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Analisis massa terhadap suatu senyawa menghasilkan 40% Ca, 48% O, dan 12% C. Senyawa tersebut mempunyai Mr 100. Bila Ar Ca= 40, O=16, dan C=12, serta tetapan Avogadro 6,0 x 1023 , maka jumlah atom ok

Analisis massa terhadap suatu senyawa menghasilkan 40% Ca, 48% O, dan 12% C. Senyawa tersebut mempunyai Mr 100.

Bila Ar Ca= 40, O=16, dan C=12, serta tetapan Avogadro 6,0 x 1023 , maka jumlah atom oksigen yang terdapat dalam 0,05 mol senyawa tersebut adalah ....

  1. 6,0 x 1023

  2. 9,0 x 1023

  3. 3,0 x 1022

  4. 4,5 x 1022

  5. 9,0 x 1022

Jawaban:

begin mathsize 14px style Dalam space penentuan space rumus space empiris comma space persen space boleh space dianggap space sebagai space massa space left parenthesis gram right parenthesis. space  Ca space equals space 40 space gram space  straight O space equals space 48 space gram space  straight C space equals space 12 space gram space space    bold Menentukan bold space bold rumus bold space bold empiris bold. space  Perbandingan space jumlah space atom space adalah space perbandingan space mol space  Ca space colon space straight O space colon space straight C space equals space mol space Ca space colon space mol space straight O space colon space mol space straight C  space space space space space space space space space space space space space space space space space equals fraction numerator gram straight space Ca over denominator Ar straight space Ca end fraction colon fraction numerator gram straight space straight O over denominator Ar straight space straight O end fraction colon fraction numerator gram straight space straight C over denominator Ar straight space straight C end fraction  space space space space space space space space space space space space space space space space space equals 40 over 40 colon 48 over 16 straight space colon 12 over 12  space space space space space space space space space space space space space space space space space equals 1 colon 3 colon 1  Rumus space empiris space colon space CaCO subscript 3  bold Menentukan bold space bold rumus bold space bold molekul space  left parenthesis Mr space CaCO subscript 3 right parenthesis straight n space equals space Mr space molekul space  left parenthesis 100 right parenthesis straight n space equals space 100 space  straight n space equals space space bold 1 bold comma bold 0 space  Jadi space rumus space molekul space equals space rumus space empiris space  Jadi space rumus space molekul space adalah space bold CaCO subscript bold 3 space space  Menentukan space jumlah space atom space straight O space dalam space 0 comma 05 space mol space senyawa space equals space mol space senyawa space straight x space jumlah space straight O 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 space space space space space space space equals space 0 comma 05 space straight x space 3 space straight x space 6 space straight x space 10 to the power of 23 space end exponent  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 space space space space space space space space space space space space space equals space bold 9 bold space bold x bold space bold 10 to the power of bold 22 bold space end exponent end style

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