Question

1. which pair of formulas illustrates the law of multiple proportions? (3pts) ch₄ and c₂h₅oh ch₄ and c₈h₁₈ feo and cuo₂ h₂o₂ and h₂so₄ 25) which of the following illustrates an increase in potential energy? (3pts)

Explanation:

The Law of Multiple Proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers.

1. For $\text{CH}_4$ and $\text{C}_2\text{H}_5\text{OH}$: They contain three elements (C, H, O), so the law does not apply.
2. For $\text{CH}_4$ and $\text{C}_8\text{H}_{18}$: These are both hydrocarbons, but they represent different ratios of H to C, not fixed mass ratios of two elements forming different simple compounds.
3. For $\text{FeO}$ and $\text{CuO}_2$: They involve different pairs of elements (Fe-O vs. Cu-O), so the law does not apply.
4. For $\text{H}_2\text{O}_2$ and $\text{H}_2\text{SO}_4$: They involve different element pairs (H-O vs. H-S-O), so the law does not apply.

Wait, correction: Re-evaluating, the first option $\text{CH}_4$ and $\text{C}_2\text{H}_5\text{OH}$ is invalid, but the correct pair that fits is $\text{CH}_4$ and $\text{C}_2\text{H}_2$ (not listed), but among the given options, the only pair with the same two elements is $\text{CH}_4$ and $\text{C}_8\text{H}_{18}$? No, no—wait, no, the Law of Multiple Proportions requires two elements forming different compounds. The only pair here with the same two elements is $\text{CH}_4$ and $\text{C}_8\text{H}_{18}$ (C and H), but actually, the correct application is when two elements form two different compounds with different ratios. Wait, no, $\text{H}_2\text{O}_2$ and $\text{H}_2\text{O}$ would fit, but $\text{H}_2\text{O}_2$ and $\text{H}_2\text{SO}_4$ do not. $\text{FeO}$ and $\text{Fe}_2\text{O}_3$ would fit, but $\text{FeO}$ and $\text{CuO}_2$ do not. $\text{CH}_4$ and $\text{C}_2\text{H}_4$ would fit, but $\text{CH}_4$ and $\text{C}_8\text{H}_{18}$ are both alkanes, with H:C ratios of 4:1 and 18:8=9:4. The fixed mass of C (12g) would combine with 4g H in $\text{CH}_4$ and $\frac{9}{2}$g H in $\text{C}_8\text{H}_{18}$, which is a ratio of 4 : 4.5 = 8:9, whole numbers. So among the given options, this is the only pair with the same two elements forming different compounds, which fits the Law of Multiple Proportions.

Wait, no, another correction: $\text{CH}_4$ and $\text{C}_2\text{H}_5\text{OH}$ have three elements, so they can't be used. $\text{FeO}$ and $\text{CuO}_2$ have different metal elements. $\text{H}_2\text{O}_2$ and $\text{H}_2\text{SO}_4$ have different non-metal elements. So the only pair with the same two elements is $\text{CH}_4$ and $\text{C}_8\text{H}_{18}$.

Answer:

$\boldsymbol{\text{CH}_4 \text{ and } \text{C}_8\text{H}_{18}}$

For question 25, the full options are not provided, so it cannot be answered. Please provide the complete options for question 25.