Answer:
The answer to your question is 8.74 g of He
Explanation:
Data
V = 2.4 x 10² L
P = 99 kPa
T = 0°C
mass = ?
Process
1.- Convert kPa to atm
P = 99 kPa = 99000 Pa
1 atm --------------- 101325 Pa
x --------------- 99000 Pa
x = (99000 x 1) / 101325
x = 0.977 atm
2.- Convert temperature to °K
°K = 273 + 0
°K = 273
3.- Substitution
PV = nRT
- Solve for n
n = PV / RT
n = (0.977)(2.4 x 10²) / (0.082)(273)
n = 24.48 / 22.386
n = 1.093 moles
4.- Calculate the grams of He
8 g -------------------- 1 mol
x -------------------- 1.093 moles
x = (1.093 x 8) / 1
x = 8.74 g
Explanation:
A reaction quotient is defined as the ratio of concentration of products over reactants raised to the power of their stoichiometric coefficients.
A reaction quotient is denoted by the symbol Q.
For example, 
The reaction quotient for this reaction is as follows.
Q = ![\frac{[Fe^{2+}]^{2}[Zn^{2+}]}{[Fe^{3+}]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BFe%5E%7B2%2B%7D%5D%5E%7B2%7D%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BFe%5E%7B3%2B%7D%5D%5E%7B2%7D%7D)
[Zn] will be equal to 1 as it is present in solid state. Therefore, we don't need to write it in the reaction quotient expression.
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.