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P a d a r u an g an 1 l i t er d i p ana s kan 1 m o l g a s N 2 ​ O 4 ​ t er ja d i re ak s i N 2 ​ O 4 ​ ( g ) ⇆ 2 N O 2 ​ ( g ) D a t a y an g d i p ero l e h : H a r g a Kc p a d a re ak s i t erse b u t a d a l ah ..

  1. K c space equals space fraction numerator open parentheses 0 comma 4 close parentheses squared over denominator open parentheses 0 comma 2 close parentheses end fraction

  2. K c space equals space fraction numerator open parentheses 0 comma 4 close parentheses squared over denominator open parentheses 0 comma 8 close parentheses end fraction

  3. K c space equals space fraction numerator open parentheses 0 comma 8 close parentheses over denominator open parentheses 0 comma 4 close parentheses squared end fraction

  4. K c space equals space fraction numerator open parentheses 0 comma 8 close parentheses over denominator open parentheses 0 comma 2 close parentheses end fraction

  5. K c space equals space fraction numerator open parentheses 0 comma 4 close parentheses over denominator open parentheses 0 comma 2 close parentheses end fraction

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M. Robo

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H a r g a space K c space d i h i t u n g space p a d a space s a a t space k e a d a a n space s e t i m b a n g. space  K c space equals space fraction numerator left square bracket N O subscript 2 right square bracket squared over denominator left square bracket N subscript 2 O subscript 4 right square bracket end fraction equals fraction numerator left square bracket 0 comma 4 right square bracket squared over denominator left square bracket 0 comma 8 right square bracket end fraction

Latihan Bab

Pendahuluan Kesetimbangan Kimia

Konsep Kesetimbangan Kimia

Konstanta Kesetimbangan (K)

Konstanta Kesetimbangan Konsentrasi (Kc)

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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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