¡¾ÌâÄ¿¡¿ôǼ×Ï㶹ËØÊÇÒ»ÖÖÖÎÁƵ¨½áʯµÄÒ©ÎºÏ³É·ÏßÈçÏÂͼËùʾ£º
ÒÑÖª£º
RCOOR'+R'OH RCOOR'+ R'OH£¨R¡¢R'¡¢R'´ú±íÌþ»ù£©
£¨1£©AÊôÓÚ·¼ÏãÌþ£¬Æä½á¹¹¼òʽÊÇ______________________¡£BÖÐËùº¬µÄ¹ÙÄÜÍÅÊÇ________________¡£
£¨2£©C¡úDµÄ·´Ó¦ÀàÐÍÊÇ___________________¡£
£¨3£©EÊôÓÚÖ¬Àà¡£½öÒÔÒÒ´¼ÎªÓлúÔÁÏ£¬Ñ¡ÓñØÒªµÄÎÞ»úÊÔ¼ÁºÏ³ÉE£¬Ð´³öÓйػ¯Ñ§·½³Ìʽ£º______________________________¡£
£¨4£©ÒÑÖª£º2E F+C2H5OH¡£FËùº¬¹ÙÄÜÍÅÓкÍ___________¡£
£¨5£©ÒÔDºÍFΪÔÁϺϳÉôǼ×Ï㶹ËØ·ÖΪÈý²½·´Ó¦£¬Ð´³öÓйػ¯ºÏÎïµÄ½á¹¹¼òʽ£º
____________________________________
¡¾´ð°¸¡¿ Ïõ»ù È¡´ú·´Ó¦ 2C2H5OH+O22CH3CHO+2H2O 2CH3CHO+O22CH3COOH C2H5OH£«CH3COOHCH3COOC2H5+H2O
¡¾½âÎö¡¿£¨1£©AÊÇ·¼ÏãÌþ£¬¸ù¾ÝCµÄ·Ö×Óʽ¿ÉÍƳöAÖÐÓ¦º¬6¸ö̼Ô×Ó£¬¼´AΪ±½£¬½á¹¹¼òʽΪ¡£A¡úB·¢ÉúÈ¡´ú·´Ó¦£¬ÒýÈë£NO2£¬Òò´ËBÖйÙÄÜÍÅÊÇÏõ»ù¡££¨2£©¸ù¾ÝCºÍDµÄ·Ö×Óʽ£¬C¡úDÊÇÓÃÁ½¸öôÇ»ùÈ¡´ú°±»ùµÄλÖ㬷¢ÉúµÄ·´Ó¦ÊÇÈ¡´ú·´Ó¦¡££¨3£©EÊôÓÚõ¥£¬Ò»¶¨Ìõ¼þÏÂÉú³ÉÒÒ´¼ºÍF£¬ÔòEÊÇÒÒËáÒÒõ¥¡£ÓÃÒÒ´¼ÖÆÈ¡ÒÒËáÒÒõ¥¿ÉÒÔÓò¿·ÖÒÒ´¼Í¨¹ýÑõ»¯ÖƱ¸ÒÒËᣬÔÙºÍÁíÒ»²¿·ÖÒÒ´¼Í¨¹ýõ¥»¯·´Ó¦µÃµ½ÒÒËáÒÒõ¥¡£·¢ÉúµÄ·´Ó¦ÊÇ2CH3CH2OH+O2 2CH3CHO+2H2O£¬2CH3CHO+O22CH3COOH£¬CH3CH2OH£«CH3COOHCH3COOCH2CH3£«H2O¡££¨4£©½áºÏÒÑ֪ת»¯£¬¸ù¾ÝÔ×ÓÊغ㣬ÍƳöFµÄ·Ö×ÓʽΪC6H10O3£¬¸ù¾ÝÐÅÏ¢£¬ÒÔ¼°ôǼ×Ï㶹ËصĽṹ¼òʽ£¬ÍƳöFµÄ½á¹¹¼òʽΪCH3CH2OOCCH2COCH3£¬³ýº¬ÓУ¬»¹º¬ÓС££¨5£©¸ù¾ÝôÇ»ùÏ㶹ËصĽṹ¼òʽ£¬ÒÔ¼°£¨2£©µÄ·ÖÎö£¬C¡¢DµÄ°±»ù¡¢ôÇ»ù·Ö±ðÔÚ±½»·µÄ¼ä룬·Ö±ðΪ¡¢¡£FµÄ½á¹¹¼òʽΪCH3CH2OOCCH2COCH3£¬FÓëD·¢ÉúÒÑÖªµÚÒ»¸ö·´Ó¦Éú³ÉÖмä²úÎï1£¬Öмä²úÎï1µÄ½á¹¹¼òʽΪ£º£¬ÔÙ·¢ÉúÒÑÖª¢ÚµÄ·´Ó¦Éú³ÉÖмä²úÎï2£º£¬È»ºó·¢ÉúÏûÈ¥·´Ó¦ÍÑË®Éú³ÉôǼ×Ï㶹ËØ¡£
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£º
¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡¡¡£©
A. ÓÃÍ×÷Ó¡Ë¢µç·ÊÇÀûÓÃÁË͵Ļ¹ÔÐÔ
B. ÀûÓÃÂÁÈÈ·´Ó¦À´º¸½Ó¸Ö¹ì
C. Ñòë¡¢²ÏË¿¡¢ÃÞ»¨µÄÖ÷Òª³É·Ö¶¼ÊÇÏËάËØ
D. ÉúÌúºÍÆÕͨ¸Ö¶¼ÊÇÌú̼ºÏ½ð
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑÖª£º2Zn(s)+O2(g)2ZnO(s) ¦¤H=701.0 kJ¡¤mol1
2Hg(l)+O2(g)2HgO(s) ¦¤H=181.6 kJ¡¤mol1
Ôò·´Ó¦Zn(s)+HgO(s)ZnO(s)+Hg(l)µÄ¦¤HΪ
A£®+519.4 kJ¡¤mol1 B£®+259.7 kJ¡¤mol1 C£®259.7 kJ¡¤mol1 D£®519.4 kJ¡¤mol1
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑÖªKMnO4ÓëŨÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2KMnO4+16HCl(Ũ)£½2MnCl2+8H2O+2KCl+5Cl2¡ü£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________________________________¡£
£¨2£©±»Ñõ»¯µÄHClÕ¼HCl×ÜÁ¿µÄ__________________________¡£
£¨3£©µ±±ê×¼×´¿öÏÂÓÐ11.2LÂÈÆøÉú³Éʱ£¬¸Ã·´Ó¦×ªÒƵĵç×ÓÊýΪ________________£¨ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£©¡£
£¨4£©15.8 g KMnO4Óë100 mL 12 mol/LŨÑÎËáÍêÈ«·´Ó¦£¨¼ÙÉèHClÎÞ»Ó·¢£©£¬ÔÚ·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄAgNO3ÈÜÒº£¬¿ÉÉú³É____________________g³Áµí¡££¨ÒªÇóд³ö¼ÆËã¹ý³Ì£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¢ñ¡¢ÓÐÏÂÁÐÎåÖÖÌþ£»¢Ù¢Ú¢ÛÒÒÍé ¢ÜÎìÍé ¢Ý£¬ÆäÖл¥ÎªÍ¬·ÖÒì¹¹ÌåµÄÊÇ____________(ÌîÐòºÅ)£¬¢ÚÓë¢ÝÖ®¼äµÄ¹ØϵΪ_____________£¬
¢Ú¢Û¢Ü¢ÝËÄÖÖÎïÖÊ°´ËüÃǵķеãÓɸߵ½µÍµÄ˳ÐòÅÅÁÐÕýÈ·µÄÊÇ_____________(ÌîÐòºÅ)
¢ò¡¢Ä³ÌþAÊÇÓлú»¯Ñ§¹¤ÒµµÄ»ù±¾ÔÁÏ£¬Æä²úÁ¿¿ÉÒÔÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½£¬AÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÈçͼËùʾµÄת»¯£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣻
(1)д³öAµÄµç×Óʽ________£¬EµÄ½á¹¹¼òʽΪ______________
(2)д³öÏÂÁз´Ó¦»¯Ñ§·½³Ìʽ£¬²¢×¢Ã÷¢Û¢Ý·´Ó¦ÀàÐÍ
¢Ú _____________________________________________
¢Û _____________________ £¬·´Ó¦ÀàÐÍ________;
¢Ý _____________________ £¬·´Ó¦ÀàÐÍ________;
(3)³ýÈ¥BÖлìÓеÄÉÙÁ¿ÔÓÖÊA£¬ËùÓõÄÊÔ¼ÁΪ___________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿±ûÏ©ÊÇʯÓÍ»¯¹¤µÄÖØÒªÔÁÏ£¬Ò»¶¨Ìõ¼þÏ¿ɷ¢ÉúÏÂÁÐת»¯
(1)AµÄ½á¹¹¼òʽΪ____________
(2)DÓë×ãÁ¿ÒÒ´¼·´Ó¦Éú³ÉEµÄ»¯Ñ§·½³ÌʽΪ_____________________________________________
(3) Óë×ãÁ¿NaOHË®ÈÜÒº·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________
(4)BÓжàÖÖͬ·ÖÒì¹¹Ì壬д³öÆäÖмÈÄÜ·¢ÉúÒø¾µ·´Ó¦ÓÖÄÜ·¢Éúõ¥»¯·´Ó¦µÄ2ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ_________________________________________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¡°84Ïû¶¾Òº¡±ÄÜÓÐЧɱÃð¼×ÐÍH1N1µÈ²¡¶¾£¬Ä³Í¬Ñ§¹ºÂòÁËһƿ¡°Íþ¶ʿ¡±ÅÆ¡°84Ïû¶¾Òº¡±£¬²¢²éÔÄÏà¹Ø×ÊÁϺÍÏû¶¾Òº°üװ˵Ã÷µÃµ½ÈçÏÂÐÅÏ¢£º¡°84Ïû¶¾Òº¡±£ºº¬25%NaClO¡¢1 000 mL¡¢ÃܶÈ1.19 g¡¤cm£3£¬Ï¡ÊÍ100±¶(Ìå»ý±È)ºóʹÓá£Í¬Ñ§²ÎÔÄ¡°Íþ¶ʿ¡±ÅÆ¡°84Ïû¶¾Òº¡±µÄÅä·½£¬ÓûÓÃNaClO¹ÌÌåÅäÖÆ480 mLº¬25% NaClOµÄÏû¶¾Òº¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ ( )
A. ÅäÖÆʱÐèÒªµÄ²£Á§ÒÇÆ÷³ýÉÕ±¡¢Á¿Í²¡¢ÈÝÁ¿Æ¿Í⻹ÐèÒ»ÖÖ²£Á§ÒÇÆ÷
B. ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»ºó£¬Ó¦ºæ¸É²ÅÄÜÓÃÓÚÈÜÒºµÄÅäÖÆ
C. ¶¨ÈÝʱˮ¼Ó¶àÁËÓ¦¸ÃÓÃÂËÖ½Îü³ö¡£
D. ÐèÒª³ÆÁ¿µÄNaClO¹ÌÌåÖÊÁ¿Îª149 g
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑÖªA¡¢B¡¢C¡¢D¡¢E¡¢FΪԪËØÖÜÆÚ±íÖÐÇ°ËÄÖÜÆÚÇÒÔ×ÓÐòÊýÒÀ´ÎÔö´óµÄÁùÖÖÔªËØ.ÆäÖÐA¡¢B¡¢C¡¢DºËµçºÉÊýÖ®ºÍΪ36£¬A¡¢CÔ×ÓµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍµÈÓÚBÔ×Ó µÄ´ÎÍâ²ãµç×ÓÊý£¬DÔ×ÓÖÊ×ÓÊýΪBÔ×ÓÖÊ×ÓÊýµÄÁ½±¶£¬EÔªËØËùÔÚÖ÷×å¾ùΪ½ðÊô£¬FµÄ¼Ûµç×ÓÊýÓëCµÄºËµçºÉÊýÏàµÈ¡£
£¨1£©ÏÂÁйØÓÚÉÏÊö¼¸ÖÖÔªËصÄ˵·¨ÕýÈ·µÄÊÇ_________¡£
a.B¡¢C¡¢DµÄÔ×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºD >C >B
b.E¡¢FµÄ×îÍâ²ãµç×ÓÊýÏàµÈ
c.A¡¢B¡¢C¡¢DËÄÖÖÔªËØÖе縺ÐԺ͵ÚÒ»µçÀëÄÜ×î´óµÄ¾ùΪB
d.BÓëCÐγɵĻ¯ºÏÎïÖпÉÄܺ¬ÓзǼ«ÐÔ¼ü
e.A¡¢C¡¢F¶¼Î»ÓÚÖÜÆÚ±íµÄ3Çø
£¨2£©Bµ¥ÖÊÓÐÁ½ÖÖͬËØÒìÐÎÌå.ÆäÖÐÔÚË®ÖÐÈܽâ¶È½Ï´óµÄÊÇ_______£¨Ìѧʽ)¡£
£¨3£©EA2ºÍA2BÈÛµã½Ï¸ßµÄÊÇ _______(Ìѧʽ£©£¬ÔÒòÊÇ_________¡£
£¨4£©DÓëB¿ÉÒÔÐγÉÁ½ÖÖ·Ö×Ó£¬ÆäÖÐDB2·Ö×ÓÖÐÐÄÔ×ÓµÄÔÓ»¯ÀàÐÍÊÇ____¡£ÏÂÁзÖ×Ó»òÀë×ÓÖÐÓëDB3½á¹¹ÏàËƵÄÊÇ____________¡£
a. NH3 b. SO32- c.NO3- d.PCl3
£¨5£©ÒÑÖªB¡¢FÄÜÐγÉÁ½ÖÖ»¯ºÏÎÆ侧°ûÈçÏÂͼËùʾ.Ôò¸ßÎÂʱ¼×Ò×ת»¯ÎªÒÒµÄÔÒòΪ ___________¡£ÈôÒÒ¾§ÌåÃܶÈΪpg/cm3£¬ÔòÒÒ¾§°ûµÄ¾§°û±ß³¤a =________nm(Óú¬PºÍNAµÄʽ×Ó±íʾ£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ä³Æø̬²»±¥ºÍÁ´ÌþCnHmÔÚÒ»¶¨Ìõ¼þÏÂÓëH2¼Ó³ÉΪCnHm+x£¬È¡CnHmºÍH2»ìºÏÆøÌå¹²60mL½øÐÐʵÑ飬·¢ÏÖËæ»ìºÏÆøÖÐH2ËùÕ¼Ìå»ýµÄ±ä»¯£¬·´Ó¦ºóµÃµ½µÄÆøÌå×ÜÌå»ýÊýÒ²²»Í¬£¬·´Ó¦Ç°»ìºÏÆøÌåÖÐH2ËùÕ¼µÄÌå»ýV(H2)ºÍ·´Ó¦ºóÆøÌå×ÜÌå»ýV(·´Ó¦ºó×Ü)µÄ¹ØϵÈçͼËùʾ(ÆøÌåÌå»ý¾ùÔÚͬÎÂͬѹϲⶨ)£®ÓÉ´Ë¿ÉÖªxµÄÊýֵΪ
A. 4 B. 3 C. 2 D. 1
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