Câu 1:

PT: \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

\(2NaHCO_3+H_2SO_4\rightarrow Na_2SO_4+2CO_2+2H_2O\)

Gọi: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\left(mol\right)\\n_{NaHCO_3}=y\left(mol\right)\end{matrix}\right.\) ⇒ 106x + 84y = 9,1 (1)

Theo PT: \(n_{CO_2}=n_{Na_2CO_3}+n_{NaHCO_3}=x+y=\dfrac{2,016}{22,4}=0,09\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,07\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na_2CO_3}=0,07.106=7,42\left(g\right)\\m_{NaHCO_3}=0,02.84=1,68\left(g\right)\end{matrix}\right.\)

Theo PT: \(n_{H_2SO_4}=n_{Na_2CO_3}+\dfrac{1}{2}n_{NaHCO_3}=0,08\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,08.98=7,84\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{7,84}{50\%}=15,68\left(g\right)\)

Câu 2:

Ta có: m _{bình 1 tăng} = m_H2O = 3,6 (g)

\(\Rightarrow n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\)

m _{bình 2 tăng} = m_CO2 = 13,2 (g)

\(\Rightarrow n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\)

PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=2n_{C_2H_4}+2n_{C_2H_2}=0,3\left(mol\right)\\n_{H_2O}=2n_{C_2H_4}+n_{C_2H_2}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,05\left(mol\right)\\n_{C_2H_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,05.28}{0,05.28+0,1.26}.100\%=35\%\\\%m_{C_2H_2}=65\%\end{matrix}\right.\)

Câu 3:

Ta có: \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(1\right)\)

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{56}{160}=0,35\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,15\left(mol\right)\\n_{C_2H_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,15.22,4}{5,6}.100\%=60\%\\\%V_{C_2H_2}=40\%\end{matrix}\right.\)

Câu3 nè  

Câu 1:

a, (1) \(CaC_2+2H_2O\rightarrow Ca\left(OH\right)_2+C_2H_2\)

(2) \(C_2H_2+H_2\underrightarrow{t^o,Pd}C_2H_4\)

b, Phần này giống phần a nhé.

c, (1) \(CO+\dfrac{1}{2}O_2\underrightarrow{t^o}CO_2\)

(2) \(CO_2+NaOH\rightarrow NaHCO_3\)

(3) \(NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\)

(4) \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

(5) \(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2+H_2\)

Câu 2:

a, \(C_2H_6+Cl_2\underrightarrow{as}HCl+C_2H_5Cl\)

b, \(CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

c, \(CH\equiv CH+2Br_2\rightarrow CHBr_2-CHBr_2\)

ngu

BT4: Hiệu suất phản ứng:

\(H=\dfrac{m_{tt}}{m_{lt}}.100\%=\dfrac{36,48}{48}.100\%=76\%\)

BT5 Khối lượng đồng thu được:

\(H=\dfrac{m_{tt}}{m_{lt}}.100\%\Rightarrow m_{tt}=\dfrac{m_{lt}.H}{100\%}=\dfrac{48.95}{100\%}=45,6\left(g\right)\)

a: \(\widehat{A}=180^0-2\cdot70^0=40^0\)

1: \(\Leftrightarrow x-2-7x+7=-1\)

=>-6x+5=-1

hay x=1(loại)

3: \(\Leftrightarrow\left(x+2\right)\left(x-1\right)-\left(x+1\right)\left(x+3\right)=4\)

\(\Leftrightarrow x^2+x-2-x^2-4x-3=4\)

=>-3x=9

hay x=-3(loại)

4: \(\Leftrightarrow x^2+2x+1-x^2+2x-1=3x\cdot\dfrac{x+1-x+1}{x+1}\)

\(\Leftrightarrow4x=\dfrac{6x}{x+1}\)

\(\Leftrightarrow4x^2+4x-6x=0\)

\(\Leftrightarrow4x^2-2x=0\)

=>2x(2x-1)=0

hay \(x\in\left\{0;\dfrac{1}{2}\right\}\)

làm giúp mình ik tại nhìu bài nhìu môn qé, nếu làm đúng mình k cho (≧∇≦)ﾉ

Lời giải:

Lời giải:

$372802=300000+70000+2000+800+2$

$430279=400000+30000+200+70+9$
$920300=900000+20000+300$

$704753=700000+4000+700+50+3$