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MakcuM [25]
3 years ago
5

A mixture of H2 and water vapor is present in a closed vessel at 20°C. The total pressure of the system is 755.0 mmHg.

Chemistry
1 answer:
sukhopar [10]3 years ago
7 0

Answer:

737.5mmHg

Explanation:

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Likurg_2 [28]

Answer:

pounds

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8 0
2 years ago
Consider the reaction H2(g) + I2(g) <---> HI(g) with an equilibrium constant of 46.3 and a reaction quotient of 525. Which
WARRIOR [948]

Answer:

The equilibrium will shift to the left to favor the reactants.

Explanation:

Remember that the reaction quotient (Qc) is derived from initial concentrations of reactants and products. Since Qc is greater than Kc, this means that initial concentrations are heavily impacted by a high product concentration ([HI]). Therefore, the reverse reaction will occur and actually create more reactants again ([H2] and [I2]). Thus, the answer is that the equilibrium will shift to the left side to favor the reactants.

8 0
3 years ago
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Musya8 [376]
The answer is C
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3 years ago
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True or False. Solutions for which water is the solvent are called aqueous solutions.
Stolb23 [73]
This is true. Water is the solvent in aqueous solutions
5 0
3 years ago
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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