a.\(C\%_{NaCl}=\dfrac{9}{91+9}.100\%=9\%\)

b.\(m_{NaCl}=0,5.58,5=29,25g\)

\(C\%_{NaCl}=\dfrac{29,25}{29,25+300}.100\%=8,88\%\)

a) \(C\%=\dfrac{9}{9+91}.100\%=9\%\)

b) \(m_{H_2O}=300.1=300\left(g\right)\)

\(C\%=\dfrac{0,5.58,5}{0,5.58,5+300}.100\%=8,88\%\)

\(1.\\ a)V_{CH_4}=1,5.24,79=37,185l\\ 2.\\ m_{ddKCl}=20+60=80g\\ C_{\%KCl}=\dfrac{20}{80}\cdot100\%=25\%\)

Sửa đề: 9,2 gam Na

\(a,n_{Na_2O}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

            0,4------------------>0,8

\(\rightarrow C_{M\left(NaOH\right)}=\dfrac{0,8}{0,5}=1,6M\)

\(b,n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

            0,4----------------->0,8

\(\rightarrow C\%_{KOH}=\dfrac{0,8.56}{362,4+37,6}.100\%=11,2\%\)

\(C\%=\dfrac{30}{170}.100\%=17,647\%\) 
\(V_{\text{dd}}=\left(30+170\right)1,1=220ml\) 
\(n_{NaCl}=\dfrac{30}{58,5}=0,513mol\)
\(C_M=\dfrac{0,513}{0,22}=0,696M\)

\(C\%_{NaCl}=\dfrac{30}{170+30}.100\%=15\%\\ C_M=C\%.\dfrac{10D}{M}=10.\dfrac{10.1,1}{58,5}=1,88M\)

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

1

\(C\%_{H_2SO_4}=\dfrac{8.100\%}{92}=8,7\%\)

2

\(C\%_{H_2SO_4}=\dfrac{8.100\%}{8+92}=8\%\)

3

\(C\%_{BaCl_2}=\dfrac{15.100\%}{15+45}=25\%\)

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