Answer:
Kc = 3.72 × 10⁶
Explanation:
Let's consider the following reaction:
NH₄HS(g) ⇄ NH₃(g) + H₂S(g)
At equilibrium, we have the following concentrations:
[NH₄HS] = 0.196 M (assuming a 1 L flask)
[NH₃] = 9.56 × 10² M
[H₂S] = 7.62 × 10² M
We can replace this data in the Kc expression.
![Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_%7B3%7D%5D%20%5Ctimes%20%5BH_%7B2%7DS%5D%20%7D%7B%5BNH_%7B4%7DHS%5D%7D%20%3D%5Cfrac%7B9.56%20%5Ctimes%2010%5E%7B2%7D%20%20%5Ctimes%207.62%20%20%5Ctimes%2010%5E%7B2%7D%7D%7B0.196%7D%20%3D3.72%20%5Ctimes%2010%5E%7B6%7D)
Answer:
During the pumping of hot air into the expansion valve.
Answer:
6. In order from left to right the balanced coeficcients are,
6, 1, 2, 3
8. 
9. In order from left to right the balanced coeficcients are,
1, 11, 7, 8
6 - one sodium atom, 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms.
V ( H2SO4) = 35 mL / 1000 => 0.035 L
M ( H2SO4) = ?
V ( NaOH ) = 25 mL / 1000 => 0.025 L
M ( NaOH ) = 0.320 M
number of moles NaOH:
n = M x V
n = 0.025 x 0.320 => 0.008 moles of NaOH
Mole ratio:
<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??
0.008 x 1 / 2 => 0.004 moles of H2SO4 :
Therefore:
M ( H2SO4) = n / V
M = 0.004 / 0.035
= 0.114 M
hope this helps!
