a)
m\(_{NaOH}=\)\(500.\frac{1}{2}=250\left(g\right)\)
\(m_{NaOH\left(20\%\right)}=\frac{250.20}{100}=50\left(g\right)\)
\(n_{NaOH}=\frac{50}{40}=1,25\left(mol\right)\)
b)
\(V_{H2SO4}=\frac{200}{1,29}=155\left(ml\right)=0,155\left(l\right)\)
n\(_{H2SO4}=0,155.5=0,775\left(mol\right)\)
Chúc bạn học tốt
Nhớ tích cho mình nhé
\(a.\\ m_{NaOH}=\frac{500.1}{2}=250\left(g\right)\\ \rightarrow n_{NaOH}=\frac{250.20}{100.40}=1,25\left(mol\right)\\ b.\\ V_{dd}=\frac{200}{1,29}=155\left(ml\right)=0,155\left(l\right)\\ n_{axit}=0,155.5=0,775\left(mol\right)\)
Mdd NaOH= 500. ½ = 250(g)
=> M NaOH = 250. 20% = 50(g)
=> n NaOH= 50/40= 1,25 mol
V dd H2SO4 = 200: 1,29 = 20000/129(ml)
= 20/129(l)
=> N H2SO4 = 20/129 .5 \(\approx\)0, 775(mol)
