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zloy xaker [14]

2 years ago

12

For the reaction N2(g)+O2(g)⇌2NO(g)N2(g)+O2(g)⇌2NO(g) classify each of the following actions by whether it causes a leftward shi

ft, a rightward shift, or no shift in the direction of the reaction.

1 answer:

bearhunter [10]2 years ago

5 0

Explanation:

Here is the complete question:

1. For the reaction

N2(g)+O2(g)⇌2NO(g)N2(g)+O2(g)⇌2NO(g)

classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the reaction.

a. halve nitrogen

b. double oxygen

c. halve oxygen

d. double nitrogen

e. double nitrogen monoxide

f. halve nitrogen monoxide

Taking note of Le Chatelier's Principle, the following is observed

1. right

2. left

3. left

4. left

5. right

6. right

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A chemistry student is given the task of analyzing the three unknown samples in the previous question. Compare the densities fro

OLEGan [10]

Answer:

It’s B

Explanation:

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2 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

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2 years ago

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Answer:
<span>D) combine light nuclei
</span>
Explanation:
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2- Fusion occurs when two light atomic nuclei combine together to form a single atom.

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2 years ago

Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these

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The formation of nitric acid from nitrogen, hydrogen, and oxygen can be written as,

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Since all the reactants are in their elemental forms, the simplified equation would be,

Hrxn = Hproduct

Substituting,

Hrxn = (-186.81 kJ/mol)(2 mols)

<em>Answer: -372.42 kJ</em>

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