`CO_2+Ba(OH)_2->BaCO_3+H_2O`.

`n_(BaCO_3)=19,7:197=0,1mol=n_(CO_2)`.

`V_(CO_2)=0,1*24,79=2,479l(đkc)`.

`V_(CO_2)=0,1*22,4=2,24l(đktc)`.

$\mathop{Zn}\limits^{0}$`+` $\mathop{HNO_3}\limits^{+5}$`->` $\mathop{Zn}\limits^{+2}(NO_3)$`+`$\mathop{N_2}\limits^{0}$`+`$\mathop{N}\limits^{-3}H_4NO_3$`+`$H_2O$

$\mathop{2N}\limits^{+5 ...

`M_(SO_x)/M_(Ne)=3,2=>M_(SO_x)=3,2*20=64g//mol`.

`M_(SO_x)=32+16x=62g//mol=>x=2`.

`=>CTHH:SO_2`.

`n_K=31,2:39=0,8mol=n_(KClO_4)`.

`m_(KClO_4)=0,8*(39+35,5+16*4)=110,8g`.

`\hat{pMn}+\hat{nMm}=140^o=\hat{mMp}`.

`=>4x+8+3x+6=140^o`.

`=>7x+14=140^o=>x=18`.

`2x^3-1=15=>2x^3=16=>x^3=8=>x=2`.

`x+16/9=y-25/16=>2+16/9=y-25/16=>y=769/144`.

`z+9/25=x+16/9=>z+9/25=2+16/9=>z=769/225`.

`B=x+y+z=2+769/144+769/225=10,758 ...

`m_(ddNaOH15%)=120*1,2=144g`.

`m_(ctNaOH15%)=144*15%=21,6g`.

`C%NaOH=(21,6+12)/(144+12)*100%~~21,54%`.

`3x=2y=>x=(2y)/3`.

`5y=3z=>z=(5y)/3`.

`8*(2y)/3+2y-4*(5y)/3=72`.

`(16y)/3+2y-(20y)/3=72`.

`1/3(16y-20y)+2y=72`.

`-4/3y+2y=72`.

`=>y=108;x=2*108:3= ...

`64^30=(4^3)^30=4^(3*30)=4^90`.

`16^44=(4^2)^44=4^(2*44)=4^88`.

Ta có `4=4` mà `90>88`.

`4^90>4^88=>64^30>16^44`.

`2^225=2^(3*75)=(2^3)^75=8^75 ...

Gốc acid `H_3PO_4` là `PO_4`.

Các nguyên tố thuộc nhóm `IIA` đều có hóa trị `II`.

Gọi `CTHH` cần tìm là:$\mathop{X_x}\limits^{II ...