·ÖÎö £¨1£©¢Ù¸ù¾ÝÍéÌþµÄÃüÃûÔÔò£ºÌ¼Á´×³ÆijÍ飬¿¿½üÖ§Á´°ÑºÅ±à£¬¼òµ¥ÔÚǰͬÏಢ£¬Æä¼äÓ¦»®Ò»¶ÌÏߣ¬£»
¢ÚÖÐ×ÇÒº¬ÓÐ̼̼˫¼üµÄ̼Á´º¬ÓÐ6¸ö̼Ô×Ó£¬¾àÀë̼̼˫¼üºÍ¼×»ù×î½üµÄ̼Ô×ÓΪ1ºÅ̼Ô×Ó£¬ÔÚ2ºÅ¡¢4ºÅ̼Ô×ÓÉϺ¬ÓÐ̼̼˫¼ü£¬Îª¶þÏ©Ìþ£»
¢ÛijȲÌþÓëH2ÍêÈ«¼Ó³ÉºóµÄ²úÎïΪCH3CH2CH£¨CH3£©C£¨CH3£©3£¬ÏàÁÚ̼Ô×ÓÉÏÿ¸ö̼Ô×ÓÈ¥µô2¸öHÔ×ÓËùµÃÎïÖʾÍÊǸÃȲÌþ£»
¢Ü¸ù¾Ý̼Ô×ӳɼü·½Ê½È·¶¨·Ö×Óʽ£¬¸ÃÎïÖÊÖÐÓм¸ÖÖÇâÔ×Ó£¬ÔòÆäÒ»ÂÈ´úÎïµÄͬ·ÖÒì¹¹Ìå¾ÍÓм¸ÖÖ£»
¢ÝÓÉÓÚ-CH3ºÍ-Br¾ùÖ»ÓÐÒ»Ìõ°ë¼ü£¬¹ÊÖ»ÄÜÁ¬ÔÚ´ËÓлúÎïµÄÍâΧ£¬ÔòÁ½¸öÔÚ´ËÓлúÎï½á¹¹µÄÄÚ²¿£»
£¨2£©Ä³ÓлúÎïAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª74£¬ÇÒºìÍâ¹âÆ×ͼÈçͼ2£¬AÖк¬ÓжԳƵļ׻ù¡¢¶Ô³ÆµÄÑǼ׻ù¡¢ÃѼü£¬
£»
£¨3£©1Ìå»ýijÌþµÄÕôÆøÍêȫȼÉÕÉú³ÉµÄCO2±ÈÉú³ÉµÄË®ÕôÆøÉÙ1Ìå»ý£¨ÔÚͬÎÂͬѹϲⶨ£©£¬ËµÃ÷HÔ×Ó¸öÊý±ÈCÔ×Ó¸öÊý¶à2£¬¼îʯ»ÒÎüÊÕ¶þÑõ»¯Ì¼ºÍË®£¬ÉèÌþµÄ·Ö×ÓʽΪCxH2x+2£¬0.1mol¸ÃÌþÍêȫȼÉÕÉú³É0.1xmolCO2¡¢£¨x+1£©¡Á0.1molË®£¬¶þÑõ»¯Ì¼ºÍË®µÄÖÊÁ¿=0.1xmol¡Á44g/mol+£¨x+1£©¡Á0.1mol¡Á18g/mol=39g£¬x=6£¬
¾Ý´ËÅжϸÃÌþµÄ·Ö×Óʽ£»Èô¸ÃÌþµÄÒ»ÂÈ´úÎïÓÐ3ÖÖ£¬ËµÃ÷¸ÃÌþ·Ö×ÓÖк¬ÓÐ3ÖÖÇâÔ×Ó£¬¾Ý´ËÅжϽṹ¼òʽ£®
½â´ð ½â£º£¨1£©¢ÙCH3CH£¨CH3£©CH£¨CH2CH3£©C£¨CH3£©3ÖÐ×ÇÒº¬ÓÐÖ§Á´×î¶àµÄ̼Á´º¬ÓÐ5¸ö̼Ô×Ó£¬¾àÀëÈ¡´ú»ù×î½üÇÒ×î¶àµÄ̼Ô×ÓΪ1ºÅ̼Ô×Ó£¬ËùÒÔ¸ÃÍéÌþÃû³ÆÊÇ2£¬2£¬4-Èý¼×»ù-3-ÒÒ»ùÎìÍ飬¹Ê´ð°¸Îª£º2£¬2£¬4-Èý¼×»ù-3-ÒÒ»ùÎìÍ飻
¢ÚÖÐ×ÇÒº¬ÓÐ̼̼˫¼üµÄ̼Á´º¬ÓÐ6¸ö̼Ô×Ó£¬¾àÀë̼̼˫¼üºÍ¼×»ù×î½üµÄ̼Ô×ÓΪ1ºÅ̼Ô×Ó£¬ÔÚ2ºÅ¡¢4ºÅ̼Ô×ÓÉϺ¬ÓÐ̼̼˫¼ü£¬ËùÒÔÆäÃû³ÆÊÇ2-¼×»ù-2£¬4-¼º¶þÏ©£¬¹Ê´ð°¸Îª£º2-¼×»ù-2£¬4-¼º¶þÏ©£»
¢ÛijȲÌþÓëH2ÍêÈ«¼Ó³ÉºóµÄ²úÎïΪCH3CH2CH£¨CH3£©C£¨CH3£©3£¬ÏàÁÚ̼Ô×ÓÉÏÿ¸ö̼Ô×ÓÈ¥µô2¸öHÔ×ÓËùµÃÎïÖʾÍÊǸÃȲÌþ£¬Ôò¸ÃȲÌþ½á¹¹¼òʽΪHC¡ÔCCH£¨CH3£©C£¨CH3£©3£¬Ãû³ÆΪ3£¬4£¬4-Èý¼×»ù-1-ÎìȲ£¬
¹Ê´ð°¸Îª£ºHC¡ÔCCH£¨CH3£©C£¨CH3£©3£»3£¬4£¬4-Èý¼×»ù-1-ÎìȲ£»
¢Ü¸ù¾Ý̼Ô×ӳɼü·½Ê½È·¶¨·Ö×ÓʽΪC9H8£¬¸Ã·Ö×ÓÖк¬ÓÐ4ÀàÇâÔ×Ó£¬ËùÒÔÆäÒ»ÂÈ´úÎïµÄͬ·ÖÒì¹¹ÌåµÄÊýĿΪ4£¬
¹Ê´ð°¸Îª£ºC9H8£»4£»
¢ÝÓÉÓÚ-CH3ºÍ-Br¾ùÖ»ÓÐÒ»Ìõ°ë¼ü£¬¹ÊÖ»ÄÜÁ¬ÔÚ´ËÓлúÎïµÄÍâΧ£¬ÔòÁ½¸öÔÚ´ËÓлúÎï½á¹¹µÄÄÚ²¿Çұ˴ËÏàÁ¬£¬Á¬½ÓºóµÃµ½½á¹¹£º£¬ËùµÃµÄ´Ë½á¹¹ÖÐËÄÌõ°ë¼üÍêÈ«µÈ¼Û£¬¹Ê½«Ò»¸ö-BrºÍÈý¸ö-CH3ÍùÉÏÁ¬£¬ËùµÃµÄ½á¹¹Ö»ÓÐÒ»ÖÖ£¬¼´£º£¨CH3£©2CHCHBrCH3£¬¹ÊËùµÃAµÄ½á¹¹¼òʽΪ£¨CH3£©2CHCHBrCH3£¬¹Ê´ð°¸Îª£º£¨CH3£©2CHCHBrCH3£»
£¨2£©ÖÊÆ×ͼ¿ÉÖª¸ÃÓлúÎïµÄÏà¶ÔÔ×ÓÖÊÁ¿Îª74£¬ºìÍâ¹âÆ×ͼÏÔʾ´æÔڶԳƵļ׻ù¡¢¶Ô³ÆµÄÑǼ׻ùºÍÃѼü¿ÉµÃ·Ö×ӵĽṹ¼òʽΪ£ºCH3CH2OCH2CH3£¬
¹Ê´ð°¸Îª£ºCH3CH2OCH2CH3£»
£¨3£©1Ìå»ýijÌþµÄÕôÆøÍêȫȼÉÕÉú³ÉµÄCO2±ÈÉú³ÉµÄË®ÕôÆøÉÙ1Ìå»ý£¨ÔÚͬÎÂͬѹϲⶨ£©£¬ËµÃ÷HÔ×Ó¸öÊý±ÈCÔ×Ó¸öÊý¶à2£¬¼îʯ»ÒÎüÊÕ¶þÑõ»¯Ì¼ºÍË®£¬ÉèÌþµÄ·Ö×ÓʽΪCxH2x+2£¬0.1mol¸ÃÌþÍêȫȼÉÕÉú³É0.1xmolCO2¡¢£¨x+1£©¡Á0.1molË®£¬¶þÑõ»¯Ì¼ºÍË®µÄÖÊÁ¿=0.1xmol¡Á44g/mol+£¨x+1£©¡Á0.1mol¡Á18g/mol=39g£¬x=6£¬
¾Ý´ËÅжϸÃÌþµÄ·Ö×ÓʽC6H14£»Èô¸ÃÌþµÄÒ»ÂÈ´úÎïÓÐ3ÖÖ£¬ËµÃ÷¸ÃÌþ·Ö×ÓÖк¬ÓÐ3ÖÖÇâÔ×Ó£¬ÔòÆä½á¹¹¼òʽΪCH3CH2C£¨CH3£©3£¬
¹Ê´ð°¸Îª£ºC6H14£»CH3CH2C£¨CH3£©3£®
µãÆÀ ±¾Ì⿼²é½Ï×ۺϣ¬Éæ¼°ÓлúÎï·Ö×ÓʽµÄÈ·¶¨¡¢ÓлúÎïÃüÃû¡¢Í¬·ÖÒì¹¹ÌåÖÖÀàÅжϵÈ֪ʶµã£¬²àÖØ¿¼²éѧÉú·ÖÎö¼ÆËãÄÜÁ¦£¬Ã÷È·ÓлúÎï½á¹¹Ìص㡢ÆäȼÉÕ¶¨ÂÉ¡¢ÃüÃûÔÔòÊǽⱾÌâ¹Ø¼ü£¬×¢Ò⣺ϩÌþÖÐÖ÷Á´²»Ò»¶¨º¬ÓÐ̼Ô×Ó¸öÊý×î¶à£¬µ«Ö÷Á´ÖÐÒ»¶¨º¬ÓÐÓйÙÄÜÍŵÄ̼Ô×Ó£¬ÌâÄ¿ÄѶȲ»´ó£®
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | Na2 C03ÈÜÒº³Ê¼îÐÔ£ºC032-+2H20?H2C03+20H- | |
B£® | NaHC03ÈÜÒºÖмӹýÁ¿Ca£¨OH£©2ÈÜÒº£ºCa2++20H-+2HC03-¨TCaC03++C032-+2H2O | |
C£® | Ư°×·ÛÈÜÒºÖÐͨÈë×ãÁ¿¶þÑõ»¯ÁòÆøÌ壺ClO-+SO2+H2O¨THC1O+HSO3- | |
D£® | ÏòNaAl02ÈÜÒºÖÐͨÈë¹ýÁ¿C02£ºAlO2-+CO2+2H2O¨TAl£¨OH£©3+HCO3- |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ³£ÎÂÏÂŨÁòËáÄÜʹÂÁ·¢Éú¶Û»¯£¬¿ÉÔÚ³£ÎÂÏÂÓÃÂÁÖÆÖü¹ÞÖüÔËŨÁòËá | |
B£® | ÌúÔÚ´¿ÑõÖÐȼÉÕ»ò¸ßÎÂϺÍË®ÕôÆø·´Ó¦¾ùÄܵõ½Fe3O4 | |
C£® | ½ðÊôÂÁÅÅÔÚ½ðÊô»î¶¯ÐÔ˳Ðò±íÖÐÇâÔªËصÄÇ°Ã棬ÂÁÓëÇ¿Ëá·´Ó¦Ò»¶¨·Å³öÇâÆø | |
D£® | ½«Ìú·Û¼ÓÈëFeCl3¡¢CuCl2»ìºÏÈÜÒºÖУ¬³ä·Ö·´Ó¦ºóÊ£ÓàµÄ¹ÌÌåÖв»Ò»¶¨ÓÐÌú |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ½ðÊôÄÆÓëË®·´Ó¦£ºNa+2H2O¨TNa++2OH-+H2¡ü | |
B£® | Óô×Ëá³ýȥˮ¹¸ÖеÄ̼Ëá¸Æ£ºCaCO3+2H+¨TCa2++H2O+CO2¡ü | |
C£® | ÁòËáÇâÄÆÈÜÒºÓëÇâÑõ»¯±µÈÜҺǡºÃ·´Ó¦³ÊÖÐÐÔ£º2H++SO42-+Ba2++2OH-¨T2H2O+BaSO4¡ý | |
D£® | µç½â±¥ºÍMgCl2ÈÜÒº£º2Cl-+2H2O$\frac{\underline{\;µç½â\;}}{\;}$2OH-+H2¡ü+Cl2¡ü |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | 0.25 | B£® | 0.1 | C£® | 0.5 | D£® | 0.05 |
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com