Answer:
The number of moles of water will be produced when 8.0 moles of ethane are burned = 24.0 mol.
Explanation:
- It is a stichiometry problem.
- The balanced equation of burning ethane is:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
- It is clear that 2.0 moles of ethane (C₂H₆) burns in 7.0 moles of oxygen and produce 4.0 moles of CO₂ and 6.0 moles of H₂O.
<em><u>Using cross multiplication:</u></em>
2.0 moles of ethane produces → 6.0 moles of H₂O, from the stichiometry.
8.0 moles of ethane produces → ??? moles of H₂O.
- ∴ The number of moles of water will be produced when 8.0 moles of ethane are burned = (6.0 x 8.0) / (2.0) = 24.0 mol.
The answer is C, they are found in the outermost layer of the atom. hope this helps, have an amazing day :)
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>
