Answer :
Milliequivalents (mEq) of iron in 160 mg of ferrous sulfate, divide 160 mg by the weight of iron in one mEq (27.95 mg/mEq). Approximately 5.72 mEq of iron is present in 160 mg of ferrous sulfate.
The subject of the question is centered on the conversion of mass into milliequivalents (mEq), specifically for ferrous sulfate (FeSO₄·7H₂O). To solve this, we firstly need to understand that 1 milliequivalent of a substance is equivalent to its atomic mass divided by the valence. The atomic mass of Fe in ferrous sulfate (FeSO₄) is 55.9.
Since the valence of iron (Fe) is 2 in ferrous sulfate, we divide 55.9 by 2 to receive 27.95 mg/mEq. Thus, 1 mEq of iron is present in 27.95 mg of ferrous sulfate.
To find the mEq in 160 mg of ferrous sulfate, we divide 160 mg by the quantity of ferrous sulfate in one mEq (27.95 mg/mEq). The result is approximately 5.72 mEq of iron. Therefore, in every 160 mg of ferrous sulfate, there are roughly 5.72 mEq of iron.
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