Hòa Tan 2 mol NaOH vào nước được 200 mol dd tính nồng độ mol
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\(a,C_{M\left(NaOH\right)}=\dfrac{0,3}{0,5}=0,6M\\ b,n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,6}{0,4}=1,5M\)
a, \(C\%_{NaOH}=\dfrac{4}{4+2,8+118,2}.100\%=3,2\%\)
\(C\%_{KOH}=\dfrac{2,8}{4+2,8+118,2}.100\%=2,24\%\)
b, \(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{0,125}=0,8\left(M\right)\)
\(n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,05}{0,125}=0,4\left(M\right)\)
\(a)m_{dd}=4+2,8+118,2=125g\\ C_{\%NaOH}=\dfrac{4}{125}\cdot100\%=3,2\%\\ C_{\%KOH}=\dfrac{2,8}{125}\cdot100\%=2,24\%\\ b)n_{NaOH}=\dfrac{4}{40}=0,1mol\\ \\ n_{KOH}=\dfrac{2,8}{56}=0,05mol\\ 125ml=0,125l\\ C_{M_{NaOH}}=\dfrac{0,1}{0,125}=0,8M\\ C_{M_{KOH}}=\dfrac{0,05}{0,125}=0,4M\)
\(m_{FeCl_2}=0,5.127=63,5g\)
\(C\%_{FeCl_2}=\dfrac{63,5}{300}.100\%=21,16\%\)
\(m_{FeCl_2}=n.M=0,5.127=63,5\left(g\right)\)
\(C_{\%_{ddFeCl_2}}=\dfrac{m_{FeCl_2}}{m_{ddFeCl_2}}.100\%=\dfrac{63,5}{300}.100\%=21,2\%\)
\(a.C_M=\dfrac{0,06}{1,5}=0,04M\\ b.C_M=\dfrac{\dfrac{400}{160}}{4}=0,625M\\ c.C_M=\dfrac{\dfrac{10,53}{58,5}}{\dfrac{450}{1,25}:1000}=0,5M\\ d.C_M=\dfrac{\dfrac{70,2}{40}}{0,5}=3,51M\\ e.C_M=\dfrac{\dfrac{42}{200}}{\dfrac{742}{1,3}:1000}=0,368M\)
a)
C% CuSO4 = 16/(16 + 184) .100% = 8%
b)
n NaOH = 20/40 = 0,5(mol)
CM NaOH = 0,5/4 = 0,125M
\(n_{CuSO_4}=\dfrac{8}{160}=0.05\left(mol\right)\)
\(C_{M_{CuSO_4}}=\dfrac{0.05}{0.2}=0.25\left(M\right)\)
nCuSO4=\(\dfrac{8}{160}=0,05mol\)
CM =\(\dfrac{0,05}{0,2}=0,25M\)
\(n_{FeCl_2}=\dfrac{25,4}{127}=0,2\left(mol\right)\\ V_{dd}=200ml=0,2l\\ \rightarrow C_{M\left(FeCl_2\right)}=\dfrac{0,2}{0,2}=1M\)
200ml chứ
\(C_M=\dfrac{2}{\left(200.22,4\right):100}=0,046M\)