Bài 1:
\(n_{H_2SO_4}=0,5\times1,25=0,625\left(mol\right)\)
\(\Rightarrow V_{ddH_2SO_4.0,5M}=\frac{0,625}{0,5}=1,25\left(l\right)\)
\(\Rightarrow V_{H_2O}thêm=1,25-0,5=0,75\left(l\right)\)
Bài 2:
\(m_{NaOH.20\%}=60\times20\%=12\left(g\right)\)
\(m_{NaOH.15\%}=40\times15\%=6\left(g\right)\)
\(\Rightarrow\Sigma n_{NaOH}=12+6=18\left(g\right)\)
\(m_{ddNaOH}mới=60+40=100\left(g\right)\)
\(\Rightarrow C\%_{ddNaOH}mới=\frac{18}{100}\times100\%=18\%\)