29: Cụ thể: n(NaOH) =0,05mol

=>C =n/v = 0,05/0,5=0,1M

30, ta có C=10*D*C%/M

M là kluong mol

Từ đó tìm dc C%=11,19%

29C

30c

Câu 29:

nNaOH= 2/40=0,05(mol)

CMddNaOH=0,05/0,5=0,1(M)

Câu 30:

C%ddNaOH(4M)= (CMddNaOH.M(NaOH) )/10D= (4.40)/(10.1,43)= 11,19%

=> Chọn C

Câu 3 : 

\(pH=-log\left[H^+\right]=-log\left(0.1\right)=1\)

Câu 4 : 

Chứa các ion : H⁺ , Cl^-

Câu 5 : 

\(n_{NaOH}=n_{HCl}=0.02\cdot0.1=0.002\left(mol\right)\)

\(\Rightarrow x=\dfrac{0.002}{0.01}=0.2\left(M\right)\)

Câu 1 : 

Bảo toàn điện tích : 

\(n_{SO_4^{2-}}=\dfrac{0.2\cdot2+0.1-0.05}{2}=0.225\left(mol\right)\)

\(m_{Muối}=0.2\cdot64+0.1\cdot39+0.05\cdot35.5+0.225\cdot96=40.075\left(g\right)\)

Câu 2 : 

\(\left[Na^+\right]=\dfrac{0.15\cdot0.5\cdot2+0.05\cdot1}{0.15+0.05}=1\left(M\right)\)

Câu 6 : 

\(pH=14+log\left[OH^-\right]=12\)

\(\Rightarrow\left[OH^-\right]=0.01\)

\(n_{NaOH}=n_{OH}=0.01\cdot2=0.02\left(mol\right)\)

\(\Rightarrow m_{NaOH\left(bđ\right)}=0.02\cdot40=0.8\left(g\right)\)

Câu 7 : 

\(n_{KOH}=0.25\cdot0.03=0.0075\left(mol\right)\)

\(n_{HCl}=0.25\cdot0.01=0.0025\left(mol\right)\)

\(\Rightarrow n_{KOH\left(dư\right)}=0.0075-0.0025=0.005\left(mol\right)\)

\(\left[OH^-\right]=\dfrac{0.005}{0.25+0.25}=0.01\)

\(pH=14+log\left[OH^-\right]=14+log\left(0.01\right)=12\)

@Lê Nguyên Hạo

@Tạ Phương Linh

@Biết để chi z bn

\(a.500ml=0,5l\\ C_{M_{Na_2CO_3}}=\dfrac{0,1}{0,5}=0,2M\\ b.m_{ddNa_2CO_3}=500.1=500g\\ m_{Na_2CO_3}=0,1.106=10,6g\\ C_{\%Na_2CO_3}=\dfrac{10,6}{500}\cdot100\%=2,12\%\)

Chọn A

a, \(K_2SO_4\rightarrow2K^++SO_4^{2-}\)

___0,5_______1______0,5_ (mol)

\(\Rightarrow\left\{{}\begin{matrix}\left[K^+\right]=\frac{1}{2}=0,5M\\\left[SO_4^{2-}\right]=\frac{0,5}{2}=0,25M\end{matrix}\right.\)

b, Ta có: \(n_{OH^-}=n_{K^+}=n_{KOH}=0,2.1=0,2\left(mol\right)\)

\(n_{H^+}=n_{Cl^-}=0,1.1=0,1\left(mol\right)\)

PT ion: \(OH^-+H^+\rightarrow H_2O\)

______0,2_____0,1_________ (mol)

⇒ OH^- dư. ⇒ n_{OH^- (dư)} = 0,1 (mol)

Dd X gồm: K⁺; Cl^- và OH^-_(dư).

\(\Rightarrow\left\{{}\begin{matrix}\left[K^+\right]=\frac{0,2}{0,3}=\frac{2}{3}M\\\left[Cl^-\right]=\frac{0,1}{0,3}=\frac{1}{3}M\\\left[OH^-\right]_{\left(dư\right)}=\frac{0,1}{0,3}=\frac{1}{3}M\end{matrix}\right.\)

c, Ta có: \(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,0005.0,5=0,00025\left(mol\right)\\n_{OH^-}=2n_{Ba\left(OH\right)_2}=2.0,0005.0,5=0,0005\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\Sigma n_{H^+}=n_{HNO_3}+n_{HCl}=1.0,1+1.0,05=0,15\left(mol\right)\\n_{NO_3^-}=n_{HNO_3}=1.0,1=0,1\left(mol\right)\\n_{Cl^-}=n_{HCl}=1.0,05=0,05\left(mol\right)\end{matrix}\right.\)

PT ion: \(OH^-+H^+\rightarrow H_2O\)

____0,0005____0,15_________ (mol)

⇒ H⁺ dư. ⇒ n_{H⁺ (dư)} = 0,1495 (mol)

Dd D gồm: Ba²⁺; NO₃^-; Cl^- và H⁺_(dư)

\(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=\frac{0,00025}{1,0005}\approx2,5.10^{-4}M\\\left[NO_3^-\right]=\frac{0,1}{1,0005}\approx0,09M\\\left[Cl^-\right]=\frac{0,05}{1,0005}\approx0,049M\\\left[H^+\right]_{\left(dư\right)}=\frac{0,1495}{1,0005}\approx0,15M\end{matrix}\right.\)

Bạn tham khảo nhé!

Mà phần c số lẻ quá, không biết đề là 0,5 ml hay 0,5 lít bạn nhỉ?

a, Ta có :

0,2l dung dịch HCl thì có nồng độ 0,1M

Suy ra 0,016l dd HCl thì có nồng độ xM
\(\rightarrow x=\frac{0,016.0,1}{0,2}=0,008\left(\frac{mol}{l}\right)\)

b, \(CM_{HCl}=\frac{0,01.0,1}{0,2}=0,005\left(mol\right)\)

\(\rightarrow n_{HCl}=0,005.0,01=0,0005\left(mol\right)\)

\(\rightarrow n_{KOH}=0,015.0,85=0,01275\left(mol\right)\)

\(PTHH:KOH+HCl\rightarrow KCl+H_2O\)

Theo phương trình : KOH dư , sau pứng tính theo HCl

\(m_{dd}=0,01+0,015=0,025\left(g\right)\)

\(CM_{KCl}=\frac{0,0005}{0,025}=0,002M\)

\(CM_{KOH}=\frac{0,01275-0,0005}{0,025}=0,508M\)

a, Ta có: C_M=\(\frac{n}{V}\)= \(\frac{0.1}{0.2}\)=0.5M

b, C%=\(\frac{mct}{mdd}\).100%=\(\frac{15}{60}\).1005=25%

a) CM CuSO4= 0.1/0.2=0.5M

b) C% NaCl= 15/ (60+15) * 100%= 20%

Mình sửa lại nhé:

a) CM CuSO4= 0.1/0.2= 0.5M

b) C% NaCl= 15/60 *100%= 25%