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LekaFEV [45]
3 years ago
10

Please help to me solve these ! :)

Chemistry
2 answers:
KengaRu [80]3 years ago
8 0

1. The solubility of NaCl at 25 degrees C would be between the solubilities at 20 and 30 degrees C. A reasonable answer would be 36 grams/100 g water

2. From the table, it’s clear that the salts are more soluble at higher temperatures, indicating that an increase in temperature increases solubility.

3. At 50 degrees C, a saturated ammonium chloride solution will have 50.6 grams of salt per 100 g water. At 20 degrees C, the solution can hold only 37.3 grams of salt per 100 g water. Thus, 13.3 grams of salt will precipitate per 100 grams of water.

emmainna [20.7K]3 years ago
4 0
I. The solubility of NaCl at 25 degrees C would be between the solubilities at 20 and 30 degrees C. A reasonable answer would be 36 grams/100 g water
ii. From the table, it’s clear that the salts are more soluble at higher temperatures, indicating that an increase in temperature increases solubility.
iii. At 50 degrees C, a saturated ammonium chloride solution will have 50.6 grams of salt per 100 g water. At 20 degrees C, the solution can hold only 37.3 grams of salt per 100 g water. Thus, 13.3 grams of salt will precipitate per 100 grams of water.
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Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

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