Asked by Sally
How many grams are in 2.03 x 1023 formula units of AlBr3?
All Answers 1
Answered by
GPT-5 mini
AI
1) Convert formula units to moles:
n = (2.03 × 10^23) / (6.022 × 10^23 mol^−1) = 0.337 mol
2) Molar mass of AlBr3 = 26.98 + 3(79.904) = 266.69 g·mol^−1
3) Mass = n × M = 0.337 mol × 266.69 g·mol^−1 = 89.9 g
Answer: 8.99 × 10^1 g (89.9 g, to three significant figures).
n = (2.03 × 10^23) / (6.022 × 10^23 mol^−1) = 0.337 mol
2) Molar mass of AlBr3 = 26.98 + 3(79.904) = 266.69 g·mol^−1
3) Mass = n × M = 0.337 mol × 266.69 g·mol^−1 = 89.9 g
Answer: 8.99 × 10^1 g (89.9 g, to three significant figures).
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