Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đáp án B
H2SO4 → 2H++ SO42-
x M 2x M
HNO3→ H++ NO3-
x M xM
HNO2 ⇌H++ NO2-
< x M
\(n_{NaOH}=0,02.2=0,04\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ a.n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,04}{2}=0,02\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,02}{0,08}=0,25\left(M\right)\\ b.\left[Na^+\right]=\dfrac{0,02.2}{0,02+0,08}=0,4\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,02+0,08}=0,2\left(M\right)\)
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 !!
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, \(\left[SO_4^{2-}\right]=\dfrac{4,5.10^{-3}}{0,05+0,15}=0,6M\)
\(\left[Na^+\right]=\dfrac{0,15.0,06}{0,05+0,15}=0,045M\)
\(\left[H^+\right]=\left[OH^-\right]=\dfrac{9.10^{-3}}{0,05+0,15}=0,045M\)
a, \(n_{H^+}=n_{Cl^-}=n_{HCl}=0,003\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\left[Cl^-\right]=\dfrac{0,003}{0,1}=0,03M\)
b, \(n_{H^+}=2n_{H_2SO_4}=0,05\left(mol\right)\Rightarrow\left[H^+\right]=\dfrac{0,05}{0,05}=1M\)
\(n_{SO_4^{2-}}=n_{H_2SO_4}=0,025\left(mol\right)\Rightarrow\left[SO_4^{2-}\right]=\dfrac{0,025}{0,05}=0,5M\)
\(n_{OH^-}=0,5.0,2+0,2.2.0,3=0,22\left(mol\right)\Rightarrow\left[OH^-\right]=\dfrac{0,22}{0,5}=0,44M\)
\(n_{Na^+}=0,5.0,2=0,1\left(mol\right)\Rightarrow\left[Na^+\right]=\dfrac{0,1}{0,5}=0,2M\)
\(n_{Ba^{2+}}=0,2.0,3=0,06\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{0,06}{0,5}=0,12M\)
\(n_{HNO_2}=\dfrac{5,64.10^{19}}{6.10^{23}}=9,4.10^{-5}\)
\(n_{NO_2^-}=\dfrac{3,6.10^{18}}{6.10^{23}}=6.10^{-6}\)
\(HNO_2⇌H^++NO_2^-\)
Ta có :
\(n_{HNO_2}=9,4.10^{-5}+6.10^{-6}=10^{-4}\)
Độ điện li \(\alpha=\dfrac{6.10^{-6}}{10^{-4}}=0,06\)
b)
\(C_{M_{HNO_2}}=\dfrac{10^{-4}}{10^{-3}}=0,1M\)
cảm ơn ạ