Q. What is the entropy change for a system that undergoes a phase transition at constant temperature?
A.
ΔS = 0
B.
ΔS = Q/T
C.
ΔS = T/Q
D.
ΔS = Q + T
Show solution
Solution
During a phase transition at constant temperature, the change in entropy is given by ΔS = Q/T, where Q is the heat absorbed or released.
Correct Answer: B — ΔS = Q/T
Learn More →
Q. What is the entropy change for an ideal gas during an isothermal expansion?
A.
Zero
B.
nR ln(Vf/Vi)
C.
nC_v ln(Tf/Ti)
D.
nC_p ln(Tf/Ti)
Show solution
Solution
The entropy change for an ideal gas during an isothermal expansion is ΔS = nR ln(Vf/Vi).
Correct Answer: B — nR ln(Vf/Vi)
Learn More →
Q. What is the entropy change for the isothermal expansion of an ideal gas from volume V1 to V2 at temperature T?
A.
R ln(V2/V1)
B.
R (V2 - V1)/T
C.
0
D.
R (V1/V2)
Show solution
Solution
The entropy change for an isothermal expansion is given by ΔS = nR ln(V2/V1). For 1 mole, ΔS = R ln(V2/V1).
Correct Answer: A — R ln(V2/V1)
Learn More →
Q. What is the entropy change for the mixing of two ideal gases at constant temperature?
A.
0
B.
R ln(2)
C.
R ln(V1/V2)
D.
R ln(V1*V2)
Show solution
Solution
The entropy change for the mixing of two ideal gases at constant temperature is ΔS = nR ln(2) for equal moles of each gas.
Correct Answer: B — R ln(2)
Learn More →
Q. What is the entropy change when 1 mole of an ideal gas is heated at constant volume from temperature T1 to T2?
A.
R ln(T2/T1)
B.
R (T2 - T1)
C.
0
D.
R (T1/T2)
Show solution
Solution
The change in entropy at constant volume is given by ΔS = nC_v ln(T2/T1). For 1 mole, ΔS = R ln(T2/T1).
Correct Answer: A — R ln(T2/T1)
Learn More →
Q. What is the entropy change when 1 mole of an ideal gas is heated at constant volume?
A.
0
B.
R ln(T2/T1)
C.
R (T2 - T1)
D.
R (T1/T2)
Show solution
Solution
The change in entropy when heating an ideal gas at constant volume is given by ΔS = nC_v ln(T2/T1). For 1 mole, it simplifies to ΔS = R ln(T2/T1).
Correct Answer: B — R ln(T2/T1)
Learn More →
Q. What is the entropy change when 1 mole of ice at 0°C is converted to water at 0°C?
A.
0 J/K
B.
R ln(2)
C.
R
D.
Positive value
Show solution
Solution
The phase change from ice to water at 0°C involves an increase in disorder, thus resulting in a positive change in entropy.
Correct Answer: D — Positive value
Learn More →
Q. What is the entropy change when 2 moles of an ideal gas are compressed isothermally from volume V2 to V1?
A.
-R ln(V1/V2)
B.
R ln(V1/V2)
C.
0
D.
R (V2 - V1)
Show solution
Solution
The change in entropy for an isothermal compression is ΔS = nR ln(V1/V2). For 2 moles, ΔS = 2R ln(V1/V2), which is negative since V1 < V2.
Correct Answer: A — -R ln(V1/V2)
Learn More →
Q. What is the equilibrium constant expression for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)?
A.
Kc = [NH3]^2 / ([N2][H2]^3)
B.
Kc = [N2][H2]^3 / [NH3]^2
C.
Kc = [NH3]^2 / [N2][H2]
D.
Kc = [N2][H2] / [NH3]^2
Show solution
Solution
The equilibrium constant Kc is given by the ratio of the concentration of products to reactants, raised to the power of their coefficients.
Correct Answer: A — Kc = [NH3]^2 / ([N2][H2]^3)
Learn More →
Q. What is the equilibrium constant expression for the reaction: 2A + B ⇌ C?
A.
[C]/([A]^2[B])
B.
[A]^2[B]/[C]
C.
[C]/[A][B]
D.
[A][B]/[C]
Show solution
Solution
The equilibrium constant K is given by the expression K = [C]/([A]^2[B]) for the reaction 2A + B ⇌ C.
Correct Answer: A — [C]/([A]^2[B])
Learn More →
Q. What is the equilibrium constant expression for the reaction: aA + bB ⇌ cC + dD?
A.
K = [C]^c [D]^d / [A]^a [B]^b
B.
K = [A]^a [B]^b / [C]^c [D]^d
C.
K = [C]^c [D]^d
D.
K = [A]^a [B]^b
Show solution
Solution
The equilibrium constant K is defined as the ratio of the concentration of products to the concentration of reactants, each raised to the power of their coefficients in the balanced equation.
Correct Answer: A — K = [C]^c [D]^d / [A]^a [B]^b
Learn More →
Q. What is the equivalent weight of H2SO4 if its molar mass is 98 g/mol?
A.
49 g
B.
98 g
C.
196 g
D.
24.5 g
Show solution
Solution
Equivalent weight = molar mass / number of equivalents = 98 g/mol / 2 = 49 g.
Correct Answer: A — 49 g
Learn More →
Q. What is the expected osmotic pressure of a 0.5 M NaCl solution at 25 °C?
A.
12.3 atm
B.
24.6 atm
C.
6.1 atm
D.
3.1 atm
Show solution
Solution
Osmotic pressure (π) can be calculated using the formula π = iCRT. For NaCl, i = 2, C = 0.5 M, R = 0.0821 L·atm/(K·mol), and T = 298 K, resulting in approximately 24.6 atm.
Correct Answer: B — 24.6 atm
Learn More →
Q. What is the formula for calculating boiling point elevation?
A.
ΔT_b = K_b * m
B.
ΔT_b = K_f * m
C.
ΔT_b = i * K_b * m
D.
ΔT_b = i * K_f * m
Show solution
Solution
The boiling point elevation is calculated using the formula ΔT_b = i * K_b * m, where i is the van 't Hoff factor.
Correct Answer: C — ΔT_b = i * K_b * m
Learn More →
Q. What is the formula for calculating the depression of freezing point?
A.
ΔTf = Kf * m
B.
ΔTf = Kb * m
C.
ΔTf = R * T
D.
ΔTf = P * V
Show solution
Solution
The depression of freezing point is calculated using the formula ΔTf = Kf * m, where Kf is the freezing point depression constant and m is the molality of the solution.
Correct Answer: A — ΔTf = Kf * m
Learn More →
Q. What is the formula for calculating the number of moles?
A.
Moles = Mass / Volume
B.
Moles = Mass x Volume
C.
Moles = Mass / Molar Mass
D.
Moles = Molar Mass / Mass
Show solution
Solution
The number of moles is calculated using the formula: Moles = Mass / Molar Mass.
Correct Answer: C — Moles = Mass / Molar Mass
Learn More →
Q. What is the freezing point depression constant (Kf) for water?
A.
1.86 °C kg/mol
B.
0.52 °C kg/mol
C.
2.00 °C kg/mol
D.
3.72 °C kg/mol
Show solution
Solution
The freezing point depression constant (Kf) for water is 1.86 °C kg/mol.
Correct Answer: A — 1.86 °C kg/mol
Learn More →
Q. What is the freezing point depression of a solution containing 2 moles of KCl in 1 kg of water?
A.
-3.72 °C
B.
-1.86 °C
C.
-2.52 °C
D.
-4.0 °C
Show solution
Solution
Freezing point depression = i * Kf * m = 3 * 1.86 * 2 = 11.16 °C, so the freezing point is lowered by 3.72 °C.
Correct Answer: A — -3.72 °C
Learn More →
Q. What is the freezing point depression of a solution containing 2 moles of KCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol)
A.
3.72 °C
B.
1.86 °C
C.
2.0 °C
D.
5.58 °C
Show solution
Solution
Freezing point depression = i * Kf * m = 3 * 1.86 * 2 = 11.16 °C.
Correct Answer: A — 3.72 °C
Learn More →
Q. What is the freezing point depression of a solution containing 2 moles of NaCl in 1 kg of water? (Kf for water = 1.86 °C kg/mol)
A.
3.72 °C
B.
1.86 °C
C.
2.72 °C
D.
5.72 °C
Show solution
Solution
Freezing point depression = i * Kf * m = 3 * 1.86 * 2 = 11.16 °C.
Correct Answer: A — 3.72 °C
Learn More →
Q. What is the freezing point depression of a solution directly proportional to?
A.
The molar mass of the solute
B.
The number of solute particles
C.
The volume of the solvent
D.
The temperature of the solvent
Show solution
Solution
Freezing point depression is directly proportional to the number of solute particles in the solution.
Correct Answer: B — The number of solute particles
Learn More →
Q. What is the freezing point depression of a solution if 0.5 mol of a non-volatile solute is dissolved in 1 kg of water? (Kf for water = 1.86 °C kg/mol)
A.
0.93 °C
B.
1.86 °C
C.
3.72 °C
D.
0.5 °C
Show solution
Solution
Freezing point depression = Kf * molality = 1.86 * 0.5 = 0.93 °C.
Correct Answer: A — 0.93 °C
Learn More →
Q. What is the freezing point of a solution containing 0.3 mol of glucose in 1 kg of water? (K_f for water = 1.86 °C kg/mol)
A.
-0.558 °C
B.
-0.558 K
C.
-1.86 °C
D.
-1.86 K
Show solution
Solution
Freezing point depression = K_f * m = 1.86 * 0.3 = 0.558 °C; Freezing point = 0 - 0.558 = -0.558 °C
Correct Answer: A — -0.558 °C
Learn More →
Q. What is the geometry of a molecule with sp3d hybridization?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Trigonal bipyramidal
Show solution
Solution
Molecules with sp3d hybridization have a trigonal bipyramidal geometry.
Correct Answer: D — Trigonal bipyramidal
Learn More →
Q. What is the Gibbs Free Energy change (ΔG) at equilibrium?
A.
ΔG < 0
B.
ΔG = 0
C.
ΔG > 0
D.
ΔG = ΔH - TΔS
Show solution
Solution
At equilibrium, the Gibbs Free Energy change (ΔG) is zero, indicating that the system is at its lowest energy state.
Correct Answer: B — ΔG = 0
Learn More →
Q. What is the Gibbs free energy change for a spontaneous process?
A.
Positive
B.
Negative
C.
Zero
D.
Undefined
Show solution
Solution
For a spontaneous process, the Gibbs free energy change is negative.
Correct Answer: B — Negative
Learn More →
Q. What is the hybridization of a molecule with a tetrahedral shape?
A.
sp
B.
sp2
C.
sp3
D.
sp3d
Show solution
Solution
A tetrahedral shape corresponds to sp3 hybridization.
Correct Answer: C — sp3
Learn More →
Q. What is the hybridization of the carbon atom in acetylene (C2H2)?
A.
sp
B.
sp2
C.
sp3
D.
dsp3
Show solution
Solution
In acetylene, each carbon atom is sp hybridized, forming a triple bond between them.
Correct Answer: A — sp
Learn More →
Q. What is the hybridization of the carbon atom in CO2?
A.
sp
B.
sp2
C.
sp3
D.
dsp3
Show solution
Solution
In CO2, the carbon atom forms two double bonds with oxygen atoms, indicating sp hybridization.
Correct Answer: A — sp
Learn More →
Q. What is the hybridization of the central atom in a molecule with a tetrahedral shape?
A.
sp
B.
sp2
C.
sp3
D.
sp3d
Show solution
Solution
A tetrahedral shape corresponds to sp3 hybridization.
Correct Answer: C — sp3
Learn More →
Showing 631 to 660 of 1494 (50 Pages)