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walter H

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'Manganese(III) fluoride, MnF3, can be prepared by the following reaction: 2MnI2(s) + 13F2(g) ? 2MnF3(s) + 4IF5(l) If the
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.033597 grams of hydrogen is combined with 1.000grams of of carbon . calculate the atomic mass of hydrogen
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3.875g MnF3
.0324*2/2=0.0346 mols MnI2 0.26315*2/13=0.020242 mols F2 F2 =LR .0202 mols MnF3x112g/mol MnF3=2.26g MnF3
mols MnI2=0.03246 mols F2=0.26315

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