Praxis 5246

Explanation

Entropy is a measure of disorder/dispersal. Condensation converts gas-phase molecules (high entropy, high disorder) into a liquid (lower entropy, more ordered arrangement), so ΔS < 0 for the system. Option B is incorrect because dissolving an ionic solid disperses ions throughout the solvent, increasing the number of accessible microstates and increasing entropy. Option C is incorrect because a solid decomposing to produce a gas greatly increases entropy; the production of CO₂(g) from a solid reactant represents a large increase in disorder. Option D is incorrect because sublimation converts a solid directly to a gas, which dramatically increases molecular disorder and entropy.

Explanation

In a galvanic (voltaic) cell, oxidation occurs at the anode, releasing electrons that travel through the external circuit. Because the anode loses electrons and drives current, it is labeled as the negative electrode in a galvanic cell. Option B is wrong because reduction occurs at the cathode, not the anode, and the anode is negative, not positive. Option C is wrong because oxidation occurs at the anode, not the cathode, and the cathode is positive in a galvanic cell. Option D is wrong because it correctly states the anode is negative but incorrectly assigns reduction to the anode rather than oxidation.

Explanation

A catalyst works by providing an alternative mechanism (reaction pathway) that has a lower activation energy than the uncatalyzed reaction. This allows a greater fraction of reactant molecules to have sufficient energy to react, thereby increasing the rate of both the forward and reverse reactions equally. The catalyst itself is not consumed in the overall reaction. (A) is incorrect because a catalyst lowers, not increases, the activation energy. (B) is incorrect because a catalyst does not change the position of equilibrium (i.e., it does not change Keq or the thermodynamic favorability); it only affects how quickly equilibrium is reached. (D) is incorrect because a catalyst does not raise the temperature of the reaction — temperature is an external condition, and increasing temperature is a separate strategy from catalysis.

Explanation

The enthalpy of vaporization corresponds to the process H₂O(l) → H₂O(g). Using Hess's law: ΔH_vap = ΔH_f°[H₂O(g)] − ΔH_f°[H₂O(l)] = (−241.8) − (−285.8) = +44.0 kJ/mol. The positive sign is expected because vaporization is endothermic (energy must be added to convert liquid to gas). Option B is wrong because it has the incorrect sign; condensation (the reverse process) releases 44.0 kJ/mol. Option C results from incorrectly adding the two formation enthalpies instead of subtracting. Option D similarly results from addition with an incorrect sign convention.

Explanation

Fe₂O₃ contains iron in the +3 oxidation state (since 2x + 3(-2) = 0, x = +3). The IUPAC Stock system name is iron(III) oxide. Choice A is incorrect because iron(II) oxide refers to FeO, where iron has a +2 oxidation state. Choice C (diiron trioxide) uses the older Greek prefix system, which is acceptable for binary nonmetal compounds but is not the preferred IUPAC name for ionic compounds containing a metal. Choice D (ferric oxide) is the classical/common name using the Latin-derived suffix system, not the preferred IUPAC name.

Explanation

Fe₂O₃ contains iron in the +3 oxidation state. Since oxygen has a −2 charge and there are three oxygen atoms (total −6), the two iron atoms must each carry a +3 charge to balance. IUPAC nomenclature for ionic compounds with variable-valence metals uses Roman numerals in parentheses to indicate the oxidation state, giving Iron(III) oxide. Option A is incomplete because it omits the oxidation state required for transition metals that form multiple ions. Option B, Iron(II) oxide, corresponds to FeO, not Fe₂O₃. Option D, Diiron trioxide, is the older Stock/additive nomenclature style using Greek prefixes, which is acceptable for binary molecular compounds but is not the preferred IUPAC ionic name for this compound.

Explanation

The molar mass of Ca₃(PO₄)₂ is calculated as follows: 3(40.08) + 2(30.97) + 8(16.00) = 120.24 + 61.94 + 128.00 = 310.18 g/mol. Option A (182.18) incorrectly accounts for only one phosphate group and miscounts oxygen atoms. Option B (278.18) results from a common error of using only 6 oxygen atoms instead of 8. Option D (422.18) results from incorrectly multiplying the entire phosphate group by 3 instead of 2, overcounting phosphorus and oxygen atoms.

Explanation

The molar mass of Ca₃(PO₄)₂ is calculated as follows: 3 Ca = 3 × 40.08 = 120.24 g/mol; 2 P = 2 × 30.97 = 61.94 g/mol; 8 O = 8 × 16.00 = 128.00 g/mol. Total = 120.24 + 61.94 + 128.00 = 310.18 g/mol. Option A (182.18) results from neglecting the subscript 2 on the phosphate group, counting only 4 oxygen atoms. Option B (278.18) results from an arithmetic error, likely using incorrect subscripts for oxygen. Option D (215.05) likely results from summing only two calcium atoms instead of three.

Explanation

Molarity = moles of solute / liters of solution. Moles of NaCl = 58.44 g ÷ 58.44 g/mol = 1.00 mol. Molarity = 1.00 mol / 2.00 L = 0.500 M. Choice B (1.00 M) would be correct if the volume were 1.00 L rather than 2.00 L — a common error of ignoring the actual volume. Choice C (2.00 M) inverts the relationship, multiplying moles by volume instead of dividing. Choice D (29.2 M) results from dividing grams by volume in mL rather than converting to moles first.

Explanation

ClO₄⁻ is the perchlorate ion. The oxyanion series for chlorine follows a systematic pattern: hypochlorite (ClO⁻), chlorite (ClO₂⁻), chlorate (ClO₃⁻), and perchlorate (ClO₄⁻). The prefix 'per-' combined with the '-ate' suffix denotes one more oxygen than the base '-ate' ion. Choice A (chlorite, ClO₂⁻) has two oxygen atoms. Choice B (chlorate, ClO₃⁻) has three oxygen atoms and is one oxygen short of perchlorate. Choice C (hypochlorite, ClO⁻) has only one oxygen atom and uses the 'hypo-' prefix to indicate the fewest oxygens in the series.