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aalyn [17]
2 years ago
5

Calculate the molar solubility of mercury (I) bromide, Hg2Br2, in 1.0 M KBr. The Ksp for Hg2Br2 is 5.6 X 10−23. (Hint: How would

the Br− concentration from the sparingly soluble compound itself compare to the Br− concentration that comes from the KBr?
Chemistry
1 answer:
Vladimir [108]2 years ago
6 0

Answer:

The correct answer is 5.6 × 10⁻²³ M.

Explanation:

As a highly soluble salt, KBr dissolves easily in water, while Hg₂Br₂ is very less soluble in comparison to KBr.

Let the solubility of Hg₂Br₂ is S mol per liter.

Therefore,

KBr (s) (1.0 M) ⇒ K⁺ (aq) (1M) + Br⁻ (aq) (1M)

Hg₂Br₂ (s) (1-S) ⇔ Hg₂⁺ (aq) (S) + 2Br⁻ (aq) (2S)

Net [Br-] = (2S + 1) M

Ksp = S (2S + 1)²

Ksp = S (4S² + 1 + 4S)

Ksp = 4S³ + S + 4S²

As the solubility is extremely less, therefore, we can ignore S² and S³. Now,

Ksp = S = 5.6 × 10⁻²³ M

Hence, the solubility of Hg₂Br₂ is 5.6 × 10⁻²³ M.

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2 years ago
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Now consider the reaction when 45.0 g NaOH have been added. What amount of NaOH is this, and what amount of FeCl3 can be consume
gayaneshka [121]

The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.

The mass of NaOH is 45 g

The molar mass of NaOH = 40 g/mol

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3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl

3 mol of NaOH react with 1 mol of FeCl₃

1.125 moles of NaOH will react with x moles of FeCl₃

x = 1.125 / 3

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The molar mass of FeCl₃ is 162 g/mol

The mass of FeCl₃ = moles × mass

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Thus, the amount of FeCl₃, which can be consumed is 60.75 g

6 0
2 years ago
10. A solution contains 130 grams of KNO3 dissolved in 100 grams of water When 3 more grams of KNO3 is added, none of it dissolv
Murljashka [212]

Answer:

Option B

Explanation:

We will check the solubility graph for potassium nitrate,  KNO 3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.

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6 0
2 years ago
2Na2O2 + 2CO2 → 2Na2CO3 + O2
ahrayia [7]

the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

we need to first find the theoretical yield

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molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

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actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

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