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)

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) \(m_{NaOH}=60.5\%=3\left(g\right)\)

\(m_{dd}=40+60=100\left(g\right)\)

\(\Rightarrow C\%\left(NaOH\right)=\dfrac{3}{100}.100\%=3\%\)

b) \(m_{dd}=100+50=150\left(g\right)\)

\(m_{KOH}=100.20\%+50.15\%=27,5\left(g\right)\)

\(\Rightarrow C\%\left(KOH\right)=\dfrac{27,5}{150}.100\%=18,33\%\)

a, m_ddsau = 40 + 60 = 100 ( g )

m_NaOH = 3 ( g )

=> \(C\%=\dfrac{3}{100}.100=3\%\)

b, - Gọi nồng độ dd thu được là X % .

- Áp đụng pp đường chéo ta có :

\(\Rightarrow\dfrac{100}{50}=2=\dfrac{X-15}{20-X}\)

=> X = 55/3 %

a)

Khối lượng của dung dịch:

\(m_{dd}=m_{ct}+m_{dm}=20+180=200\left(g\right)\)

Nồng độ phần trăm của dung dịch:

\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{20}{200}.100\%=10\%\)

b) đề sai nha bạn

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

=> \(C_M=\dfrac{0,06}{0,25}=0,24M\)

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

1. \(C\%_{NaOH}=\dfrac{60}{300}.100\%=20\%\)

2. \(m_{HCl}=150.12\%=18\left(g\right)\)

3. \(m_{ddNa_2CO_3}=\dfrac{20}{15\%}=\dfrac{400}{3}\left(g\right)\)