18-¢ñ ÒÒËáÏãÀ¼õ¥ÊÇÓÃÓÚµ÷ÅäÄÌÓÍ¡¢±ùä¿ÁܵÄʳÓÃÏ㾫£¬ÆäºÏ³É·´Ó¦µÄ»¯Ñ§·½³ÌʽÈçÏ£º

ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨_____£©

A£®¸Ã·´Ó¦ÊôÓÚÈ¡´ú·´Ó¦

B£®ÒÒËáÏãÀ¼õ¥µÄ·Ö×ÓʽΪC10H8O4

C£®FeCl3ÈÜÒº¿ÉÓÃÓÚÇø±ðÏãÀ¼ËØÓëÒÒËáÏãÀ¼õ¥

D£®ÒÒËáÏãÀ¼õ¥ÔÚ×ãÁ¿NaOHÈÜÒºÖÐË®½âµÃµ½ÒÒËáºÍÏãÀ¼ËØ

18-¢òÓà A£¨CH2=CH2£©ºÍ D(HOOCCH=CHCH=CHCOOH)ºÏ³É¸ß·Ö×ÓP£¬ÆäºÏ³É·ÏßÈçÏ£º

ÒÑÖª£º¢Ù

¢Úõ¥Óë´¼¿É·¢ÉúÈçÏÂõ¥½»»»·´Ó¦£º

£¨1£©BµÄÃû³ÆΪ_________________£¬DÖйÙÄÜÍŵÄÃû³ÆΪ_________________________¡£

£¨2£©CµÄ·Ö×ÓʽÊÇC2H6O2£¬·´Ó¦¢ÚµÄÊÔ¼ÁºÍ·´Ó¦Ìõ¼þÊÇ____________________________¡£

£¨3£©FµÄ½á¹¹¼òʽÊÇ__________________________¡£

£¨4£©·´Ó¦¢ÞµÄ»¯Ñ§·½³ÌʽÊÇ____________________________________¡£

£¨5£©GµÄÒ»ÖÖͬ·ÖÒì¹¹ÌåG'Ϊ¼×Ëáõ¥¡¢ºË´Å¹²ÕñÇâÆ×ÓÐ3ÖÖ·åÇÒ1mol¸ÃÓлúÎïËáÐÔÌõ¼þÏÂË®½â²úÎïÄÜÓë2 mol NaOH·´Ó¦¡£G'µÄ½á¹¹¼òʽΪ___________________¡£

£¨6£©ÒÔ¶Ô±½¶þ¼×´¼¡¢¼×´¼ÎªÆðʼԭÁÏ£¬Ñ¡ÓñØÒªµÄÎÞ»úÊÔ¼ÁºÏ³ÉG£¬Ð´³öºÏ³É·Ïߣ¨Óýṹ¼òʽ±íʾÓлúÎÓüýÍ·±íʾת»¯¹Øϵ£¬¼ýÍ·ÉÏ×¢Ã÷ÊÔ¼ÁºÍ·´Ó¦Ìõ¼þ£©¡£___________________

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄêºÓ±±Ê¡¸ß¶þÏÂѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ ( )

A£®³£Î³£Ñ¹Ï£¬16 g ÓÉO2ºÍO3×é³ÉµÄ»ìºÏÆøÌåËùº¬µç×ÓÊýΪ8NA

B£®ÊµÑéÊÒ·Ö±ðÓÃKClO3ºÍH2O2ÖÆÈ¡3.2g O2ʱ£¬×ªÒƵĵĵç×ÓÊý¾ùΪ0.4NA

C£®25¡æʱ£¬1 L pH£½1µÄÏ¡ÁòËáÖÐÔ¼º¬2NA¸öH+

D£®±ê×¼×´¿öÏ£¬22.4 L SO3Ëùº¬·Ö×ÓÊýΪNA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìÌì½òÊиßÈýÏÂѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

Èç±íËùʾΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬²ÎÕÕÔªËآ١«¢áÔÚ±íÖеÄλÖã¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

×å

ÖÜÆÚ

IA

0

1

¢Ù

¢òA

¢óA

¢ôA

¢õA

¢öA

¢÷A

2

¢à

¢á

¢Ú

¢Û

3

¢Ü

¢Ý

¢Þ

¢ß

£¨1£©¢Û¡¢¢Ü¡¢¢ßµÄÔ­×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ_________£¨ÓÃÔªËØ·ûºÅ±íʾ£©¡£

£¨2£©ÏÂÁÐÊÂʵÄÜ˵Ã÷¢ÚÔªËصķǽðÊôÐԱȢÞÔªËصķǽðÊôÐÔÇ¿µÄÊÇ__________¡£

a.¢ÚµÄµ¥ÖÊÓë¢ÞÔªËصļòµ¥Ç⻯ÎïÈÜÒº·´Ó¦£¬ÈÜÒº±ä»ë×Ç

b.ÔÚÑõ»¯»¹Ô­·´Ó¦ÖУ¬1mol¢Úµ¥ÖʱÈ1mol¢Þµ¥Öʵõç×Ó¶à

c.¢ÚºÍ¢ÞÁ½ÔªËصļòµ¥Ç⻯ÎïÊÜÈȷֽ⣬ǰÕߵķֽâζȸߡ£

£¨3£©¢Ù¡¢¢ÚÁ½ÖÖÔªËØ°´Ô­×Ó¸öÊýÖ®±ÈΪ1£º1×é³ÉµÄ³£¼ûҺ̬»¯ºÏÎÔÚËáÐÔÈÜÒºÖÐÄܽ«Fe2+ Ñõ»¯£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ ___________________¡£

£¨4£© ÒÑÖªÖÜÆÚ±íÖдæÔÚ¶Ô½ÇÏàËƹæÔò£¬Èçî루Be£©ÓëÂÁ»¯Ñ§ÐÔÖÊÏàËÆ£¬¢àµÄÑõ»¯Îï¡¢ÇâÑõ»¯ÎïÒ²ÓÐÁ½ÐÔ£¬Ð´³ö¢àµÄÇâÑõ»¯ÎïÓë¢ÜµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦µÄ»¯Ñ§·½³Ìʽ _______________________¡£

£¨5£©ÒÑÖªW+X=Y+Z£¨·´Ó¦ÐèÒª¼ÓÈÈ£¬£©£¬W¡¢X¡¢Y¡¢Z·Ö±ðÊÇÓÉ¢Ù¢Ú¢áÈýÖÖÔªËØÐγɵÄËÄÖÖ10µç×ÓÁ£×Ó£¨W¡¢XΪÀë×Ó£¬Y¡¢ZΪ·Ö×Ó£©£¬Ð´³ö¸Ã»¯Ñ§·½³Ìʽ_________________¡£

£¨6£©ÓɱíÖÐÔªËØÐγɵÄÎïÖÊ¿É·¢ÉúÈçͼÖеķ´Ó¦£¬ÆäÖÐB¡¢C¡¢GÊǵ¥ÖÊ£¬BΪ»ÆÂÌÉ«ÆøÌ壬 DÈÜÒºÏÔ¼îÐÔ¡£

¢Ùд³öDÈÜÒºÓëG·´Ó¦µÄÀë×Ó·½³Ìʽ______________________¡£

¢Úд³ö¼ìÑéAÈÜÒºÖÐÈÜÖʵÄÒõÀë×ӵķ½·¨____________________¡£

¢Û³£ÎÂÏ£¬Èôµç½â1L0.1mol/LµÄAÈÜÒº£¬Ò»¶Îʱ¼äºó²âµÃÈÜÒºpHΪ12£¨ºöÂÔÈÜÒºÌå»ý±ä»¯£©£¬Ôò¸Ãµç½â¹ý³ÌÖÐתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª£º________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄêɽÎ÷Ê¡¸ßÒ»3ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÔªËØR¡¢X¡¢T¡¢Z¡¢QÔÚÔªËØÖÜÆÚ±íÖеÄÏà¶ÔλÖÃÈçϱíËùʾ,ÆäÖÐRµ¥ÖÊÔÚ°µ´¦ÓëH2¾çÁÒ»¯ºÏ²¢·¢Éú±¬Õ¨¡£ÔòÏÂÁÐÅжÏÕýÈ·µÄÊÇ(¡¡¡¡)

¢Ù·Ç½ðÊôÐÔ:Z<T<X ¢ÚRÓëQµÄµç×ÓÊýÏà²î26

¢ÛÆø̬Ç⻯ÎïÎȶ¨ÐÔ:R<T<Q ¢Ü×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄËáÐÔ:T>Q

A. ¢Ù¢Û B. ¢Ù¢Ü C. ¢Ú¢Ü D. ¢Ú¢Û

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄêɽÎ÷Ê¡¸ßÒ»3ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁйØÓÚ¼î½ðÊôÔªËغͱËصÄ˵·¨ÖдíÎóµÄÊÇ(¡¡¡¡)

A. ËæºËµçºÉÊýµÄÔö¼Ó,¼î½ðÊôÔªËغͱËصÄÔ­×Ӱ뾶¶¼Öð½¥Ôö´ó

B. ¼î½ðÊôÔªËØÖÐ,ï®Ô­×Óʧȥ×îÍâ²ãµç×ÓµÄÄÜÁ¦×îÈõ;±ËØÖÐ,·úÔ­×ӵõç×ÓÄÜÁ¦×îÇ¿

C. äåµ¥ÖÊÓëË®µÄ·´Ó¦±ÈÂȵ¥ÖÊÓëË®µÄ·´Ó¦¸ü¾çÁÒ

D. ¼ØÓëË®µÄ·´Ó¦±ÈÄÆÓëË®µÄ·´Ó¦¸ü¾çÁÒ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºº£ÄÏÊ¡¡¢ÎIJýÖÐѧ2017½ì¸ßÈýÏÂѧÆÚÁª¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

A¡¢B¡¢C¡¢D¡¢EÎåÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ÆäÖÐC¡¢D¡¢EͬÖÜÆÚ£¬A¡¢CͬÖ÷×壬B¡¢EͬÖ÷×壬BÔªËصÄÔ­×Ó×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄÈý±¶£¬ÓÖÖªAµ¥ÖÊÊÇÃܶÈ×îСµÄÆøÌå¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÔªËØCÔÚÖÜÆÚ±íÖеÄλÖÃ______________________________¡£

£¨2£© A¡¢C¡¢EÒÔÔ­×Ó¸öÊý±È1¡Ã1¡Ã1Ðγɻ¯ºÏÎïX£¬Æäµç×ÓʽΪ_________________¡£

£¨3£© B¡¢E¶ÔÓ¦¼òµ¥Ç⻯ÎïÎȶ¨ÐԵĴóС˳ÐòÊÇ£¨Ó÷Ö×Óʽ±íʾ£© ________________¡£

£¨4£©ÈôDÊǷǽðÊôÔªËØ£¬Æäµ¥ÖÊÔÚµç×Ó¹¤ÒµÖÐÓÐÖØÒªÓ¦Óã¬Çëд³öÆäÑõ»¯ÎïÈÜÓÚÇ¿¼îÈÜÒºµÄÀë×Ó·½³Ìʽ£º___________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºº£ÄÏÊ¡¡¢ÎIJýÖÐѧ2017½ì¸ßÈýÏÂѧÆÚÁª¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐ×°ÖûòʵÑé²Ù×÷ÕýÈ·µÄÊÇ

A. ¢ÙÓÃpHÊÔÖ½²âijÈÜÒºµÄËá¼îÐÔ B. ¢Ú̽¾¿Ñõ»¯ÐÔ£ºKMnO4>Cl2>I2

C. ¢ÛÎüÊÕ°±ÆøÖÆ°±Ë® D. ¢ÜÖк͵ζ¨ÊµÑé

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2016-2017ѧÄêËÄ´¨Ê¡³É¶¼Êб±ºþУÇø¸ß¶þ3ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐÓ뻯ѧ¸ÅÄîÓйصÄ˵·¨»ò»¯Ñ§ÓÃÓï²»ÕýÈ·µÄÊÇ£¨ £©

A. ÓÃʯī×÷µç¼«µç½âNH4ClÈÜÒºµÄÑô¼«·´Ó¦Ê½Îª£º2Cl£­£­2e£­£½Cl2¡ü

B. ÓÉAl¡¢Cu¡¢Å¨ÏõËá×é³ÉµÄÔ­µç³Ø£¬ÆäÕý¼«·´Ó¦Ê½Îª£ºNO3£­£«2H£«£«e£­= NO2¡ü£«H2O

C. ÓÉAl¡¢Mg¡¢NaOH×é³ÉµÄÔ­µç³Ø£¬Æ为¼«·´Ó¦Ê½Îª£ºAl£­3e£­+3OH£­ = Al(OH)3¡ý

D. Mg(OH)2¹ÌÌåÔÚÈÜÒºÖдæÔÚÈܽâƽºâ£º Mg(OH)2(s) Mg2+(aq) + 2OH£­(aq)£¬¸Ã¹ÌÌå¿ÉÈÜÓÚNH4ClÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2017½ìÕã½­Ê¡¸ßÈý¡°³¬¼¶È«ÄÜÉú¡±3ÔÂÁª¿¼£¨Ñ¡¿¼¿ÆÄ¿£©»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ò´ðÌâ

¸ßÎÂÏÂÓÃH2»¹Ô­CuClÖƱ¸»îÐÔÍ­£¬·´Ó¦Ô­ÀíÈçÏ£º

2Cu(s)+Cl2 (g)2CuCl(s) ¡÷H1=-36 kJ¡¤mol-1 ¢Ù

H2(g)+2CuCl( s)=2Cu(s)+2HCl(g) ¡÷H2 ¢Ú

ÓйØÎïÖʵļüÄÜÊý¾ÝÈçÏÂ±í£º

ÎïÖÊ

H2

Cl2

HCl

¼üÄÜ(kJ¡¤mol-1)

436

243

432

(1)Çó¡÷H2=_______ kJ¡¤mol-1¡£

(2)¾­²â¶¨·´Ó¦¢ÚÖƱ¸»îÐÔÍ­µÄ·´Ó¦Ç÷ÊÆ´ó£¬Ô­ÒòÊÇ_______________¡£

(3)ÔÚijζÈÏ£¬·´Ó¦¢Ù´ïµ½Æ½ºâ״̬£¬ÔÚtlʱ£¬Ôö¼Óѹǿµ½Ô­À´µÄ2±¶£¨CuµÄÁ¿×ã¹»£©£¬ÔÚͼÖл­³öCl2Ũ¶ÈµÄ±ä»¯Ç÷ÊÆÏß¡£_______________

(4)°×É«²»ÈÜÓÚË®µÄCuCl¿ÉÒÔÓɵç½â·¨ÖƵã¬ÈçͼËùʾ£º

¢Ù×°ÖÃÖÐÓõĽ»»»Ä¤Îª_____________¡£

A.ÑôÀë×Ó½»»»Ä¤ B£®ÒõÀë×Ó½»»»Ä¤ C£®ÖÊ×Ó½»»»Ä¤ D£®ÇâÑõ¸ùÀë×Ó½»»»Ä¤

¢ÚÑô¼«µÄµç¼«·´Ó¦Ê½Îª_________________¡£

(5)ÒÑÖªCuCl¿ÉÈܽâÓÚÏ¡ÏõËᣬд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_________________¡£

(6)¸ù¾ÝÒÑѧ֪ʶд³öÖÆÈ¡CuClµÄÒ»ÖÖ·½·¨£¬Óû¯Ñ§·½³Ìʽ±íʾ£º_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸