Q. What is the molecular orbital configuration of the F2 molecule?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)⁴
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)¹
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)³(π*2p)²
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Solution
The correct configuration for F2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)².
Correct Answer: A — (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
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Q. What is the molecular orbital configuration of the O2 molecule?
A.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
B.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)²
C.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)¹(π*2p)¹
D.
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)¹(π*2p)²
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Solution
The correct configuration for O2 is (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹.
Correct Answer: A — (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p)²(π2p)²(π*2p)¹
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Q. What is the molecular shape of a molecule with the formula AX3E?
A.
Trigonal planar
B.
Tetrahedral
C.
Trigonal pyramidal
D.
Bent
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Solution
AX3E indicates three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape.
Correct Answer: C — Trigonal pyramidal
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Q. What is the molecular shape of BF3 according to VSEPR theory?
A.
Bent
B.
Trigonal planar
C.
Tetrahedral
D.
Octahedral
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Solution
BF3 has three bonding pairs and no lone pairs, resulting in a trigonal planar shape.
Correct Answer: B — Trigonal planar
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Q. What is the molecular shape of NH3 according to VSEPR theory?
A.
Linear
B.
Trigonal planar
C.
Tetrahedral
D.
Trigonal pyramidal
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Solution
NH3 has three bonding pairs and one lone pair, resulting in a trigonal pyramidal shape.
Correct Answer: D — Trigonal pyramidal
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Q. What is the Nernst equation used for?
A.
Calculating pH
B.
Determining cell potential
C.
Finding molarity
D.
Measuring temperature
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Solution
The Nernst equation is used to determine the cell potential under non-standard conditions.
Correct Answer: B — Determining cell potential
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Q. What is the normality of a solution containing 1 mole of H2SO4 in 1 liter of solution?
A.
1 N
B.
2 N
C.
0.5 N
D.
4 N
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Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 provides 2 equivalents, so N = 2 moles / 1 L = 2 N.
Correct Answer: B — 2 N
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Q. What is the normality of a solution containing 2 moles of H2SO4 in 1 liter of solution?
A.
2 N
B.
4 N
C.
1 N
D.
0.5 N
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Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 provides 2 equivalents, so 2 moles × 2 = 4 N.
Correct Answer: B — 4 N
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Q. What is the normality of a solution containing 3 moles of H2SO4 in 2 liters of solution? (H2SO4 is a diprotic acid)
A.
3 N
B.
6 N
C.
1.5 N
D.
1 N
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Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 has 2 equivalents per mole, so 3 moles = 6 equivalents. Normality = 6 equivalents / 2 L = 3 N.
Correct Answer: B — 6 N
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Q. What is the normality of a solution containing 4 moles of H2SO4 in 2 liters of solution?
A.
4 N
B.
8 N
C.
2 N
D.
1 N
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Solution
Normality (N) = equivalents of solute / liters of solution. H2SO4 has 2 equivalents, so 4 moles = 8 equivalents. N = 8 eq / 2 L = 4 N.
Correct Answer: B — 8 N
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Q. What is the normality of a solution containing 4 moles of H2SO4 in 2 liters of solution? (H2SO4 is a diprotic acid)
A.
4 N
B.
2 N
C.
8 N
D.
1 N
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Solution
Normality (N) = number of equivalents / liters of solution. H2SO4 has 2 equivalents per mole, so 4 moles = 8 equivalents. Normality = 8 equivalents / 2 L = 4 N.
Correct Answer: C — 8 N
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Q. What is the normality of a solution that is 1 M in H2SO4?
A.
1 N
B.
2 N
C.
0.5 N
D.
4 N
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Solution
Normality (N) = Molarity (M) x number of equivalents. H2SO4 has 2 acidic protons, so 1 M x 2 = 2 N.
Correct Answer: B — 2 N
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Q. What is the number of atoms in 2 moles of aluminum (Al)? (2044)
A.
6.022 x 10^23
B.
1.2044 x 10^24
C.
3.011 x 10^23
D.
12.044 x 10^24
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Solution
Number of atoms = moles x Avogadro's number = 2 moles x 6.022 x 10^23 atoms/mole = 1.2044 x 10^24 atoms.
Correct Answer: B — 1.2044 x 10^24
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Q. What is the number of atoms in 2 moles of CaCO3?
A.
6.022 x 10^23
B.
1.2044 x 10^24
C.
3.011 x 10^23
D.
1.8066 x 10^24
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Solution
Each CaCO3 has 5 atoms (1 Ca, 1 C, 3 O). Number of atoms = moles x atoms per molecule x Avogadro's number = 2 moles x 5 x 6.022 x 10^23 = 1.206 x 10^24 atoms.
Correct Answer: D — 1.8066 x 10^24
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Q. What is the number of atoms in 2 moles of Na2SO4?
A.
6.022 x 10^23
B.
1.2044 x 10^24
C.
1.2044 x 10^25
D.
3.011 x 10^23
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Solution
Na2SO4 has 2 Na + 1 S + 4 O = 7 atoms. Number of atoms = moles x atoms per mole = 2 moles x 7 atoms = 14 atoms = 1.2044 x 10^24 atoms.
Correct Answer: B — 1.2044 x 10^24
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Q. What is the number of atoms in 2 moles of NaCl? (2044)
A.
6.022 x 10^23
B.
1.2044 x 10^24
C.
1.2044 x 10^25
D.
3.011 x 10^23
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Solution
Each NaCl unit has 2 atoms (Na and Cl). Therefore, 2 moles of NaCl contain 2 x 6.022 x 10^23 x 2 = 1.2044 x 10^24 atoms.
Correct Answer: B — 1.2044 x 10^24
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Q. What is the number of moles in 10 grams of Na?
A.
0.43
B.
0.22
C.
0.5
D.
0.1
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Solution
Molar mass of Na = 23 g/mol. Number of moles = mass / molar mass = 10 g / 23 g/mol = 0.43 moles.
Correct Answer: A — 0.43
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Q. What is the number of moles in 10 grams of NaOH?
A.
0.25
B.
0.5
C.
0.75
D.
1
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Solution
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Number of moles = mass/molar mass = 10 g / 40 g/mol = 0.25 moles.
Correct Answer: B — 0.5
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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 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 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
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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 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 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 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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