Trong nicotin có: \(\left\{{}\begin{matrix}m_C=\dfrac{162.74,07\%}{100\%}=120\left(g\right)\\m_H=\dfrac{162.8,64\%}{100\%}=14\left(g\right)\\m_N=\dfrac{162.17,28\%}{100\%}=28\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_C=\dfrac{120}{12}=10\left(mol\right)\\n_H=\dfrac{14}{1}=14\left(mol\right)\\n_N=\dfrac{28}{14}=2\left(mol\right)\end{matrix}\right.\)

Vậy CTHH của nicotin là \(C_{10}H_{14}N_2\)

Giups mik vs ạ

Gọi \(\left\{{}\begin{matrix}n_{N_2}=x\left(mol\right)\\n_{H_2}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{28x+2y}{x+y}=12,4\)

\(\Leftrightarrow12,4x+12,4y-28x-2y=0\Leftrightarrow-15,6x=-10,4y\\ \Leftrightarrow\dfrac{x}{y}=-\dfrac{10,4}{-15,6}=\dfrac{2}{3}\)

Giả sử \(\left\{{}\begin{matrix}n_{N_2}=2\left(mol\right)\\n_{H_2}=3\left(mol\right)\end{matrix}\right.\)

\(n_{khí.giảm}=n_{NH_3}=\dfrac{2}{3}n_{H_2}.H\%=\dfrac{2}{3}.3.40\%=0,8\left(mol\right)\)

\(n_Y=2+3-0,8=4,2\left(mol\right)\)

\(m_Y=m_X=28.2+2.3=62\left(g\right)\)

\(\overline{M_Y}=\dfrac{62}{4,2}=14,76g/mol\)

\(n_{N_2}=a;n_{H_2}=b\\ PT:\dfrac{1}{2}N_2+\dfrac{3}{2}H_2-Fe,t^{^0}->NH_3\\ M_X=\dfrac{28a+2b}{a+b}=12,4\\ a=1,5b\\ Sau.pư:\\ M_Y=\dfrac{2.0,6b+28\left(a-\dfrac{0,4b}{3}\right)+17.0,4b}{0,6b+a-\dfrac{0,4b}{3}+0,4b}=19,55\)

Dung dịch A thể tích bao nhiêu? Nếu không có thì không cho đáp số.

Tổng quát:

\(CO_3^{2-}+2H^+->CO_2+H_2O\\ n_{CO_2}=\dfrac{0,672}{22,4}=0,03mol=n_{H^+}.2=2n_{Cl^-}=n_{CO_3^{2-}}\\ m_{muối}=10-60.0,03+2.35,5.0,03=10,33g\)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{NaOH}=0,3.1=0,3\left(mol\right)\)

Xét \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,3}{0,25}=1,2\)

=> Phản ứng tạo 2 muối: muối axit và muối trung hòa.

Gọi \(\left\{{}\begin{matrix}n_{NaHCO_3}=x\left(mol\right)\\n_{Na_2CO_3}=y\left(mol\right)\end{matrix}\right.\)

\(CO_2+NaOH\rightarrow NaHCO_3\)

x<-------x<---------x

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

y<------2y<----------y

Từ PTHH trên có hệ phương trình: \(\left\{{}\begin{matrix}x+y=0,25\\x+2y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,05\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaHCO_3}=0,2.84=16,8\left(g\right)\\m_{Na_2CO_3}=0,05.106=5,3\left(g\right)\end{matrix}\right.\)

Gọi kim loại cần tìm là R và \(n_R=x\left(mol\right)\)

\(RO+H_2SO_4\rightarrow RSO_4+H_2O\)

x----->x---------->x

\(m_{dd.H_2SO_4}=\dfrac{98x.100\%}{4,9\%}=2000x\left(g\right)\)

\(m_{dd}=m_{RO}+m_{dd.H_2SO_4}=\left(R+16\right)x+2000x=Rx+2016x\left(g\right)\)

\(C\%_{RSO_4}=\dfrac{\left(R+96\right).x.100\%}{Rx+2016x}=5,88\%\)

\(\Leftrightarrow\dfrac{Rx+96x}{Rx+2016x}=0,0588\\ \Leftrightarrow0,0588Rx+118,5408x-Rx-96x=0\\ \Leftrightarrow-0,9412Rx=-22,5408x\\ \Leftrightarrow R=-\dfrac{22,5408x}{-0,9412x}=24\left(g/mol\right)\)

Vậy kim loại là Mg

(Có quy thì quy 1 mol ra rồi làm cách hệt mình, làm xong bài rồi mới thấy dòng cuối của bạn: )

H2SO4 4,9% nhé 

Gọi a, b, c, x, y, z là số p, n, e trong X, Y

\(2a+b=52\\ 2a-b=16\\ a=17=p_X=e_X\\ b=18=n_X\\ X:Argon\\ 2x+y=58\\ 2x-y=18\\ x=19=p_Y=e_Y\\ y=20=n_Y\\ Y:Potassium\text{/}Kali\)

Argon: 

Potassium:

\(C_4H_{10}O\\ \)

CH3-CH2-CH2-CH2-OH: Butan-1-ol

CH3-CH(CH3)CH2OH: 2-metylpropan-1-ol

CH3-CH2-CH(OH)-CH3: Butan-2-ol

CH3-C(OH)(CH3)-CH3:  2-metylpropan -2- ol

CH3-CH2-CH2-O-CH3:1- metoxypropan

CH3-CH(CH3)-O-CH3: 2 - metoxypropan

CH3-CH2-O-CH2-CH3 : etoxyetan

\(C_4H_9Br\)

CH3-CH2-CH2-CH2Br: 1-bromobutan

CH3-CH(CH3)-CH2Br: 1-bromo-2-metylpropan

CH3-CBr(CH3)-CH3: 2-bromo-2-metylpropan

CH3-CH2-CHBr-CH3: 2-bromobutan

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,2------->0,1--------->0,1

Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư

Trong dung dịch D có:

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

b

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,35<---------0,35

\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)

\(a.C_M=\dfrac{0,06}{1,5}=0,04M\\ b.C_M=\dfrac{\dfrac{400}{160}}{4}=0,625M\\ c.C_M=\dfrac{\dfrac{10,53}{58,5}}{\dfrac{450}{1,25}:1000}=0,5M\\ d.C_M=\dfrac{\dfrac{70,2}{40}}{0,5}=3,51M\\ e.C_M=\dfrac{\dfrac{42}{200}}{\dfrac{742}{1,3}:1000}=0,368M\)