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Sebanyak 11,6 gram senyawa hidrat Na subscript 2 SO subscript 4 space straight x space straight H subscript 2 straight O dipanaskan sampai terbentuk Na subscript 2 SO subscript 4 sebanyak 7,1 gram, menurut reaksi Na subscript 2 SO subscript 4 space straight x space straight H subscript 2 straight O subscript left parenthesis straight s right parenthesis end subscript rightwards arrow Na subscript 2 SO subscript 4 left parenthesis straight s right parenthesis end subscript plus space space straight x space straight H subscript 2 straight O subscript left parenthesis straight g right parenthesis end subscript. Jika Ar Na = 23 ; S =  32 ; O = 16 ; H = 1, rumus senyawa kristal tersebut adalah ...

  1. Na subscript 2 SO subscript 4 space. space straight H subscript 2 straight O

  2. Na subscript 2 SO subscript 4 space. space 2 space straight H subscript 2 straight O

  3. Na subscript 2 SO subscript 4 space. space 3 space straight H subscript 2 straight O

  4. Na subscript 2 SO subscript 4 space. space 4 space straight H subscript 2 straight O

  5. Na subscript 2 SO subscript 4 space. space 5 space straight H subscript 2 straight O

S. Utari

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

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Pembahasan

Na subscript 2 SO subscript 4 space straight x space straight H subscript 2 straight O subscript left parenthesis straight s right parenthesis end subscript rightwards arrow Na subscript 2 SO subscript 4 left parenthesis straight s right parenthesis end subscript plus space space straight x space straight H subscript 2 straight O subscript left parenthesis straight g right parenthesis end subscript

11,6 gram                           7,1 gram

Maka massa x H2O adalah 11,6 gram – 7,1 gram = 4,5 gram.

Untuk mengetahui x, dimana x adalah mol dari H2O maka menggunakan perbandingan

straight x equals fraction numerator mol space straight H subscript 2 straight O over denominator mol space Na subscript 2 SO subscript 4 end fraction    mol space straight H subscript 2 straight O space equals space fraction numerator gr space straight H subscript 2 straight O over denominator Mr space straight H subscript 2 straight O end fraction equals fraction numerator 4 comma 5 space gram over denominator 18 end fraction equals 0 comma 25 space mol    mol space Na subscript 2 SO subscript 4 space equals fraction numerator gr space Na subscript 2 SO subscript 4 over denominator Mr space Na subscript 2 SO subscript 4 end fraction equals fraction numerator 7 comma 1 space gram over denominator 142 end fraction equals 0 comma 05 space mol    Jadi comma space  straight x equals fraction numerator mol space straight H subscript 2 straight O over denominator mol space Na subscript 2 SO subscript 4 end fraction space equals fraction numerator 0 comma 25 over denominator 0 comma 05 end fraction space equals space 5

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