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112 mL of hydrogen combines with 56 mL o...

`112 mL` of hydrogen combines with `56 mL` of oxygen of form water. When `224 mL` of hydrogen is passes over hand cupric oxide, the cupric oxide loses. `0.160 g` of weight. All volumes are measured at `STP`. Show that the result agrees with the law of constant composition `(22.4 L` hydrogen and oxygen at `STP` weigh, respectively, `2g` and `32 g`)

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To solve the problem, we will analyze the reaction between hydrogen and oxygen to form water and verify the law of constant composition using the given data. ### Step 1: Analyze the Reaction The reaction between hydrogen (H₂) and oxygen (O₂) to form water (H₂O) can be represented as: \[ 2H_2 + O_2 \rightarrow 2H_2O \] From the problem, we know: - 112 mL of hydrogen combines with 56 mL of oxygen. ...
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