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a. Từ công thức : C%=\(\frac{mct}{mdd}\).100 => mct=\(\frac{C\%.mdd}{100}\) hay mKOH= \(\frac{10.28}{100}\)=2.8(g)
➩ nKOH= \(\frac{2.8}{56}\)=0.05(mol)
b. mdd=144+36=180 (g)
C%=\(\frac{36.100}{180}\)=20%
c. nNaOH=\(\frac{0.8}{40}\)=0.02(mol) , 80ml=0.08l
Từ công thức CM=\(\frac{n}{V}\) hay CM=\(\frac{0.02}{0.08}\)=0.25M
good luck!!!
a)
\(C\%_{dd.KOH}=\dfrac{7,5}{7,5+42,5}.100\%=15\%\)
b) \(n_{HNO_3}=\dfrac{1,26}{63}=0,02\left(mol\right)\Rightarrow C_{M\left(dd.HNO_3\right)}=\dfrac{0,02}{0,016}=1,25M\)
a) \(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(C_M=\dfrac{0,1}{0,2}=0,5M\)
b) \(n_{K_2O}=\dfrac{2,82}{94}=0,03\left(mol\right)\)
PTHH: K2O + H2O --> 2KOH
0,03------------->0,06
=> \(C_M=\dfrac{0,06}{0,25}=0,24M\)
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\ C_M=\dfrac{0,1}{0,2}.=0,5M\)
\(n_{K_2O}=\dfrac{2,82}{94}=0,03\left(mol\right)\\ C_M=\dfrac{0,03}{0,25}=0,12M\)
Câu 2:
Ta có: mNaOH = 400 x 20% = 80 (g)
=> nNaOH (mới) = \(\dfrac{80}{40}\) = 2 (mol)
=> CM của dd mới = \(\dfrac{2}{4}\) = 0,5M
Cau 1:
Theo de bai ta co
VCuSO4=250ml=0,25 l
\(\rightarrow\) nCuSO4=CM.V=1,5.0,25=0,375 mol
Ta co
n\(_{CuSO4.5H2O}=n_{CuSO4}=0,375\left(mol\right)\)
\(\Rightarrow m_{CuSO4.5H2O}=n.M=0,375.250=93,75\left(g\right)\)
a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)
b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)
c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)
d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)
e, \(m_{NaCl}=150.60\%=90\left(g\right)\)
f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)
g, \(n_{NaOH}=120.20\%=24\left(g\right)\)
Gọi: nNaOH (thêm vào) = a (g)
\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)
\(a,m_{KOH}=\dfrac{28.10}{100}=2,8\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ b,C\%=\dfrac{36}{144+36}.100\%=20\%\\ c, n_{NaOH}=\dfrac{0,8}{40}=0,02\left(mol\right)\\ \rightarrow C_{M\left(NaOH\right)}=\dfrac{0,02}{0,08}=0,25M\)
\(a,m_{KOH}=\dfrac{28.10}{100}=2,8\left(g\right)\\ n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ C\%=\dfrac{36}{36+144}.100\%=20\%\\ C_M=\dfrac{0,8}{0,08}=10M\)