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Untuk reaksi P + Q + R → hasil, diperoleh data sebagai berikut. Persamaan laju untuk reaksi tersebut adalah ...

Untuk reaksi P + Q + R → hasil, diperoleh data sebagai berikut.
 

 
 

Persamaan laju untuk reaksi tersebut adalah ...space

  1. italic r equals italic k italic space open square brackets P close square brackets to the power of italic 2 open square brackets Q close square brackets space  

  2. italic r equals italic k italic space open square brackets P close square brackets squared open square brackets Q close square brackets open square brackets R close square brackets space

  3. italic r equals italic k italic space open square brackets P close square brackets open square brackets Q close square brackets open square brackets R close square brackets space

  4. italic r equals italic k italic space open square brackets P close square brackets open square brackets Q close square brackets squared open square brackets R close square brackets space

  5. italic r equals italic k italic space open square brackets P close square brackets open square brackets Q close square brackets open square brackets R close square brackets squared space

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Jawaban terverifikasi

Jawaban

dapat disimpulkan jawaban yang tepat adalah E.

dapat disimpulkan jawaban yang tepat adalah E. 

Pembahasan

Pembahasan
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Orde reaksi adalah bilangan pangkat pada persamaan laju reaksi. Persamaan laju reaksinya Jadi, dapat disimpulkan jawaban yang tepat adalah E.

Orde reaksi adalah bilangan pangkat pada persamaan laju reaksi.


italic r equals italic k italic middle dot open square brackets A close square brackets to the power of x open square brackets B close square brackets to the power of y  
 

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell orde space reaksi space terhadap space R end cell row cell fraction numerator r 1 over denominator r 2 end fraction end cell equals cell k over k open parentheses fraction numerator P 1 over denominator P 2 end fraction close parentheses to the power of x open parentheses fraction numerator Q 1 over denominator Q 2 end fraction close parentheses to the power of y open parentheses fraction numerator R 1 over denominator R 2 end fraction close parentheses to the power of z end cell row cell fraction numerator 0 comma 1 over denominator 0 comma 025 end fraction end cell equals cell open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses to the power of x open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses to the power of y open parentheses fraction numerator 0 comma 1 over denominator 0 comma 05 end fraction close parentheses to the power of z end cell row 4 equals cell 2 to the power of z end cell row z equals 2 row blank blank blank row blank blank cell orde space reaksi space terhadap space Q end cell row cell fraction numerator r 2 over denominator r 3 end fraction end cell equals cell k over k open parentheses fraction numerator P 2 over denominator P 3 end fraction close parentheses to the power of x open parentheses fraction numerator Q 2 over denominator Q 3 end fraction close parentheses to the power of y open parentheses fraction numerator R 2 over denominator R 3 end fraction close parentheses to the power of z end cell row cell fraction numerator 0 comma 025 over denominator 0 comma 05 end fraction end cell equals cell open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses to the power of x open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of y open parentheses fraction numerator 0 comma 05 over denominator 0 comma 05 end fraction close parentheses to the power of z end cell row cell 0 comma 5 end cell equals cell 0 comma 5 to the power of y end cell row y equals 1 row blank blank blank row blank blank cell orde space reaksi space terhadap space P end cell row cell fraction numerator r 1 over denominator r 2 end fraction end cell equals cell k over k open parentheses fraction numerator P 1 over denominator P 2 end fraction close parentheses to the power of x open parentheses fraction numerator Q 1 over denominator Q 2 end fraction close parentheses to the power of y open parentheses fraction numerator R 1 over denominator R 2 end fraction close parentheses to the power of z end cell row cell fraction numerator 0 comma 1 over denominator 0 comma 025 end fraction end cell equals cell open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses to the power of x open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses to the power of 1 open parentheses fraction numerator 0 comma 1 over denominator 0 comma 05 end fraction close parentheses squared end cell row 4 equals cell 1 to the power of x cross times 1 cross times 2 squared end cell row x equals 1 end table 
 

Persamaan laju reaksinya italic r equals italic k italic middle dot open square brackets P close square brackets open square brackets Q close square brackets open square brackets R close square brackets squared 
 

Jadi, dapat disimpulkan jawaban yang tepat adalah E. 

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