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Vilka [71]
4 years ago
12

Please help me! This is too hard!?

Chemistry
1 answer:
Annette [7]4 years ago
5 0

Answer: 59.24 atm

Explanation:

Given that:

Original Volume of gas V1 = 2.7L

Temperature T1 = 42.7°C

Convert Celsius to Kelvin

(42.7°C + 273 = 315.7K)

Pressure P1 = 684.9 torr

Final Volume V2 = 0.14 L

Final temperature T2 = 803.1°C

Convert Celsius to Kelvin

(803.1°C + 273 = 1076.1K)

Final pressure = ?

Then, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(684.9 torr x 2.7L)/315.7K = (P2 x 0.14L)/1076.1K

1849.23/315.7 = 0.14P2/1076.1

Then, cross multiply

1849.23 x 1076.1 = 315.7 x 0.14P2

1989956.403 = 44.198P2

Divide both sides by 44.198

1989956.403/44.198 = 44.198P2/44.198

45023.67 torr = P2

Now, convert the pressure in torr to atmosphere

If 760 torr = 1 atm

45023.67 torr = Z atm

Then, cross multiply

760 torr x Z = 45023.67 torr x 1 atm

Z = 45023.67 torr / 760 torr

Z = 59.24 atm

Thus, the new pressure of the gas will be 59.24 atm

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12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

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2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

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3.0 atm      0              0

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(3.0-2p)     p               3p

Equilibrium partial pressure of nitrogen gas = p = 0.90 atm

The expression of a pressure equilibrium constant will be given by :

K_p=\frac{p_{N_2}\times (p_{H_2})^3}{(p_{NH_3})^2}

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=\frac{0.90 atm\times (3\times 0.90 atm)^3}{(3.0-2\times 0.90 atm)^2}

K_p=12.30\approx 12.

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

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When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+
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Answer:

D) 65.7%

Explanation:

Based on the reaction:

2H2(g)+O2(g)⟶2H2O(l)

<em>2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.</em>

<em />

To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.

<em>Theoretical yield:</em>

Moles of 5.58g H₂:

5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂

As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:

2.768 moles H₂O ₓ (18.015g / mol) =

49.86g H₂O is theoretical yield

<em>Percent yield:</em>

Percent yield = Actual yield / Theoretical yield ₓ 100

32.8g H₂O / 49.86g ₓ 100 =

65.7% is percent yield of the reaction

<h3>D) 65.7% </h3>

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4 years ago
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