º£Ë®ÊǾ޴óµÄ×ÊÔ´±¦¿â£¬º£Ë®µ­»¯¼°Æä×ÛºÏÀûÓþßÓÐÖØÒªÒâÒå¡£

Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©ÂȼҵÖ÷ÒªÒÔʳÑÎΪԭÁÏ¡£ÎªÁ˳ýÈ¥´ÖÑÎÖеÄCa2+¡¢Mg2+¡¢SO42£­¼°Äàɳ£¬¿É½«´ÖÑÎÈÜÓÚË®£¬È»ºó½øÐÐÏÂÁвÙ×÷£¬ÕýÈ·µÄ²Ù×÷˳ÐòÊÇ             ¡£
¢Ù¹ýÂË£»¢Ú¼Ó¹ýÁ¿µÄNaOHÈÜÒº£»¢Û¼ÓÊÊÁ¿µÄÑÎË᣻¢Ü¼Ó¹ýÁ¿µÄNa2CO3ÈÜÒº£»¢Ý¼Ó¹ýÁ¿µÄBaCl2ÈÜÒº
a£®¢Ú¢Ý¢Ü¢Ù¢Û       b£®¢Ù¢Ü¢Ú¢Ý¢Û       c        d£®¢Ý¢Ú¢Ü¢Ù¢Û
£¨2£©ÔÚʵÑéÊÒÖпÉÒÔÓÃÝÍÈ¡µÄ·½·¨ÌáÈ¡ä壬¿ÉÑ¡ÓõÄÊÔ¼ÁÊÇ________________£¬ËùÓÃÖ÷ÒªÒÇÆ÷µÄÃû³ÆÊÇ____________________¡£
£¨3£©²½Öè¢ñÖÐÓÃÁòËáËữ¿ÉÌá¸ßCl2ÀûÓÃÂʵÄÔ­ÒòÊÇ                       ¡£
£¨4£©²½ÖèII·´Ó¦µÄÀë×Ó·½³Ìʽ__________________________________________¡£
£¨5£©º£Ë®ÌáäåÕôÁó¹ý³ÌÖУ¬Î¶ÈÓ¦¿ØÖÆÔÚ80~90¡æ£¬Î¶ȹý¸ß»ò¹ýµÍ¶¼²»ÀûÓÚÉú²ú £¬Çë½âÊÍÔ­Òò                                                                ¡£
£¨6£©Mg(OH)2³ÁµíÖлìÓÐCa(OH)2£¬¿ÉÑ¡ÓÃ__________ÈÜÒº½øÐÐÏ´µÓ³ýÈ¥¡£ÈçÖ±½Ó¼ÓÈÈMg(OH)2µÃµ½MgO£¬ÔÙµç½âÈÛÈÚMgOÖƽðÊôþ£¬ÕâÑù¿É¼ò»¯ÊµÑé²½Ö裬Äã_______£¨Ñ¡ÌͬÒ⡱£¬¡°²»Í¬Ò⡱£©¸Ã˵·¨£¬ÀíÓÉÊÇ                                  ¡£
£¨1£©AD
£¨2£©CCl4£¨»ò±½£©£»·ÖҺ©¶·
£¨3£©Ëữ¿ÉÒÖÖÆCl2 ¡¢Br2ÓëË®·´Ó¦
£¨4£©Br2+SO2+2H2O¡ú4H++SO+2Br¡ª£¨2·Ö£©
£¨5£©Î¶ȹý¸ß£¬´óÁ¿Ë®ÕôÆøËæË®Åųý³ö£¬äåÕôÆøÖÐË®Ôö¼Ó£»Î¶ȹýµÍ£¬äå²»ÄÜÍêÈ«Õô³ö£¬ÎüÊÕÂʵ͡££¨2·Ö£©
£¨6£©MgCl2¡£²»Í¬Ò⣻MgOÈÛµãºÜ¸ß£¬ÈÛÈÚʱºÄÄܸߣ¬Ôö¼ÓÉú²ú³É±¾¡££¨2·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©³ýÔÓʱΪÁË°ÑÔÓÖʳý¾»³ýÔÓÊÔ¼ÁÐè¹ýÁ¿£¬³ýÈ¥´ÖÑÎÖеÄCa2+¡¢Mg2+¡¢SO42£­·Ö±ðÑ¡ÓÃNa2CO3ÈÜÒº¡¢NaOHÈÜÒº¡¢ BaCl2ÈÜÒº£¬Ç°Á½ÖÐÊÔ¼Á¿ÉÒÔÓÃÊÊÁ¿µÄÑÎËá³ýÈ¥£¬ BaCl2ÈÜÒºÖ»ÄÜ·ÅÔÚ¼ÓÈëNa2CO3ÈÜÒº²½ÖèÇ°Ã棬ÓÃÆä³ýÈ¥£¬×¢Òâ¼ÓÑÎËáÇ°¹ýÂ˳ýÈ¥³Áµí£¬¹Ê²Ù×÷˳ÐòΪ¢Ú¢Ý¢Ü¢Ù¢Û »ò¢Ý¢Ú¢Ü¢Ù¢Û£»
£¨2£©ÝÍÈ¡ÊÔ¼Á²»ÄÜÓëË®»¥ÈÜÇÒ±»ÌáÈ¡µÄÎïÖÊÔÚÊÔ¼ÁÖеÄÈܽâ¶ÈÒªÔ¶´óÓÚÔÚË®ÖеÄÈܽâ¶È£¬ÊµÑéÔÚ·ÖҺ©¶·ÖнøÐУ»
£¨3£©Cl2 ¡¢Br2ÓëË®·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬Ëữ¿ÉÒÖÖÆCl2 ¡¢Br2ÓëË®·´Ó¦£»
£¨5£©Î¶ȹý¸ß£¬´óÁ¿Ë®ÕôÆøËæË®Åųý³ö£¬äåÕôÆøÖÐË®Ôö¼Ó£»Î¶ȹýµÍ£¬äå²»ÄÜÍêÈ«Õô³ö£¬ÎüÊÕÂʵ͡£
£¨6£©Ñ¡ÓÃMgCl2ÈÜÒº£¬Ê¹Ca(OH)2ת»¯ÎªMg(OH)2£¬Í¬Ê±Ï´È¥ÔÓÖÊ£»MgOÈÛµãºÜ¸ß£¬ÈÛÈÚʱºÄÄܸߣ¬Ôö¼ÓÉú²ú³É±¾£¬¹Ê²»Óõç½âÈÛÈÚMgOÖƽðÊôþ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ÂÈËáþ[Mg(ClO3)2]³£ÓÃ×÷´ßÊì¼Á¡¢³ý²Ý¼ÁµÈ£¬ÊµÑéÊÒÖƱ¸ÉÙÁ¿Mg(ClO3)2¡¤6H2OµÄÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢Ù±¿éÖ÷Òª³É·ÖΪMgCl2¡¤6H2O£¬º¬ÓÐMgSO4¡¢FeCl2µÈÔÓÖÊ¡£
¢ÚËÄÖÖ»¯ºÏÎïµÄÈܽâ¶È(S )ËæζÈ(T )±ä»¯ÇúÏßÈçͼËùʾ¡£

£¨1£©¹ýÂËËùÐèÒªµÄÖ÷Òª²£Á§ÒÇÆ÷ÓР          ¡£
£¨2£©¼ÓÈëBaCl2µÄÄ¿µÄÊÇ        £¬¼ÓMgOºó¹ýÂËËùµÃÂËÔüµÄÖ÷Òª³É·ÖΪ            ¡£
£¨3£©¼ÓÈëNaClO3±¥ºÍÈÜÒººó·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                             
£¨4£©²úÆ·ÖÐMg(ClO3)2¡¤6H2Oº¬Á¿µÄ²â¶¨£º
²½Öè1£º×¼È·³ÆÁ¿3.50 g²úÆ·Åä³É100 mLÈÜÒº¡£
²½Öè2£ºÈ¡10.00 mLÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë10.00 mLÏ¡ÁòËáºÍ20 .00mL 1.000 mol¡¤L£­1µÄFeSO4ÈÜÒº£¬Î¢ÈÈ¡£
²½Öè3£ºÀäÈ´ÖÁÊÒΣ¬ÓÃ0.100 mol¡¤L£­1 K2Cr2O7 ÈÜÒºµÎ¶¨Ê£ÓàµÄFe2£«ÖÁÖյ㣬´Ë¹ý³ÌÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºCr2O72£­£«6Fe2£«£«14H£«£½2Cr3£«£«6Fe3£«£«7H2O¡£
²½Öè4£º½«²½Öè2¡¢3Öظ´Á½´Î£¬Æ½¾ùÏûºÄK2Cr2O7   ÈÜÒº15.00 mL¡£
¢Ùд³ö²½Öè2Öз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                        £»
¢Ú²úÆ·ÖÐMg(ClO3)2¡¤6H2OµÄÖÊÁ¿·ÖÊýΪ          ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

×ÛºÏÀûÓú£Ë®¿ÉÒÔÖƱ¸Ê³ÑΡ¢´¿¼î¡¢½ðÊôþµÈÎïÖÊ£¬ÆäÁ÷³ÌÈçÏÂͼËùʾ£º

£¨1£©·´Ó¦¢Ù¡«¢ÝÖУ¬ÊôÓÚÑõ»¯»¹Ô­·´Ó¦µÄÊÇ   £¨Ìî±àºÅ£©¡£
£¨2£©Ð´³ö·´Ó¦¢ÚµÄÀë×Ó·½³Ìʽ   ¡£
£¨3£©XÈÜÒºÖеÄÖ÷ÒªÑôÀë×ÓÊÇNa+ºÍ   ¡£
£¨4£©´ÖÑÎÖк¬ÓÐNa2SO4¡¢MgCl2¡¢CaCl2µÈ¿ÉÈÜÐÔÔÓÖÊ£¬ÎªÖƵô¿¾»µÄNaCl¾§Ì壬²Ù×÷ÈçÏ£º
¢ÙÈܽ⣻¢ÚÒÀ´Î¼ÓÈë¹ýÁ¿µÄBaCl2ÈÜÒº¡¢NaOHÈÜÒº¡¢Na2CO3ÈÜÒº£»¢Û  £»¢Ü¼ÓÊÊÁ¿ÑÎË᣻¢Ý    ¡££¨Ç벹ȫȱÉÙµÄʵÑé²½Ö裩
£¨5£©¼ìÑé´¿¼îÑùÆ·ÖÐÊÇ·ñº¬NaClӦѡÓõÄÊÔ¼ÁÊÇ        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉϳ£Óú¬ÉÙÁ¿SiO2¡¢Al2O3µÄ¸õÌú¿ó£¨FeOCr2O3£©Ò±Á¶¸õ£¬¼òÒªÁ÷³ÌÈçÏ£º

£¨1£©Íê³ÉÏÂÁл¯Ñ§·½³Ìʽ£¨ÔÚºáÏßÌîдÎïÖʵĻ¯Ñ§Ê½¼°ÏµÊý£©£º
2FeO¡¤Cr2O3£«4Na2CO3£«7NaNO34Na2CrO4£«Fe2O3£«4CO2£«______________¡£
£¨2£©²Ù×÷¢Ù°üÀ¨¹ýÂËÓëÏ´µÓ£¬ÔÚʵÑéÊÒÖнøÐÐÏ´µÓ³ÁµíµÄ²Ù×÷__________________________________¡£²Ù×÷¢Ú¿ÉÑ¡ÓõÄ×°Ö㨲¿·Ö¼Ð³Ö×°ÖÃÂÔÈ¥£©ÊÇ________£¨ÌîÐòºÅ£©

£¨3£©Ð´³öÄܹ»Íê³É²Ù×÷¢ÛµÄÏà¹Ø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____________________¡£
£¨4£©»¯Ñ§ÐèÑõÁ¿£¨COD£©¿É¶ÈÁ¿Ë®ÔâÊÜÓлúÎïÎÛȾµÄ³Ì¶È¡£ÔÚÇ¿Ëá²¢¼ÓÈȵÄÌõ¼þÏ£¬ÓÃK2Cr2O7×öÇ¿Ñõ»¯¼Á´¦ÀíË®Ñù£¬²¢²â¶¨ÏûºÄµÄK2Cr2O7µÄÁ¿£¬È»ºó»»Ëã³ÉÏ൱ÓÚO2µÄº¬Á¿³ÆΪ»¯Ñ§ÐèÑõÁ¿£¨ÒÔmg/L¼Æ£©¡£»¯Ñ§ÐËȤС×é²â¶¨Ä³Ë®ÑùµÄ»¯Ñ§ÐèÑõÁ¿£¨COD£©¹ý³ÌÈçÏ£º
I.È¡amLË®ÑùÖÃÓÚ׶ÐÎÆ¿£¬¼ÓÈë10£®00mL 0.2500 mol/LµÄK2Cr2O7ÈÜÒº£»
II£®¡­¡­¡£
III£®¼Óָʾ¼Á£¬ÓÃc molµÄÁòËáÑÇÌúï§[ £¨NH4£©2Fe£¨SO4£©]µÎ¶¨£¬ÖÕµãʱÏûºÄb mL£¨´Ë²½ÖèµÄÄ¿µÄÊÇÓÃFe2£«°Ñ¶àÓàµÄCr2O72£­×ª»¯Îª£©Cr3£«¡£
¢ÙIÖÐÁ¿È¡K2Cr2O7ÈÜÒºµÄÒÇÆ÷ÊÇ_____________£»
¢Ú¼ÆËã¸ÃË®ÑùµÄ»¯Ñ§ÐèÑõÁ¿Ê±ÐèÓõ½ÏÂÁйØϵ£ºÒª³ýÈ¥1molCr2O72£­ ÐèÏûºÄ___molFe2£«£¬1molCr2O72£­Ï൱ÓÚ_______ molO2¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖª£ºBaSO4(s) + 4C(s)4CO(g) + BaS(s)¹¤ÒµÉÏÒÔÖؾ§Ê¯¿ó£¨Ö÷Òª³É·ÖBaSO4£¬ÔÓÖÊΪFe2O3¡¢SiO2£©ÎªÔ­ÁÏ£¬Í¨¹ýÏÂÁÐÁ÷³ÌÉú²úÂÈ»¯±µ¾§Ì壨BaCl2¡¤nH2O£©¡£

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²»ÈÜÎïAµÄ»¯Ñ§Ê½ÊÇ_________£»ÈôÔÚʵÑéÊÒ½øÐбºÉÕʱ£¬Ëù²úÉúµÄÆøÌåµÄ´¦Àí·½·¨ÊÇ       
a.ÓÃNaOHÈÜÒºÎüÊÕ       b.ÓÃŨÁòËáÎüÊÕ      c.µãȼ
£¨2£©Óõ¥Î»Ìå»ýÈÜÒºÖÐËùº¬µÄÈÜÖÊÖÊÁ¿ÊýÀ´±íʾµÄŨ¶È½ÐÖÊÁ¿-Ìå»ýŨ¶È£¬¿ÉÒÔÓÃg/L±íʾ,ÏÖÓÃ38%µÄŨÑÎËáÅäÖƺ¬ÈÜÖÊ109.5g/LµÄÏ¡ÑÎËá500mL,ËùÐèÒªµÄ²£Á§ÒÇÆ÷³ýÁ˲£Á§°ô»¹ÓР                          ¡£
£¨3£©³Áµí·´Ó¦ÖÐËù¼ÓµÄÊÔ¼ÁR¿ÉÒÔÊÇÏÂÁÐÊÔ¼ÁÖеĠ               
a.NaOHÈÜÒº    b.BaO¹ÌÌå      c.°±Ë®       d.Éúʯ»Ò
Ö¤Ã÷³ÁµíÒѾ­ÍêÈ«µÄ·½·¨ÊÇ________________________________________________________¡£
£¨4£©Éè¼ÆÒ»¸öʵÑéÈ·¶¨²úÆ·ÂÈ»¯±µ¾§Ì壨BaCl2¡¤nH2O£©ÖеÄnÖµ£¬ÍêÉÆÏÂÁÐʵÑé²½Ö裺
¢Ù³ÆÁ¿ÑùÆ·¢Ú_______ ¢ÛÖÃÓÚ_________£¨ÌîÒÇÆ÷Ãû³Æ£©ÖÐÀäÈ´ ¢Ü³ÆÁ¿ ¢ÝºãÖزÙ×÷¡£
ºãÖزÙ×÷ÊÇÖ¸____________________________________________                   _£»
µÚ¢Û²½ÎïÆ·Ö®ËùÒÔ·ÅÔÚ¸ÃÒÇÆ÷ÖнøÐÐʵÑéµÄÔ­ÒòÊÇ                                  ¡£
£¨5£©½«Öؾ§Ê¯¿óÓë̼ÒÔ¼°ÂÈ»¯¸Æ¹²Í¬±ºÉÕ£¬¿ÉÒÔÖ±½ÓµÃµ½ÂÈ»¯±µ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌΪ
BaSO4+ 4C+CaCl24CO + CaS+ BaCl2¡£ÇëÄãÍêÉÆÏÂÁдӱºÉÕºóµÄ¹ÌÌåÖзÖÀëµÃµ½ÂÈ»¯±µ¾§ÌåµÄʵÑéÁ÷³ÌµÄÉè¼Æ£¨ÒÑÖªÁò»¯¸Æ²»ÈÜÓÚË®£¬Ò×ÈÜÓÚÑÎËᣩ¡£

£¨·½¿òÄÚÌîд²Ù×÷Ãû³Æ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ijͭ¿óʯÖÐÍ­ÔªËغ¬Á¿½ÏµÍ£¬ÇÒº¬ÓÐÌú¡¢Ã¾¡¢¸ÆµÈÔÓÖÊÀë×Ó¡£Ä³Ð¡×éÔÚʵÑéÊÒÖÐÓýþ³ö-ÝÍÈ¡·¨ÖƱ¸ÁòËáÍ­£º

£¨1£©²Ù×÷1µÄÃû³ÆΪ           ¡£²Ù×÷2Óõ½µÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­ºÍ                     
£¨2£©¡°½þ³ö¡±²½ÖèÖУ¬ÎªÌá¸ßÍ­µÄ½þ³öÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÓР                            
£¨3£©±È½Ï²Ù×÷2֮ǰÓë²Ù×÷3Ö®ºóµÄÈÜÒº£¬ËµÃ÷Á½²½ÖèÖ÷ҪĿµÄÊÇ                                                                              ¡£
£¨4£©È¡ÉÙÁ¿ËùµÃÈÜÒºA£¬µÎ¼Ó          £¨ÌîÎïÖÊÃû³Æ£©ÈÜÒººó³ÊºìÉ«£¬ËµÃ÷ÈÜÒºÖдæÔÚFe3+£¬¼ìÑéÈÜÒºÖл¹´æÔÚFe2+µÄ·½·¨ÊÇ                  £¨×¢Ã÷ÊÔ¼Á¡¢ÏÖÏ󣩣¨²»¿¼Âdzý×¢Ã÷ÍâµÄÆäËüÔÓÖʸÉÈÅ£©
£¨5£©Óõζ¨·¨²â¶¨CuSO4¡¤5H2OµÄº¬Á¿¡£È¡a gÊÔÑùÅä³É100 mLÈÜÒº£¬È¡20.00mLÓÃc mol /L µÎ¶¨¼Á(H2Y2¨C£¬µÎ¶¨¼Á²»ÓëÔÓÖÊ·´Ó¦)µÎ¶¨ÖÁÖյ㣬ÏûºÄµÎ¶¨¼ÁbmL.
µÎ¶¨·´Ó¦£ºCu2+ + H2Y2¨CCuY2¨C+ 2H+¡£ÔòCuSO4¡¤5H2OÖÊÁ¿·ÖÊýµÄ±í´ïʽÊÇ       ¡£
£¨6£©ÏÂÁвÙ×÷»áµ¼ÖÂCuSO4¡¤5H2Oº¬Á¿µÄ²â¶¨½á¹ûÆ«¸ßµÄÊÇ_____________¡£
A£®µÎ¶¨ÁÙ½üÖÕµãʱ£¬ÓÃÏ´Æ¿ÖеÄÕôÁóˮϴÏµζ¨¹Ü¼â×ì¿ÚµÄ°ëµÎ±ê×¼ÒºÖÁ׶ÐÎÆ¿ÖÐ
B£®µÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó£¬Ö±½Ó×¢Èë´ý²âÒº£¬È¡20.00mL½øÐеζ¨
C£®µÎ¶¨Ç°£¬µÎ¶¨¹Ü¼â¶ËÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐËÄ×éÈÜÒº£¬²»ÓÃÆäËûÊÔ¼Á¾ÍÄܼø±ð¿ªÀ´µÄÊÇ£¨   £©
¢ÙAgNO3ÈÜÒº¡¢Ï¡°±Ë®
¢ÚNaAlO2  KHCO3  NaCl  NaHSO4
¢ÛHNO3¡¢Na2SO3¡¢Na2SO4¡¢BaCl2
¢ÜAlCl3¡¢NaAlO2
A£®¢ÙB£®¢Ù¢Ú¢ÜC£®¢Ù¢Ú¢ÛD£®È«²¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÖÓÐÈý×éÈÜÒº£º¢ÙÆûÓͺÍÂÈ»¯ÄÆÈÜÒº ¢Ú39£¥µÄÒÒ´¼ÈÜÒº ¢ÛÂÈ»¯Äƺ͵¥ÖÊäåµÄË®ÈÜÒº£¬·ÖÀëÒÔÉϸ÷»ìºÏÒºµÄÕýÈ··½·¨ÒÀ´ÎÊÇ
A£®·ÖÒº¡¢ÝÍÈ¡¡¢ÕôÁóB£®ÝÍÈ¡¡¢ÕôÁó¡¢·ÖÒº
C£®·ÖÒº¡¢ÕôÁó¡¢ÝÍÈ¡D£®ÕôÁó¡¢ÝÍÈ¡¡¢·ÖÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÃÏÂͼËùʾװÖüìÑéÒÒϩʱ²»ÐèÒª³ýÔÓµÄÊÇ(   )

 
ÒÒÏ©µÄÖƱ¸
ÊÔ¼ÁX
ÊÔ¼ÁY
A
CH3CH2BrÓëNaOHÒÒ´¼ÈÜÒº¹²ÈÈ
H2O
KMnO4ËáÐÔÈÜÒº
B
CH3CH2BrÓëNaOHÒÒ´¼ÈÜÒº¹²ÈÈ
H2O
Br2µÄCCl4ÈÜÒº
C
C2H5OHÓëŨH2SO4¼ÓÈÈÖÁ170 ¡æ
NaOHÈÜÒº
KMnO4ËáÐÔÈÜÒº
D
C2H5OHÓëŨH2SO4¼ÓÈÈÖÁ170 ¡æ
NaOHÈÜÒº
Br2µÄCCl4ÈÜÒº
 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸