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nosovavalera
@nosovavalera
September 2021
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в избыток воды внесено 355 г оксида фосфора (5). Какая масса (в граммах) продукта образовалась?
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naugros
P2O5+2H2o=2H3PO4
M(P2O5)=31*/2+16*5=142
M(H3PO4)=98
355*2*98/142=490
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Answers & Comments
M(P2O5)=31*/2+16*5=142
M(H3PO4)=98
355*2*98/142=490