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Sebanyak 24,6 gram kristal garam left parenthesis MgSO subscript 4. xH subscript 2 straight O right parenthesis dipanaskan sehingga menghasilkan 12 gram MgSO subscript 4 menurut reaksi:

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

Rumus kristal garam adalah .... (Ar Mg = 24; S = 32; O = 16; H = 1)

  1. MgSO subscript 4 space. space 3 space straight H subscript 2 straight O

  2. MgSO subscript 4 space. space 4 space straight H subscript 2 straight O

  3. MgSO subscript 4 space. space 5 space straight H subscript 2 straight O

  4. MgSO subscript 4 space. space 6 space straight H subscript 2 straight O

  5. MgSO subscript 4 space. space 7 space straight H subscript 2 straight O

S. Hidayati

Master Teacher

Mahasiswa/Alumni Universitas Indonesia

Jawaban terverifikasi

Pembahasan

Massa hidrat left parenthesis MgSO subscript 4. xH subscript 2 straight O right parenthesis adalah 24,6 gram, ketika dipanaskan ada massa hilang, massa yang hilang ini merupakan massa straight H subscript 2 straight O. Kemudian massa tersisa adalah 12 gram ini merupakan massa anhidrat (garam murni).

Massa space straight H subscript 2 straight O space space equals space massa space hidrat space – space massa space anhidrat space  space space space space space space space space space space space space space space space space space space space space space space equals space 24 comma 6 space gram space – space 12 space gram space  space space space space space space space space space space space space space space space space space space space space space space equals space 12 comma 6 space gram

Kemudian, untuk menentukan jumlah straight H subscript 2 straight O maka hitung perbandingan mol dari straight H subscript 2 straight O dan garam.

mol space MgSO subscript 4 space colon space Mol space straight H subscript 2 straight O equals fraction numerator massa space MgSO subscript 4 over denominator Mr space MgSO subscript 4 end fraction space colon space fraction numerator massa space straight H subscript 2 straight O over denominator Mr space straight H subscript 2 straight O end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 12 space straight g over denominator 120 end fraction space colon space fraction numerator 12 comma 6 space straight g over denominator 18 end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0 comma 1 space colon space 0 comma 7  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 space colon space 7

       Sehingga X adalah 7

Maka rumus hidratnya = bold italic M bold italic g bold italic S bold italic O subscript bold 4 bold space bold. bold space bold 7 bold space bold H subscript bold 2 bold O

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