mNH3= 5,1gr volume larutan= 1gr mNH4Cl=53,5 mrNH3= 17 mrNH4Cl=53,5 gr kb= 10^-5 Ditanya: PH?
Kimia
nabilamustaqim1
Pertanyaan
mNH3= 5,1gr
volume larutan= 1gr
mNH4Cl=53,5
mrNH3= 17
mrNH4Cl=53,5 gr
kb= 10^-5
Ditanya: PH?
volume larutan= 1gr
mNH4Cl=53,5
mrNH3= 17
mrNH4Cl=53,5 gr
kb= 10^-5
Ditanya: PH?
1 Jawaban
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1. Jawaban Iksaladrian
PH=mrNH1CJ=55,7gr
PH=PH4VU:mrNH4CI