\(n_{NaOH}=\frac{10.40}{100.40}=0,1\left(mol\right)\)
\(PTHH:NaOH+HCl\rightarrow NaCl+H_2O\)
\(\rightarrow m_{ddHCl}=\frac{100.\left(36,5.0,1\right)}{13}=28,08\left(g\right)\)
\(C\%_M=\frac{0,1.58,5}{28,08+40}.100\%=8,59\left(\%\right)\)
m\(_{NaOH}=\)\(\frac{40.10}{100}=\left(g\right)\)
n\(_{NaOH}=\frac{4}{40}=0,1\left(mol\right)\)
NaOH +HCl ---> NaCl +H2O
Theo pthh
n\(_{HCl}=n_{NaOH}=0,1mol\)
m\(_{HCl}=0,1.36,5=3,65\left(g\right)\)
m=\(\frac{3,65.100}{13}=28,08\left(g\right)\)
Theo pthh
n\(_{NaCl}=n_{NaOH}=0,1mol\)
m\(_{NaCl}=0,1.58,5=5,85\left(g\right)\)
m\(_{dd}=28,08+40=68,08\left(g\right)\)
C%=\(\frac{5,85}{68,08}.100\%=5.=8,59\left(g\right)\)
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