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zlopas [31]
3 years ago
15

15.24. Calcium hydroxide, also known as slaked lime, is used in industrial processes in which low concentrations of base are req

uired. Only 0.16 g of Ca(OH)2 dissolves in 100 mL of water at 25°C. What is the concentration of hydroxide ions in 250 mL of a solution containing the maximum amount of dissolved calcium hydroxide?​
Chemistry
1 answer:
Ivahew [28]3 years ago
7 0

The concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

We'll begin by calculating the number of mole of in 0.16 g of Ca(OH)₂. This can be obtained as follow:

Mass of Ca(OH)₂ = 0.16 g

Molar mass of Ca(OH)₂ = 40 + 2[16 + 1] = 74 g/mol

<h3>Mole of Ca(OH)₂ =? </h3>

Mole = mass / molar mass

Mole of Ca(OH)₂ = 0.16 / 74

<h3>Mole of Ca(OH)₂ = 0.00216 mole </h3>

  • Next, we shall determine the molarity of the stock solution of Ca(OH)₂.

Mole of Ca(OH)₂ = 0.00216 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

<h3>Molarity of Ca(OH)₂ =? </h3>

Molarity = mole / Volume

Molarity of Ca(OH)₂ = 0.00216 / 0.1

<h3>Molarity of Ca(OH)₂ = 0.0216 M</h3>

  • Next, we shall determine the molarity of the diluted solution. This can be obtained as follow:

Volume of stock solution (V₁) = 100 mL

Molarity of stock solution (M₁) = 0.0216 M

Volume of diluted solution (V₂) = 250 mL

<h3>Molarity of diluted solution (M₂) =?</h3>

<h3>M₁V₁ = M₂V₂</h3>

0.0216 × 100 = M₂ × 250

2.16 = M₂ × 250

Divide both side by 250

M₂ = 2.16 / 250

<h3>M₂ = 0.00864 M</h3>

Thus, the molarity of the diluted solution is 0.00864 M

  • Finally, we shall determine the concentration of the hydroxide ions, OH¯ in the diluted solution. This can be obtained as follow:

Ca(OH)₂(aq) —> Ca²⁺(aq) + 2OH¯(aq)

From the balanced equation above,

1 mole of Ca(OH)₂ contains 2 moles of OH¯

Therefore,

0.00864 M Ca(OH)₂ will contain =  2 × 0.00864 = 0.01728 M OH¯

Thus, the concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

Learn more: brainly.com/question/11471182

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The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this
Semmy [17]

Answer:

The molar concentration of Cu^{2+} ions in the given amount of sample is 4.73\times 10^{-5}M

Explanation:

Given that,

Mass of sample = 21.5 g

0.07 % (m/m) of copper (II) sulfate in plant fertilizer

This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present

So, in 20 g of plant fertilizer

,=\frac{0.07}{100}\times 21.5}\\=0.0151g of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Mass of solute (copper (II) sulfate) = 0.0151 g

Molar mass of copper (II) sulfate = 159.6 g/mol

Volume of solution = 2.0 L

\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M

The chemical equation for the ionization of copper (II) sulfate follows:

CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = 4.73\times 10^{-5}M

Hence, the molar concentration of Cu^{2+} ions in the given amount of sample is 4.73\times 10^{-5}M

4 0
3 years ago
Can someone please please help me with this? I will mark you as brainlist? I only have two more questions but anyways
DENIUS [597]

Answer:

its 20 L

Explanation:

7 0
3 years ago
Read 2 more answers
A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water
S_A_V [24]

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

8 0
3 years ago
The following equilibrium is formed when copper and bromide ions are placed in a solution:
JulsSmile [24]

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

3 0
3 years ago
Phosphorus that is stored in organisms
Mumz [18]

Explanation:

B. Recycles slowly

And it also depends on the organism

3 0
3 years ago
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