Q. If x + 4 = 10, what is the value of x?
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Solution
Subtracting 4 from both sides gives x = 6.
Correct Answer: B — 6
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Q. If ΔG is negative for a reaction, what can be inferred about the reaction?
A.
The reaction is at equilibrium.
B.
The reaction is spontaneous.
C.
The reaction is non-spontaneous.
D.
The reaction requires energy input.
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Solution
A negative ΔG indicates that the reaction is spontaneous.
Correct Answer: B — The reaction is spontaneous.
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Q. If ΔG is negative for a reaction, what can be inferred?
A.
The reaction is non-spontaneous.
B.
The reaction is at equilibrium.
C.
The reaction is spontaneous.
D.
The reaction requires energy input.
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Solution
A negative ΔG indicates that the reaction is spontaneous under the given conditions.
Correct Answer: C — The reaction is spontaneous.
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Q. If ΔG is negative, what does it indicate about the reaction?
A.
Reaction is at equilibrium
B.
Reaction is spontaneous
C.
Reaction is non-spontaneous
D.
Reaction requires energy input
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Solution
A negative ΔG indicates that the reaction is spontaneous.
Correct Answer: B — Reaction is spontaneous
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Q. If ΔG is positive, what can be inferred about the reaction?
A.
The reaction is spontaneous.
B.
The reaction is at equilibrium.
C.
The reaction is non-spontaneous.
D.
The reaction will proceed in reverse.
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Solution
If ΔG is positive, the reaction is non-spontaneous under the given conditions.
Correct Answer: C — The reaction is non-spontaneous.
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Q. If ΔH = 100 kJ and ΔS = 0.2 kJ/K, what is ΔG at 298 K?
A.
100 kJ
B.
96 kJ
C.
104 kJ
D.
90 kJ
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Solution
ΔG = ΔH - TΔS = 100 kJ - 298 K * 0.2 kJ/K = 100 kJ - 59.6 kJ = 40.4 kJ.
Correct Answer: B — 96 kJ
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Q. If ΔH is negative and ΔS is positive, what can be said about ΔG?
A.
ΔG is always positive.
B.
ΔG is always negative.
C.
ΔG can be positive or negative depending on temperature.
D.
ΔG is zero.
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Solution
If ΔH is negative and ΔS is positive, ΔG will always be negative, indicating that the reaction is spontaneous at all temperatures.
Correct Answer: B — ΔG is always negative.
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Q. In a binary solution of A and B, if component A has a higher vapor pressure than component B, what can be inferred about the solution?
A.
It will show positive deviation from Raoult's Law.
B.
It will show negative deviation from Raoult's Law.
C.
It will have a lower boiling point than pure A.
D.
It will have a higher boiling point than pure B.
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Solution
If component A has a higher vapor pressure than component B, the solution will show positive deviation from Raoult's Law.
Correct Answer: A — It will show positive deviation from Raoult's Law.
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Q. In a binary solution of A and B, if the mole fraction of A is 0.6, what is the mole fraction of B?
A.
0.4
B.
0.6
C.
1.0
D.
0.2
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Solution
The mole fraction of B can be calculated as 1 - mole fraction of A = 1 - 0.6 = 0.4.
Correct Answer: A — 0.4
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Q. In a binary solution of A and B, if the vapor pressure of A is 80 mmHg and that of B is 40 mmHg, what is the total vapor pressure when both are present?
A.
40 mmHg
B.
80 mmHg
C.
120 mmHg
D.
100 mmHg
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Solution
The total vapor pressure is the sum of the partial pressures: P_total = P_A + P_B = 80 mmHg + 40 mmHg = 120 mmHg.
Correct Answer: D — 100 mmHg
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Q. In a binary solution of A and B, if the vapor pressure of A is 80 mmHg and that of B is 40 mmHg, what is the total vapor pressure of the solution if the mole fraction of A is 0.75?
A.
60 mmHg
B.
70 mmHg
C.
80 mmHg
D.
90 mmHg
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Solution
Total vapor pressure = P_A + P_B = (0.75 * 80) + (0.25 * 40) = 60 + 10 = 70 mmHg.
Correct Answer: D — 90 mmHg
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Q. In a binary solution of A and B, if the vapor pressure of pure A is 150 mmHg and pure B is 50 mmHg, what is the total vapor pressure when the mole fraction of A is 0.4?
A.
90 mmHg
B.
120 mmHg
C.
150 mmHg
D.
100 mmHg
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Solution
Total vapor pressure = (X_A * P_A^0) + (X_B * P_B^0) = (0.4 * 150) + (0.6 * 50) = 60 + 30 = 90 mmHg.
Correct Answer: B — 120 mmHg
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Q. In a binary solution of A and B, if the vapor pressure of pure A is 80 mmHg and that of pure B is 40 mmHg, what is the vapor pressure of the solution if the mole fraction of A is 0.6?
A.
64 mmHg
B.
72 mmHg
C.
80 mmHg
D.
56 mmHg
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Solution
Using Raoult's Law, the vapor pressure of the solution = (0.6 * 80 mmHg) + (0.4 * 40 mmHg) = 64 mmHg.
Correct Answer: B — 72 mmHg
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Q. In a binary solution of A and B, if the vapor pressure of pure A is 80 mmHg and pure B is 40 mmHg, what is the vapor pressure of the solution if the mole fraction of A is 0.6?
A.
64 mmHg
B.
72 mmHg
C.
80 mmHg
D.
56 mmHg
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Solution
Using Raoult's Law, the vapor pressure of the solution = (0.6 * 80 mmHg) + (0.4 * 40 mmHg) = 64 mmHg.
Correct Answer: B — 72 mmHg
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Q. In a binary solution of A and B, if the vapor pressure of pure A is 80 mmHg and pure B is 40 mmHg, what is the vapor pressure of component A if the mole fraction of A is 0.6?
A.
48 mmHg
B.
64 mmHg
C.
80 mmHg
D.
32 mmHg
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Solution
Using Raoult's Law, the vapor pressure of A in the solution is 0.6 * 80 mmHg = 48 mmHg.
Correct Answer: B — 64 mmHg
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Q. In a binary solution of A and B, if the vapor pressure of pure A is 80 mmHg and that of pure B is 40 mmHg, what is the total vapor pressure when the mole fraction of A is 0.6?
A.
64 mmHg
B.
72 mmHg
C.
80 mmHg
D.
88 mmHg
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Solution
Using Raoult's Law, the total vapor pressure = (0.6 * 80 mmHg) + (0.4 * 40 mmHg) = 48 mmHg + 16 mmHg = 64 mmHg.
Correct Answer: B — 72 mmHg
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Q. In a binary solution of A and B, if the vapor pressure of pure A is 80 mmHg and pure B is 40 mmHg, what is the total vapor pressure when the mole fraction of A is 0.5?
A.
60 mmHg
B.
70 mmHg
C.
80 mmHg
D.
50 mmHg
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Solution
Total vapor pressure = P_A + P_B = (0.5 * 80) + (0.5 * 40) = 40 + 20 = 60 mmHg.
Correct Answer: B — 70 mmHg
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Q. In a chemical reaction, if the enthalpy change is positive, the reaction is classified as:
A.
Exothermic
B.
Endothermic
C.
Isothermal
D.
Adiabatic
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Solution
A positive enthalpy change indicates that the reaction absorbs heat, classifying it as endothermic.
Correct Answer: B — Endothermic
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Q. In a chemical reaction, if the enthalpy of products is less than that of reactants, what can be concluded?
A.
The reaction is endothermic
B.
The reaction is exothermic
C.
The reaction is at equilibrium
D.
The reaction is spontaneous
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Solution
If the enthalpy of products is less, the reaction releases heat, indicating it is exothermic.
Correct Answer: B — The reaction is exothermic
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Q. In a closed system at equilibrium, if the pressure is increased, what will happen to the equilibrium position?
A.
Shift to the side with more moles of gas
B.
Shift to the side with fewer moles of gas
C.
No change
D.
Increase in temperature
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Solution
Increasing the pressure will shift the equilibrium position to the side with fewer moles of gas.
Correct Answer: B — Shift to the side with fewer moles of gas
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Q. In a closed system at equilibrium, if the pressure is increased, what will happen to the equilibrium position if there are more moles of gas on the reactant side?
A.
Shift to the right
B.
Shift to the left
C.
No change
D.
Increase in temperature
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Solution
Increasing the pressure will shift the equilibrium to the side with fewer moles of gas, which in this case is the right side.
Correct Answer: B — Shift to the left
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Q. In a closed system at equilibrium, if the volume is decreased, what will happen to the equilibrium position?
A.
Shift to the side with more moles of gas
B.
Shift to the side with fewer moles of gas
C.
No change
D.
Increase in pressure
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Solution
Decreasing the volume increases the pressure, and the equilibrium will shift to the side with fewer moles of gas to counteract the change.
Correct Answer: B — Shift to the side with fewer moles of gas
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Q. In a closed system, if the temperature increases, what happens to the entropy?
A.
It decreases
B.
It increases
C.
It remains constant
D.
It becomes zero
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Solution
In a closed system, an increase in temperature generally leads to an increase in entropy, as the molecular motion becomes more chaotic.
Correct Answer: B — It increases
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Q. In a closed system, if the volume is decreased, what will happen to the equilibrium of the reaction 2A(g) + B(g) ⇌ 3C(g)?
A.
Shift to the left
B.
Shift to the right
C.
No change
D.
Depends on temperature
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Solution
Decreasing the volume increases the pressure, and the equilibrium will shift towards the side with fewer moles of gas, which is the left side.
Correct Answer: A — Shift to the left
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Q. In a closed system, if the volume of gas is halved at constant temperature, what happens to the pressure?
A.
Halved
B.
Doubled
C.
Remains the same
D.
Increases four times
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Solution
According to Boyle's law, if the volume is halved, the pressure is doubled.
Correct Answer: B — Doubled
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Q. In a concentration cell, the potential difference arises due to:
A.
Different temperatures
B.
Different concentrations
C.
Different pressures
D.
Different materials
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Solution
In a concentration cell, the potential difference arises due to different concentrations of the same species.
Correct Answer: B — Different concentrations
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Q. In a cyclic process, the net work done is equal to the:
A.
Change in internal energy
B.
Heat added to the system
C.
Heat removed from the system
D.
Net heat transfer
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Solution
In a cyclic process, the net work done is equal to the heat added to the system.
Correct Answer: B — Heat added to the system
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Q. In a Daniell cell, which metal is oxidized?
A.
Copper
B.
Zinc
C.
Lead
D.
Silver
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Solution
In a Daniell cell, zinc is oxidized.
Correct Answer: B — Zinc
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Q. In a dynamic equilibrium, which of the following statements is true?
A.
The rates of the forward and reverse reactions are equal
B.
The concentrations of reactants and products are equal
C.
The reaction has stopped
D.
The system is at rest
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Solution
In a dynamic equilibrium, the rates of the forward and reverse reactions are equal, but the concentrations of reactants and products may not be equal.
Correct Answer: A — The rates of the forward and reverse reactions are equal
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Q. In a first-order reaction, if the concentration of the reactant is halved, what happens to the half-life?
A.
It doubles
B.
It remains the same
C.
It is halved
D.
It becomes zero
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Solution
In a first-order reaction, the half-life is independent of the concentration of the reactant, so it remains the same.
Correct Answer: B — It remains the same
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