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Analisis terhadap suatu senyawa menghasilkan rumus...

Analisis terhadap suatu senyawa menghasilkan rumus empiris NaCO2 (Ar Na = 23, O = 16, dan C = 12).

Jika 1 mol senyawa ini mempunyai massa 134 g dan bilangan Avogadro L = 6,02 × 1023

maka jumlah atom karbon dalam 0,1 mol senyawa ini adalah ….

  1. 1,2 × 1023

  2. 6,02 × 1023

  3. 3,01 × 1023

  4. 3,01 × 1022
     

  5. 6,02 × 1020

Jawaban:

begin mathsize 14px style Mr space senyawa space equals space gram over mol equals fraction numerator 134 over denominator 1 straight space end fraction equals space 134 space straight g divided by mol  Menentukan space rumus space molekul space colon  left parenthesis Mr space NaCO subscript 2 right parenthesis straight n equals space Mr space Senyawa space  67 space straight n space equals space 134 space  straight n space equals space 2 space  Sehingga space rumus space molekul space equals space left parenthesis NaCO subscript 2 right parenthesis 2 space equals space Na subscript 2 straight C subscript 2 straight O subscript 4 space  Menentukan space jumlah space atom space straight C space dalam space senyawa space equals space mol space senyawa space straight x space jumlah space straight C space straight x space straight L 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 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 straight x space 2 space straight x space 6 comma 02 space straight x space 10 to the power of 23  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 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 bold 1 bold comma bold 204 bold space bold x bold space bold 10 to the power of bold 23  end style

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