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\(b.n_{NaOH\left(tổng\right)}=0,4.0,5+\dfrac{100.1,33.20\%}{40}=0,865\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\left(sau\right)\right]=\dfrac{0,865}{0,4+0,1}=1,73\left(M\right)\\ c.n_{HCl}=0,05.0,12=0,006\left(mol\right)\\ n_{HNO_3}=0,15.0,1=0,015\left(mol\right)\\ \left[H^+\right]=\dfrac{0,006+0,015}{0,05+0,15}=0,105\left(M\right)\\ \left[NO^-_3\right]=\dfrac{0,015}{0,05+0,15}=0,075\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,006}{0,05+0,15}=0,03\left(M\right)\)
\(d.n_{H_2SO_4}=0,4.0,05=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ f.n_{KOH}=\dfrac{20.1,31.32\%}{56}=\dfrac{131}{875}\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,08.1=0,08\left(mol\right)\\ \left[OH^-\right]=\dfrac{\dfrac{131}{875}+0,08.2}{0,02+0,08}=\dfrac{542}{175}\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,08}{0,02+0,08}=0,8\left(M\right)\)
\(\left[K^+\right]=\dfrac{\dfrac{131}{875}}{0,02+0,08}=\dfrac{262}{175}\left(M\right)\)
Ok, để thử coi chứ tui ngu hóa thấy mồ :(
a/ \(n_{NaOH}=0,2.0,1=0,02\left(mol\right)\)
\(NaOH\rightarrow Na^++OH^-\)
\(n_{Na^+}=n_{OH^-}=0,02\left(mol\right)\)
\(\Rightarrow C_{MNa^+}=\frac{0,02}{0,4+0,1}=0,04\left(mol/l\right)\)
\(n_{Ba\left(OH\right)_2}=0,3.0,4=0,12\left(mol\right)\)
\(Ba\left(OH\right)_2=Ba^{2+}+2OH^-\)
\(\Rightarrow n_{OH^-}=0,24\left(mol\right);n_{Ba^{2+}}=0,12\left(mol\right)\)
\(\Rightarrow C_{MBa^{2+}}=\frac{0,12}{0,5}=0,24\left(mol/l\right)\)
\(n_{OH^-}=0,02+0,24=0,26\left(mol\right)\)
\(\Rightarrow C_{MOH^-}=\frac{0,26}{0,5}=0,52\left(mol/l\right)\)
b/ \(n_{HCl}=0,2V\left(mol\right)\)
\(\Rightarrow n_{H^+}=n_{Cl^-}=0,2V\)
\(\Rightarrow C_{MCl^-}=\frac{0,2V}{2V}=0,1\left(mol/l\right)\)
\(n_{H_2SO_4}=0,3V\left(mol\right)=\frac{n_{H^+}}{2}=n_{SO_4^{2-}}\)
\(\Rightarrow C_{MSO_4^{2-}}=\frac{0,3V}{2V}=0,15\left(mol/l\right)\)
\(n_{H^+}=0,2V+0,6V=0,8V\left(mol\right)\)
\(\Rightarrow C_{MH^+}=\frac{0,8V}{2V}=0,4\left(mol/l\right)\)
Bác nào hảo tâm giúp em mấy câu còn lại chớ đến đây thì em chịu chết òi :(
\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\\ \left[HCl\right]=\dfrac{0,2}{0,2}=1\left(M\right)\\ \left[H_2SO_4\right]=\dfrac{0,1}{0,2}=0,5\left(M\right)\\ \left[SO_4^{2-}\right]=\left[H_2SO_4\right]=0,5\left(M\right)\\ \left[Cl^-\right]=\left[HCl\right]=1\left(M\right)\\ \left[H^+\right]=1+0,5.2=2\left(M\right)\)
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, \(n_{H^+}=2n_{H_2SO_4}=0,05\left(mol\right)\Rightarrow\left[H^+\right]=\dfrac{0,05}{0,05}=1M\)
\(n_{SO_4^{2-}}=n_{H_2SO_4}=0,025\left(mol\right)\Rightarrow\left[SO_4^{2-}\right]=\dfrac{0,025}{0,05}=0,5M\)
\(n_{H_2SO_4\left(2M\right)}=0,15.2=0,3\left(mol\right)\)
\(n_{H_2SO_4\left(3M\right)}=0,15.3=0,45\left(mol\right)\)
\(n_{H_2SO_4\left(B\right)}=0,3+0,45=0,75\left(mol\right)\)
\(\Rightarrow C_{M_{ddB}}=\dfrac{0,75}{0,2}=3,75M\)
\(n_{H_2SO_4\left(tổng\right)}=0,15.2+0,05.3=0,45\left(mol\right)\\ V_{ddH_2SO_4\left(tổng\right)}=150+50=200\left(ml\right)=0,2\left(l\right)\\ C_{MddH_2SO_4\left(sau\right)}=C_{MddB}=\dfrac{0,45}{0,2}=2,25\left(M\right)\)
Ta có: \(n_{H^+}=2n_{H_2SO_4}+n_{HCl}=2\cdot0,02+0,07=0,11\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{0,11}{0,35+0,05}=0,275\left(M\right)\) \(\Rightarrow\left[OH^-\right]=\dfrac{10^{-14}}{\left[H^+\right]}\approx3,64\cdot10^{-14}\left(M\right)\)
\(\Rightarrow pH=-log\left(0,275\right)\approx0,56\)
*Môi trường axit và làm quỳ tím hóa đỏ
\(n_{H_2SO_4}=0,05.0,4=0,02\left(mol\right)\\ n_{HCl}=0,2.0,35=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ pH=-log\left[H^+\right]=-log\left[0,275\right]=0,56\\ \Rightarrow Qùy.hóa.đỏ\)
\(n_{H^+}=n_{HCl}+n_{HCl}=0,15\cdot0,2+0,35\cdot0,04=0,044mol\)
\(C_M=\dfrac{0,044}{0,15+0,35}=\dfrac{0,044}{0,5}=0,088M\)
\(n_{H_2SO_4}=0,05.0,4=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H_2SO_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[HCl\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ \Rightarrow\left[H^+\right]=0,05.2+0,175.1=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=0,05\left(M\right)\\ \left[Cl^-\right]=0,175\left(M\right)\)