Question

BŁAGAM O POMOC oblicz masy cząsteczkowe wypisanych wodorotlenków KOH, Al(OH)3, Ba(OH)2, Fe(OH)2, Fe(OH)3, AgOH, Mn(OH)2, CsOH, bardzo prosze z obliczeniami daje naj

Answer 1

KOH

40u + 16u+ 1u = 57u

 

Al(OH)3

27u + 48u + 3u = 78u

 

Ba(OH)2

137u + 32u + 2u = 171u

 

Fe(OH)2

56u + 32u + 2u = 90u

 

AgOH

108u + 16u + 1u = 125u

 

Mn(OH)2

55u + 32u + 2u = 89u

 

CsOH 

133u + 16u + 1u = 150u

 

 

Answer 2

mKOH=39u+16u+1u=56u

mAl(OH)₃=27u+3*16u+3*1u=27u+48u+3u=78u

mBa(OH)₂=137u+2*16u+2*1u=137u+32u+2u=171u

mFe(OH)₂=56u+2*16u+2*1u=56u+32u+2u=90u

mFe(OH)₃=56u+3*16u+3*1u=56u+48u+3u=107u

mAgOH=108u+16u+1u=125u

mMn(OH)₂=55u+2*16u+2*1u=55u+32u+2u=89u

mCsOH=133u+16u+1u=150u