Answer:
[H₂] = 1.61x10⁻³ M
Explanation:
2H₂S(g) ⇋ 2H₂(g) + S₂(g)
Kc = 9.30x10⁻⁸ = ![\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_%7B2%7D%5D%5E2%5BS_%7B2%7D%5D%7D%7B%5BH_%7B2%7DS%5D%5E2%7D)
First we <u>calculate the initial concentration</u>:
0.45 molH₂S / 3.0L = 0.15 M
The concentrations at equilibrium would be:
[H₂S] = 0.15 - 2x
[H₂] = 2x
[S₂] = x
We <u>put the data in the Kc expression and solve for x</u>:


We make a simplification because x<<< 0.0225:

x = 8.058x10⁻⁴
[H₂] = 2*x = 1.61x10⁻³ M
That it’s the power house and only hold protons and neutrons.
Answer:
- The first picture attached is the diagram that accompanies the question.
- The<u> second picture attached</u> is the diagram with the answer.
Explanation:
In the box on the left there are 8 Cl⁻ ions and 8 Na⁺ ions.
The dissociaton equation for NaCl(aq) is:
- NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)
The dissociation equation for CaCl₂ (aq) is:
- CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)
A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.
Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half) and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).
Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.