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

31: B

32. D

0,1M

0, 1M

\(n_{K2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\)

Pt : \(K_2O+H_2O\rightarrow2KOH|\)

         1          1               2

       0,25                      0,5

\(n_{KOH}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)

\(C_{M_{ddKOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

⇒ Chọn câu : B 

 Chúc bạn học tốt

\(N_{K_{2^O}}\)=\(\dfrac{23,5}{94}=0,25mol\)

\(K_{2^O}\)+\(H_{2^O}\)=>\(2KOH\)

\(0,25->0,5mol\)

->\(C_{M\left(A\right)}\)=\(\dfrac{n}{v}=\dfrac{0,5}{0,5}=1M\)

=>B

ai giúp mình với, đang cần gấp lắm ạ

\(n_{Cu^{2+}}=n_{SO_4^{2-}}=n_{CuSO_4}=0,5\left(mol\right)\)

\(\Rightarrow\left[Cu^{2+}\right]=\left[SO_4^{2-}\right]=\dfrac{0,5}{2}=0,25M\)

Câu 1  :

a) n Na2O = 3,1/62 = 0,05(mol)

$Na_2O + H_2O \to 2NaOH$

Theo PTHH : n NaOH = 2n Na2O = 0,1(mol)

=> CM NaOH = 0,1/2 = 0,05M

Câu 2 : 

Coi n KOH = 1(mol)

=> V dd KOH = 1/2 = 0,5(lít) = 500(ml)

=> mdd KOH = D.V = 500.1,43 = 715(gam)

=> C% KOH = 1.56/715  .100% = 7,83%

1. Ta có : \(n_{Na_2O}=\dfrac{m}{M}=0,05mol\)

\(PTHH:Na_2O+H_2O\rightarrow2NaOH\)

Theo PTHH: \(n_{NaOH}=2n_{Na_2O}=0,1mol\)

\(\Rightarrow C_{MNaOH}=\dfrac{n}{V}=0,05M\)

2. - Gọi số lít KOH là a lít

\(\Rightarrow m_{dd}=D.V=1430a\left(g\right)\)

Mà \(n_{KOH}=C_M.V=2amol\)

\(\Rightarrow m_{KOH}=n.M=112a\left(g\right)\)

\(\Rightarrow C\%=\dfrac{m}{m_{dd}}.100\%=\dfrac{112a}{1430a}.100\%=~7,83\%\)

1)

$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,2(mol)$
$m_{dd} = 6,2 + 193,8 = 200(gam) \Rightarrow C\%_{NaOH} = \dfrac{0,2.40}{200}.100\% = 4\%$

2) 

$n_{K_2O} = \dfrac{23,5}{94} = 0,25(mol)$
$K_2O + H_2O \to 2KOH$
$n_{KOH} = 2n_{K_2O} = 0,5(mol) \Rightarrow C_{M_{KOH}} = \dfrac{0,5}{0,5} = 1M$

3) $n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} =0,8M$

4)

$Na_2SO_3 + 2HCl \to 2NaCl +S O_2 + H_2O$
Theo PTHH : 

$n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)$
$V_{SO_2} = 0,1.22,4 = 2,24(lít)$

5) $n_{CaO} = \dfrac{5,6}{56} = 0,1(mol)$

$CaO + 2HCl \to CaCl_2 + H_2O$

Theo PTHH : 

$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$

Bài 1 : 

\(n_{Na2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

Pt : \(Na_2O+H_2O\rightarrow2NaOH|\)

           1          1              2

          0,1                      0,2

\(n_{NaOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{NaOH}=0,2.40=8\left(g\right)\)

\(m_{ddspu}=6,2+193,8=200\left(g\right)\)

\(C_{ddNaOH}=\dfrac{8.100}{200}=4\)⁰/₀

 Chúc bạn học tốt

$n_{BaCl_2} = \dfrac{41,6}{208} = 0,2(mol)$
$C_{M_{BaCl_2}} = 0,2 : 0,5 = 0,4M$

$BaCl_2 \to Ba^{2+} + 2Cl^-$

$[Ba^{2+}] = 0,4M ; [Cl^-] = 0,4.2 = 0,8M$

ai giúp mình với ạ