Answer: 48.6 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_6H_6}\times \Delta H_{C_6H_6})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_6H_6%7D%5Ctimes%20%5CDelta%20H_%7BC_6H_6%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-6534.0=[(12\times -393.5)+(6\times -285.8)]-[(15\times 0)+(2\times \Delta H_{C_6H_6})]](https://tex.z-dn.net/?f=-6534.0%3D%5B%2812%5Ctimes%20-393.5%29%2B%286%5Ctimes%20-285.8%29%5D-%5B%2815%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelta%20H_%7BC_6H_6%7D%29%5D)
Therefore, the enthalpy change for benzene is 48.6 kJ/mol
The correct answer is 9.7 grams because mg are 1,000 difference to grams.
Answer:
This is False
Explanation:
A chemical change is when the matter is completely changed.
When you have water and boil it to be a gas, it is still the same matter, just in a different form.
I hope this helps :)
What elem eat is this? That has three orbitals? Usually the first orbital will contain only two electrons. The second orbital would contain four, the third would contain 16
Answer:
t = 5.7634 s
Explanation:
- A → Pdts
- - rA = K (CA)∧α = - δCA/δt
∴ T = 400°C
∴ α = 1 ....first-order
∴ CAo = 0.950 M
∴ CA = 0.300 M
⇒ t = ?
⇒ - δCA/δt = K*CA
⇒ - ∫δCA/CA = K*∫δt
⇒ Ln (CAo/CA) = K*t
⇒ t = Ln(CAo/CA) / K
⇒ t = (Ln(0.950/0.300)) / (0.200 s-1)
⇒ t = 1.1527 / 0.200 s-1
⇒ t = 5.7634 s
