If it takes 526 J of energy to warm 7.40g of water by 17 degrees celsius, how much energy would be needed to warm 7.40g of water
2 answers:
Answer:
Heat required = 1702 J
Explanation:
The amount of heat energy (Q) required to raise the temperature of water of mass (m) is given as:
where c = specific heat capacity
ΔT =change in temperature
It is given that:
When ΔT = 17 C, Q = 526 J and m = 7.40 g
i.e.
The amount of heat required to increase the temperature of 7.40 g water by 55 C would be:
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Answer:
4.41
Explanation:
Step 1: Write the balanced equation
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
Step 2: Calculate the respective concentrations
Step 3: Make an ICE chart
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
I 0.100 0.300 0 0
C -x -3x +x +x
E 0.100-x 0.300-3x x x
Step 4: Find the value of x
Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396
Step 5: Find the concentrations at equilibrium
[CO] = 0.100-x = 0.100-0.0396 = 0.060 M
[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M
[CH₄] = x = 0.0396 M
[H₂O] = x = 0.0396 M
Step 6: Calculate the equilibrium constant (Kc)
- Density=3g/mL
- Volume=0.8mL