Answer: 600 kJ
-
Explanation:
C₃H₈ (g) + 5 O₂ (g) =============== 3 CO₂ (g) + 4 H₂O (l)
Δ⁰Hf kJ/mol -104 0 -393.5 -285.8
Δ⁰Hcomb C₃H₈ = 3(-393.5) + 4 (-285.80) - (-104) kJ/mol
Δ⁰Hcomb = 2219.70 kJ/mol
n= m /MW MW c₃H₈ = 44.1 g/mol
n= 12 g/44.1 g/mol = 0.27 mol
then for 12 g the heat released will be
0.27 mol x 2219.70 kJ/mol = 600 KJ
<u>Answer:</u> The value of 
for the reaction at 690 K is 0.05
<u>Explanation:</u>
We are given:
Initial pressure of 
= 1.0 atm
Total pressure at equilibrium = 1.2 atm
The chemical equation for the decomposition of phosgene follows:
Initial: 1 - -
At eqllm: 1-x x x
We are given:
Total pressure at equilibrium = [(1 - x) + x+ x]
So, the equation becomes:
![[(1 - x) + x+ x]=1.2\\\\x=0.2atm](https://tex.z-dn.net/?f=%5B%281%20-%20x%29%20%2B%20x%2B%20x%5D%3D1.2%5C%5C%5C%5Cx%3D0.2atm)
The expression for for above equation follows:
Putting values in above equation, we get:
Hence, the value of for the reaction at 690 K is 0.05
An atomic mass unit is defined as precisely 1/12 the mass of an atom of carbon-12.
At almost the opposite point on the Earth's surface, the "P" waves reappear. The shadow zone exists because the waves are refracted as they pass through the boundary between the mantle and the core and are diverted from their original paths.
Given:
P1 = 13.0 atm
T1 = 20 °C
T2 = 102 °C
Required:
P2 of oxygen
Solution:
At constant volume,
we can apply Gay-Lussac’s law of pressure and temperature relationship
P1/T1=P2/T2
(13.0 atm) / (20 °C)
= P2 / (102 °C)
P2 = 66.3 atm
The answer is not in the choices given.
