3 moles of H2 = 1 mole of N2

3 moles of H2 = 2 moles of NH3

Explanation:

The balanced equation for the reaction is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Thus, we can say that:

1 mole of N2 = 3 moles of H2

1 mole of N2 = 2 moles of NH3

3 moles of H2 = 2 moles of NH3

Thus, considering the options given above, the right answers to the question are:

2 moles NH3 = 1 mole of N2

3 moles of H2 = 1 mole of N2

3 moles of H2 = 2 moles of NH3

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For the reaction: 3 H2(g) + N2(g) <--> 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [
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The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

Read more on equilibrium constant;

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What happens when samples of matter interact?
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2 years ago
You are given 10ml (M) 20 Naoh solution in a conical flask and asked to titrate with (M) 20 Hcl and (M) 20 H2so4 separately. cal
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Answer:

n_{HCl}=0.2molHCl\\n_{H_2SO_4}=0.1molH_2SO_4

Explanation:

Hello!

In this case, since the reactions between NaOH and the acids are:

NaOH+HCl\rightarrow NaCl+H_2O\\\\2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

Whereas we can see the 1:1 and 2:1 mole ratios between NaOH and HCl and H2SO4 respectively. In such a way, at the equivalence point we realize that:

n_{HCl}=n_{NaOH}=V_{NaOH}M_{NaOH}=0.01L*20mol/L=0.2molHCl\\\\2n_{H_2SO_4}=n_{NaOH}\\\\n_{H_2SO_4}=\frac{1}{2} V_{NaOH}M_{NaOH}=\frac{0.01L*20mol/L}{2} =0.1molH_2SO_4

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