What amount of the substance is contained in 945 mg of calcium phosphate Ca3 (PO4) 2?
| September 9, 2021 | education
Given:
m (Ca3 (PO4) 2) = 945mg
Find: m (Ca3 (PO4) 2)
Solution:
n = m / M
n (Ca3 (PO4) 2) = 0.945 / 310 = 0.003 mol = 3 mmol
Answer: 3 mmol.
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