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KingMack1136
20.03.2020 •
Chemistry
A mixture consisting of 0.250 M N2(g) and 0.500 M H2(g) reaches equilibrium according to the equation N2(g) + 3 H2(g) → 2 NH3(g). At equilibrium, the concentration of ammonia is 0.150 M. Calculate the concentration of H2(g) at equilibrium.
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Ответ:
The concentration of H2 at the equilibrium is 0.275 M
Explanation:
Step 1: Data given
Molarity of N2 = 0.250 M
Molarity of H2 = 0.500 M
At equilibrium, the concentration of ammonia is 0.150 M
Step 2: The balanced equation
N2(g) + 3 H2(g) → 2 NH3(g)
Step 3: The initial concentrations
[N2] = 0.250 M
[H2] = 0.500 M
[NH3] = 0 M
Step 4: The concentrations at the equilibrium
[N2] = 0.250 - X M
[H2] = 0.500 - 3X M
[NH3] = 2X M = 0.150
X = 0.150 / 2 = 0.075
[N2] = 0.250 - 0.075 M = 0.175 M
[H2] = 0.500 - 3X M = 0.275 M
[NH3] = 2X M = 0.150
Step 5: Calculate Kc
Kc = [NH3]² / [N2][H2]³
Kc = (0.150²) / (0.175 * 0.275³)
Kc = 1.70
The concentration of H2 at the equilibrium is 0.275 M
Ответ:
The mass of sample is 21.6 grams.
Explanation:
We are given:
Volume of cylinder without object,
= 10 mL
Volume of cylinder with object,
= 18 mL
Volume of object =![V_2-V_1=18-10=8mL](/tpl/images/0226/7313/e8999.png)
To calculate volume of a substance, we use the equation:
We are given:
Density of sample (aluminum) = 2.7 g/mL
Volume of sample (aluminum) = 8 mL
Putting values in above equation, we get:
Hence, the mass of sample is 21.6 grams.