How much energy is required to raise 10 grams of water by 10 degrees Celsius? (in Joules)
2 answers:
<u>Answer:</u> The amount of heat required by water is 418.6 J
<u>Explanation:</u>
To calculate the heat absorbed by water, we use the equation:
where,
q = heat absorbed
m = mass of water = 10 g
c = heat capacity of water = 4.186 J/g°C
= change in temperature = 10°C
Putting values in above equation, we get:
Hence, the amount of heat required by water is 418.6 J
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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>
<em />
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