Q. What is the number of moles in 100 grams of NaOH?
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Solution
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Number of moles = mass/molar mass = 100 g / 40 g/mol = 2.5 moles.
Correct Answer: A — 1
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Q. What is the number of moles in 180 grams of glucose (C6H12O6)?
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Solution
Molar mass of C6H12O6 = 6*12 + 12*1 + 6*16 = 180 g/mol. Number of moles = mass/molar mass = 180 g / 180 g/mol = 1 mole.
Correct Answer: A — 1
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Q. What is the number of moles in 44 grams of CO2?
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Solution
Molar mass of CO2 = 12 + 16*2 = 44 g/mol. Number of moles = mass/molar mass = 44 g / 44 g/mol = 1 mole.
Correct Answer: A — 1
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Q. What is the number of moles in 5 liters of a 2 M NaCl solution?
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Solution
Moles = Molarity x Volume = 2 moles/L x 5 L = 10 moles.
Correct Answer: A — 2.5
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Q. What is the number of proper subsets of the set E = {a, b}?
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Solution
The total number of subsets is 2^n = 2^2 = 4. Proper subsets exclude the set itself, so there are 4 - 1 = 3 proper subsets.
Correct Answer: B — 3
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Q. What is the number of significant figures in 0.0001230?
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Solution
The significant figures are 1, 2, 3, and the trailing zero counts.
Correct Answer: C — 5
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Q. What is the number of significant figures in the measurement 0.00456?
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Solution
The significant figures are 4, 5, and 6. Leading zeros are not counted.
Correct Answer: B — 3
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Q. What is the number of subsets of the empty set?
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Solution
The empty set has exactly one subset, which is itself (∅).
Correct Answer: B — 1
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Q. What is the number of subsets of the set F = {x, y, z}?
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Solution
The number of subsets of a set with n elements is 2^n. Here, n = 3, so 2^3 = 8.
Correct Answer: D — 8
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Q. What is the number of subsets of the set H = {1, 2, 3, 4, 5}?
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Solution
The number of subsets of a set with n elements is 2^n. Here, n = 5, so 2^5 = 32.
Correct Answer: A — 32
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Q. What is the number of subsets of the set H = {x, y, z, w, v}?
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Solution
The number of subsets of a set with n elements is 2^n. Here, n = 5, so the number of subsets is 2^5 = 32.
Correct Answer: A — 32
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Q. What is the number of subsets of the set {1, 2, 3, 4}?
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Solution
The number of subsets of a set with n elements is 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.
Correct Answer: C — 16
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Q. What is the number of subsets of the set {a, b, c, d}?
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Solution
The number of subsets of a set with n elements is 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.
Correct Answer: B — 8
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Q. What is the Nyquist rate for a signal with a maximum frequency of 1 kHz? (2000)
A.
1 kHz
B.
2 kHz
C.
500 Hz
D.
4 kHz
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Solution
Nyquist rate = 2 * f_max = 2 * 1000 Hz = 2000 Hz = 2 kHz.
Correct Answer: B — 2 kHz
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Q. What is the Nyquist rate for a signal with a maximum frequency of 5 kHz?
A.
5 kHz
B.
10 kHz
C.
15 kHz
D.
20 kHz
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Solution
Nyquist rate = 2 * f_max = 2 * 5 kHz = 10 kHz.
Correct Answer: B — 10 kHz
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Q. What is the Nyquist rate in a communication system?
A.
The minimum sampling rate to avoid aliasing
B.
The maximum data rate achievable
C.
The frequency of the carrier wave
D.
The bandwidth of the communication channel
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Solution
The Nyquist rate is the minimum sampling rate required to avoid aliasing, which is twice the highest frequency of the signal.
Correct Answer: A — The minimum sampling rate to avoid aliasing
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Q. What is the Nyquist rate in communication systems?
A.
The minimum sampling rate to avoid aliasing
B.
The maximum frequency of the signal
C.
The bandwidth of the communication channel
D.
The maximum data rate of a channel
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Solution
The Nyquist rate is the minimum sampling rate required to avoid aliasing, which is twice the maximum frequency of the signal.
Correct Answer: A — The minimum sampling rate to avoid aliasing
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Q. What is the Nyquist rate in the context of signal sampling?
A.
The minimum sampling rate to avoid aliasing
B.
The maximum frequency of the signal
C.
The bandwidth of the signal
D.
The average power of the signal
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Solution
The Nyquist rate is defined as twice the maximum frequency of the signal, which is the minimum sampling rate required to avoid aliasing.
Correct Answer: A — The minimum sampling rate to avoid aliasing
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Q. What is the Nyquist rate?
A.
The minimum sampling rate to avoid aliasing
B.
The maximum frequency of a signal
C.
The bandwidth of a communication channel
D.
The rate of data transmission in a channel
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Solution
The Nyquist rate is the minimum sampling rate required to avoid aliasing, which is twice the maximum frequency of the signal.
Correct Answer: A — The minimum sampling rate to avoid aliasing
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Q. What is the orbital speed of a satellite in a circular orbit at a height h above the Earth's surface?
A.
sqrt(GM/R^2)
B.
sqrt(GM/(R+h)^2)
C.
sqrt(GM/(R-h)^2)
D.
sqrt(GM/R)
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Solution
The orbital speed v of a satellite in a circular orbit is given by v = sqrt(GM/(R+h)^2), where G is the gravitational constant, M is the mass of the Earth, R is the radius of the Earth, and h is the height of the satellite above the Earth's surface.
Correct Answer: B — sqrt(GM/(R+h)^2)
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Q. What is the orbital speed of a satellite in a low Earth orbit (LEO) approximately 2000 km above the Earth's surface? (2000)
A.
7.9 km/s
B.
11.2 km/s
C.
5.0 km/s
D.
3.0 km/s
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Solution
The orbital speed of a satellite in low Earth orbit is approximately 7.9 km/s.
Correct Answer: A — 7.9 km/s
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Q. What is the order of the first bright fringe in a single-slit diffraction pattern?
A.
m = 0
B.
m = 1
C.
m = 2
D.
m = 3
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Solution
In single-slit diffraction, the central maximum corresponds to m = 0, which is the first bright fringe.
Correct Answer: A — m = 0
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Q. What is the order of the reaction if the rate constant has the unit L/mol·s?
A.
Zero order
B.
First order
C.
Second order
D.
Third order
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Solution
If the rate constant has the unit L/mol·s, the reaction is second order.
Correct Answer: C — Second order
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Q. What is the osmotic pressure of a 0.2 M NaCl solution at 25 °C? (R = 0.0821 L atm/(K mol))
A.
4.92 atm
B.
2.46 atm
C.
1.23 atm
D.
0.61 atm
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Solution
Osmotic pressure = iCRT = 2 * 0.2 * 0.0821 * 298 = 4.92 atm (i = 2 for NaCl)
Correct Answer: A — 4.92 atm
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Q. What is the osmotic pressure of a solution containing 0.2 moles of solute in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))
A.
4.92 atm
B.
1.64 atm
C.
0.82 atm
D.
2.46 atm
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Solution
Osmotic pressure = (n/V)RT = (0.2/1) * 0.0821 * 298 = 4.92 atm.
Correct Answer: A — 4.92 atm
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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 1 liter of water at 25 °C?
A.
12.3 atm
B.
24.6 atm
C.
6.1 atm
D.
3.1 atm
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Solution
Osmotic pressure (π) = iCRT = 1 * 0.5 * 0.0821 * 298 = 12.3 atm.
Correct Answer: A — 12.3 atm
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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 1 liter of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))
A.
12.3 atm
B.
0.5 atm
C.
1.0 atm
D.
2.5 atm
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Solution
Osmotic pressure (π) = nRT/V = (0.5 mol)(0.0821 L·atm/(K·mol))(298 K) = 12.3 atm.
Correct Answer: A — 12.3 atm
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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 1 L of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))
A.
12.3 atm
B.
10.2 atm
C.
8.2 atm
D.
6.1 atm
Show solution
Solution
Osmotic pressure (π) = nRT/V = (0.5 moles)(0.0821 L·atm/(K·mol))(298 K) = 12.3 atm.
Correct Answer: A — 12.3 atm
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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 1 L of water at 25 °C? (R = 0.0821 L·atm/(K·mol))
A.
12.3 atm
B.
1.23 atm
C.
0.5 atm
D.
2.5 atm
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Solution
Osmotic pressure (π) = nRT/V = (0.5 moles * 0.0821 * 298) / 1 = 12.3 atm.
Correct Answer: B — 1.23 atm
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Q. What is the osmotic pressure of a solution containing 0.5 moles of glucose in 2 liters of solution at 25 °C? (R = 0.0821 L·atm/(K·mol))
A.
6.13 atm
B.
12.26 atm
C.
3.07 atm
D.
1.54 atm
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Solution
Osmotic pressure (π) = nRT/V = (0.5 moles)(0.0821 L·atm/(K·mol))(298 K) / 2 L = 6.13 atm.
Correct Answer: A — 6.13 atm
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