The student originally has 252 grams of water in this experiment.
LAW OF CONSERVATION OF MASS:
- The law of conservation of mass explains that matter (mass) can neither be created nor destroyed but can be changed from one form to another.
- This means that in a chemical reaction, the sum of the masses of the reactants must equate to the total mass of product(s).
- According to this question, a student conducts an experiment to separate water into hydrogen and oxygen. The student collects 28.0 g of hydrogen and 224.0 g of oxygen.
- Since hydrogen and oxygen are the constituent elements of water, the sum of their masses must equate the mass of water.
- Therefore, 224g of oxygen + 28g of hydrogen = 252g of water.
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Answer:
0.0119
Explanation:
There was a part missing. I think this is the whole question:
<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>
A (aq) ⇌ 2B (aq) + C(aq)
<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>
First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.
A (aq) ⇌ 2B (aq) + C (aq)
I 0.0510 0 0
C -x +2x +x
E 0.0510-x 2x x
Since the concentration at equilibrium of A is 0.0153 M, we get
We can use the value of x to calculate the concentrations at equilibrium.
![[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\](https://tex.z-dn.net/?f=%5BA%5De%20%3D%200.0153%20M%20%5C%5C%5BB%5De%20%3D%202x%20%3D%202%280.0357%29%20%3D%200.0714%20M%20%5C%5C%5BC%5De%20%3D%20x%20%3D%200.0357%20M%20%5C%5C)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
![Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BB%5D%5E%7B2%7D%20%20%5Ctimes%20%5BC%5D%7D%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B0.0714%5E%7B2%7D%20%20%5Ctimes%200.0357%7D%7B0.0153%7D%20%3D%200.0119)
The equilibrium constant for this reaction at equilibrium is 0.0119.
You can learn more about equilibrium here: brainly.com/question/4289021
Answer: 0.0944 gram of H2
Explanation:
Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:
2.0 atm x (973 K / 298 K) = 6.53 atm
Where
Kc = Kp because the moles of product equals the moles of reactants.
At equilibriuim, the amounts are
P(H2) = 6.53 - x
P(CO2) = 6.53 - x
P(H2O) = x
P(CO) = x
Kc = Kp = .534 = (x)(x) / [(6.53 - x)(6.53 - x)]
Take the square root of each side
(.534)^0.5 = x / (6.53 - x)
x = 0.731 (6.53 - x)
x = 4.77 - 0.731x
1.731x = 4.77
x = 4.77 / 1.731 = 2.76 atm
P(H2) at equilibriuim = 6.53 - 2.76 = 3.77 atm
P(CO2) at equilibrium = 6.53 - 2.76 = 3.77 atm
PV = nRT
n = PV/RT = [(3.77 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0472 mol H2
0.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g
Answer: option B. 0.020 mole
Explanation:Please see attachment for explanation
