1.

 \(C_{12}H_{22}O_{11}+H_2O\rightarrow C_6H_{12}O_6+C_6H_{12}O_6\)

\(C_6H_{12}O_6\xrightarrow[xt:enzim]{t^o:30-35^o}2C_2H_5OH+2CO_2\)

\(C_2H_5OH+O_2\underrightarrow{xt:men\left(giấm\right)}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_4đặc,t^o}CH_3COOC_2H_5+H_2O\)

1.

 C12H22O11+H2O→C6H12O6+C6H12O6C12H22O11+H2O→C6H12O6+C6H12O6

C6H12O6to:30−35o−−−−−→xt:enzim2C2H5OH+2CO2C6H12O6→xt:enzimto:30−35o2C2H5OH+2CO2

C2H5OH+O2xt:men(giấm)−−−−−−−−−−−−→CH3COOH+H2OC2H5OH+O2xt:men(giấm)→CH3COOH+H2O

CH3COOH+C2H5OHH2SO4đặc,to−−−−−−−−−−→CH3COOC2H5+H2O

Bài 1:

a. \(p=dh=10300\cdot36=370800\left(N/m^2\right)\)

b. \(160cm^2=0,016m^2\)

\(p=\dfrac{F}{S}\Rightarrow F=S\cdot p=0,016\cdot370800=5932,8\left(N\right)\)

Bài 2:

Ta có: \(p_2>p_1\left(1165000>875000\right)\)

\(\Rightarrow\) Tàu đăng lặn xuống, vì khi càng xuống áp suất lại càng tăng.

\(\left\{{}\begin{matrix}p'=dh'\Rightarrow h'=\dfrac{p'}{d}=\dfrac{875000}{10300}\approx84,9\left(m\right)\\p''=dh''\Rightarrow h''=\dfrac{p''}{d}=\dfrac{1165000}{10300}\approx113\left(m\right)\end{matrix}\right.\)

Bài 3:

\(20cm=0,2m\)

\(p=dh=8000\cdot\left(1,8-0,2\right)=12800\left(N/m^2\right)\)

Bài 4:

\(p=dh=10000\cdot2,8=28000\left(N/m^2\right)\)

\(150cm^2=0,015m^2\)

\(p=\dfrac{F}{S}\Rightarrow F=S\cdot p=0,015\cdot28000=420\left(N\right)\)

Hiệu điện thế hai đầu cuộn thứ cấp:

\(\dfrac{U_1}{U_2}=\dfrac{N_1}{N_2}\Rightarrow\dfrac{12}{U_2}=\dfrac{500}{50}\)

\(\Rightarrow U_2=1,2V\)

Câu 1.

\(CH_3COOH+NaOH\rightarrow CH_4\uparrow+Na_2CO_3\)

\(2CH_4\xrightarrow[LLN]{1500^oC}C_2H_2+2H_2\uparrow\)

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

a)\(C_2H_4+HBr\rightarrow C_2H_5Br\)

b)\(C_2H_4+HCl\underrightarrow{as}C_2H_5Cl\)