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\(a)\ CH_3-CH=CH-CH_3 + H_2 \xrightarrow{t^o,Ni} CH_3-CH_2-CH_2-CH_3\\ b)\ CH_2=CH-CH_3 + Br_2 \to CH_2Br-CHBr-CH_3\\ c)\ CH_2=C(CH_3)-CH_3 + HBr \to CH_3-CBr(CH_3)-CH_3\\ d)\ CH_2=CH-CH_2-CH_3 + H_2O \xrightarrow{H^+,t^o} CH_3-CH(OH)-CH_2-CH_3\\ e)\ CH_3-CH=CH-CH_3 + HBr \to CH_3-CH_2-CHBr-CH_3\\ f)\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ g)\ nCH_2=CH_2 \xrightarrow{t^o,p,xt} (-CH_2-CH_2-)_n\\ h)\ nCH_2=CH-CH_3 \xrightarrow{t^o,p,xt} (-CH_2-CH(CH_3)-)_n\)
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\(n_{C_4H_6} = n.n_{cao\ su} = n.\dfrac{1000}{54n} = \dfrac{500}{27}(kmol)\\ n_{C_2H_5OH} = 2.\dfrac{1}{60\%}.n_{C_4H_6} = \dfrac{5000}{81}(kmol)\\ n_{C_6H_{12}O_6} = \dfrac{1}{2}.\dfrac{1}{80\%}.n_{C_2H_5OH} = \dfrac{3125}{81}(kmol)\\ m_{gỗ} = \dfrac{m_{C_6H_{12}O_6}}{35\%} = \dfrac{\dfrac{3125}{81}.180}{35\%} = 19841,25(kg)\)
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CaCO3 \(\underrightarrow{t^o}\) CaO + CO2
CaO + H2O \(\rightarrow\) Ca(OH)2
Ca(OH)2 + CO2 \(\rightarrow\) CaCO3 + H2O
Chọn đáp án A
$CH_3-CH_2-CH_2-CH_2OH \xrightarrow{H_2SO_4,t^o} CH_3-CH_2-CH=CH_2 + H_2O$
$CH_3-CH_2-CH=CH_2 + HBr \to CH_3-CH_2-CHBr-CH_3$
$CH_3-CH_2-CHBr-CH_3 + KOH \xrightarrow{xt} CH_3-CH_2-CHOH-CH_3 + KBr$