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Boron forms a series of compounds with hydrogen, all with the general formula By Hy. X B₂H₁ (s) + excess O₂(g) → B2O3(s) + / H₂O(g) · If 0.293 g of one of these compounds gives 0.835 g of B₂03 when burned in excess O2, what is its empirical formula? Empirical formula:

Boron forms a series of compounds with hydrogen, all with the general formula By Hy. X B₂H₁ (s) + excess O₂(g) → B2O3(s) + / H₂O(g) · If 0.293 g of one of these compounds gives 0.835 g of B₂03 when burned in excess O2, what is its empirical formula? Empirical formula:

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Boron forms a series of compounds with hydrogen, all with the general formula By Hy. Y B₂H₁ (s) + excess O₂(g) → – B₂O3(s) + H₂O(g) - = If 0.293 g of one of these compounds gives 0.835 g of B203 when burned in excess O2, what is its empirical formula? Empirical formula:
Transcribed Image Text:Boron forms a series of compounds with hydrogen, all with the general formula By Hy. Y B₂H₁ (s) + excess O₂(g) → – B₂O3(s) + H₂O(g) - = If 0.293 g of one of these compounds gives 0.835 g of B203 when burned in excess O2, what is its empirical formula? Empirical formula:
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