A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. Determine the
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A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. Determine the empirical formula of phosphorus selenide.
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Given that 452 mg of phosphorus reacts with selenium to form 1316 mg of phosphorus selenide we can d...View the full answer
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