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\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=\dfrac{m_1}{23}+m_2-\dfrac{m_1}{46}=\dfrac{m_1}{46}+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{m_1}{46}+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=m_1+m_2-\dfrac{m_1}{23}=\dfrac{22}{23}m_1+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{22}{23}m_1+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
\(n_{Na}=\dfrac{2.3}{23}=0.1\left(mol\right)\)
\(m_{NaOH\left(10\%\right)}=100\cdot10\%=10\left(g\right)\)
\(n_{NaOH\left(10\%\right)}=\dfrac{10}{40}=0.25\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(0.1......................0.1..........0.05\)
\(\sum n_{NaOH}=0.25+0.1=0.35\left(mol\right)\)
\(m_{NaOH}=0.35\cdot40=14\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=2.3+100-0.05\cdot2=102.2\left(g\right)\)
\(C\%_{NaOH}=\dfrac{14}{102.2}\cdot100\%=13.7\%\)
\(V_{dd}=\dfrac{102.2}{1.05}=97.33\left(ml\right)=0.0973\left(l\right)\)
\(C_{M_{NaOH}}=\dfrac{0.35}{0.0973}=3.6\left(M\right)\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,04\left(mol\right)\Rightarrow m_1=m_{Zn}=0,04.65=2,6\left(g\right)\)
\(n_{HCl}=2n_{H_2}=0,08\left(mol\right)\Rightarrow m_{HCl}=0,08.36,5=2,92\left(g\right)\)
\(\Rightarrow m_2=m_{ddHCl}=\dfrac{2,92}{14,6\%}=20\left(g\right)\)
b, Ta có: m dd sau pư = mZn + m dd HCl - mH2 = 22,52 (g)
\(n_{ZnCl_2}=n_{H_2}=0,04\left(mol\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,04.136}{22,52}.100\%\approx24,16\%\)
\(n_{SO_3}=\dfrac{20}{80}=0,25\left(mol\right)\\ PTHH:SO_3+H_2O\rightarrow H_2SO_4\\ Mol:0,25\rightarrow0,25\rightarrow0,25\\ C_{MH_2SO_4}=\dfrac{0,25}{0,5}=0,5M\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\\ Mol:0,25\leftarrow0,25\\ m_{Mg}=0,25.24=6\left(g\right)\)
a, Gọi \(m_{NaCl\left(thêm\right)}=a\left(g\right)\)
\(m_{NaCl\left(bđ\right)}=5\%.100=5\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{5+a}{100+a}.100\%=5,5\%\\ \Leftrightarrow a=0,53\left(g\right)\)
b, \(m_{NaCl}=58,5.5,5\%=3,2175\left(g\right)\\ n_{NaCl}=\dfrac{3,2175}{58,5}=0,055\left(mol\right)\)
PTHH: NaCl + AgNO3 ---> AgCl↓ + NaNO3
0,055-->0,055------>0,055---->0,055
\(m_{AgCl}=0,055.143,5=7,8925\left(g\right)\\ m_{ddY}=58,5+200-7,8925=250,6075\left(g\right)\\ \Rightarrow C\%_{NaNO_3}=\dfrac{0,055.85}{250,6075}.100\%=1,87\%\)
Ta có: \(n_{Na_2CO_3}=n_{Na_2CO_3.10H_2O}=\dfrac{38,61}{286}=0,135\left(mol\right)\)
m dd sau pư = 38,61 + 256 = 294,61 (g)
\(\Rightarrow C\%_{Na_2CO_3}=\dfrac{0,135.106}{294,61}.100\%\approx4,86\%\)
Có: \(V_{ddsaupư}=\dfrac{294,61}{1,156}\approx254,85\left(ml\right)\approx0,255\left(l\right)\)
\(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{0,135}{0,255}\approx0,53M\)
Bạn tham khảo nhé!
Gọi số mol của Na2CO3 là a (mol) \(\Rightarrow n_{H_2O\left(phân.tử\right)}=10a\left(mol\right)\)
\(\Rightarrow106a+18\cdot10a=38,61\) \(\Leftrightarrow a=0,135\left(mol\right)\)
\(\Rightarrow C\%_{Na_2CO_3}=\dfrac{0,135\cdot106}{38,61+256}\cdot100\%\approx4,86\%\)
Mặt khác: \(V_{ddNa_2CO_3}=\dfrac{38,61+256}{1,156}\approx254,41\left(ml\right)\) \(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{0,135}{0,25441}\approx0,53\left(M\right)\)
a, \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)
\(m_{NaOH}=0,2.40=8\left(g\right)\)
b, \(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)
\(c,C\%=\dfrac{6}{200}.100\%=3\%\)
\(m_{NaCl}=\dfrac{200.8}{100}=16\left(g\right)\)
Khối lượng CuSO4 có trong m gam tinh thể : \(\frac{160}{250}\)m = 0,64(g)
Khối lượng CuSO4 trong V ml dung dịch CuSO4 c% ((khối lượng riêng bằng d g/ml) là : \(\frac{V.d.c}{100}\) = 0,01 V.d.c (g)
Khối lượng dung dịch X bằngv : m+V.d (g)
Nồng độ phần trăm của dung dịch X:
\(\frac{0,64m+0,01V.d.c}{m+V.d}.100\%=\frac{64m+V.d.c}{m+V.d}\left(\%\right)\)
Phần c thì làm ntn ạ