How many grams of carbon dioxide are produced from the combustion of 250.G of ethane, C3H8 in the reaction C3H8 +5O2 - 3CO2 +4H2
1 answer:
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Answer:
b)15.0°C
Explanation:
Specific Heat of Water=4.2 J/g°C
This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.
∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.
80×4.2 J=336 J
Total Energy Provided=1680 J
The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}
Temperature increase=
=5°C
Initial Temperature =10°C
Final Temperature=Initial + Increase in Temperature
=10+5=15°C
Answer : The correct option is, (B) 0.11 M
Solution :
First we have to calculate the concentration and
.
The given equilibrium reaction is,
Initially 0.70 0.70 0
At equilibrium (0.70-x) (0.70-x) x
The expression of will be,
Now put all the given values in the above expression, we get:
By solving the term x, we get
From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.
Thus, the concentration of at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of at equilibrium = x = 0.59 M
Therefore, the concentration of at equilibrium is 0.11 M