Answered

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What is the percent of Boron in Sr3(BO3)2

Sagot :

Answer:

5.68%

Explanation:

Percent Composition of Boron = Atomic Mass of Boron / Molar mass of Sr3(BO3)2  * 100%

Atomic mass of B = 10.811 g/mol

Molar mass of Sr3(BO3)2 = 380.4784 g/mol

Percent composition = 2 (10.811) / 380.4784  * 100

Percent composition = 21.622 / 380.4784  * 100

Percent composition = 5.68%

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