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D = mdd/V ---> mdd = D.V = 1,28.200 = 256 gam. ---> mCaCl2 = mdd.C%/100 = 256.30/100 = 76,8gam.
a) 300 ml nước ~ 300g nước
m dd = 25+300 = 325(g)
n CaCl2.6H2O = n CaCl2 = 25/219 = 0.11 (mol)
m CaCl2 = 0.11*111 = 12.21(g)
C%dd = 12.21/325*100% = 3.76%
V dd = 325/1.08 = 300.93(ml) = 0.3(l)
CM = 0.11/0.3 = 0.37M
\(n_{CaCl_2.6H_2O}=\frac{25}{219}\left(mol\right)\)
Ta có: \(n_{CaCl_2}=n_{CaCl_2.6H_2O}=\frac{25}{219}\left(mol\right)\)
\(\Rightarrow m_{CaCl_2}=\frac{25}{219}\times111=12,67\left(g\right)\)
\(m_{H_2O}=300\times1=300\left(g\right)\)
Ta có: \(m_{ddA}=25+300=325\left(g\right)\)
\(C\%_{CaCl_2}mới=\frac{12,67}{325}\times100\%=3,9\%\)
Ta có: \(V_{ddA}=\frac{325}{1,08}=300,93\left(ml\right)=0,30093\left(l\right)\)
\(\Rightarrow C_{M_{CaCl_2}}=\frac{25}{219}\div0,30093=0,38\left(M\right)\)
ncacl2.6H2O = 25/219 (mol) => mCaCl2 = \(\frac{925}{73}\)(g)
mA = 300.1,08=324 (g)
=> C% = \(\frac{\frac{925}{73}}{324}.100\%\approx3,9\%\)
CM = \(\frac{\frac{25}{219}}{0,3}\)=0,381%
Nồng độ \(CaCl_2\) trong \(CaCl_2.6H_2O\) là:
\(C\%=\dfrac{M_{CaCl_2}}{M_{CaCl_2.6H_2O}}\cdot100\%=\dfrac{111}{219}\cdot100\%=50,68\%\)
Sơ đồ chéo:
\(CaCl_2.6H_2O\) \(m_1\) 50,68 40
40
\(H_2O\) \(m_2\) 0 10,68
\(\Rightarrow\dfrac{m_1}{m_2}=\dfrac{40}{10,68}\) (*)
\(m_{dd}=V\cdot D=10\cdot1,395=13,95g=m_1+m_2\)
\(\Rightarrow m_2=13,95-m_1\) Thay vào (*) ta được:
\(\Rightarrow\dfrac{m_1}{13,95-m_1}=\dfrac{40}{10,68}\Rightarrow m_1=11,01g\)
\(\Rightarrow m_2=13,95-11,01=2,94g\Rightarrow n_{H_2O}=0,163mol\)
\(V_{H_2O}=0,163\cdot22,4=3,65l\)
\(m_{dd} = 25 + 300 = 325(gam)\\ n_{CaCl_2} = \dfrac{325.3,9\%}{111} = \dfrac{169}{1480}\\ \Rightarrow n_{CaCl_2.nH_2O} = n_{CaCl_2} = \dfrac{169}{1480}(mol)\\ \Rightarrow (111 + 18n).\dfrac{169}{1480} = 25\\ \Rightarrow n = 6\)
CTPT tinh thể : \(CaCl_2.6H_2O\)
mdd=25+300=325(gam)nCaCl2=325.3,9%111=1691480⇒nCaCl2.nH2O=nCaCl2=1691480(mol)⇒(111+18n).1691480=25⇒n=6mdd=25+300=325(gam)nCaCl2=325.3,9%111=1691480⇒nCaCl2.nH2O=nCaCl2=1691480(mol)⇒(111+18n).1691480=25⇒n=6
CTPT tinh thể : CaCl2.6H2OCaCl2.6H2O
$n_{Na_2CO_3} = n_{Na_2CO_3.10H_2O} = \dfrac{28,6}{286} = 0,1(mol)$
$C_{M_{Na_2CO_3}} = \dfrac{0,1}{0,2} = 0,5M$
$m_{dd} = D.V = 200.1,05 = 210(gam)$
$C\%_{Na_2CO_3} = \dfrac{0,1.106}{210}.100\% = 5,05\%$
300ml nước ~ 300g nước
m dd=25+300=325(g)
n CaCl2.6H2O= n CaCl2=25/219=0.11(mol)
m CaCl2=0.11*111=12.21(g)
C%dd=12.21/325*100%=3.76%
V dd= 325/1.08=300.93(ml)=0.3(l)
CM= 0.11/0.3=0.37M