Q. If 'PEN' is coded as '16-5-14', what is the code for 'PAPER'? (2022)
A.
16-1-16-5-18
B.
16-1-16-5-17
C.
15-1-16-5-18
D.
16-1-15-5-18
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Solution
P=16, A=1, P=16, E=5, R=18; thus PAPER = 16-1-16-5-18
Correct Answer: A — 16-1-16-5-18
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Q. If 'PENCIL' is coded as 'QFODJM', what is the code for 'ERASER'? (2023)
A.
FSBTFQ
B.
FSBTEQ
C.
FSCUFR
D.
FSCUHQ
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Solution
Each letter is shifted by +1. E->F, R->S, A->B, S->T, E->F, R->Q. So, ERASER = FSBTFQ.
Correct Answer: A — FSBTFQ
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Q. If 'RAT' is coded as '18-1-20', what is the code for 'BAT'? (2021)
A.
2-1-20
B.
2-1-19
C.
2-1-21
D.
2-1-22
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Solution
B=2, A=1, T=20; thus BAT = 2-1-20
Correct Answer: A — 2-1-20
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Q. If 'SUN' is coded as '19-21-14', what is the code for 'MOON'? (2023)
A.
13-15-15-14
B.
13-15-15-13
C.
13-15-15-12
D.
13-15-15-15
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Solution
M=13, O=15, O=15, N=14; thus MOON = 13-15-15-14
Correct Answer: A — 13-15-15-14
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Q. If 'TABLE' is coded as 'UBMDF', how is 'CHAIR' coded? (2021)
A.
DIBJS
B.
DIBJT
C.
DIBJR
D.
DIBJQ
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Solution
Each letter is shifted by +1. C->D, H->I, A->B, I->J, R->S. So, CHAIR = DIBJS.
Correct Answer: A — DIBJS
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Q. If 'TABLE' is coded as 'UBMDF', what is the code for 'CHAIR'? (2021)
A.
DIBJS
B.
DIBJR
C.
DIBJT
D.
DIBJU
Show solution
Solution
Each letter is shifted by +1. C -> D, H -> I, A -> B, I -> J, R -> S. So, CHAIR = DIBJS.
Correct Answer: A — DIBJS
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Q. If 'WATER' is coded as 'XBTFS', how is 'FIRE' coded? (2021)
A.
GJTF
B.
GJTE
C.
GJTD
D.
GJTC
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Solution
Each letter is shifted by +1. F->G, I->J, R->S, E->F. So, FIRE = GJTF.
Correct Answer: A — GJTF
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Q. If 'XYZ' in base-4 equals 27 in decimal, what is the value of 'X'?
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Solution
In base-4, 'XYZ' = 4^2*X + 4^1*Y + 4^0*Z = 27. The only valid combination is X=3, Y=3, Z=3.
Correct Answer: C — 3
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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
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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
Show solution
Solution
[H+] = 10^(-pH) = 10^(-2.87) ≈ 0.001 M.
Correct Answer: B — 0.01 M
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Q. If 0.1 M of a strong acid is mixed with 0.1 M of a strong base, what will be the resulting pH?
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Solution
Strong acid and strong base neutralize each other, resulting in a neutral solution with pH = 7.
Correct Answer: B — 7
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Q. If 0.1 M of a weak acid has a pH of 4.0, what is the Ka of the acid?
A.
1 x 10^-4
B.
1 x 10^-5
C.
1 x 10^-6
D.
1 x 10^-7
Show solution
Solution
Using the formula Ka = [H+]^2 / [HA], where [H+] = 10^(-4) M and [HA] = 0.1 M, we find Ka = (10^-4)^2 / 0.1 = 1 x 10^-5.
Correct Answer: B — 1 x 10^-5
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Q. If 0.5 mol of a non-volatile solute is dissolved in 1 kg of water, what is the vapor pressure lowering? (Vapor pressure of pure water = 23.76 mmHg)
A.
1.88 mmHg
B.
2.88 mmHg
C.
3.88 mmHg
D.
4.88 mmHg
Show solution
Solution
Vapor pressure lowering = (n_solute / n_solvent) * P°_solvent = (0.5 / 55.5) * 23.76 = 1.88 mmHg
Correct Answer: A — 1.88 mmHg
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Q. If 0.5 moles of a gas occupy 11.2 liters at STP, what is the molar volume of the gas?
A.
22.4 L
B.
11.2 L
C.
5.6 L
D.
44.8 L
Show solution
Solution
At STP, 1 mole of gas occupies 22.4 L. 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 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
Show solution
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
Show solution
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
Show solution
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
Show solution
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
Show solution
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 are dissolved in 1 liter of water, what is the concentration of NaCl in the solution?
A.
0.5 M
B.
1 M
C.
2 M
D.
0.25 M
Show solution
Solution
Concentration (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.
Correct Answer: A — 0.5 M
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Q. If 0.5 moles of NaCl are dissolved in 1 liter of water, what is the concentration of NaCl?
A.
0.5 M
B.
1 M
C.
2 M
D.
0.25 M
Show solution
Solution
Concentration (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.
Correct Answer: A — 0.5 M
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Q. If 0.5 moles of NaCl are dissolved in 1 liter of water, what is the molarity of the solution?
A.
0.5 M
B.
1 M
C.
2 M
D.
0.25 M
Show solution
Solution
Molarity (M) = moles of solute / liters of solution = 0.5 moles / 1 L = 0.5 M.
Correct Answer: A — 0.5 M
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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
Show solution
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 20°C to 100°C, how much heat is absorbed? (Specific heat of water = 4.2 J/g°C)
A.
3360 J
B.
4000 J
C.
4200 J
D.
4800 J
Show solution
Solution
Q = m*c*ΔT = 1000 g * 4.2 J/g°C * (100°C - 20°C) = 4200 J.
Correct Answer: C — 4200 J
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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
Show solution
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 L of a 2 M solution is diluted to 3 L, what is the new molarity of the solution?
A.
0.67 M
B.
1 M
C.
1.5 M
D.
2 M
Show solution
Solution
Using the dilution formula M1V1 = M2V2, we have 2 M * 1 L = M2 * 3 L, thus M2 = 2/3 = 0.67 M.
Correct Answer: A — 0.67 M
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Q. If 1 L of a 3 M solution is diluted to 2 L, what is the new molarity?
A.
1.5 M
B.
3 M
C.
6 M
D.
0.5 M
Show solution
Solution
Using the dilution formula M1V1 = M2V2, we have 3 M × 1 L = M2 × 2 L. Thus, M2 = 1.5 M.
Correct Answer: A — 1.5 M
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Q. If 1 liter of a 2 M solution is diluted to 3 liters, what is the new molarity?
A.
0.67 M
B.
1 M
C.
1.5 M
D.
2 M
Show solution
Solution
Using the dilution formula M1V1 = M2V2, (2 M)(1 L) = M2(3 L) => M2 = 2/3 = 0.67 M.
Correct Answer: A — 0.67 M
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Q. If 1 mol of NaCl is dissolved in 1 kg of water, how many particles are present in solution?
Show solution
Solution
NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl produces 2 mol of particles in solution.
Correct Answer: B — 2
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Q. If 1 mol of NaCl is dissolved in 1 kg of water, what is the expected van 't Hoff factor (i)?
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Solution
NaCl dissociates into 2 ions (Na+ and Cl-), so the van 't Hoff factor i = 2.
Correct Answer: B — 2
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