C373 General Chemistry I with Lab
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Free C373 General Chemistry I with Lab Questions
The hybridization and geometry of the central atom in [Co(NH₃)₆]³⁺ are:
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sp³d², octahedral
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d²sp³, octahedral
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sp³, tetrahedral
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dsp², square planar
Explanation
[Co(NH₃)₆]³⁺ is a coordination complex with six ligands around the central Co³⁺ ion. The coordination number is 6, which corresponds to octahedral geometry. The central atom uses d²sp³ hybrid orbitals to accommodate six bonding pairs in an octahedral arrangement.
Correct Answer:
d²sp³, octahedral
The correct configuration for Cu²⁺ is:
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[Ar] 4s¹ 3d⁹
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[Ar] 4s⁰ 3d⁹
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[Ar] 4s² 3d⁷
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[Ar] 4s⁰ 3d¹⁰
Explanation
A neutral copper atom has the electron configuration [Ar] 4s¹ 3d¹⁰ due to its half-filled s orbital and completely filled d subshell providing stability. When it forms a Cu²⁺ ion, it loses two electrons. Electrons are removed first from the 4s orbital and then from the 3d orbital because 4s electrons are higher in energy once filled. Removing two electrons (one from 4s and one from 3d) results in the configuration [Ar] 3d⁹, or equivalently written as [Ar] 4s⁰ 3d⁹.
Correct Answer:
[Ar] 4s⁰ 3d⁹
A 4.00 L sample of N₂ at 300 K and 2.00 atm is heated to 600 K at constant pressure. New volume?
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2.00 L
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4.00 L
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6.00 L
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8.00 L
Explanation
Using Charles’s Law (V₁/T₁ = V₂/T₂) at constant pressure, V₂ = V₁ × (T₂/T₁). Substituting values: V₂ = 4.00 L × (600 K / 300 K) = 4.00 × 2 = 8.00 L. Therefore, the volume doubles when the temperature is doubled at constant pressure.
Correct Answer:
8.00 L
In the reaction 3NO₂ + H₂O → 2HNO₃ + NO, the oxidation number of nitrogen changes from:
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+4 to +5 and +2
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+5 to +4
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+4 to +2 only
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+4 to +5 only
Explanation
In NO₂, nitrogen has an oxidation number of +4. In HNO₃, nitrogen is +5, and in NO, nitrogen is +2. This reaction is a disproportionation reaction, where a single species (NO₂) is simultaneously oxidized to HNO₃ and reduced to NO. Therefore, nitrogen changes from +4 to both +5 and +2.
Correct Answer:
+4 to +5 and +2
Rutherford’s gold foil experiment provided evidence for:
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Electrons orbiting in fixed paths
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A dense, positive nucleus
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Neutrons in the nucleus
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Quantized energy levels
Explanation
Rutherford’s gold foil experiment involved directing alpha particles at a thin sheet of gold foil. Most particles passed through, but a few were deflected at large angles. This showed that atoms are mostly empty space, with a small, dense, positively charged center that repelled the alpha particles. This discovery led to the nuclear model of the atom, identifying the nucleus as the core where most of the atom’s mass and positive charge reside.
Correct Answer:
A dense, positive nucleus
Which molecule is an exception to the octet rule with only 6 electrons?
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BF₃
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PF₅
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SF₆
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XeF₄
Explanation
Boron trifluoride (BF₃) is a well-known exception to the octet rule. In this molecule, boron forms three single bonds with fluorine atoms, resulting in only six electrons around the boron atom. Because boron lacks sufficient valence electrons to complete an octet and does not have available d orbitals for expansion, it remains electron-deficient. In contrast, PF₅, SF₆, and XeF₄ all involve central atoms that have expanded octets due to available d orbitals in period 3 or higher.
Correct Answer:
BF₃
A 0.500 g sample of a compound containing C, H, and O produces 1.100 g CO₂ and 0.450 g H₂O on combustion. Its molar mass is 60.0 g/mol. Molecular formula?
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CH₃COOH
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C₂H₅OH
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CH₂O
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C₃H₆O
Explanation
First, determine moles of C from CO₂: 1.100 g ÷ 44.01 g/mol ≈ 0.0250 mol C. Moles of H from H₂O: 0.450 g ÷ 18.02 g/mol ≈ 0.0250 mol H₂, which is 0.0500 mol H. Mass of C + H = (0.0250 × 12.01) + (0.0500 × 1.008) ≈ 0.300 + 0.050 = 0.350 g. Oxygen mass = 0.500 − 0.350 = 0.150 g → 0.150 ÷ 16.00 ≈ 0.009375 mol O. The empirical formula ratio: C: 0.0250, H: 0.0500, O: 0.009375 → divide by 0.009375: C ≈ 2.67, H ≈ 5.33, O ≈ 1 → closest whole-number ratio ≈ C₃H₆O. The molar mass matches 60.0 g/mol, confirming the molecular formula is C₃H₆O.
Correct Answer:
C₃H₆O
As you move left to right across period 3, what happens to electronegativity?
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Decreases
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Remains constant
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Increases
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First increases, then decreases
Explanation
Electronegativity refers to an atom’s ability to attract electrons in a chemical bond. Across a period from left to right, the number of protons in the nucleus increases while the atomic radius decreases due to stronger attraction between the nucleus and electrons. This stronger pull causes atoms to attract bonding electrons more effectively, so electronegativity increases from left to right across period 3.
Correct Answer:
Increases
Which species is isoelectronic with Kr?
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Br⁻
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Se²⁻
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Rb⁺
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All of the above
Explanation
Isoelectronic species have the same number of electrons. Krypton (Kr) has 36 electrons. Bromine (Z=35) gains one electron to form Br⁻, selenium (Z=34) gains two electrons to form Se²⁻, and rubidium (Z=37) loses one electron to form Rb⁺. In each case, the ion ends up with 36 electrons, the same as krypton. Hence, all three ions—Br⁻, Se²⁻, and Rb⁺—are isoelectronic with Kr.
Correct Answer:
All of the above
Which gas effuses fastest at the same T and P?
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He
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Ne
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Ar
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Kr
Explanation
According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. He (helium) has the smallest molar mass (4 g/mol) compared with Ne (20 g/mol), Ar (40 g/mol), and Kr (84 g/mol). Therefore, helium effuses the fastest under the same conditions of temperature and pressure.
Correct Answer:
He
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