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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 :
\(\left[Na^+\right]=\dfrac{0.15\cdot0.5\cdot2+0.05\cdot1}{0.15+0.05}=1\left(M\right)\)
a) Ta có: \(n_{Al\left(NO_3\right)_3}=\dfrac{4,26}{213}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^+}=0,02\left(mol\right)\\n_{NO_3^-}=0,06\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[Al^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\\left[NO_3^-\right]=\dfrac{0,06}{0,1}=0,6\left(M\right)\end{matrix}\right.\)
b) Ta có: \(\left[Na^+\right]=0,1+0,02\cdot2+0,3=0,304\left(M\right)\)
c) Bạn xem lại đề !!
Đáp án C
nMgSO4= 0,1 mol; nAl2(SO4)3= 0,1 mol; Vdd= 0,5 lít
CMgSO4=0,2M; CMAl2(SO4)3= 0,2M
MgSO4 → Mg2++ SO42-
0,2M 0,2M
Al2(SO4)3→ 2Al3++ 3SO42-
0,2M 0,6M
[SO42-] = 0,8M
a,nA=\(\dfrac{18,25}{36,5}\)=0,5(mol)
nB=\(\dfrac{10,95}{36,5}\)=0,3(mol)
→nC=0,3+0,5=0,8(mol)
→CM(C)=\(\dfrac{0,8}{2}\)=0,4M
b,CM(A)=\(\dfrac{0,5}{V1}\)
CM(B)=\(\dfrac{0,3}{V2}\)
→\(\dfrac{0,5}{V1}\)=\(\dfrac{0,3}{V2}\)=0,8
=>V1=0,625 l
=>V2=0,375 l
=>CmV1=\(\dfrac{0,5}{0,625}\)=0,8M
=>CmV2=\(\dfrac{0,3}{0,375}\)=0,8M
\(a,n_A=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ n_B=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
\(\rightarrow n_C=0,3+0,5=0,8\left(mol\right)\\ \rightarrow C_{M\left(C\right)}=\dfrac{0,8}{2}=0,4M\)
\(b,C_{M\left(A\right)}=\dfrac{0,5}{V_1}\\ C_{M\left(B\right)}=\dfrac{0,3}{V_2}\\ \rightarrow\dfrac{0,5}{V_1}:\dfrac{0,3}{V_2}=0,8\\ \rightarrow\dfrac{0,5}{V_1}=\dfrac{0,24}{V_2}=\dfrac{0,5+0,24}{V_1+V_2}=\dfrac{0,74}{2}=0,37\\ \rightarrow\left\{{}\begin{matrix}V_1=\dfrac{0,5}{0,34}=1,4\left(l\right)\\V_2=\dfrac{0,24}{0,34}=0.6\left(l\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}C_{M\left(A\right)}=\dfrac{0,5}{1,4}=0,36M\\C_{M\left(B\right)}=\dfrac{0,5}{0,6}=0,83M\end{matrix}\right.\)
Đáp án C
n N a 2 S O 4 = 0,02 mol; nNaCl=0,5 mol; nNa+= 0,02.2+0,5= 0,54 mol
[Na+]= 0,54/(0,117+0,171+0,212)= 1,08M