¡¾ÌâÄ¿¡¿ÒÑÖªA¡¢B¡¢C¡¢DÊÇÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄËÄÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬AµÄÖÜÆÚÊýµÈÓÚÆäÖ÷×åÐòÊý£¬BÔ­×ӵļ۵ç×ÓÅŲ¼Îªnsnnpn£¬DÊǵؿÇÖк¬Á¿×î¶àµÄÔªËØ¡£EÊǵÚËÄÖÜÆÚpÇøµÄÔªËØÇÒ×îÍâ²ãÖ»ÓÐ2¶Ô³É¶Ôµç×Ó£¬FÊÇ29ºÅÔªËØ¡£

£¨1£©B¡¢C¡¢DÈýÔªËصÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪ £¨ÓÃÔªËØ·ûºÅ±íʾ£©

£¨2£©BD32-ÖÐÐÄÔ­×ÓÔÓ»¯¹ìµÀµÄÀàÐÍΪ________ÔÓ»¯£»CA4+µÄ¿Õ¼ä¹¹ÐÍΪ______________¡£

£¨3£©»ù̬EÔ­×ӵļ۵ç×ÓÅŲ¼Í¼______________________________¡£

£¨4£©1mol BC£­Öк¬ÓЦмüµÄÊýĿΪ______________¡£

£¨5£©±È½ÏD¡¢EÔªËØ×î¼òµ¥Ç⻯ÎïµÄ·Ðµã¸ßµÍ£º £¨Óû¯Ñ§Ê½±íʾ£©¡£

£¨6£©C¡¢FÁ½ÔªËØÐγɵÄij»¯ºÏÎïµÄ¾§°û½á¹¹ÈçͼËùʾ£¬¶¥µãΪCÔ­×Ó¡£Ôò¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇ £¬CÔ­×ÓµÄÅäλÊýÊÇ ¡£ÈôÏàÁÚCÔ­×ÓºÍFÔ­×Ó¼äµÄ¾àÀëΪa cm£¬°¢·üÙ¤µÂÂÞ³£ÊýΪNA£¬Ôò¸Ã¾§ÌåµÄÃܶÈΪ________________g£¯cm3£¨Óú¬a¡¢NAµÄ·ûºÅ±íʾ£©¡£

¡¾´ð°¸¡¿£¨1£©N£¾O£¾C

£¨2£©sp2 ÕýËÄÃæÌå

£¨3£©

£¨4£©2NA

£¨5£©H2O£¾H2Se

£¨6£©Cu3N 6 103/4a3NA

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º¸ù¾ÝÌâ¸øÐÅÏ¢£¬ A¡¢B¡¢C¡¢DÊÇÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄËÄÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬AµÄÖÜÆÚÊýµÈÓÚÆäÖ÷×åÐòÊý£¬BÔ­×ӵļ۵ç×ÓÅŲ¼Îªnsnnpn£¬DÊǵؿÇÖк¬Á¿×î¶àµÄÔªËØ£¬ÔòAΪH£¬BΪ̼£¬DΪÑõ£¬ÔòCΪµªÔªËØ£¬EÊǵÚËÄÖÜÆÚµÄpÇøÔªËØÇÒ×îÍâ²ãÖ»ÓÐ2¶Ô³É¶Ôµç×Ó£¬EΪSe£¬FÔªËصÄÔ­×ÓÐòÊýΪ29£¬FΪͭ¡£Ôò

£¨1£©Í¬ÖÜÆÚ×Ô×óÏòÓÒµÚÒ»µçÀëÄÜÖð½¥¼õС£¬µ«NÔªËصÄ2p¹ìµÀµç×Ó´¦ÓÚ°ë³äÂú״̬£¬Îȶ¨ÐÔÇ¿£¬µÚÒ»µçÀëÄÜ´óÓÚÑõÔªËØ£¬ÔòB¡¢C¡¢DÈýÔªËصÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪN£¾O£¾C£»

£¨2£©CO32-ÖÐÐÄÔ­×Ó̼ԭ×ӵļ۲ãµç×Ó¶ÔÊý£½3+£¬Òò´ËÆäÔÓ»¯¹ìµÀµÄÀàÐÍΪsp2ÔÓ»¯£»NH4+µÄ¿Õ¼ä¹¹ÐÍΪÕýËÄÃæÌå¡£

£¨3£©SeµÄÔ­×ÓÐòÊýÊÇ34£¬Ôò¸ù¾ÝºËÍâµç×ÓÅŲ¼¹æÂÉ¿ÉÖª»ù̬SeÔ­×ӵļ۵ç×ÓÅŲ¼Í¼Îª£»

£¨4£©CN-Àë×ÓÖÐC¡¢N¼äΪC¡ÔN£¬1mol CN£­Öк¬ÓЦмüµÄÊýĿΪ2NA£»

£¨5£©Ë®·Ö×Ó¼ä´æÔÚÇâ¼ü£¬ÔòÇ⻯Îï·ÐµãΪH2O£¾H2Se£»

£¨6£©¸ù¾Ý¾§°ûµÄ½á¹¹¿ÉÖªºÚÇòÊÇCu£¬Æä¸öÊýÊÇ£¬°×ÇòÊÇN£¬Æä¸öÊýΪ£¬ËùÒÔ»¯ºÏÎïµÄ»¯Ñ§Ê½Cu3N£»¶¥µãµÄN3-£¨1/8£©¶ÔÓ¦µÄÈýÌõÀâÉÏÓÐ3¸öCu+£¨1/4£©£¬¹ÊN3-£ºCu+=1/8£º3/4=1£º6£¬ÆäÅäλÊýÊÇ6¡£¾§°ûµÄÌå»ýÊÇV=£¨2a£©3cm3£¬ÖÊÁ¿Îª206/NA£¬¹ÊÃܶÈΪ103/4a3NA¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¾ßÓÐÒÔϽṹµÄÔ­×Ó£¬Ò»¶¨ÊôÓÚÖ÷×åÔªËصÄÊÇ£¨ £©

A. ×îÍâ²ãÓÐ3¸öδ³É¶Ôµç×ÓµÄÔ­×Ó

B. ×îÍâ²ãµç×ÓÅŲ¼Îªns2µÄÔ­×Ó

C. ´ÎÍâ²ãÎÞδ³É¶Ôµç×ÓµÄÔ­×Ó

D. ×îÍâ²ãÓÐ8¸öµç×ÓµÄÔ­×Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁз´Ó¦ÖÐÊôÓÚÎüÈÈ·´Ó¦µÄÊÇ£¨ £©

A£®Ã¾ÓëÑÎËá·´Ó¦·Å³öÇâÆø

B£®ÇâÑõ»¯ÄÆÓëÑÎËáµÄ·´Ó¦

C£®ÁòÔÚ¿ÕÆø»òÑõÆøÖÐȼÉÕ

D£®Ba(OH£©28H2OÓëNH4Cl·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Óлú²£Á§ÊÇÒ»ÖÖÖØÒªµÄËÜÁÏ£¬ËüÊÇÓɵ¥ÌåA£¨ CH2=C£¨CH3£©COOCH3 £©Í¨¹ý¼Ó¾Û·´Ó¦ÖƵã¬ÒÑÖªµ¥ÌåA¿É·¢ÉúÈçͼËùʾµÄת»¯¹Øϵ£º

ÒÑÖª£ºÒÒËáÒÒõ¥ÔÚÏ¡H2SO4Öз¢ÉúË®½â£º

CH3C18OOCH2CH3 + H2O CH3C18OOH + CH3CH2OH

£¨1£©Óлú²£Á§µÄ½á¹¹¼òʽÊÇ £®

£¨2£©B·Ö×ÓÖк¬ÓеĹÙÄÜÍÅÊÇ £¨ÌîÃû³Æ£©£®

£¨3£©ÓÉBת»¯ÎªCµÄ·´Ó¦ÊôÓÚ £¨ÌîÐòºÅ£©£®

¢ÙÑõ»¯·´Ó¦ ¢Ú»¹Ô­·´Ó¦ ¢Û¼Ó³É·´Ó¦ ¢ÜÈ¡´ú·´Ó¦

£¨4£©CµÄÒ»ÂÈ´úÎïDÓÐ Öֽṹ£¨¡ªCOOHÉϵÄÇâ²»»áºÍCl2·¢ÉúÈ¡´ú·´Ó¦£©£®

£¨5£©ÓÉAÉú³ÉBµÄ»¯Ñ§·½³ÌʽÊÇ £®

£¨6£©CH3CH2OHÊÇCH3OHµÄͬϵÎÊÔд³öCH3CH2OH ÓëO2ÔÚÍ­×ö´ß»¯¼Á²¢¼ÓÈȵÄÌõ¼þÏ·¢Éú·´Ó¦µÄ»¯Ñ§·´Ó¦·½³Ìʽ£º £®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Óлú»¯ºÏÎïGÊǺϳÉάÉúËØÀàÒ©ÎïµÄÖмäÌ壬ÆäºÏ³É·ÏßÈçÏ£º

ÆäÖÐA¡«F·Ö±ð´ú±íÒ»ÖÖÓлú»¯ºÏÎºÏ³É·ÏßÖв¿·Ö²úÎï¼°·´Ó¦Ìõ¼þÒÑÂÔÈ¥ÒÑÖª£º

GΪ£»

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©GµÄ·Ö×Óʽ_____________£»DÖйÙÄÜÍŵÄÃû³ÆÊÇ_________¡£

£¨2£©µÚ¢Ú²½·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________¡£

£¨3£©µÚ¢Û²½·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________¡£

£¨4£©Ð´³öFµÄ½á¹¹¼òʽ_____________¡£

£¨5£©µÚ¢Ù~¢Þ²½·´Ó¦ÖÐÊôÓڼӳɷ´Ó¦µÄÓÐ___________________£»ÊôÓÚÈ¡´ú·´Ó¦µÄÓÐ

__________________________¡££¨Ìî²½Öè±àºÅ£©

£¨6£©Í¬Ê±Âú×ãÏÂÁÐÌõ¼þµÄEµÄͬ·ÖÒì¹¹ÌåÓÐ_____________ÖÖ¡£

¢ÙÖ»º¬Ò»ÖÖ¹ÙÄÜÍÅ£»

¢ÚÁ´×´½á¹¹ÇÒÎÞ¡ªO¡ªO¡ª£»

¢ÛºË´Å¹²ÕñÇâÆ×Ö»ÓÐ2×é·å¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÎªÁËÖÎÀí·ÏË®ÖÐCr2O72-µÄÎÛȾ£¬³£ÏȼÓÈëÊÔ¼Áʹ֮±äΪCr3+£¬¸ÃÊÔ¼ÁΪ

A. NaOHÈÜÒº B. FeCl3ÈÜÒº C. Ã÷·¯ D. Na2SO3ºÍH2SO4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÀûÓÃÏÂͼװÖòⶨÖкÍÈȵÄʵÑé²½ÖèÈçÏ£º

¢ÙÁ¿È¡50mL 0.25mol/L H2SO4ÈÜÒºµ¹ÈëСÉÕ±­ÖУ¬²âÁ¿Î¶ȣ»

¢ÚÁ¿È¡50mL 0.55mol/L NaOHÈÜÒº£¬²âÁ¿Î¶ȣ»

¢Û½«NaOHÈÜÒºµ¹ÈëСÉÕ±­ÖÐ,»ìºÏ¾ùÔȺó²âÁ¿»ìºÏҺζȡ£Çë»Ø´ð£º

£¨1£©ÈçÓÒͼËùʾ£¬ÒÇÆ÷AµÄÃû³ÆÊÇ_________ ______£»

£¨2£©NaOHÈÜÒºÉÔ¹ýÁ¿µÄÔ­Òò ____________________________¡£

£¨3£©¼ÓÈëNaOHÈÜÒºµÄÕýÈ·²Ù×÷ÊÇ_______£¨Ìî×Öĸ£©¡£

A£®Ñز£Á§°ô»ºÂý¼ÓÈë B£®Ò»´ÎѸËÙ¼ÓÈë C£®·ÖÈý´Î¼ÓÈë

£¨4£©Ê¹ÁòËáÓëNaOHÈÜÒº»ìºÏ¾ùÔȵÄÕýÈ·²Ù×÷ÊÇ ________ __________________¡£

£¨5£©ÉèÈÜÒºµÄÃܶȾùΪ1g¡¤cm-3,ÖкͺóÈÜÒºµÄ±ÈÈÈÈÝc=4.18 J¡¤£¨g¡¤¡æ£©-1£¬Çë¸ù¾ÝʵÑéÊý¾Ýд³ö¸ÃÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ_______________________________

£¨6£©ÉÏÊöʵÑéÊýÖµ½á¹ûÓë57.3 kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔ­Òò¿ÉÄÜÊÇ£¨Ìî×Öĸ£©

a£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î

b£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±­ÖÐ

c£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζÈ

£¨7£©ÔõÑù²ÅÄÜÈ·±£¶ÁÈ¡»ìºÏÒºµÄ×î¸ßζȣ¿_______ ____________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁвÄÁϵÄÌØÐÔ¼°ÓÃ;˵·¨´íÎóµÄÊÇ(

A£®¸ß´¿¹èÓÃÓÚÖÆÔì¹âµ¼ÏËά£¬ÊµÏÖ¹âµçÐźÅת»¯

B£®Ê¯Ó¢ÖÐÎÞɫ͸Ã÷µÄ¾§Ìå¾ÍÊÇͨ³£Ëù˵µÄË®¾§£¬ÆäÖ÷Òª³É·ÖÊǶþÑõ»¯¹è

C£®¹âµ¼ÏËάµ¼¹âµÄÄÜÁ¦ºÜÇ¿£¬ÊǷdz£ºÃµÄͨѶ²ÄÁÏ

D£®¹è½º¶à¿×£¬Îü¸½Ë®ÄÜÁ¦Ç¿£¬¿ÉÒÔÓÃ×÷´ß»¯¼ÁµÄÔØÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ ÏÂÁÐÓйؼîÍÁ½ðÊôïÈ£¨Sr£©ÔªËص¥Öʼ°Æ仯ºÏÎïµÄÐðÊöÖУ¬ÕýÈ·µÄÊÇ£¨ £©

A. ïÈÄÜÓëË®·´Ó¦£¬µ«·´Ó¦»ºÂý

B. ÏõËáïÈÊÇÀë×Ó»¯ºÏÎ²»ÈÜÓÚË®

C. ÇâÑõ»¯ïȼîÐÔÈõÓÚÇâÑõ»¯Ã¾µÄ¼îÐÔ

D. ½ðÊôïȵ¥ÖÊÒø°×É«£¬µ¼µçÐÔÁ¼ºÃ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸