¡¾ÌâÄ¿¡¿¼×õ¥³ý²ÝÃÑÊÇÒ»ÖÖ¹âºÏ×÷ÓÃÒÖÖƼÁ£¬Äܱ»Ò¶Æ¬½Ï¿ìµØÎüÊÕ£¬µ«ÔÚÖ²ÎïÌåÄÚ´«µ¼ËٶȽÏÂý£¬ËüÊÇÑ¿Ç°³ý²Ý¼Á£¬Ö÷ÒªÓÃÓڴ󶹳ý²ÝµÈ¡£¹¤ÒµÉÏͨ³£ÓÃÌþA½øÐкϳɣ¬ÆäºÏ³É·ÏßÈçÏ£º

ÒÑÖª£ºRCH=O+H2O(R´ú±íÌþ»ù)

(1)ÊÔ¼Á¢ÙΪ£º_____________£¬ÊÔ¼Á¢ÚΪ___________¡£

(2)д³öAµÄ½á¹¹¼òʽ_____________¡£

(3)д³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º·´Ó¦¢Ù____________¡£·´Ó¦¢Ý____________¡£

(4)ÓÉÓÚ±½»·ºÍ²àÁ´»ùÍŵÄÏ໥ӰÏ죬ÐÂÒý½ø±½»·µÄ»ùÍÅÔÚ±½»·ÉÏÈ¡´úµÄλÖÃÓÉÔ­ÓлùÍžö¶¨£¬È磺±½·Ó·Ö×ÓÖС°¡±Ê¹±½»·ÉÏ_________Ñ¡±àºÅÌî¿Õ£¬ÏÂͬµÄHÔ­×ÓÈÝÒ×±»È¡´ú£»¸ù¾ÝÌâÄ¿ÐÅÏ¢¿ÉÖª¡°¡±Ê¹±½»·ÉÏ________µÄHÔ­×ÓÈÝÒ×±»È¡´ú¡£

ÁÚλ ¼äλ ¶Ôλ

(5)·´Ó¦¢ÚÐëÔÚ±¥ºÍÈÜÒºÖнøÐУ¬ÈôÔÚNaOHÈÜÒºÖнøÐУ¬Ôò»á½øÒ»²½Ë®½â£¬ÊÔд³öÔÚ×ãÁ¿NaOHÈÜÒºÖÐÍêÈ«Ë®½âµÄ»¯Ñ§·´Ó¦·½³Ìʽ_________________

¡¾´ð°¸¡¿ÂÈÆø ¼×´¼ +HNO3+H2O b

¡¾½âÎö¡¿

¸ù¾ÝAÔÚ¹âÕÕÌõ¼þÏÂÉú³É£¬¿ÉÖªAΪ£¬ÔÙ½áºÏDºÍÉú³É¿ÉÖªDΪ£»CÔÚÏõËᣬÁòËá×÷ÓÃϵĵ½D£¬¿ÉÖªCΪ£º£¬ÔÙÓÉC·´ÍÆBΪ£¬ÊÔ¼ÁΪ¼×´¼¡£

ÓÉ·ÖÎö¿ÉÖªÊÔ¼ÁΪÂÈÆø£¬ÊÔ¼ÁΪ¼×´¼£»

¹Ê´ð°¸Îª£ºÎªÂÈÆø£»Îª¼×´¼£»

ÓÉ·ÖÎö¿ÉÖªAµÄ½á¹¹¼òʽ£»

¹Ê´ð°¸Îª£º£»

ÓÉ·ÖÎö¿ÉÖª·½³ÌʽΪ£º£»+HNO3+H2O£»

¹Ê´ð°¸Îª£º£»+HNO3+H2O£»

ÓÉÓÚ±½»·ºÍ²àÁ´»ùÍŵÄÏ໥ӰÏ죬ÐÂÒý½ø±½»·µÄ»ùÍÅÔÚ±½»·ÉÏÈ¡´úµÄλÖÃÓÉÔ­ÓлùÍžö¶¨£¬È磺±½·Ó·Ö×ÓÖС°¡±Ê¹±½»·ÉϵÄÁÚ¶Ôλ±äµÃ»îÆã¬ÓÉ·´Ó¦¿ÉÖª¡°¡±Ê¹±½»·ÉϼäλµÄHÔ­×ÓÈÝÒ×±»È¡´ú£»

¹Ê´ð°¸Îª£ºac£»b£»

·´Ó¦ÈôÔÚNaOHÈÜÒºÖнøÐлáÒ»²½Éú³É±½¼×ËáÄÆ£¬·½³ÌʽΪ£º+4NaOH¡ú£»

¹Ê´ð°¸Îª£º+4NaOH¡ú¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚ¼Ûµç×ÓÅŲ¼Ê½Îª3s23p4µÄÁ£×ÓÃèÊöÕýÈ·µÄÊÇ£¨ £©

A.ËüµÄÔªËØ·ûºÅΪO

B.ËüµÄºËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p4

C.Ëü¿ÉÓëH2Éú³ÉҺ̬»¯ºÏÎï

D.Æäµç×ÓÅŲ¼Í¼Îª

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ( )

A.·´Ó¦2Mg£«CO22MgO£«C ¦¤H£¼0´Óìرä½Ç¶È¿´£¬¿É×Ô·¢½øÐÐ

B.ÔÚÃܱÕÈÝÆ÷·¢Éú¿ÉÄæ·´Ó¦£º2NO(g)£«2CO(g)N2(g)£«2CO2(g) ¦¤H£½£­113.0 kJ/mol£¬´ïµ½Æ½ºâºó£¬±£³ÖζȲ»±ä£¬ËõСÈÝÆ÷Ìå»ý£¬ÖØдﵽƽºâºó£¬¦¤H±äС

C.ÒÑÖª£ºKsp(AgCl)£½1.8¡Á10£­10£¬Ksp(Ag2CrO4)£½2.0¡Á10£­12£¬½«µÈÌå»ýŨ¶ÈΪ1.0¡Á10£­4mol/LµÄAgNO3ÈÜÒºµÎÈ뵽Ũ¶È¾ùΪ1.0¡Á10£­4mol/LµÄKClºÍK2CrO4µÄ»ìºÏÈÜÒºÖвúÉúÁ½ÖÖ²»Í¬³Áµí£¬ÇÒAg2CrO4³ÁµíÏȲúÉú

D.¸ù¾ÝHClOµÄKa£½3.0¡Á10£­8£¬H2CO3µÄKa1£½4.3¡Á10£­7£¬ Ka2£½5.6¡Á10£­11£¬¿ÉÍƲâÏàͬ״¿öÏ£¬µÈŨ¶ÈµÄNaClOÓëNa2CO3ÈÜÒºÖУ¬pHÇ°ÕßСÓÚºóÕß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚÓлú»¯ºÏÎïµÄ˵·¨´íÎóµÄÊÇ£¨ £©

A.¼×ÍéºÍÂÈÆøÔÚ¹âÕÕÌõ¼þÏ·´Ó¦µÄ²úÎïÓÐ5ÖÖ

B.2¡ªÒÒ»ù¡ª1£¬3¡ª¶¡¶þÏ©·Ö×ӵļüÏßʽ£º

C.·Ö×ÓʽΪC5H12O2µÄ¶þÔª´¼µÄÖ÷Á´Ì¼Ô­×ÓÊýΪ3µÄ½á¹¹ÓÐ2ÖÖ

D.Ϊ¼ìÑéÂȱûÍéÖеÄÂÈÔªËØ£¬¿É½«ÂȱûÍéÓëNaOHÈÜÒº¹²Èȼ¸·ÖÖÓºó£¬ÀäÈ´£¬µÎ¼ÓAgNO3ÈÜÒº£¬¹Û²ìÊÇ·ñÓа×É«³ÁµíÉú³É

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¶«½ú¡¶»ªÑô¹úÖ¾ÄÏÖÐÖ¾¡·¾íËÄÖÖÒÑÓйØÓÚ°×Í­µÄ¼ÇÔØ£¬ÔÆÄÏÄø°×Í­(Í­ÄøºÏ½ð)ÎÄÃ÷ÖÐÍ⣬ÔøÖ÷ÒªÓÃÓÚÔì±Ò£¬Òà¿ÉÓÃÓÚÖÆ×÷·ÂÒøÊÎÆ·¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÄøÔªËØ»ù̬ԭ×ӵļ۵ç×Ó¹ìµÀ±íʾʽ___£¬3dÄܼ¶ÉϵÄδ³É¶ÔµÄµç×ÓÊýΪ___¡£

(2)ÁòËáÄøÈÜÓÚ°±Ë®ÐγÉ[Ni(NH3)6]SO4À¶É«ÈÜÒº¡£

¢Ù[Ni(NH3)6]SO4ÖÐÒõÀë×ÓµÄÁ¢Ìå¹¹ÐÍÊÇ___¡£

¢ÚÔÚ[Ni(NH3)6]2+ÖÐNi2+ÓëNH3Ö®¼äÐγɵĻ¯Ñ§¼ü³ÆΪ___£¬Ìṩ¹Âµç×ӶԵijɼüÔ­×ÓÊÇ___¡£

¢Û°±µÄ·Ðµã_______(¡°¸ßÓÚ¡±»ò¡°µÍÓÚ¡±)ì¢(PH3)£¬Ô­ÒòÊÇ______¡£

(3)µ¥ÖÊÍ­¼°Äø¶¼ÊÇÓÉ__¼üÐγɵľ§Ì壺ԪËØͬÓëÄøµÄµÚ¶þµçÀëÄÜ·Ö±ðΪ£ºICu=1959kJ/mol£¬INi=1753kJ/mol£¬ICu>INiµÄÔ­ÒòÊÇ___¡£

(4)ijÄø°×Í­ºÏ½ðµÄÁ¢·½¾§°û½á¹¹ÈçͼËùʾ¡£

¢Ù¾§°ûÖÐÍ­Ô­×ÓÓëÄøÔ­×ÓµÄÊýÁ¿±ÈΪ___¡£

¢ÚÈôºÏ½ðµÄÃܶÈΪdg/cm3£¬¾§°û²ÎÊýa=____nm¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐͼʾÓë¶ÔÓ¦µÄÐðÊöÏà·ûµÄÊÇ

ͼ1 ͼ2 ͼ3 ͼ4

A. ͼ1±íʾ0.1molMgCl2¡¤6H2OÔÚ¿ÕÆøÖгä·Ö¼ÓÈÈʱ¹ÌÌåÖÊÁ¿Ëæʱ¼äµÄ±ä»¯

B. ͼ2±íʾÓÃ0.1000 mol¡¤L¡¥lNaOHÈÜÒºµÎ¶¨25.00 mLCH3COOHµÄµÎ¶¨ÇúÏߣ¬Ôòc(CH3COOH)£½0.0800 mol¡¤L¡¥1

C. ͼ3±íʾºãκãÈÝÌõ¼þÏ£¬2NO2(g)N2O4(g)ÖУ¬¸÷ÎïÖʵÄŨ¶ÈÓëÆäÏûºÄËÙÂÊÖ®¼äµÄ¹Øϵ£¬ÆäÖн»µãA¶ÔÓ¦µÄ״̬Ϊ»¯Ñ§Æ½ºâ״̬

D. ͼ4±íʾ³£ÎÂÏ£¬Ï¡ÊÍHA¡¢HBÁ½ÖÖËáµÄÏ¡ÈÜҺʱ£¬ÈÜÒºpHËæ¼ÓË®Á¿µÄ±ä»¯£¬Ôò³£ÎÂÏ£¬NaAÈÜÒºµÄpHСÓÚͬŨ¶ÈµÄNaBÈÜÒºµÄpH

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÖÓÐÁùÖÖÔªËØ£¬ÆäÖÐA¡¢B¡¢C¡¢D¡¢EΪ¶ÌÖÜÆÚÖ÷Òª×åÔªËØ£¬FΪµÚËÄÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£Çë¸ù¾ÝÏÂÁÐÏà¹ØÐÅÏ¢£¬»Ø´ðÎÊÌâ¡£

AÔªËØÐγɵÄÎïÖÊÖÖÀà·±¶à£¬ÆäÐγɵÄÒ»ÖÖ¹ÌÌåµ¥Öʹ¤ÒµÉϳ£ÓÃ×÷Çи¾ß

BÔªËØÔ­×ӵĺËÍâpµç×ÓÊý±Èsµç×ÓÊýÉÙ1

CÔªËØ»ù̬ԭ×Óp¹ìµÀÓÐÁ½¸öδ³É¶Ôµç×Ó

DÔ­×ӵĵÚÒ»ÖÁµÚËĵçÀëÄÜ·Ö±ðÊÇ£º

¢ñ1=738kJ¡¤mol-1£»¢ñ2=1451kJ¡¤mol-1£»¢ñ3=7733kJ¡¤mol-1£»¢ñ4=10540kJ¡¤mol-1£»

EÔ­×ÓºËÍâËùÓÐp¹ìµÀÈ«Âú»ò°ëÂú

FÔÚÖÜÆÚ±íµÄµÚ8×ÝÁÐ

£¨1£©Ä³Í¬Ñ§¸ù¾ÝÉÏÊöÐÅÏ¢£¬ÍƶÏA»ù̬ԭ×ӵĺËÍâ×î×ÓÅŲ¼ÎªÁË£¬¸ÃͬѧËù»­µÄµç×ÓÅŲ¼Í¼Î¥±³ÁË___________¡£

£¨2£©BÔªËصĵ縺ÐÔ_____£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©CÔªËصĵ縺ÐÔ¡£

£¨3£©CÓëDÐγɵĻ¯ºÏÎïËùº¬ÓеĻ¯Ñ§¼üÀàÐÍΪ_____________¡£

£¨4£©E»ù̬ԭ×ÓÖÐÄÜÁ¿×î¸ßµÄµç×Ó£¬Æäµç×ÓÔÆÔÚ¿Õ¼äÓÐ__________¸ö·½Ïò¡£

£¨5£©ÏÂÁйØÓÚFÔ­×ӵļ۲ãµç×ÓÅŲ¼Í¼ÕýÈ·µÄÊÇ___________¡£

a. b.

c. d.

£¨6£©»ù̬F3+Àë×ÓºËÍâµç×ÓÅŲ¼Ê½Îª_____________¡£¹ýÁ¿µ¥ÖÊFÓëBµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄÏ¡ÈÜÒºÍêÈ«·´Ó¦£¬Éú³ÉBCÆøÌ壬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________¡£

£¨7£©ÔªËØÍ­ÓëÄøµÄµÚ¶þµçÀëÄÜ·Ö±ðΪ£ºICu=1959kJ¡¤mol-1,INi=1753kJ¡¤mol-1,ICu£¾INiÔ­ÒòÊÇ__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³ÖÐѧ»¯Ñ§ÐËȤС×éΪÁ˵÷²éµ±µØijһºþ²´µÄË®ÖÊÎÛȾÇé¿ö£¬ÔÚ×¢Èëºþ²´µÄ3¸öÖ÷ҪˮԴµÄÈë¿Ú´¦²É¼¯Ë®Ñù£¬²¢½øÐÐÁË·ÖÎö£¬¸ø³öÁËÈçÏÂʵÑéÐÅÏ¢£ºÆäÖÐÒ»´¦Ë®Ô´º¬ÓÐA¡¢BÁ½ÖÖÎïÖÊ£¬Ò»´¦º¬ÓÐC¡¢DÁ½ÖÖÎïÖÊ£¬Ò»´¦º¬ÓÐEÎïÖÊ£¬A¡¢B¡¢C¡¢D¡¢EΪÎåÖÖ³£¼û»¯ºÏÎ¾ùÓÉϱíÖеÄÀë×ÓÐγɣº

ÑôÀë×Ó

K+ Na+ Cu2+ Al3+

ÒõÀë×Ó

SO42£­ HCO3- NO3- OH£­

ΪÁ˼ø±ðÉÏÊö»¯ºÏÎï¡£·Ö±ðÍê³ÉÒÔÏÂʵÑ飬Æä½á¹ûÊÇ£º

¢Ù½«ËüÃÇÈÜÓÚË®ºó£¬DΪÀ¶É«ÈÜÒº£¬ÆäËû¾ùΪÎÞÉ«ÈÜÒº£»

¢Ú½«EÈÜÒºµÎÈëµ½CÈÜÒºÖгöÏÖ°×É«³Áµí£¬¼ÌÐøµÎ¼Ó£¬³ÁµíÈܽ⣻

¢Û½øÐÐÑæÉ«·´Ó¦£¬Ö»ÓÐB¡¢CΪ×ÏÉ«£¨Í¸¹ýÀ¶É«îܲ£Á§£©£»

¢ÜÔÚ¸÷ÈÜÒºÖмÓÈëÏõËá±µÈÜÒº£¬ÔÙ¼Ó¹ýÁ¿Ï¡ÏõËᣬAÖзųöÎÞÉ«ÆøÌ壬C¡¢DÖж¼ÄܲúÉú°×É«³Áµí£º

¢Ý½«B¡¢DÁ½ÈÜÒº»ìºÏ£¬Î´¼û³Áµí»òÆøÌåÉú³É¡£

¸ù¾ÝÉÏÊöʵÑéÌî¿Õ£º

£¨1£©Ð´³öC¡¢D¡¢µÄ»¯Ñ§Ê½£ºC_______£¬D______¡£

£¨2£©½«º¬1 mol AµÄÈÜÒºÓ뺬l mol EµÄÈÜÒº·´Ó¦ºóÕô¸É£¬½öµÃµ½Ò»ÖÖ»¯ºÏÎÇëд³öAÓëE·´Ó¦µÄÀë×Ó·½³Ìʽ£º_______________¡£

£¨3£©ÔÚAÈÜÒºÖмÓÈëÉÙÁ¿³ÎÇåʯ»ÒË®£¬ÆäÀë×Ó·½³ÌʽΪ_____________________¡£

£¨4£©C³£ÓÃ×÷¾»Ë®¼Á£¬ÓÃÀë×Ó·½³ÌʽºÍÊʵ±ÎÄ×Ö˵Ã÷Æ侻ˮԭÀí______________________¡£

£¨5£©ÈôÏòº¬ÈÜÖÊ0.5 molµÄCÈÜÒºÖÐÖðµÎ¼ÓÈëBa(OH)2ÈÜÒº£¬Éú³É³ÁµíÖÊÁ¿×î´óΪ__________g¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»Ø´ðÏÂÁйØÓÚNO¡¢NO2µÄÎÊÌ⣺

(1)Æû³µÅÅÆø¹ÜÉÏ×°Óд߻¯×ª»¯Æ÷¿É¼õÉÙβÆø¶Ô»·¾³µÄÎÛȾ£¬Æû³µÎ²ÆøÖеÄÓк¦ÆøÌåCOºÍNO·´Ó¦¿Éת»¯ÎªÎÞº¦ÆøÌåÅÅ·Å£¬Ð´³öÏà¹Ø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º___________________

(2)¹¤ÒµÉú²úÖÐÀûÓð±Ë®ÎüÊÕSO2ºÍNO2£¬Ô­ÀíÈçͼËùʾ£º

NO2±»ÎüÊÕºóÉú³ÉµÄï§ÑÎÊÇ____________(Ìѧʽ)£»ÎªÖ¤Ã÷ÈÜÒºÖÐNH4+µÄ´æÔÚ£¬¿ÉÏòÉÙÁ¿ï§ÑÎÈÜÒºÖмÓÈë___________ÈÜÒº¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸