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\(n_{Na_2CO_3}=n_{Na_2CO_3\cdot10H_2O}=\dfrac{57.2}{106+18\cdot10}=0.2\left(mol\right)\)
\(C_{M_{Na_2CO_3}}=\dfrac{0.2}{0.4}=0.5\left(M\right)\)
\(m_{Na_2CO_3}=0.2\cdot106=21.2\left(g\right)\)
\(m_{dd}=400\cdot1.05=420\left(g\right)\)
\(C\%_{Na_2CO_3}=\dfrac{21.2}{420}\cdot100\%=5.04\%\)
\(n_{Na_2CO_3.10H_2O}=\dfrac{28,6}{286}=0,1\left(mol\right)\)
=> nNa2CO3 = 0,1(mol)
=> \(C_M=\dfrac{0,1}{0,2}=0,5M\)
mdd sau pư = 1,05.200 = 210 (g)
=> \(C\%=\dfrac{0,1.106}{210}.100\%=5,05\%\)
\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
=> \(C_M=\dfrac{0,2}{0,3}=0,667M\)
\(m_{dd}=300.1,05=315\left(g\right)\)
=> \(C\%=\dfrac{21,2}{315}.100\%=6,73\%\)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(m_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c, \(n_{HCl}=2n_{Mg}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
Ta có : \(n_{CaCl_2.6H_2O}=n_{CaCl_2}=\dfrac{5,475}{219}=0,025\left(mol\right)\)
=> CM CaCl2= \(\dfrac{0,025}{0,1}=0,25M\)
a) Gọi KL cần tìm là X
nHCl=\(\frac{5,6}{22,4}\)=0,25
PTHH: X + HCl \(\rightarrow\) XCl2 + H2
0,25 0,5 0,25 0,25
\(\Rightarrow\)mX = \(\frac{16.25}{0,25}\)=65g ( Zn )
b) mHCl= \(0,5.36,5\)=18.25g
mdd= \(\frac{18.25}{0,1825}\)=100g
Cm = \(\frac{0,5}{\frac{0,1}{0,2}}\)=6 mol/l
c) C% = 0,25.(65+71)/(100+16,25-0,5).100=29.73%
$n_{Na_2CO_3} = n_{Na_2CO_3.10H_2O} = \dfrac{28,6}{286} = 0,1(mol)$
$C_{M_{Na_2CO_3}} = \dfrac{0,1}{0,2} = 0,5M$
$m_{dd} = D.V = 200.1,05 = 210(gam)$
$C\%_{Na_2CO_3} = \dfrac{0,1.106}{210}.100\% = 5,05\%$
\(m_{dd}\)=1,05.200=210 g
=>C%dd =\(\dfrac{28,6}{210}\) .100% =13,62%
Mặt khác : 200ml=0,2l
Mct=23.2+12+16.3+10.(1.2+16)=286 (M nguyên tử khối )
=>nct=\(\dfrac{28,6}{286}\) =0,1 mol
=>CM=\(\dfrac{0,1}{0,2}\) =0,5M