Answer: The equilibrium concentration of hydrogen gas is 0.0269 M
Explanation:
The chemical reaction follows the equation:

At t = 0 0.044M 0.044M 0.177M
At
(0.044-x)M (0.044-x)M (0.177+x)M
The expression for
for the given reaction follows:
![K_c=\frac{[HI]^2}{[H_2]\times [I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5Ctimes%20%5BI_2%5D%7D)
We are given:

Putting values in above equation, we get:


Hence, the equilibrium concentration of hydrogen gas is (0.044-x) M =(0.044-0.0171) M= 0.0269 M
To solve this, we can use two equations.
t1/2 = ln 2 / λ = 0.693 / λ
where, t1/2 is half-life and λ is the decay constant.
t1/2 = 10 min = 0.693 / λ
Hence, λ = 0.693 / 10 min - (1)
Nt = Nο e∧(-λt)
Nt = amount of atoms at t =t time
Nο= initial amount of atoms
t = time taken
by rearranging the equation,
Nt/Nο = e∧(-λt) - (2)
From (1) and (2),
Nt/Nο = e∧(-(0.693 / 10 min) x 20 min)
Nt/Nο = 0.2500
Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%
= (Nt/Nο ) x 100%
= 0.2500 x 100%
= 25.00%
Hence, Percentage of remaining nuclei is 25.00%
Answer:
Q = -33.6kcal .
Explanation:
Hello there!
In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:

Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:

Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:

Best regards!