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a) Ta có: \(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,1}{0,5}=0,2\left(M\right)=\left[Na^+\right]=\left[Cl^-\right]\)
b) Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=0,4\left(M\right)\\\left[OH^-\right]=0,8\left(M\right)\end{matrix}\right.\)
c) Ta có: \(n_{H_2SO_4}=0,025\cdot2=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,125+0,025}\approx0,33\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,66\left(M\right)\\\left[SO_4^{2-}\right]=0,33\left(M\right)\end{matrix}\right.\)
a, \(\left[Ca^{2+}\right]=\dfrac{0,15.0,5}{0,15+0,05}=0,375M\)
\(\left[Na^+\right]=\dfrac{0,05.2}{0,15+0,05}=0,5M\)
\(\left[Cl^-\right]=\dfrac{0,15.2.0,5+0,05.2}{0,15+0,05}=1,25M\)
Đáp án B
Na2O+ H2O→ 2NaOH
0,02 0,02+0,02
NaOH+ HCl→ NaCl + H2O
0,02 0,02
2NaOH dư+ H2SO4→ Na2SO4+ 2H2O
0,02 0,01
Do đó a=mNa2O= 0,02.62= 1,24 gam
a. Xem như nấc 2 của H2SO4 điện li hoàn toàn
\(H_2SO_4\rightarrow2H^++SO_4^{2-}\)
nH2SO4=0,05(mol)
nH+=0,1(mol) nSO42=0,05(mol)
\(\left[H^+\right]=\dfrac{0,1}{0,2}=0,5M\)
\(\left[SO^{2-}_4\right]=\dfrac{0,05}{0,2}=0,25M\)
b. \(HCl\rightarrow H^++Cl^-\)
Đặt mdd=100(g) ⇒ \(n_{HCl}=\dfrac{100.7,3\%}{36,5.100\%}=0,2\left(mol\right)\)
⇒\(V=\dfrac{100}{1,25}=80\left(ml\right)=0,08\left(l\right)\)
\(\left[H^+\right]=\left[Cl^-\right]=\dfrac{0,2}{0,08}=2,5M\)
c. \(CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}\)
nCuSO4.5H2O=nCu2+=nSO42-=0,05(mol)
\(\left[Cu^{2+}\right]=\left[SO_4^{2-}\right]=\dfrac{0,05}{0,5}=0,1M\)
a, \(n_{H^+}=n_{OH^-}=9.10^{-3}\left(mol\right)\Rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{\dfrac{9.10^{-3}}{2}}{0,05}=0,09M\)
b, \(\left[SO_4^{2-}\right]=\dfrac{4,5.10^{-3}}{0,05+0,15}=0,6M\)
\(\left[Na^+\right]=\dfrac{0,15.0,06}{0,05+0,15}=0,045M\)
\(\left[H^+\right]=\left[OH^-\right]=\dfrac{9.10^{-3}}{0,05+0,15}=0,045M\)
Những pt ion này bạn nên nhớ khi làm dạng toán HNO3.
\(3Cu+8H^++2NO_3^-\rightarrow3Cu^{2+}+2NO+4H_2O\)
0,3 \(\rightarrow\)0,8\(\rightarrow\) 0,2 \(\rightarrow\) 0,3\(\rightarrow\) 0,2
\(3Fe^{2+}+4H^++NO^-_3\rightarrow3Fe^{3+}+NO+2H_2O\)
0,6 \(\rightarrow\) 0,8 \(\rightarrow\) 0,2 \(\rightarrow\) 0,6 \(\rightarrow\) 0,2
\(\underrightarrow{BTe:}\) \(3n_{NO}=2n_{Fe}+2n_{Cu}\rightarrow n_{NO}=0,4\Rightarrow V_{NO}=8,96l\)
a) Ta có: \(\left\{{}\begin{matrix}\left[Cu^{2+}\right]=C_{M_{Cu\left(NO_3\right)_2}}=0,3\left(M\right)\\\left[NO_3^-\right]=2C_{M_{Cu\left(NO_3\right)_2}}=0,6\left(M\right)\end{matrix}\right.\)
b) Ta có: \(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,5\left(M\right)\\\left[SO_4^{2-}\right]=0,25\left(M\right)\end{matrix}\right.\)