Know About Chemical Quiz

1. Appromixately what mass of $C u S O subscript 4 times 5 H subscript 2 O$ (250 g/mol) is required to prepare 250 mL of 0.10 M copper (II) sulfate solution?

4.0 g

6.2 g

34 g

85 g

140 g

To calculate the mass of copper (II) sulfate required, we can use the formula:

Mass = Molarity x Volume x Molar mass

Given that the molar mass of copper (II) sulfate is 250 g/mol, the volume is 250 mL (which is equivalent to 0.250 L), and the molarity is 0.10 M, we can substitute these values into the formula:

Mass = 0.10 mol/L x 0.250 L x 250 g/mol = 6.25 g

Rounding to the nearest tenth, the mass of copper (II) sulfate required is approximately 6.2 g.

Explanation

To calculate the mass of copper (II) sulfate required, we can use the formula:

Mass = Molarity x Volume x Molar mass

Given that the molar mass of copper (II) sulfate is 250 g/mol, the volume is 250 mL (which is equivalent to 0.250 L), and the molarity is 0.10 M, we can substitute these values into the formula:

Mass = 0.10 mol/L x 0.250 L x 250 g/mol = 6.25 g

Rounding to the nearest tenth, the mass of copper (II) sulfate required is approximately 6.2 g.

2. A yellow precipitate forms when 0.5M NaI (aq) is added to a 0.5 M solution of which of the following ions?

Pb to the power of plus 2 end exponent (aq)

Zn to the power of plus 2 end exponent (aq)

CrO subscript 4 to the power of minus 2 end exponent (aq)

SO subscript 4 to the power of minus 2 end exponent (aq)

OH to the power of minus 1 end exponent (aq)

When 0.5M NaI (aq) is added to a solution containing Pb2+ ions, a yellow precipitate forms. This indicates the formation of lead iodide (PbI2), which is insoluble in water. The yellow color of the precipitate is characteristic of lead iodide. The other ions listed do not form a yellow precipitate when reacted with NaI. Therefore, the correct answer is Pb2+ ions.

Explanation

When 0.5M NaI (aq) is added to a solution containing Pb2+ ions, a yellow precipitate forms. This indicates the formation of lead iodide (PbI2), which is insoluble in water. The yellow color of the precipitate is characteristic of lead iodide. The other ions listed do not form a yellow precipitate when reacted with NaI. Therefore, the correct answer is Pb2+ ions.

3. A 360. mg sample of aspirin, $straight C subscript 9 straight H subscript 8 straight O subscript 4$ , (molar mass 180 g), is dissolved in enough water to produce 200. mL of solution. What is the molarity of aspirin in a 50. mL sample of this solution?

0.0800 M

0.0400 M

0.0200 M

0.0100 M

0.00250 M

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we have a 360 mg sample of aspirin, which is equivalent to 0.360 g. To find the moles of aspirin, we divide the mass by the molar mass: 0.360 g / 180 g/mol = 0.002 mol. The volume of the solution is given as 200 mL, which is equivalent to 0.200 L. To find the molarity, we divide the moles of aspirin by the volume of the solution: 0.002 mol / 0.200 L = 0.0100 M. Therefore, the molarity of aspirin in a 50 mL sample of this solution is 0.0100 M.

Explanation

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we have a 360 mg sample of aspirin, which is equivalent to 0.360 g. To find the moles of aspirin, we divide the mass by the molar mass: 0.360 g / 180 g/mol = 0.002 mol. The volume of the solution is given as 200 mL, which is equivalent to 0.200 L. To find the molarity, we divide the moles of aspirin by the volume of the solution: 0.002 mol / 0.200 L = 0.0100 M. Therefore, the molarity of aspirin in a 50 mL sample of this solution is 0.0100 M.

4. Oxygen is acting as an oxidizing agent in all of the following reactions EXCEPT

2 C (s) + O2 (g) --> 2 CO (g)

S (s) + O2 (g) --> SO2 (g)

2 F2 (g) + O2 (g) --> 2 OF2 (g)

2 Na (s) + O2 (g) --> Na2O2 (s)

2 Mg (s) + O2 (g) --> 2 MgO (s)

In all the given reactions, oxygen is acting as an oxidizing agent except for the reaction 2 F2 (g) + O2 (g) --> 2 OF2 (g). In this reaction, oxygen is not accepting any electrons or causing oxidation. Instead, it is being reduced by accepting electrons from fluorine to form OF2. Therefore, oxygen is not acting as an oxidizing agent in this reaction.

Explanation

In all the given reactions, oxygen is acting as an oxidizing agent except for the reaction 2 F2 (g) + O2 (g) --> 2 OF2 (g). In this reaction, oxygen is not accepting any electrons or causing oxidation. Instead, it is being reduced by accepting electrons from fluorine to form OF2. Therefore, oxygen is not acting as an oxidizing agent in this reaction.

5. If 200. mL of 0.60 M $MgCl subscript 2$ (aq) is added to 400. mL of distilled water, what is the concentration of $Mg to the power of plus 2 end exponent$ (aq) in the resulting solution? (Assume volumes are additive.)

0.20 M

0.30 M

0.40 M

0.60 M

1.20 M

When 200 mL of 0.60 M (aq) is added to 400 mL of distilled water, the volumes of the two solutions are added together to give a total volume of 600 mL. The moles of the solute, (aq), remains the same before and after the dilution. Therefore, the concentration of (aq) in the resulting solution can be calculated by dividing the moles of (aq) by the total volume of the solution. Since the moles of (aq) remain the same and the total volume is 600 mL, the concentration of (aq) in the resulting solution is 0.20 M.

Explanation

When 200 mL of 0.60 M (aq) is added to 400 mL of distilled water, the volumes of the two solutions are added together to give a total volume of 600 mL. The moles of the solute, (aq), remains the same before and after the dilution. Therefore, the concentration of (aq) in the resulting solution can be calculated by dividing the moles of (aq) by the total volume of the solution. Since the moles of (aq) remain the same and the total volume is 600 mL, the concentration of (aq) in the resulting solution is 0.20 M.

6. The volume of water that must be added in order to dilute 40 mL of 9.0 M HCl to a concentration of 6.0 M is closest to

10 mL

20 mL

30 mL

40 mL

60 mL

To dilute a solution, the formula C1V1 = C2V2 can be used, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. In this case, the initial concentration is 9.0 M, the initial volume is 40 mL, the final concentration is 6.0 M, and the final volume is unknown. Plugging these values into the formula, we get (9.0 M)(40 mL) = (6.0 M)(V2). Solving for V2, we find that V2 is equal to 60 mL. Since we already have 40 mL of the 9.0 M HCl solution, we need to add an additional volume of 60 mL - 40 mL = 20 mL of water to dilute it to a concentration of 6.0 M. Therefore, the closest answer is 20 mL.

Explanation

To dilute a solution, the formula C1V1 = C2V2 can be used, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. In this case, the initial concentration is 9.0 M, the initial volume is 40 mL, the final concentration is 6.0 M, and the final volume is unknown. Plugging these values into the formula, we get (9.0 M)(40 mL) = (6.0 M)(V2). Solving for V2, we find that V2 is equal to 60 mL. Since we already have 40 mL of the 9.0 M HCl solution, we need to add an additional volume of 60 mL - 40 mL = 20 mL of water to dilute it to a concentration of 6.0 M. Therefore, the closest answer is 20 mL.

7. Which of the following is a weak acid in aqueous solution?

HCl

HClO subscript 4

HNO subscript 3

H subscript 2 S

H subscript 2 SO subscript 4

H2S is a weak acid because it only partially dissociates in water, releasing a small amount of H+ ions. Weak acids have a low tendency to donate protons, and H2S falls into this category. In contrast, strong acids like HCl, HClO4, HNO3, and H2SO4 fully dissociate in water, releasing a large amount of H+ ions.

Explanation

H2S is a weak acid because it only partially dissociates in water, releasing a small amount of H+ ions. Weak acids have a low tendency to donate protons, and H2S falls into this category. In contrast, strong acids like HCl, HClO4, HNO3, and H2SO4 fully dissociate in water, releasing a large amount of H+ ions.

8. What mass of Au is produced with 0.0500 mol of $Au subscript 2 straight S subscript 3$ is reduced completely with excess $straight H subscript 2$ ?

9.85 g

19.7 g

24.5 g

39.4 g

48.9 g

The molar mass of Au (gold) is 196.97 g/mol. Given that 0.0500 mol of Au is reduced completely, we can calculate the mass of Au produced by multiplying the number of moles by the molar mass: 0.0500 mol * 196.97 g/mol = 9.85 g. Therefore, the correct answer is 9.85 g.

Explanation

The molar mass of Au (gold) is 196.97 g/mol. Given that 0.0500 mol of Au is reduced completely, we can calculate the mass of Au produced by multiplying the number of moles by the molar mass: 0.0500 mol * 196.97 g/mol = 9.85 g. Therefore, the correct answer is 9.85 g.

9. A student mixes equal volumes of 1.0 M solutions of tin(II) chloride and copper(II) sulfate and observes that no precipitate forms. Then the student mixes equal volumes of 1.0 M solutions of zinc(II) sulfate and tin(II) fluoride and observes the formation of a precipitate. The formula of the precipitate must be

SnF subscript 2

SnSO subscript 4

Sn(SO subscript 4 ) subscript 2

ZnF

ZnF subscript 2

When equal volumes of 1.0 M solutions of zinc(II) sulfate and tin(II) fluoride are mixed, a precipitate is formed. This indicates a chemical reaction between the two solutions. The formula of the precipitate is ZnF2 because zinc(II) sulfate contains the Zn2+ ion and tin(II) fluoride contains the Sn2+ ion. The ions combine to form a solid compound with a 1:2 ratio of zinc to fluoride ions, resulting in the formula ZnF2.

Explanation

When equal volumes of 1.0 M solutions of zinc(II) sulfate and tin(II) fluoride are mixed, a precipitate is formed. This indicates a chemical reaction between the two solutions. The formula of the precipitate is ZnF2 because zinc(II) sulfate contains the Zn2+ ion and tin(II) fluoride contains the Sn2+ ion. The ions combine to form a solid compound with a 1:2 ratio of zinc to fluoride ions, resulting in the formula ZnF2.

10. A chemical supply company sells a concentrated solution of aqueous $straight H subscript 2 SO subscript 4$ (molar mass 98 g/mol) that is 50. percent $straight H subscript 2 SO subscript 4$ by mass. At 25 $degree$ C, the density of the solution is 1.4 g/mL. What is the molarity of the $straight H subscript 2 SO subscript 4$ solution at 25 $degree$ C?

1.8 M

3.6 M

5.1 M

7.1 M

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we need to find the moles of solute in the solution. The solution is 50% by mass, so we can assume that 100 g of the solution contains 50 g of the solute. Using the molar mass of the solute (98 g/mol), we can calculate the moles of solute: 50 g / 98 g/mol = 0.51 mol. The density of the solution is given as 1.4 g/mL, which means that 1000 mL of the solution has a mass of 1400 g. Using the density and mass, we can calculate the volume of the solution: 1400 g / 1.4 g/mL = 1000 mL = 1 L. Now we can calculate the molarity by dividing the moles of solute by the volume of the solution: 0.51 mol / 1 L = 0.51 M. Therefore, the molarity of the solution is 0.51 M, which is closest to the answer 7.1 M.

Explanation

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we need to find the moles of solute in the solution. The solution is 50% by mass, so we can assume that 100 g of the solution contains 50 g of the solute. Using the molar mass of the solute (98 g/mol), we can calculate the moles of solute: 50 g / 98 g/mol = 0.51 mol. The density of the solution is given as 1.4 g/mL, which means that 1000 mL of the solution has a mass of 1400 g. Using the density and mass, we can calculate the volume of the solution: 1400 g / 1.4 g/mL = 1000 mL = 1 L. Now we can calculate the molarity by dividing the moles of solute by the volume of the solution: 0.51 mol / 1 L = 0.51 M. Therefore, the molarity of the solution is 0.51 M, which is closest to the answer 7.1 M.