Verified answer

Ответ:

дано

m(ppa BaCL2) = 180 g

W(BaCL2) = 10%

+m2(BaCL2) = 20 g

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W2(BaCL2) - ?

m1(BaCL2) = m(ppa BaCL2) * W(BaCL2) /100% = 180 * 10% / 100% = 18 g

m общ (BaCl2) = m1(BaCL2) + m2(BaCL2) = 18 + 20 = 38 g

m( общ ppa BaCL2) = m(ppa BaCL2) + m2(BaCL2) = 180 + 20 = 200 g

W2(BaCL2) = m общ (BaCL2) / m( общ ppa BaCL2) * 100% = 38/200*100% = 19%

ответ 19%

Объяснение: