The specific heat capacity of gold is 0.128 J/g C. How much would be needed to warm 250.0 grams of gold from 25.0 C to 100.0 C.
1 answer:
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Answer:
0.36 M
Explanation:
There is some info missing. I think this is the complete question.
<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible: </em>
<em>N₂(g) +O₂(g) ⇄ 2 NO(g) </em>
<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>
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Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:
[N₂] = 0.30 mol / 0.250 L = 1.2 M
[NO] = 0.70 mol / 0.250 L = 2.8 M
We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.
N₂(g) +O₂(g) ⇄ 2 NO(g)
I 1.2 0 2.8
C +x +x -2x
E 1.2+x x 2.8 - 2x
The equilibrium constant (K) is:
Solving for x, the positive one is x = 0.3601 M
[O₂] = 0.3601 M ≈ 0.36 M