¡¾ÌâÄ¿¡¿Ê¯ÓÍÊǹ¤ÒµµÄѪҺ£¬ÓëÎÒÃǵÄÉú²ú¡¢Éú»îϢϢÏà¹Ø¡£

£¨1£©ÒÒÏ©ÊÇʯÓÍ»¯¹¤ÖØÒªµÄ»ù´¡Ô­ÁÏ¡£ д³öÒÒÏ©µÄµç×Óʽ____________

£¨2£©·Ö×ÓʽΪC5H12µÄijÌþ£¬·Ö×ÓÖк¬ÓÐ4¸ö¼×»ù£¬¸ÃÌþµÄ½á¹¹¼òʽΪ_______________________¡£

£¨3£©ÓëÒÒÏ©»¥ÎªÍ¬ÏµÎïµÄÊÇ_______ ¡£(Ñ¡Ìî±àºÅ)

a. CH3CH=CH2 b. CH2=CHCH=CH2

c. CH¡ÔCH d. CH3CH3

£¨4£©¾ÛÒÒÏ©°²È«ÎÞ¶¾£¬¿ÉÓÃÓÚÖÆʳƷ°ü×°´ü¡£¾ÛÒÒÏ©µÄ½á¹¹¼òʽΪ ________________ ¡£

£¨5£©ÒÒȲÈý¾Û¿ÉµÃµ½±½»ò¶þÒÒÏ©»ùÒÒȲ (CH2=CH-C-C-CH=CH2) ¡£¼ø±ð±½ºÍ¶þÒÒÏ©»ùÒÒȲ¿ÉÓõÄÊÔ¼ÁÊÇ__________________¡£

¡¾´ð°¸¡¿ÂÔ a äåË®»òËáÐÔ¸ßÃÌËá¼ØÈÜÒº

¡¾½âÎö¡¿

¸ù¾ÝÒÒϩΪ¹²¼Û»¯ºÏÎ¼°CÔ­×ÓºÍHÔ­×Ó×îÍâ²ã¿É´ïµ½µÄµç×ÓÊýд³öÒÒÏ©µÄµç×Óʽ£»

¸ù¾ÝÌþµÄ·Ö×ÓʽºÍº¬¼×»ùµÄ¸öÊýд³ö½á¹¹¼òʽ£»

¸ù¾ÝͬϵÎïÐÔÖÊÀàËÆ£¬·Ö×ÓʽÏà²în¸öCH2£¬½áºÏÑ¡ÏîÖÐÎïÖʵĹÙÄÜÍÅÅжϣ»

¸ù¾Ý¾ÛÒÒÏ©ÊÇÓÉÒÒϩͨ¹ý¼Ó¾ÛµÃµ½µÄд³ö½á¹¹¼òʽ£»

ÕÒ³öÓë±½²»·´Ó¦£¬Óë¶þÒÒÏ©»ùÒÒȲ·´Ó¦ÇÒÓÐÑÕÉ«µÄÎïÖʼ´¿É¼ø±ð´ËÁ½ÖÖÎïÖÊ¡£

£¨1£©ÒÒϩΪ¹²¼Û»¯ºÏÎCÔ­×Ó×îÍâ²ã¿É´ïµ½8¸öµç×Ó£¬HÔ­×Ó×îÍâ²ã¿É´ïµ½2¸öµç×Ó£¬ÒÒÏ©ÕýÈ·µÄµç×ÓʽΪ£º£»

£¨2£©¸ù¾ÝÌþµÄ·Ö×ÓʽºÍº¬¼×»ùµÄ¸öÊý¿ÉÖª£¬¸ÃÌþÖ»ÄÜΪÐÂÎìÍ飬½á¹¹¼òʽΪ£»

£¨3£©¸ù¾ÝͬϵÎïµÄ¶¨Òå¿ÉÖªÓëÒÒÏ©»¥ÎªÍ¬ÏµÎïµÄÖ»ÄÜÊÇa£»

£¨4£©¾ÛÒÒϩΪÒÒϩͨ¹ý¼Ó¾Û·´Ó¦Éú³ÉµÄ£¬ÆäÕýÈ·µÄ½á¹¹¼òʽΪ£º£»

£¨5£©¸ù¾Ý±½²»ÓëäåË®£¨»ò¸ßÃÌËá¼ØÈÜÒº£©·´Ó¦¶ø¶þÒÒÏ©»ùÒÒȲ¿ÉÓëäåË®£¨»ò¸ßÃÌËá¼ØÈÜÒº£©·´Ó¦£¬Ê¹äåË®£¨»ò¸ßÃÌËá¼ØÈÜÒº£©ÍÊÉ«¿ÉÖª¿ÉÓÃäåË®£¨»ò¸ßÃÌËá¼ØÈÜÒº£©¼ø±ð±½ºÍ¶þÒÒÏ©»ùÒÒȲ£»ÆäÖн«äåË®£¨»ò¸ßÃÌËá¼ØÈÜÒº£©¼ÓÈëµ½±½ÖÐʱ£¬Äܹ۲쵽µÄÏÖÏóÊÇÈÜÒº·Ö²ã£¬±½²ãÔÚÉÏ£¬Ë®ÔÚÏ£¬Õñµ´£¬¾²Öú󣬱½²ã³Êºì×ØÉ«£¬Ë®²ãÑÕÉ«±ädz£¨»òÈÜÒº·Ö²ã£¬±½²ãÔÚÉÏ£¬Ë®ÔÚÏ£¬Õñµ´£¬¾²Öú󣬱½²ãÎÞÉ«£¬Ë®²ãÈÔ³Ê×ÏÉ«£©¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ

A.Ôھƾ«µÆ¼ÓÈÈÌõ¼þÏ£¬Na2CO3¡¢NaHCO3¹ÌÌ嶼ÄÜ·¢Éú·Ö½â

B.Cl2ÊÇÒ»ÖÖÓж¾ÆøÌ壬²»¿ÉÓÃÓÚ×ÔÀ´Ë®µÄɱ¾úÏû¶¾

C.SiO2¼ÈÄܺÍNaOHÈÜÒº·´Ó¦ÓÖÄܺÍÇâ·úËá·´Ó¦£¬ËùÒÔÊÇÁ½ÐÔÑõ»¯Îï

D.Na2SiO3Ë®ÈÜÒºË׳ÆË®²£Á§£¬ÊÇÖƱ¸¹è½ººÍľ²Ä·À»ð¼ÁµÈµÄÔ­ÁÏ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª·´Ó¦£º2SO2(g)£«O2(g)2SO3(g)¦¤H<0¡£Ä³Î¶ÈÏ£¬½« 2 mol SO2 ºÍ 1 mol O2 ÖÃÓÚ 10L ÃܱÕÈÝÆ÷ÖУ¬·´Ó¦´ïƽºâºó£¬SO2 µÄƽºâת»¯ÂÊ(¦Á)ÓëÌåϵ×Üѹǿ(p)µÄ¹ØϵÈçͼ¼×Ëùʾ¡£ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

¼×ÒÒ±û

A.ÓÉͼ¼×Íƶϣ¬B µã SO3µÄƽºâŨ¶ÈΪ 0.3molL1

B.ÔÚͼ¼×ÖУ¬ÔÚ´ËζÈÏ£¬C µã ¦Ô Õý£¼¦Ô Äæ

C.´ïµ½Æ½ºâºó£¬±£³ÖÌå»ý²»±ä£¬³äÈ뺤Æø£¬Ñ¹Ç¿Ôö´ó£¬Ôò·´Ó¦ËÙÂʱ仯ͼÏñ¿ÉÒÔÓÃͼÒÒ±íʾ

D.ѹǿΪ 0.50 MPa ʱ£¬²»Í¬Î¶ÈÏ SO2 µÄƽºâת»¯ÂÊÓëʱ¼ä¹ØϵÈçͼ±û£¬Ôò T2>T1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»¯Ñ§ÓëÉú»îÃÜÇÐÏà¹Ø¡£

I.K2Cr2O7ÔøÓÃÓÚ¼ì²â˾»úÊÇ·ñ¾Æºó¼ÝÊ»£ºCr2O72-(³ÈÉ«)+CH3CH2OH¡úCr3+(ÂÌÉ«)+CH3COOH(δÅäƽ£©

£¨1£©»ù̬CrÔ­×ӵļ۵ç×Ó¹ìµÀ±í´ïʽΪ__¡£

£¨2£©CH3COOH·Ö×ÓÖÐËùº¬ÔªËصĵ縺ÐÔÓÉ´óµ½Ð¡µÄ˳ÐòΪ__£¬Ì¼Ô­×ӵĹìµÀÔÓ»¯ÀàÐÍΪ__£¬Ëùº¬¦Ò¼üÓë¦Ð¼üµÄÊýÄ¿Ö®±ÈΪ__¡£

£¨3£©ÒÑÖªCr3+µÈ¹ý¶ÉÔªËØË®ºÏÀë×ÓµÄÑÕÉ«Èç±íËùʾ£º

Àë×Ó

Sc3+

Cr3+

Fe2+

Zn2+

Ë®ºÏÀë×ÓµÄÑÕÉ«

ÎÞÉ«

ÂÌÉ«

dzÂÌÉ«

ÎÞÉ«

Çë¸ù¾ÝÔ­×ӽṹÍƲâSc3+¡¢Zn2+µÄË®ºÏÀë×ÓΪÎÞÉ«µÄÔ­ÒòΪ__¡£

II.ZnCl2ŨÈÜÒº³£ÓÃÓÚ³ýÈ¥½ðÊô±íÃæµÄÑõ»¯ÎÀýÈçÓëFeO·´Ó¦¿ÉµÃFe[Zn(OH)Cl2]2ÈÜÒº¡£

£¨4£©Fe[Zn(OH)Cl2]2ÈÜÒºÖв»´æÔÚµÄ΢Á£¼ä×÷ÓÃÁ¦ÓÐ__(ÌîÑ¡Ïî×Öĸ)£»

A.Àë×Ó¼ü B.¹²¼Û¼ü C.½ðÊô¼ü D.Åäλ¼ü E.·¶µÂ»ªÁ¦ F.Çâ¼ü

ÈÜÒºÖÐ[Zn(OH)Cl2]-µÄ½á¹¹Ê½Îª__¡£

III.пÊÇÈËÌå±ØÐèµÄ΢Á¿ÔªËØÖ®Ò»£¬Æä¶Ñ»ý·½Ê½Èçͼ1£¬¾§°û½á¹¹Èçͼ2¡£

£¨5£©Ð¿µÄ¶Ñ»ý·½Ê½Îª__£¬ÅäλÊýΪ__¡£

£¨6£©Èôпԭ×ӵİ뾶Ϊapm£¬°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪNA£¬Ôòп¾§ÌåµÄÃܶÈΪ___g/cm3(Óú¬aµÄ´úÊýʽ±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁС°½âÊÍ»ò½áÂÛ¡±Ó롰ʵÑé²Ù×÷¼°ÏÖÏó¡±²»Ïà·ûµÄÒ»×éÊÇ

ÐòºÅ

ʵÑé²Ù×÷¼°ÏÖÏó

½âÊÍ»ò½áÂÛ

A

ŨÁòËáµÎµ½Ö½ÕÅÉÏ£¬Ö½±äºÚ

ŨÁòËáÓÐÍÑË®ÐÔ

B

Ïò×ÏɫʯÈïÈÜÒºÖмÓÈëÂÈË®£¬ÈÜÒºÏȱäºì£¬ËæºóÍÊÉ«

ÂÈË®Öк¬ÓÐËáÐÔÎïÖʺÍ

Ư°×ÐÔÎïÖÊ

C

ÏòijÈÜÒºÖмÓÈëÏ¡ÑÎËᣬ²úÉúÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÆøÌå

¸ÃÈÜÒºÖÐÒ»¶¨ÓÐCO32-

D

ÏòijÈÜÒºÖмÓÈëŨNaOHÈÜÒº£¬¼ÓÈÈ£¬²úÉúÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌå

¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓÐNH

A.AB.BC.CD.D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ

A. ´×ËáÈܽâË®¹¸ÖеÄCaCO3£º CaCO3 + 2H+= Ca2++ H2O + CO2¡ü

B. ¶èÐԵ缫µç½â±¥ºÍMgCl2ÈÜÒº£º Mg2++2Cl£­ + 2H2O Mg(OH)2¡ý + H2¡ü + Cl2¡ü

C. ±½·ÓÄÆÈÜÒºÖÐͨÈëÉÙÁ¿µÄCO2£º+H2O+CO2¡ú+

D. ÓÃÒø°±ÈÜÒº¼ìÑéÒÒÈ©ÖеÄÈ©»ù£ºCH3CHO£«£«2OH£­ CH3COONH4£«H2O£«2Ag¡ý£«3NH3¡ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿°²È«ÊÇ˳Àû½øÐÐʵÑé¼°±ÜÃâÉ˺¦µÄ±£ÕÏ.ÏÂÁÐʵÑé²Ù×÷ÕýÈ·ÇÒ²»ÊÇ´ÓʵÑ鰲ȫ½Ç¶È¿¼ÂǵÄÊÇ£¨ £©

A. ²Ù×÷¢Ù£ºÊ¹ÓÃÉÔ½þÈëÒºÃæϵĵ¹¿Û©¶·¼ìÑéÇâÆøµÄ´¿¶È

B. ²Ù×÷¢Ú£ºÊ¹ÓÃCCl4ÝÍÈ¡äåË®ÖеÄäåʱ£¬Õñµ´ºóÐè´ò¿ª»îÈûʹ©¶·ÄÚÆøÌå·Å³ö

C. ²Ù×÷¢Û£ºÎüÊÕ°±Æø»òÂÈ»¯ÇâÆøÌå²¢·ÀÖ¹µ¹Îü

D. ²Ù×÷¢Ü£ºÓÃʳָ¶¥×¡Æ¿Èû£¬ÁíÒ»Ö»ÊÖÍÐסƿµ×£¬°ÑÆ¿µ¹Á¢£¬¼ì²éÈÝÁ¿Æ¿ÊÇ·ñ©ˮ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓлúÎïGÊǺϳÉÐÂÅ©Ò©µÄÖØÒªÖмäÌå¡£ÒÔ»¯ºÏÎïAΪԭÁϺϳɻ¯ºÏÎïGµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©»¯ºÏÎïGÖк¬Ñõ¹ÙÄÜÍŵÄÃû³ÆΪ________¡£

£¨2£©·´Ó¦D¡úEµÄ·´Ó¦ÀàÐÍΪ________¡£

£¨3£©»¯ºÏÎïBµÄ·Ö×ÓʽΪC7H6Cl2£¬BµÄ½á¹¹¼òʽΪ______¡£

£¨4£©Ð´³öͬʱÂú×ãÏÂÁÐÌõ¼þµÄGµÄÒ»ÖÖͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º______¡£

¢ÙÄÜ·¢ÉúÒø¾µ·´Ó¦£»

¢ÚºË´Å¹²ÕñÇâÆ×ÏÔʾÇâÔ­×ӵķåÖµ±ÈΪ3¡Ã2¡Ã2¡Ã1¡£

£¨5£©ÇëÒÔ»¯ºÏÎïFºÍCH2(COOC2H5)2ΪԭÁÏÖƱ¸£¬Ð´³öÖƱ¸µÄºÏ³É·ÏßÁ÷³Ìͼ£¨ÎÞ»úÊÔ¼ÁÈÎÓ㬺ϳɷÏßÁ÷³ÌͼʾÀý¼û±¾ÌâÌâ¸É£©¡£

__________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ê³´×£¨Ö÷Òª³É·ÖCH3COOH £©¡¢´¿¼î£¨Na2CO3 £©ºÍСËÕ´ò£¨NaHCO3 £©¾ùΪ¼ÒÍ¥³ø·¿Öг£ÓõÄÎïÖÊ¡£ÒÑÖª£ºCH3COOH¡¢H2CO3¡¢HNO2µÄµçÀë³£Êý£¨25¡æ£©·Ö±ðΪKa=1.8¡Á10-5£»Ka1=4.3¡Á10-7¡¢Ka2=5.6¡Á10-11£»Ka=5.0¡Á10-4Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÅÒ»¶¨Î¶ÈÏ£¬Ïò0.1mol/LCH3COOH ÈÜÒºÖмÓÈëÉÙÁ¿CH3COONa ¾§Ìåʱ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ____£¨Ìî´úºÅ¡££©

a£®ÈÜÒºµÄpHÔö´ó b£®CH3COOHµÄµçÀë³Ì¶ÈÔö´ó

c£®ÈÜÒºµÄµ¼µçÄÜÁ¦¼õÈõ d£®ÈÜÒºÖÐc(OH-)¡¤c(H+)²»±ä

¢Æ25¡æʱ£¬ÏòCH3COOHÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄNaHCO3£¬ËùµÃ»ìºÏÒºµÄpH=6£¬Ôò»ìºÏÒºÖÐ:

c(CH3COO-)/c(CH3COOH)=____

¢Ç³£ÎÂÏ£¬½«20mL 0.1mol/L CH3COOHÈÜÒº ºÍ20mL 0.1mol/LHNO2 ÈÜÒº·Ö±ðÓë 20mL 0.1mol/LNaHCO3ÈÜÒº»ìºÏ£¬ÊµÑé²âµÃ²úÉúµÄÆøÌåÌå»ý£¨V£©Ëæʱ¼ä£¨t£©µÄ±ä»¯ÈçͼËùʾ£¬Ôò±íʾCH3COOHÈÜÒºµÄÇúÏßÊÇ_______£¨ÌîдÐòºÅ£©£»

¢ÈÌå»ý¾ùΪ100mL pH=2µÄCH3COOHÓëÒ»ÔªËáHX£¬¼ÓˮϡÊ͹ý³ÌÖÐpHÓëÈÜÒºÌå»ýµÄ¹ØϵÈçͼËùʾ£¬ÔòHXµÄµçÀëƽºâ³£Êý______£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©CH3COOHµÄµçÀëƽºâ³£Êý¡£

¢É25¡æʱ£¬½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ´×ËáÓ백ˮ»ìºÏºó£¬ÈÜÒºµÄpH=7£¬ÔòNH3¡¤H2OµÄµçÀë³£ÊýKb =___________

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸