First, find the total moles of $$CO_2$$ produced using the ideal gas law: $$n = PV/RT = (1.00 \text{ atm} \times 0.985 \text{ L}) / (0.08206 \text{ L atm/mol K} \times 298 \text{ K}) = 0.0403 \text{ mol} CO_2$$. Let x be the mass of $$NaHCO_3$$ and y be the mass of $$Na_2CO_3$$. We have two equations: 1) $$x + y = 4.00$$ and 2) Moles of $$CO_2$$ from $$NaHCO_3$$ + Moles of $$CO_2$$ from $$Na_2CO_3$$ = $$0.0403$$. The reactions are $$NaHCO_3 \rightarrow CO_2$$ (1:1) and $$Na_2CO_3 \rightarrow CO_2$$ (1:1). So, $$(x/84.0) + (y/106.0) = 0.0403$$. Substitute $$y = 4.00 - x$$ into the second equation: $$(x/84.0) + ((4.00 - x)/106.0) = 0.0403$$. Solving for x gives $$x = 1.68$$ g. Mass percent of $$NaHCO_3$$ = ($$1.68 \text{ g} / 4.00 \text{ g}$$) * 100% = 42.0%.

Ethanol and water are miscible, meaning they mix uniformly at the molecular level to form a solution, which is a homogeneous mixture. (A) is incorrect because the components are not chemically bonded in a mixture. (B) is incorrect because the mixture is uniform, not heterogeneous. (D) is incorrect because the ratio is fixed for this specific mixture, but it can be varied, which is a characteristic of mixtures, not compounds.

First, find the partial pressure of $$C_2H_2$$: $$P_{C2H2} = P_{total} - P_{H2O} = 1.00 \text{ atm} - 0.035 \text{ atm} = 0.965 \text{ atm}$$. Next, find moles of $$C_2H_2$$ using the ideal gas law: $$n = PV/RT = (0.965 \text{ atm} \times 0.520 \text{ L}) / (0.08206 \text{ L atm/mol K} \times 300 \text{ K}) = 0.0204 \text{ mol}$$. The reaction is $$CaC_2(s) + 2H_2O(l) \rightarrow C_2H_2(g) + Ca(OH)_2(aq)$$. The mole ratio of $$CaC_2$$ to $$C_2H_2$$ is 1:1. Mass of $$CaC_2$$ = $$0.0204 \text{ mol} \times 64.1 \text{ g/mol} = 1.307 \text{ g}$$. Percent purity = ($$1.307 \text{ g} / 1.80 \text{ g}$$) * 100% = 72.6%.

The mass percent of Fe in FeO is $$(55.85/71.85) \times 100% = 77.73%$$. The mass percent of Fe in $$Fe_2O_3$$ is $$(2 \times 55.85 / 159.70) \times 100% = 69.94%$$. Let x be the mass fraction of FeO in the mixture. The weighted average gives: $$x(0.7773) + (1-x)(0.6994) = 0.720$$. Expanding: $$0.7773x + 0.6994 - 0.6994x = 0.720$$, so $$0.0779x = 0.0206$$, and $$x = 0.264 = 26.4%$$. Choice (A) is closest at 27.8%. Choice (B) assumes equal masses of each oxide. Choice (C) represents the mass percent of $$Fe_2O_3$$. Choice (D) is the overall iron percentage.

The loss of mass is due to the evolution of $$CO_2$$ gas from the decomposition of $$CaCO_3$$. Mass of $$CO_2$$ lost = $$\10.0 \text{ g} - 7.8 \text{ g} = 2.2 \text{ g}$$. Moles of $$CO_2$$ = $$2.2 \text{ g} / 44.0 \text{ g/mol} = 0.050 \text{ mol}$$. The reaction is $$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$$. The mole ratio of $$CaCO_3$$ to $$CO_2$$ is 1:1. So, moles of $$CaCO_3$$ = $$0.050 \text{ mol}$$. Mass of $$CaCO_3$$ = $$0.050 \text{ mol} \times 100.1 \text{ g/mol} = 5.0 \text{ g}$$. Initial mass percent of $$CaCO_3$$ = ($$5.0 \text{ g} / 10.0 \text{ g}$$) * 100% = 50.0%. (A) is the mass percent of $$CO_2$$ lost. (C) is the mass percent of the final residue. (D) incorrectly assumes the entire sample decomposed.

The change in properties (loss of magnetism, change in solubility) indicates that a chemical reaction has occurred, forming a new substance (iron sulfide) with properties different from the original components. The initial sample was a mixture of elements. (A) is likely incorrect, as a mixture of powders is typically heterogeneous. (B) describes a physical change, but the evidence points to a chemical change. (C) is incorrect because the original sample was a mixture of elements, not a compound.

Assume a total mass, for example, 100 g. Then there are 50 g of He and 50 g of Ne. Moles of He = $$\50 \text{ g} / 4.00 \text{ g/mol} = 12.5 \text{ mol}$$. Moles of Ne = $$\50 \text{ g} / 20.2 \text{ g/mol} = 2.48 \text{ mol}$$. Total moles = $$\12.5 + 2.48 = 14.98 \text{ mol}$$. Mole fraction of He = Moles of He / Total moles = $$\12.5 / 14.98 = 0.834$$. (B) incorrectly assumes equal moles instead of equal masses. (A) is the mole fraction of neon. (D) is based on a calculation error.

This question tests the skill of calculating mole fraction in a gas mixture. Mole fraction is calculated as moles of component divided by total moles. The total moles = 1.0 mol CH₄ + 4.0 mol H₂ = 5.0 mol. Therefore, mole fraction of CH₄ = 1.0 mol / 5.0 mol = 0.20. Choice A (0.80) represents the mole fraction of H₂ rather than CH₄, a common mistake when students calculate for the wrong component. Always double-check which component's mole fraction is being requested in the problem.

This question tests the skill of calculating mass percent composition in a solution. Mass percent is found using: mass percent = (mass of solute / total mass) × 100%. The total mass = 15.0 g ethanol + 35.0 g water = 50.0 g. Therefore, mass percent of ethanol = (15.0 g / 50.0 g) × 100% = 30.0%. Choice A (70.0%) represents the mass percent of water, not ethanol, which is a common error when students confuse which component they're calculating. Remember to identify clearly which component's mass percent is being asked for before calculating.

This question tests the skill of calculating mass percent composition in a liquid mixture. To find mass percent: (mass of component / total mass) × 100%. The total mass = 30.0 g acetone + 20.0 g water = 50.0 g. Therefore, mass percent of water = (20.0 g / 50.0 g) × 100% = 40.0%. Choice A (60.0%) represents the mass percent of acetone rather than water, a common error when students calculate for the wrong component. Always verify which component's mass percent is requested before performing the calculation.