a) \(m_{NaOH}=\dfrac{60.20}{100}=12\left(g\right)\)
\(C\%_{dd.sau.khi.pha}=\dfrac{12}{60+40}.100\%=12\%\)
b) \(C\%_{dd.sau.khi.pha}=\dfrac{12+12}{60+12}.100\%=33,33\%\)
60g dd 20% có 60.20%=12g NaOH, 60-12=48g H2O
a, Pha thêm 40g H2O , ta có 88g H2O
→C%NaOH=12.100:88=12,64%
b)
tan thêm 12 g
=>m NaOH=24g
=>C%=\(\dfrac{24}{60+12}100\)=33,33%
\(m_{NaOH}=\dfrac{60.20}{100}=12\left(g\right)\\ a,m_{\text{dd}}=60+40=100\left(g\right)\\ C\%=\dfrac{12}{100}.100\%=12\%\\ b,m_{NaOH}=12+12=24\left(g\right)\\ C\%=\dfrac{24}{60}.100\%=40\%\)