Chỉ cần ra đáp án là được ạ

Đáp án:

Giải thích các bước giải:

`a.`

 `m_{HCl(15%)}={60.15}/100=9(g)`

`m_{HCl(25%)}={40.25}/100=10(g)`

`C%_{X}={9+10}/{60+40} .100%=19%`

`b.`

`m_{H_2SO_4(10%)}={30.10}/100=3(g)`

`m_{H_2SO_4(20%)}={20.30}/100=6(g)`

`C%_{Y}={3+6}/{30+20} .100%=18%`

`c.`

`m_{NaOH(5%)}={40.5}/100=2(g)`

`m_{NaOH(20%)}={20.20}/100=4(g)`

`C%_{Z}={2+4}/{40+20} .100%=10%`

`d.`

`m_{Ba(OH)_2 (20%)}={25.20}/100=5(g)`

`m_{Ba(OH)_2 (35%)}={15.35}/100=5,25(g)`

`C%_T={5+5,25}/{25+15} .100% =25,645%`

Tham khảo :

$a,$

$C\%(X)=\dfrac{\dfrac{60 \cdot 15}{100}+\dfrac{40 \cdot 25}{100}}{60+40} \cdot 100\%=19\%$

$b,$

$C\%(Y)=\dfrac{\dfrac{30 \cdot 10}{100}+\dfrac{20 \cdot 30}{100}}{20+30} \cdot 100\%=18\%$

$c,$

$C\%(Z)=\dfrac{\dfrac{40 \cdot 5}{100}+\dfrac{20 \cdot 20}{100}}{40+20} \cdot 100\%=10\%$

$d,$

$C\%(T)=\dfrac{\dfrac{25 \cdot 20}{100}+\dfrac{15 \cdot 35}{100}}{25+15} \cdot 100\%=25,625\%$