<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
Ka x Kb = Kw Kw = 1 x 10⁻¹⁴
(1.3 x 10⁻⁷) x Kb = 1 x 10⁻¹⁴
Kb = ( 1 x 10⁻¹⁴ ) / ( 1.3 x 10⁻⁷)
Kb = 7.7 x 10⁻⁸
Answer D
hope this helps!
Answer: O 10.1 is your answer! Please mark me as brainlyest!?
Answer:
Explanation:
Hello,
In this case, the combustion of methane is shown below:
And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:
Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:
Best regards.
<span>greater dispersion forces is in fact correct</span>
