`2NaBr+Cl_2->2NaCl+Br_2`

Tăng giảm khối lượng:

`n_{NaBr(pứ) }=(10, 3-8, 075) /(103-58,5) =0, 05(mol) `

`=> m_{NaBr}=0, 05.103=5, 15g`

`n_{MnO_2}=(17, 4) /(87) =0, 2(mol) ;n_{HCl}=0, 5.2=1(mol) ;n_{NaOH}=0, 25.2=0, 5(mol) `

$MnO_2+4HCl\xrightarrow{t°}Cl_2+MnCl_2+2H_2O$

`0,2  -> 0,8   -> 0,2`< ...

CTCT: `CH_2OH-[CHOH]_3-CHO`

PTHH:

`CH_2OH-[CHOH]_3-CHO+Br_2+H_2O->CH_2OH-[CHOH]_3-COOH+2HBr`

$CH_2OH-[CHOH]_3-CHO+2AgNO_3+3NH_3+H_2O\xrightarrow{t°} CH_2OH-[CHOH]_3-COON ...

a) BTKL:`m_X+m_{NaOH}=m_{rắn}+m_{glycerol}`

`=> m_{rắn}=13, 26+0, 5.0,1.40-1, 38=13, 88g`

$b) (RCOO) _3C_3H_5+3NaOH\xrightarrow{t°}3RCOONa+C_3H_5(OH) _3$

`n_X=n_{glycero ...

$(COOCH_3)_2+2KOH\xrightarrow{t°}(COOK) _2+2CH_3OH$

$HCOO-CH_2-COOCH_3+2KOH\xrightarrow{t°} HCOOK+HO-CH_2COOK+CH_3OH$

$HCOO-CH_2-OOC-CH_3+2KOH\xrightarrow{t°}HCOOK+HCHO+H_ ...

Sau khi trộn: `C_{Fe^{3+}}=10^{-4}(M) ;C_{NaF}=0, 02M;C_{KSCN}=0,01M`

Bảo toàn nồng độ đầu:

`C_{Fe^{3+}}=[Fe^{3+}]+[FeOH^{2+}]+Fe(SCN)_i^{3-i} + +FeF_i^{3-i}`

`=> 10^ ...

`2Al+3Cu^{2+}->2Al^{3+}+3Cu`

`Al+3Ag^{+}->Al^{3+}+3Ag`

`Mg+Cu^{2+}->Mg^{2+}+Cu`

`Mg+2Ag^{+}->Mg^{2+}+2Ag`

`Fe+Cu^{2+}->Fe^{2+}+Cu`

`Fe+3Ag^{+}-> ...

Sau khi trộn:`C_{H_2SO_4}=0, 05(M) ;C_{Pb^{2+}}=0, 025(M) ;C_{Ca^{2+}}=0, 005(M) `

`H_2SO_4->H^{+}+HSO_4^-`

`0,05-> 0,05 -> 0,05(M) `

`Ca^{2+}+HSO_4^{-}->CaSO_4+H^+ ...

b) PT Nernst :

`E_1=E^{0}_{AgCl/Ag, Cl^-}-(0, 0592) /2.log(([H^+]^2[Cl^-]^2)/p_{H_2}) => E^{0}_{AgCl/Ag, Cl^-}=0, 587+0, 0592.log(0, 001^2}=0, 2318(V)` 

`E_2=E^{0}_{AgCl/Ag, C ...

Câu 1:

`C_6H_6`:benzene

`C_6H_5CH_3`:methyl benzene hay toluene

Câu 2:

Dãy đồng đẳng benzene:`C_nH_{2n-6}(n>=6) `