a)

C% muối =\(\dfrac{25}{25+35+190}100=10\%\)

C% đường =\(\dfrac{35}{25+35+190}.100=14\%\)

b)

200ml=0,2, x(ml)=0,001x(l)

nNaOH=0,2×3=0,6(mol)

→CMNaOH 1,5M=0,6\0,2+0,001x=1,5M

       →x=200

a)

C% CuSO4 = 16/(16 + 184)  .100% = 8%
b)

n NaOH = 20/40 = 0,5(mol)

CM NaOH = 0,5/4 = 0,125M

\(a.\)

\(m_{dd_{CuSO_4\:}}=16+184=200\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{16}{200}\cdot100\%=8\%\)

\(b.\)

\(n_{NaOH}=\dfrac{20}{40}=0.5\left(mol\right)\)

\(C_{M_{NaOH}}=\dfrac{0.5}{4}=0.125\left(M\right)\)

a)   \(m_{dmCUSO4}\) =  16+184 = 200g 

C% = \(\dfrac{16}{200}\)  x 100% =8 % 

b) \(n_{NaOH}\) = \(\dfrac{20}{40}\) = 0,5 (Mol)

\(C_M\) = \(\dfrac{0.5}{4}\) = 0,125 (M)  Vì 4000ml= 4l

\(a,C_{M\left(NaOH\right)}=\dfrac{0,3}{0,5}=0,6M\\ b,n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,6}{0,4}=1,5M\)

a) \(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

 \(C_M=\dfrac{0,1}{0,2}=0,5M\)

b) \(n_{K_2O}=\dfrac{2,82}{94}=0,03\left(mol\right)\)

PTHH: K₂O + H₂O --> 2KOH

            0,03------------->0,06

=> \(C_M=\dfrac{0,06}{0,25}=0,24M\)

\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\ C_M=\dfrac{0,1}{0,2}.=0,5M\) 
 \(n_{K_2O}=\dfrac{2,82}{94}=0,03\left(mol\right)\\ C_M=\dfrac{0,03}{0,25}=0,12M\)

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: n_{NaOH (thêm vào)} = a (g)

\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)

a)

$BaO + H_2O \to Ba(OH)_2$
$n_{Ba(OH)_2} = n_{BaO} = \dfrac{30,6}{153} = 0,2(mol)$
$C_{M_{Ba(OH)_2}} = \dfrac{0,2}{0,2} = 1M$

b)

$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 2n_{Ba(OH)_2} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{0,5} = 0,8(lít)$

$n_{FeCl_2} = \dfrac{50,8}{127} = 0,4(mol)$
$V_{dd} = 200(ml) = 0,2(lít)$
$C_{M_{FeCl_2}} = \dfrac{0,4}{0,2} = 2M$

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

a) \(n_{K_2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\)

\(n_{KOH}=2n_{K_2O}=0,5\left(mol\right)\)

\(m_{KOH}=0,5.56=28\left(g\right)\)

b) \(C_{M_{ddKOH}}=\dfrac{n}{V}=\dfrac{0,5}{0,4}=1,25M\)