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Dalam volume 2 liter dipanaskan 1 mol gas S O 3 ​ sehingga terjadi kesetimbangan: Data yang diperoleh sebagai berikut: Harga Kc pada keseimbangan tersebut adalah ...

Dalam volume 2 liter dipanaskan 1 mol gas sehingga terjadi kesetimbangan:

Data yang diperoleh sebagai berikut:

Harga Kc pada keseimbangan tersebut adalah ...

  1. Kc space equals space fraction numerator left square bracket 0 comma 5 divided by 2 right square bracket over denominator left square bracket 0 comma 5 divided by 2 right square bracket left square bracket 0 comma 25 divided by 2 right square bracket end fraction

  2. Kc space equals space fraction numerator left square bracket 0 comma 5 divided by 2 right square bracket left square bracket 0 comma 5 divided by 2 right square bracket over denominator left square bracket 0 comma 25 divided by 2 right square bracket end fraction

  3. Kc space equals space fraction numerator left square bracket 0 comma 5 right square bracket squared left square bracket 0 comma 25 right square bracket over denominator left square bracket 0 comma 5 right square bracket squared end fraction

  4. Kc space equals space fraction numerator left square bracket 0 comma 5 right square bracket over denominator left square bracket 0 comma 5 right square bracket end fraction

  5. Kc space equals space fraction numerator left square bracket 0 comma 5 divided by 2 right square bracket squared left square bracket 0 comma 25 divided by 2 right square bracket over denominator left square bracket 0 comma 5 divided by 2 right square bracket squared end fraction

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S. Utari

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

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space space space space space space space space 2 SO subscript 3 space left parenthesis straight g right parenthesis space left right arrow space 2 space SO subscript 2 space left parenthesis straight g right parenthesis space plus space straight O subscript 2 space left parenthesis straight g right parenthesis space  straight S space equals space 0 comma 50 space mol space space space space space 0 comma 50 space mol space space space space 0 comma 25 space mol    Dalam space volume space 2 space liter  Kc space equals space fraction numerator italic left square bracket p r o d u k italic right square bracket to the power of k o e f i s i e n end exponent over denominator italic left square bracket r e a k t a n italic right square bracket to the power of k o e f i s i e n end exponent end fraction space equals fraction numerator space left square bracket SO subscript 2 right square bracket squared space straight x space left square bracket straight O subscript 2 right square bracket to the power of 1 over denominator space left square bracket SO subscript 3 right square bracket squared end fraction space equals space fraction numerator left square bracket 0 comma 5 divided by 2 right square bracket squared left square bracket 0 comma 25 divided by 2 right square bracket over denominator left square bracket 0 comma 5 divided by 2 right square bracket squared end fraction

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Jika molaritas zat dalam reaksi kesetimbangan: A ( g ) + 2 B ( g ) ⇄ 2 C ( g ) adalah [ A ] = 2 , 4 × 1 0 − 2 M ; [ B ] = 4 , 6 × 1 0 − 3 M ; [ C ] = 6 , 2 × 1 0 − 3 M , maka hitung nilai tetapa...

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