Answer:
11.0 L
Explanation:
The equation for this reaction is given as;
2H2 + O2 --> 2H2O
2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O
At STP;
1 mol = 22.4 L
This means;
44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O
In this reaction, the limiting reactant is H2 as O2 is in excess.
The relationship between H2 and H2O;
44.8 L = 44.8 L
11.0 L would produce x
Solving for x;
x = 11 * 44.8 / 44.8
x = 11.0 L
Answer:
The value of Kc for the reaction is 3.24
Explanation:
A reversible chemical reaction, indicated by a double arrow, occurs in both directions: reagents transforming into products (
direct reaction) and products transforming back into reagents (inverse reaction)
Chemical Equilibrium is the state in which direct and indirect reactions have the same reaction rate. Then taking into account the rate constant of a direct reaction and its inverse the chemical constant Kc is defined.
Being:
aA + bB ⇔ cC + dD
where a, b, c and d are the stoichiometric coefficients, the equilibrium constant with the following equation:
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also raised to their stoichiometric coefficients.
Then, in the reaction 3A(g) + 2B(g) ⇔ 2C(g), the constant Kc is:
![Kc=\frac{[C]^{2} }{[A]^{3} *[B]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7B2%7D%20%7D%7B%5BA%5D%5E%7B3%7D%20%2A%5BB%5D%5E%7B2%7D%20%7D)
where:
- [A]= 0.855 M
- [B]= 1.23 M
- [C]= 1.75 M
Replacing:

Solving you get:
Kc=3.24
<u><em>The value of Kc for the reaction is 3.24</em></u>
1 kPa = 7.5 mmHg so 7.0 mmHg / 7.5 mmHg x 1 kPa = .93 kPa
101.3 kPa = 1 atm so 10 kPa / 101.3 kPa x 1 atm = .0987 atm
1 kPa = 7.5 mmHg so 15 kPa x 7.5 mmHg / 1 kPa = 112.5 mmHg
Answer:
b. 4/3
Explanation:
Given data
- Final pressure: P₂ = 3 P₁
- Final temperature: T₂ = 4 T₁
We can find by what factor will the volume of the sample change using the combined gas law.

B. It keeps the warm inside