Elements X and Y form compounds X_2Y_3 and X_ | vedclass

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(C) Let the atomic mass of $X$ be $x$ and $Y$ be $y$. For $X_2Y_3$: $0.2 \ mol$ weighs $32.0 \ g$,so $1 \ mol$ weighs $32.0 / 0.2 = 160 \ g/mol$. Thus

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$56.0, 16.0$

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(C) Let the atomic mass of $X$ be $x$ and $Y$ be $y$. For $X_2Y_3$: $0.2 \ mol$ weighs $32.0 \ g$,so $1 \ mol$ weighs $32.0 / 0.2 = 160 \ g/mol$. Thus,$2x + 3y = 160$ --- $(1)$ For $X_3Y_4$: $0.4 \ mol$ weighs $92.8 \ g$,so $1 \ mol$ weighs $92.8 / 0.4 = 232 \ g/mol$. Thus,$3x + 4y = 232$ --- $(2)$ Multiplying $(1)$ by $4$ and $(2)$ by $3$: $8x + 12y = 640$ $9x + 12y = 696$ Subtracting the equations: $x = 56$. Substituting $x = 56$ in $(1)$: $2(56) + 3y = 160 \implies 112 + 3y = 160 \implies 3y = 48 \implies y = 16$. Therefore,the atomic masses are $X = 56$ and $Y = 16$.

Elements $X$ and $Y$ form compounds $X_2Y_3$ and $X_3Y_4$ respectively. If $0.2 \ mol$ of $X_2Y_3$ weighs $32.0 \ g$ and $0.4 \ mol$ of $X_3Y_4$ weighs $92.8 \ g$,then the atomic masses of $X$ and $Y$ are respectively:

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