Therefore, 1 mole<span> of gold weighs </span>196.9665<span> grams. So, in 2.8 grams of gold we would have:</span>
(2.8 gram)(1 mole/196.9665<span> gram) = 0.0142 mole.</span>
From Avogadro's number, we know that there are approximately 6.02 x 1023<span>atoms/mole.</span>
If points E and D are midpoints of triangle ABC, then the perimeter of the triangle is:AB + BC + CA = 48, or:
3 t + 3 t + 4 t + 4 t + 7 t + 6 = 4821 t = 48 - 621 t = 42t = 42 : 21t = 2Answer: A. 2.
The ansewr is A i would guess
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)