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Pertanyaan

Dari hasil percobaan reaksi

begin mathsize 14px style 2 N O open parentheses italic g close parentheses and 2 H subscript 2 open parentheses italic g close parentheses yields N subscript 2 open parentheses italic g close parentheses and 2 H subscript 2 O open parentheses italic l close parentheses end style 

diperoleh data sebagai berikut:
 

 space 
 

Tingkat reaksi untuk reaksi di atas adalah ...space

  1. 1,0space

  2. tetapspace

  3. 2,0space

  4. 2,5space

  5. 3,0space

A. Tri

Master Teacher

Mahasiswa/Alumni Universitas Pertamina

Jawaban terverifikasi

Jawaban

dapat disimpulkan jawaban yang tepat adalah E.space

Pembahasan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator v 3 over denominator v 4 end fraction end cell equals cell fraction numerator italic k open square brackets N O close square brackets subscript 3 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y over denominator italic k open square brackets N O close square brackets subscript 4 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y end fraction end cell row cell fraction numerator italic t 4 over denominator t 3 end fraction end cell equals cell fraction numerator italic k open square brackets N O close square brackets subscript 3 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y over denominator italic k open square brackets N O close square brackets subscript 4 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y end fraction end cell row cell 1 fourth end cell equals cell open square brackets 0 comma 2 close square brackets to the power of x over open square brackets 0 comma 4 close square brackets to the power of x end cell row cell 1 fourth end cell equals cell 1 half to the power of x end cell row x equals 2 row blank blank blank row cell fraction numerator v 1 over denominator v 2 end fraction end cell equals cell fraction numerator italic k open square brackets N O close square brackets subscript 1 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y over denominator italic k open square brackets N O close square brackets italic 2 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y end fraction end cell row cell fraction numerator italic t italic 2 over denominator t 1 end fraction end cell equals cell fraction numerator italic k open square brackets N O close square brackets subscript 1 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y over denominator italic k open square brackets N O close square brackets subscript italic 2 to the power of italic x open square brackets H subscript 2 close square brackets to the power of italic y end fraction end cell row cell fraction numerator 3 comma 2 over denominator 9 comma 6 end fraction end cell equals cell open square brackets 0 comma 1 close square brackets to the power of y over open square brackets 0 comma 3 close square brackets to the power of y end cell row cell 1 third end cell equals cell 1 third to the power of y end cell row x equals 1 row blank blank blank row blank blank blank row blank blank blank end table end style 
 

Orde total 2+1=3.

Jadi, dapat disimpulkan jawaban yang tepat adalah E.space

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