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Pertanyaan

Pada reaksi 100 mL larutan Pb left parenthesis NO subscript 3 right parenthesis subscript 2 0,4 M dan 100 mL larutan straight K subscript 2 CO subscript 3 0,4 M jika Ksp PbCO subscript 3 space end subscript equals space 7 comma 4 space. space 10 to the power of negative 4 end exponent massa zat yang mengendap adalah ...

(Ar Pb = 207, C = 12, O = 16, K = 39)

  1. 56,67 gram

  2. 10,68 gram

  3. 5,34 gram

  4. 2,67 gram

  5. 1,068 gram

S. Utari

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

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Pembahasan

space space space space space space space space space space space space space Pb left parenthesis NO subscript 3 right parenthesis subscript 2 space space space space plus space space space space space space space space straight K subscript 2 CO subscript 3 space space space space space space space space space space space space space rightwards arrow space space PbCO subscript 3 space plus space 2 space KNO subscript 3 space  straight M space equals space 0 comma 4 space straight M space straight x space 100 space mL space space space space space space 0 comma 4 space straight M space straight x space 100 space mL space space space space space space space space space space space space minus space space space space space space space space space space space space space space space minus space  space space space space equals space 40 space mmol space space space space space space space space space space space space space space space space space equals space 40 space mmol space space  bottom enclose straight R space equals space 40 space mmol space space space space space space space space space space space space space space space space space space space space space space 40 space mmol space space space space space space space space space space space space space space space space space space space 40 space mmol space space space space space space 80 space mmol space space end enclose space  straight S space equals space space space space space space space minus 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 minus 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 40 space mmol space space space space space 80 space mmol    Karena space diketahui space Ksp space PbCO subscript 3 comma space maka space massa space yang space mengendap space kemungkinan space besar space adalah space PbCO subscript 3. space  Mmol space sisa space untuk space PbCO subscript 3 space adalah space 40 space mmol.    Mol space PbCO subscript 3 equals fraction numerator gram space PbCO subscript 3 over denominator Mr end fraction  4 space straight x space 10 to the power of negative 2 end exponent space mol equals fraction numerator gram space PbCO subscript 3 over denominator 267 end fraction  gram space PbCO subscript 3 equals 4 space straight x space 10 to the power of negative 2 end exponent space mol space straight x space 267 equals 10 comma 68 space gram

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