Answer:
<h3>the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
Explanation:
Amount of HBr dissociated
2HBr(g) ⇆ H2(g) + Br2(g)
Initial Changes 2.15 0 0 (mol)
- 0.789 + 0.395 + 0.395 (mol)
At equilibrium 1.361 0.395 0.395 (mole)
Concentration 1.361 / 1 0.395 / 1 0.395 / 1
at equilibrium (mole/L)
![K_c=\frac{[H_2][Br_2]}{[HBr]^2} \\\\=\frac{(0.395)(0.395)}{(1.361)^2} \\\\=\frac{0.156025}{1.852321} \\\\=0.084](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5BBr_2%5D%7D%7B%5BHBr%5D%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B%280.395%29%280.395%29%7D%7B%281.361%29%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B0.156025%7D%7B1.852321%7D%20%5C%5C%5C%5C%3D0.084)
<h3>Therefore, the equilibrium constant of the decomposition of hydrogen bromide is 0.084</h3>
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The name "flint" is often the preferred name used for material with a very fine grain size and a slightly higher luster. These "fine-grained" materials break with greater predictability and produce a sharper edge. So the middle answer
Answer:
Explanation:
Just as context, write the chemical equation and the mole ratios
1) <u>Balanced chemical equation</u>:
- CuO (s) + H₂SO₄ (aq) → CuSO₄ (aq) + H₂O (l)
2) <u>Therotetical (stoichiometric) mole ratios</u>:
- 1 mol CuO : 1 mol H₂SO₄ : 1 mol CuSO₄ : 1 mol H₂O
You can calculate the percent yield from the amount of CuSO₄ obtained and the theoretical yield
3) <u>Percent yield</u>
Percent yield = (actual yield / theoretical yield)×100
- Theoretical yiedl (given): 3.19 moles CuSO₄
- Actual yield (given): 2.50 moles CuSO₄
Substitute the values in the formula:
- Percent yield = (2.50 moles CuSO₄ / 3.19 moles CuSO₄)×100 = 78.4%
