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Q. How much heat is required to raise the temperature of 250 g of water from 25°C to 75°C? (Specific heat of water = 4.2 J/g°C) (2020)

A. 5250 J
B. 4200 J
C. 2500 J
D. 1000 J

Solution

Heat required = mass × specific heat × change in temperature = 250 g × 4.2 J/g°C × (75°C - 25°C) = 5250 J.

Correct Answer: A — 5250 J

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Q. How much heat is required to raise the temperature of 500 g of water from 25°C to 75°C? (2020)

A. 10000 J
B. 5000 J
C. 20000 J
D. 15000 J

Solution

Q = mcΔT = 500 g * 4.2 J/g°C * (75 - 25)°C = 10000 J.

Correct Answer: D — 15000 J

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Q. How much work is done when a force of 10 N moves an object 5 m in the direction of the force?

A. 25 J
B. 50 J
C. 75 J
D. 100 J

Solution

Work = Force * Distance = 10 N * 5 m = 50 J.

Correct Answer: B — 50 J

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Q. How much work is done when a force of 15 N moves an object 3 m in the direction of the force?

A. 45 J
B. 30 J
C. 15 J
D. 60 J

Solution

Work done = Force × Distance = 15 N × 3 m = 45 J

Correct Answer: A — 45 J

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Q. How much work is done when a force of 30 N moves an object 4 m in the direction of the force?

A. 120 J
B. 90 J
C. 60 J
D. 30 J

Solution

Work = Force * Distance = 30 N * 4 m = 120 J.

Correct Answer: A — 120 J

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Q. Identify the oxidizing agent in the reaction: 2H2 + O2 → 2H2O. (2023)

A. H2
B. O2
C. H2O
D. None of the above

Solution

O2 is the oxidizing agent as it gains electrons and is reduced to H2O.

Correct Answer: B — O2

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Q. If 0.1 M acetic acid (CH3COOH) is 1% ionized, what is the concentration of H+ ions?

A. 0.001 M
B. 0.01 M
C. 0.1 M
D. 0.2 M

Solution

Ionization = 1% of 0.1 M = 0.001 M, so [H+] = 0.001 M.

Correct Answer: B — 0.01 M

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Q. If 0.1 M acetic acid has a pH of 2.87, what is the concentration of H+ ions?

A. 0.001 M
B. 0.01 M
C. 0.1 M
D. 0.5 M

Solution

[H+] = 10^(-pH) = 10^(-2.87) ≈ 0.001 M.

Correct Answer: B — 0.01 M

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Q. If 0.5 moles of a gas occupy 12 liters at STP, what is the molar volume of the gas? (2023)

A. 22.4 L
B. 24 L
C. 12 L
D. 6 L

Solution

At STP, 1 mole of any gas occupies 22.4 L. Therefore, 0.5 moles occupy 0.5 x 22.4 L = 11.2 L, confirming the molar volume is 22.4 L.

Correct Answer: A — 22.4 L

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Q. If 0.5 moles of a gas occupy 12.2 liters at STP, what is the molar volume of the gas? (2023)

A. 22.4 L
B. 24 L
C. 12.2 L
D. 6.1 L

Solution

At STP, 1 mole of gas occupies 22.4 liters. Therefore, the molar volume is 22.4 L.

Correct Answer: A — 22.4 L

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Q. If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.52 °C kg/mol) (2021)

A. 0.26 °C
B. 0.52 °C
C. 1.04 °C
D. 0.78 °C

Solution

Boiling point elevation = i * Kb * m = 1 * 0.52 * 0.5 = 0.26 °C, so change = 0.52 °C.

Correct Answer: C — 1.04 °C

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Q. If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2021)

A. 0.5 P0
B. 0.25 P0
C. 0.75 P0
D. 0.1 P0

Solution

Vapor pressure lowering = X_solute * P0 = (0.5 / (0.5 + 55.5)) * P0 ≈ 0.25 P0.

Correct Answer: B — 0.25 P0

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Q. If 0.5 moles of a non-volatile solute is dissolved in 1 kg of water, what is the expected vapor pressure lowering? (2020)

A. 0.5 P0
B. 0.25 P0
C. 0.75 P0
D. 0.1 P0

Solution

Vapor pressure lowering = (moles of solute / moles of solvent) * P0 = (0.5 / 55.5) * P0 ≈ 0.25 P0.

Correct Answer: B — 0.25 P0

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Q. If 0.5 moles of NaCl is dissolved in 1 liter of water, what is the molality of the solution? (2022)

A. 0.5 m
B. 1 m
C. 2 m
D. 0.25 m

Solution

Molality (m) = moles of solute / kg of solvent. Assuming 1 L of water = 1 kg, molality = 0.5 moles / 1 kg = 0.5 m.

Correct Answer: A — 0.5 m

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Q. If 1 kg of water is heated from 25°C to 75°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C) (2021)

A. 21000 J
B. 42000 J
C. 84000 J
D. 105000 J

Solution

Q = mcΔT = (1000 g)(4.2 J/g°C)(50°C) = 210000 J.

Correct Answer: B — 42000 J

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Q. If 1 kg of water is mixed with 1 kg of sugar, what is the expected change in boiling point? (2023)

A. No change
B. Increase
C. Decrease
D. Depends on the sugar type

Solution

The addition of sugar (a non-volatile solute) will raise the boiling point of water due to boiling point elevation.

Correct Answer: B — Increase

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Q. If 1 mole of a gas occupies 22.4 liters at STP, how many liters will 0.5 moles occupy? (2023)

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 5.6 L

Solution

At STP, 1 mole of gas occupies 22.4 L. Therefore, 0.5 moles will occupy 0.5 x 22.4 L = 11.2 L.

Correct Answer: A — 11.2 L

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Q. If 1 mole of a non-volatile solute is dissolved in 1 kg of water, what is the expected freezing point depression? (2019)

A. 1.86 °C
B. 3.72 °C
C. 0.93 °C
D. 2.0 °C

Solution

Freezing point depression (ΔTf) = i * Kf * m = 1 * 1.86 °C kg/mol * 1 mol/kg = 1.86 °C.

Correct Answer: A — 1.86 °C

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Q. If 10 g of glucose (C6H12O6) is dissolved in 200 g of water, what is the mass percentage of the solution? (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: B — 5.00%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of sugar in the solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / total mass of solution) x 100. Total mass = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the solution? (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 20.00%

Solution

Mass percent = (mass of solute / mass of solution) × 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) × 100 = 4.76%.

Correct Answer: B — 5.00%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percentage of the sugar solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the sugar solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / mass of solution) × 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) × 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percentage of the solution? (Molar mass of sugar = 342 g/mol) (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percentage = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percentage = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: B — 5.00%

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Q. If 10 g of sugar (C12H22O11) is dissolved in 200 g of water, what is the mass percent of the sugar solution? (2023)

A. 4.76%
B. 5.00%
C. 10.00%
D. 2.50%

Solution

Mass percent = (mass of solute / mass of solution) x 100. Mass of solution = 10 g + 200 g = 210 g. Mass percent = (10 g / 210 g) x 100 = 4.76%.

Correct Answer: A — 4.76%

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Q. If 10 g of sugar is dissolved in 100 g of water, what is the mass percent of sugar in the solution? (2023)

A. 10%
B. 5%
C. 20%
D. 15%

Solution

Mass percent = (mass of solute / total mass of solution) x 100. Total mass = 10 g + 100 g = 110 g. Mass percent = (10 g / 110 g) x 100 = 9.09%, approximately 10%.

Correct Answer: A — 10%

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Q. If 10 g of sugar is dissolved in 100 g of water, what is the mass percentage of sugar in the solution? (2023)

A. 10%
B. 5%
C. 20%
D. 15%

Solution

Mass percentage = (mass of solute / mass of solution) × 100. Mass of solution = 10 g + 100 g = 110 g. Mass percentage = (10 g / 110 g) × 100 = 9.09%, approximately 10%.

Correct Answer: A — 10%

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Q. If 10 grams of NaCl is dissolved in 500 mL of water, what is the concentration in g/L? (2020)

A. 20 g/L
B. 10 g/L
C. 5 g/L
D. 15 g/L

Solution

Concentration = (10 g / 0.5 L) = 20 g/L.

Correct Answer: A — 20 g/L

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Q. If 10 grams of NaCl is dissolved in 500 mL of water, what is the mass/volume percent concentration? (2020) 2020

A. 2%
B. 5%
C. 10%
D. 20%

Solution

Mass/volume % = (mass of solute / volume of solution) * 100 = (10 g / 500 mL) * 100 = 2%.

Correct Answer: B — 5%

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