Answer:
1.03 atm
Explanation:
Primero <u>convertimos 21 °C y 37 °C a K</u>:
- 21 °C + 273.16 = 294.16 K
- 37 °C + 273.16 = 310.16 K
Una vez tenemos las temperaturas absolutas, podemos resolver este problema usando la<em> ley de Gay-Lussac</em>:
En este caso:
Colocando los datos:
- 294.16 K * P₂ = 310.16 K * 0.98 atm
Y <u>despejando P₂</u>:
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
Answer : The mass of sucrose added to water will be, 189.0 grams.
Explanation :
As we are given that 9 % solution (mass per volume) that means 9 grams of sucrose present in 100 mL volume of solution.
Total given volume of solution = 2.1 L = 2100 mL (1 L = 1000 mL)
Now we have to determine the mass of sucrose in solution.
As, 100 mL of solution contains 9 grams of sucrose
So, 2100 mL of solution contains
grams of sucrose
Therefore, the mass of sucrose added to water will be, 189.0 grams.
Answer:
Hello is there a specific aliment you have in mind If so I can explain and tell you the answer.