Giải mik câu 5
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Bài 5 :
Độ tan của dd KNO3 bão hòa là :
\(S=\dfrac{m_{ct}}{m_{dm}}.100\%=\dfrac{60}{190}.100\approx31,58\)
Bài 6 :
\(C\%_{NaCl}=\dfrac{5}{125}.100\%=4\%\)
C5:
nC2H5OH = 8,2 / 46 = 0,2 (mol)
C2H5OH + 3O2-- (t^o)-- > 2CO2 + 3H2O
VCO2 = 0,4 . 22,4 = 8,96 (l)
VO2 = 0,6.22,4=13,44 (l)
$a) n_{Fe_2O_3} = \dfrac{32}{160} = 0,2(mol)$
$b) m_{O_2} = 0,12.32 = 3,84(gam)$
$c) n_{CuSO_4} = \dfrac{1,2.6^{23}}{6.10^{23}} = 0,2(mol)$
$m_{hh} = 0,12.342 + 0,2.160 = 73,04(GAM)$
$d) n_{CO_2} = \dfrac{3,6.10^{23}}{6.10^{23}} = 0,6(mol)$
$V_{CO_2} = 0,6.22,4 = 13,44(lít)$
$e) n_{CO_2} = 0,5(mol) \Rightarrow V_{hh} = (0,5 + 0,5).22,4 = 22,4(lít)$
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