Match each state of matter to its description.

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

Ivahew [28]

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

Mole = mass / molar mass

Mole of Ca(OH)₂ = 0.16 / 74

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>

0.0216 × 100 = M₂ × 250

2.16 = M₂ × 250

Divide both side by 250

M₂ = 2.16 / 250

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

7 0

3 years ago

Bismuth oxide reacts with carbon to form bismuth metal: bi2o3(s) + 3c(s) → 2bi(s) + 3co(g) when 689 g of bi2o3 reacts with exces

ExtremeBDS [4]

Using the answer from the first part, we know that 2.957 moles of bismuth have formed. Moreover, the molar ratio between bismuth and carbon monoxide is:

2 : 3

Using the method of ratios,

2 : 3
2.957 : CO

CO = (3 * 2.957) / 2
CO = 4.4355

4.436 moles of carbon monoxide will be formed

8 0

3 years ago

Read 2 more answers

A compound is found to contain 38.65 % carbon, 16.25 % hydrogen, and 45.09 % nitrogen by mass. what is the empirical formula for

iren [92.7K]

The molar mass of carbon is 12, hydrogen is 1, and nitrogen is 14, hence the ratio are:

C = 38.65 / 12 = 3.22

H = 16.25 / 1 = 16.25

N = 45.09 / 14 = 3.22

Divide the three by the lowest ratio which is 3.22:

C = 3.22 / 3.22 = 1

H = 16.25 / 3.22 = 5

N = 3.22 / 3.22 = 1

So the empirical formula is:

CHN

5 0

3 years ago

2) Write a hypothesis for the Observation and question below. Be sure to use the proper “If…, then…” format in your answer. Iden

Anna [14]

Answer:

If one astronaut used more force, then that one would be faster than the other. The independent variable is force. The dependent variable is speed.

Explanation:

Force is what is to be changed. Speed is what is being measured.

IV = Changed/factor DV=Measured/Changes by factor

4 0

3 years ago

In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 1

weeeeeb [17]

Answer:

A. 3.36g of Na.

B. 14.62g of Ba3(PO4)2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2

Next, we shall determine the mass of Ba that reacted and the mass of Na and Ba3(PO4)2 produced from the equation.

This is illustrated below:

Molar Mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 6 x 23 = 138g

Molar mass of Ba3(PO4)2 = (3 x 137) + 2[31 + (4x16)] = 411 + 2[31 + 64] = 601g/mol

Mass of Ba3(PO4)2 from the balanced equation = 1 x 601 = 601g

Summary:

From the balanced equation above,

411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.

A. Determination of the mass of Na produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 138g of Na.

Therefore, 10g of Ba will react to produce = (10 x 138)/411 = 3.36g of Na.

Therefore, 3.36g of Na is produced.

B. Determination of the mass of Ba3(PO4)2 produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 601g of Ba3(PO4)2.

Therefore, 10g of Ba will react to produce = (10 x 601)/411 = 14.62g of Ba3(PO4)2.

Therefore, 14.62g of Ba3(PO4)2 is produced.

7 0

3 years ago