Solving this chemistry is a little bit hard because the question didn't give some important detailed.
So first, there are a couple problems with your question.
We will just need to know which direction will it proceed to reach equilibrium.
Your expression for Kc (and Qc ) for the reaction should be:
Kc = [C] / [A] [B]^2
You have not provided a value for Kc, so a value of Qc tells you absolutely nothing. Qc is only valuable in relation to a numerical value for Kc. If Qc = Kc, then the reaction is at equilibrium. If Q < K, the reaction will form more products to reach equilibrium, and if Q > Kc, the reaction will form more reactants.
Answer:
N2
Explanation:
Rate of effusion is defined by Graham's Law:
(Rate 1/Rate 2) = (sqrt (M2)/ sqrt (M1))
(Where M is the molar mass of each substance. )
Molar Mass of oxygen, O2, is 32 (M1).
Rate of effusion of O2 to an unknown gas is .935(Rate 1).
Rate 2 is unknown so put 1.
Solve for x (M2).
.935/1 = sqrt x/ sqrt32
.935 x sqrt 32 = sqrt x
5.29 = sq rt x
5.29^2 = 27.975 = 28
N2 has a molar mass of 28 so it is the correct gas.
<u>Answer:</u> The equilibrium partial pressure of chlorine gas is 0.360 atm
<u>Explanation:</u>
For the given chemical equation:
The expression of 
for above reaction follows:
We are given:
Putting values in above equation, we get:
Hence, the equilibrium partial pressure of chlorine gas is 0.360 atm
Answer: 22g of chlorine would be needed to carry out this synthesis reaction
Explanation:
A synthesis reaction is one in which two or more than two elements combine together to forma single product.
The atoms present in the reactants are found on the product side. According to the law of conservation of mass, the number of atoms on both sides of the arrow must be same as the total mass must be conserved.
15 grams of sodium reacts with 22 grams of chlorine to yield 37 grams of sodium chloride. Thus 22g of chlorine would be needed to carry out this synthesis reaction.
Answer:
The products are: KCl03 and H20.
Explanation:
The reaction between HC03 (chloric acid) and KOH (potassium hydroxide) is:
HC03 + KOH ----> KCl03 (KCl03 and H20) + H20 (water)
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<em>The reaction is of the double displacement type (in this case parts of the reagents are exchanged, producing two generating new compounds).</em>
