Answer:
Kc = [CH₄] / [H₂]²
Kp = [CH₄] / [H₂]² * (0.082*T)^-1
Explanation:
Equilibrium constant, Kc, is defined as the ratio of the concentrations of the products over the reactants. Also, each concentration of product of reactant is powered to its coefficient.
<em>Pure solids and liquids are not taken into account in an equilibrium</em>
Thus, for the reaction:
C(s)+ 2H₂(g) ⇌ CH₄(g)
Equilibrium constant is:
<h3>Kc = [CH₄] / [H₂]²</h3>
Now, using the formula:
Kp = Kc* (RT)^Δn
<em>Where R is gas constant (0.082atmL/molK), T is the temperature of the reaction and Δn is difference in coefficients of gas products - coefficients of gas reactants (1 - 2= -1)</em>
Replacing:
<h3>Kp = [CH₄] / [H₂]² * (0.082*T)^-1</h3>
<em />
Answer:
The insect exerts no force on the windshield, and the windshield strikes the insect with a large force.
Explanation:
:#D
Answer:
0.221M
Explanation:
From the question ,
The Molarity of AgNO₂ = 0.310 M
Hence , the concentration of Ag⁺ = 0.301 M
The volume of AgNO₂ = 250 mL
and,
The Molarity of Sodium chromate = 0.160 M
The volume of Sodium chromate = 100 mL.
As the solution is mixed the final volume becomes ,
250mL +100mL = 350mL
Now, using the formula , to find the final molarity of the mixture ,
M₁V₁ ( initial ) = M₂V₂ ( final )
substituting the values , in the above equation ,
0.310M * 250ml = M₂ * 350ml
M₂ = 0.221M
Hence , the concentration of the silver in the final solution = 0.221M
I think the answer is B. I really don't remember but im pretty sure its that one.
Explanation:
Since, it is given that carbon dioxide is completely removed by absorption with NaOH. And, pressure inside the container is 0.250 atm.
For Kr = 0.250 atm and pressure 
will be calculated as follows.
= (0.708 - 0.250) atm
= 0.458 atm
Now, we will calculate the mole fraction as follows.
= 0.646
Kr = 
= 0.353
Now, we will convert into gram fraction as follows.
= 28.424
Kr = 
= 29.57
Therefore, total mass is calculated as follows.
Total mass = (28.424 + 29.57)
= 57.994
Hence, the percentage of and Kr are calculated as follows.
= 49%
Kr = 
= 51%
Hence, amount of and Kr present i mixture is as follows.
in mixture = 
= 17.15 g
Kr = 
= 17.85 g
Thus, we can conclude that 17.15 g of is originally present and 17.85 g of Kr is recovered.
