Answer:
Explanation:
Hello,
In this case, by knowing the given reference reactions, one could rearrange them as follows:
Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):
Consequently, the equilibrium constant is computed as:
![Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6](https://tex.z-dn.net/?f=Kp%3D%5Cfrac%7B%5BN_2%5D%5BO_2%5D%7D%7B%5BNO%5D%5E2%7D%20%2A%20%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%5E2%7D%20%3DKp_2%2AKp_3%3D4.35x10%5E%7B18%7D%2A7.056x10%5E%7B-13%7D%3D3.07x10%5E6)
Best regards.
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
Cell theory was discovered from the scientist
Hello there.
Thomson's atomic model is best described by which of the following statements?
A nucleus with electrons moving around it like planets.
Answer:10,0000 years
Explanation:
If it takes 2000 years to weather 1cm of limestone
It will take 5*2000 years to weather 5cm of limestone. This gives us 10000 years.
