ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ä¿Ç°ÊÀ½çÉϹ㷺²ÉÓð±ºÍ¶þÑõ»¯Ì¼¸ßθßѹÏÂÖƱ¸ÄòËØ£¬Ö÷Òª·´Ó¦ÎªÁ½²½£º

µÚÒ»²½£ºÉú³É°±»ù¼×Ëáï§

µÚ¶þ²½£º°±»ù¼×Ëáï§ÍÑË®Éú³ÉÄòËØ

2NH3(l)£«CO2(g)H2NCOONH4(°±»ù¼×Ëáï§) (l) H1

H2NCOONH4(l)H2O(l)£«H2NCONH2(l)

H2

¿ìËÙ·ÅÈÈ

ÂýËÙÎüÈÈ

£¨1£©Ð´³öÖƱ¸ÄòËصÄ×Ü·´Ó¦»¯Ñ§·½³Ìʽ£º____________________________£¬¸Ã·´Ó¦ÈÈΪ£¬Ôò______________£¨¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©¡£

£¨2£©ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________________¡£

A.ÀûÓöþÑõ»¯Ì¼ÖƱ¸ÄòËØÊǼõ»ºÎÂÊÒЧӦµÄÓÐЧ·½·¨

B.µÚ¶þ²½·´Ó¦¸ßÎÂÌõ¼þÏÂ×Ô·¢½øÐÐ

C.Ìá¸ßͶÁÏÖеÄˮ̼±ÈÓÐÀûÓÚÄòËصÄÉú³É

D.µÚÒ»²½·´Ó¦µÄ»î»¯ÄÜ´óÓÚµÚ¶þ²½·´Ó¦

£¨3£©ÄòËØÉú²ú¹ý³ÌÖÐת»¯ÂÊͨ³£ÓöþÑõ»¯Ì¼×ª»¯ÂÊÀ´±íʾ£¬µ±¶þÑõ»¯Ì¼ÆðʼŨ¶ÈΪʱ£¨£¬Ë®Ì¼±È=0.5£©£¬ÄòËØƽºâת»¯ÂÊË氱̼±ÈµÄ±ä»¯Èç±íËùʾ£º

°±Ì¼±È/Ũ¶È±È

2.95

3.10

3.20

3.50

ÄòËØƽºâת»¯ÂÊ/%

56.4

57.5

57.9

60.0

Ìá¸ß°±Ì¼±ÈÓÐÀûÓÚÉú³ÉÄòËØ£¬Ö÷ÒªÓÐÁ½¸öÔ­Òò£ºÒ»ÊÇÔö´ó°±ÆøŨ¶ÈÓÐÀûÓÚ·´Ó¦ÕýÏòÒƶ¯£»¶þÊÇ_____________¡£°±Ì¼±ÈΪ3.50ʱ£¬Çó¸Ã״̬ϵÄÖƱ¸ÄòËØ×Ü·´Ó¦Æ½ºâ³£ÊýK=___________.

£¨4£©Ò»¶¨Ìõ¼þÏ£¬ÔÚͼÖлæÖÆ°±»ù¼×Ëá泥¨£©ÔÚ·´Ó¦¹ý³ÌÖÐÎïÖʵÄÁ¿Óëʱ¼äµÄ¹Øϵͼ¡£

_______________

£¨5£©Í¨¹ýÖ±½ÓÄòËØȼÁϵç³Ø×°Öã¬ÊµÏÖÁË¡°ÄòËØÄÜ¡±µÄÀûÓã¬ÇÒ²úÉúÎÞÎÛȾµÄ²úÎд³ö¸º¼«·´Ó¦£º____________________________.

¡¾´ð°¸¡¿ ´óÓÚ AB ¹ýÁ¿°±ÆøÓëµÚ¶þ²½·´Ó¦Éú³ÉµÄË®½áºÏ£¬´Ù½øƽºâÕýÏòÒƶ¯ 0.13 £¨³öÏÖÏÈÉÏÉýºóϽµÔÙÎȶ¨Ç÷ÊƼ´µÃ·Ö£©

¡¾½âÎö¡¿

£¨1£©ÒÀ¾Ý¸Ç˹¶¨ÂÉ·ÖÎö£»

£¨2£©¸ù¾Ý·´Ó¦×Ô·¢ÐÔ¡¢·´Ó¦ËÙÂʺͻ¯Ñ§Æ½ºâµÄ»ù±¾Ô­Àí×÷´ð£»

£¨3£©´ÓŨ¶ÈÓ°Ï컯ѧƽºâµÄ½Ç¶È¿¼ÂÇ£»¸ù¾ÝÌâÒ⣬ÁгöÈý¶Îʽ£¬¸ù¾Ýƽºâ³£Êý±í´ïʽÁÐʽ¼ÆË㣻

£¨4£©ÒÀ¾ÝÁ½²½·´Ó¦·ÖÎö×÷´ð£»

£¨5£©ÓÉÌâÄ¿ÐÅÏ¢¿ÉÖªÔ­µç³Ø¸º¼«·¢ÉúÄòËصÄÑõ»¯·´Ó¦£¬ÔÚ¼îÐÔÌõ¼þϽøÐУ¬¹ÊÉú³É̼Ëá¸ù¡£

£¨1£©ÓÉÌâÒâ¿ÉÖªÁ½²½·´Ó¦Ïà¼Ó¼´Îª×Ü·´Ó¦£¬¹ÊÖƱ¸ÄòËصÄ×Ü·´Ó¦»¯Ñ§·½³ÌʽΪ£º£»¸ù¾Ý¸Ç˹¶¨ÂÉ£¬£¬ÔòÓУ¬¹Ê´ð°¸Îª£º£»´óÓÚ£»

£¨2£©A. ÓÃÓÚÉú²úÄòËØÊÇÏûºÄ¶þÑõ»¯Ì¼µÄÓÐЧ;¾¶£¬AÏîÕýÈ·£»

B. µÚ¶þ²½·´Ó¦£¬£¬¸ßÎÂ×Ô·¢£¬BÏîÕýÈ·£»

C. Ë®Êǵڶþ²½·´Ó¦²úÎÌá¸ßˮ̼±È²»ÀûÓÚÌá¸ßÄòËصÄÉú³É£¬CÏî´íÎó£»

D. µÚÒ»²½·´Ó¦¿ìËÙ£¬µÚ¶þ²½·´Ó¦ÂýËÙ£¬¹ÊµÚÒ»²½·´Ó¦»î»¯ÄÜСÓÚµÚ¶þ²½·´Ó¦£¬DÏî´íÎó£»

¹Ê´ð°¸ÎªAB¡£

£¨3£©¹ýÁ¿°±Æø½áºÏµÚ¶þ²½·´Ó¦Éú³ÉµÄË®£¬´Ù½øƽºâÕýÏòÒƶ¯£»ÓÉÌâÄ¿ÖеÄˮ̼±ÈºÍ°±Ì¼±È£¬½áºÏÄòËØƽºâת»¯ÂÊ£¬¼ÙÉ迪ʼ£¬

¿ªÊ¼ 7 2 0

ת»¯ 2.4 1.2 1.2

ƽºâ 4.6 0.8 1.2£¬

Òòˮ̼±ÈΪ0.5£¬ÆðʼˮÕôÆøµÄŨ¶ÈΪ2mol/L0.5=1mol/L£¬¹ÊË®ÕôÆøµÄƽºâŨ¶ÈΪ1.2mol/L+1mol/L=2.2mol/L£¬Ôòƽºâ³£Êý£»

£¨4£©µÚÒ»²½·´Ó¦¿ìËÙÉú³É°±»ù¼×Ëá泥¬µÚ¶þ²½·´Ó¦ÏûºÄ½ÏÂý£¬¹Ê°±»ù¼×Ëáï§ÏÈÀÛ»ýºó´ïµ½Æ½ºâŨ¶È£¬¹Ê´ð°¸Îª£º£¨³öÏÖÏÈÉÏÉýºóϽµÔÙÎȶ¨Ç÷ÊƼ´µÃ·Ö£©£»

£¨5£©½áºÏ¸º¼«ÄòËزúÉúµªÆøºÍ̼Ëá¸ùÀë×Ó£¬Æäµç¼«·´Ó¦Ê½Îª£º¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ã¾ºÏ½ð¼°Ã¾µÄ»¯ºÏÎïÔÚÉú²ú¡¢Éú»îÖÐÓÐ׏㷺µÄÓ¦Óá£

(1)þÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ__________________¡£

(2)ÓÃË®ÂÈþʯ(Ö÷Òª³É·ÖΪMgCl2¡¤6H2O)ÖƱ¸½ðÊôþµÄ¹Ø¼üÁ÷³ÌÈçÏ£º

¸Ã¹¤ÒÕÖпÉÑ­»·Ê¹ÓõÄÎïÖÊÓÐ______________¡£

(3)´¢Çâ²ÄÁÏMg(AlH4)2ÔÚ110¡«200 ¡æµÄ·´Ó¦ÎªMg(AlH4)2=MgH2£«2Al£«3H2¡ü£¬Ã¿×ªÒÆ6 molµç×ÓÉú³ÉÇâÆøµÄÎïÖʵÄÁ¿Îª________mol¡£

(4)¼îʽ̼ËáþÃܶÈС£¬ÊÇÏð½ºÖÆÆ·µÄÓÅÁ¼ÌîÁÏ£¬¿ÉÓø´ÑÎMgCO3¡¤(NH4)2CO3¡¤2H2O×÷Ô­ÁÏÖƱ¸¡£ÖƱ¸¹ý³ÌÖУ¬ÐèÒªÓõ½Â±Ë®(ÂÈ»¯Ã¾ÈÜÒº)¡£Ä³¿ÆÑÐС×éÓóÁµíµÎ¶¨·¨·ÖÎö²úÆ·ÖÐCl£­µÄº¬Á¿£¬³ÆÈ¡6.100 0 g²úÆ·ÓÃÊÊÁ¿ÏõËáÈܽ⣬¾­Ï¡Ê͵Ȳ½Öè×îÖÕÅäµÃ500 mLµÄÈÜÒº¡£

a£®×¼È·Á¿È¡25.00 mL ´ý²âÒº£¬ÓÃ0.100 0 mol/L AgNO3±ê×¼ÒºµÎ¶¨£¬µÎ¶¨Ç°ºóµÎ¶¨¹ÜÖеÄÒºÃæ¶ÁÊýÈçͼËùʾ£¬ÔòµÎ¶¨¹ý³ÌÖÐÏûºÄ±ê×¼ÒºµÄÌå»ýΪ________mL¡£

b.

AgCl

AgBr

AgI

Ag2CrO4

Ksp

2¡Á10£­10

5.4¡Á10£­13

8.3¡Á10£­17

2¡Á10£­12

ÑÕÉ«

°×

µ­»Æ

Ȯ

שºì

²ÎÕÕÉϱíÊý¾Ý¼°ÐÅÏ¢·ÖÎö£¬µÎ¶¨Ê±¿ÉÒÔ×÷ָʾ¼ÁµÄÊÇ________(ÌîÊý×ÖÐòºÅ)¡£

¢ÙCaCl2¡¡¡¡¢ÚNaBr¡¡¡¡¢ÛNaI¡¡¡¡¢ÜK2CrO4

¡¾ÌâÄ¿¡¿ºÏÀíÀûÓÃ×ÊÔ´£¬¼ÓÇ¿»·¾³±£»¤£¬½µµÍ̼ÅÅ·Å£¬ÊµÊ©µÍ̼¾­¼ÃÊǽñºó¾­¼ÃÉú»îÖ÷Á÷¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÏÂÁдëÊ©²»ÀûÓÚÓÐЧ¼õÉÙ¶þÑõ»¯Ì¼µÄÊÇ______(Ìî×Öĸ)¡£

a.Ö²Ê÷ÔìÁÖ£¬±£»¤É­ÁÖ£¬±£»¤Ö²±»

b.¼Ó´ó¶ÔúºÍʯÓ͵Ŀª²É£¬²¢¹ÄÀøʹÓÃÒº»¯Ê¯ÓÍÆø

c.´óÁ¦·¢Õ¹·çÄÜ¡¢Ë®Á¦¡¢³±Ï«ÄÜ·¢µçºÍºËµç£¬´óÁ¦ÍÆÐÐÌ«ÑôÄܵÄ×ۺϿª·¢

d.ÍƹãʹÓýÚÄܵƺͽÚÄܵçÆ÷£¬Ê¹Óÿյ÷ʱÏļ¾Î¶Ȳ»ÒËÉèÖùýµÍ£¬¶¬Ìì²»Ò˹ý¸ß

(2)¿Æѧ¼ÒÖÂÁ¦ÓÚ¶þÑõ»¯Ì¼µÄ¡°×éºÏת»¯¡±¼¼ÊõÑо¿£¬È罫CO2ºÍH2ÒÔ1¡Ã4µÄÎïÖʵÄÁ¿Ö®±È»ìºÏͨÈë·´Ó¦Æ÷£¬ÔÚÊʵ±Ìõ¼þÏ·´Ó¦¿É»ñµÃÒ»ÖÖÖØÒªÄÜÔ´¡£ÇëÍê³ÉÒÔÏ»¯Ñ§·½³Ìʽ£ºCO2+4H2_____+2H2O¡£

(3)ÓÃCO2ºÏ³ÉȼÁϼ״¼(CH3OH)ÊÇ̼¼õÅŵÄз½Ïò¡£ÏÖ½øÐÐÈçÏÂʵÑ飺ijζÈÏÂÔÚÌå»ýΪ1LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë2molCO2ºÍ6molH2£¬·¢Éú·´Ó¦£ºCO2(g)+3H2(g)CH3OH(g)+H2O(g)¡£ÏÖ²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊv(H2)=___¡£

(4)ÒÔKOHΪµç½âÖʵļ״¼È¼Áϵç³Ø×Ü·´Ó¦Îª2CH3OH+3O2+4KOH=2K2CO3+6H2O¡£Í¨Èë¼×´¼µÄµç¼«ÎªÈ¼Áϵç³ØµÄ______(Ìî¡°Õý¡±»ò¡°¸º¡±)¼«£¬Õý¼«·´Ó¦Ê½Îª_______¡£

¡¾ÌâÄ¿¡¿¶þÑõ»¯Ì¼ÀûÓþßÓÐÊ®·ÖÖØÒªµÄÒâÒ壬¿Æѧ¼ÒÓÐÒÔϼ¸¸öÉèÏë¡£

(1)ÓÃÌ«ÑôÄܽ«CO2ת»¯³ÉO2ºÍC(ʯīϩ£©£¬ÆäÉèÏëÈçͼ£º

ÔòÖØÕûϵͳ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________¡£

(2)¶þÑõ»¯Ì¼ºÍÇâÆøÔÚ´ß»¯¼Á×÷ÓÃÏ¿ÉÖÆÈ¡µÍ̼ϩÌþ¡£ÔÚÒ»ÃܱÕÈÝÆ÷ÖзֱðͶÈë1molCO2¡¢3molH2£¬·¢Éú·´Ó¦£º2CO2(g)+6H2(g)C2H4(g)+ 4H2O(g) ¡÷H£»ÔÚ²»Í¬Î¶ÈÏ£¬Óô«¸Ð¼¼Êõ²â³öƽºâʱH2µÄÎïÖʵÄÁ¿±ä»¯¹ØϵÈçͼËùʾ¡£

¢ÙÆäËüÌõ¼þ²»±ä£¬ÆðʼʱÈô°´lmolCO2¡¢2molH2½øÐÐͶÁÏ£¬CO2ת»¯Âʽ«________(Ìî¡°Ôö´ó¡±¡¢¡° ¼õС¡±»ò¡°²»±ä¡±£©£»

¢Ú¡÷H_____0£¨Ìî¡°>¡±¡°<¡±¡° ²»ÄÜÈ·¶¨¡±£©¡£

¢ÛÈô²âÊÔÖÐÌåϵÄÚÎÞÑõÆø²úÉú£¬ÊÔ½áºÏͼʾÍƶÏÈÈÎȶ¨ÐÔC2H4 _____H2O £¨Ìî¡°>¡±¡°<¡±¡° ²»ÄÜÈ·¶¨¡±£©¡£

(3)Óð±Ë®ÎüÊÕCO2ÖÆ»¯·Ê(NH4HCO3)

¢ÙÒÑÖª£ºNH3¡¤H2O(aq)NH4+(aq)+OH-(aq) ¡÷H1=akJ/mol

CO2(g)+H2O(l)H2CO3(aq) ¡÷H2=bkJ/mol

H2CO3(aq)+OH-(aq)HCO3-(aq)+H2O(l) ¡÷H3=ckJ/mol

ÔòÀûÓÃNH3 H2OÎüÊÕCO2ÖƱ¸NH4HCO3µÄÈÈ»¯Ñ§·½³ÌʽΪ______________________£»

¢ÚÒÑÖª³£ÎÂÏÂÏà¹ØÊý¾ÝÈç±í£º

Kb(NH3¡¤H2O)

2¡Á10-5kJ/mol

Ka1(H2CO3)

4¡Á10-7kJ/mol

Ka2(H2CO3)

4¡Á10-11kJ/mol

Ôò·´Ó¦NH4++HCO3-+H2ONH3 H2O+H2CO3µÄƽºâ³£ÊýK=___________¡£

¡¾ÌâÄ¿¡¿µþµª»¯¼Ø£¨£©ÄÜ´Ùʹ×÷Îï»òÄÑÓÚÃÈ·¢µÄÖÖ×Ó·¢Óý¡£Éè¼ÆÈçÏÂʵÑéÖƱ¸µþµª»¯¼Ø²¢²â¶¨Æä´¿¶È£º

I.ÖƱ¸

²½Öè1£ºÖƱ¸ÑÇÏõËᶡõ¥£¨£©

·´Ó¦×°ÖÃÈçͼ1£¨¼Ð³Ö×°ÖÃÂÔÈ¥£©£¬ÏòÉÕ±­ÖÐÒÀ´Î¼ÓÈëÏ¡ÁòËá¡¢¶¡´¼¡¢ÑÇÏõËáÄÆÈÜÒº£¬´ý·´Ó¦ÍêÈ«ºó£¬·ÖÀë³öÉϲãÓÍ×´ÎÓú͵ĻìºÏÈÜҺϴµÓÈý´Î£¬¾­¸ÉÔïºó±¸Óá£

²½Öè2£ºÖƱ¸µþµª»¯¼Ø

·´Ó¦×°ÖÃÈçͼ2£¨¼Ð³Ö¼°¼ÓÈÈ×°Ö÷ȥ£©£¬ÏòÒÇÆ÷AÖмÓÈëÒÒ´¼ÈÜÒº¡¢µÄÁª°±£¨£©¡¢ÑÇÏõËᶡõ¥£¬ÕôÆûÔ¡¼ÓÈÈ£¬·´Ó¦ÍêÈ«ºó£¬µþµª»¯¼Ø¼´³Áµí³öÀ´£¬±ùÔ¡ÀäÈ´£¬¹ýÂË£¬ÏÈÓÃÎÞË®ÒÒ´¼Ï´µÓ£¬ÔÙÓÃÎÞË®ÒÒÃÑÏ´µÓ£¬ÔÚ¿ÕÆøÖÐÓÚ¸ÉÔï¡£

Ïà¹ØÎïÖÊÐÔÖÊÈçÏ£º

ÎïÖÊ

ÑÕÉ«¡¢×´Ì¬

·Ðµã£¨¡æ£©

ÈܽâÐÔ

ÎÞÉ«¾§Ìå

ÊÜÈÈÒ×·Ö½â

Ò×ÈÜÓÚË®£¬Î¢ÈÜÓÚÒÒ´¼£¬²»ÈÜÓÚÒÒÃÑ

ÎÞÉ«ÒºÐÝ

118

΢ÈÜÓÚË®£¬ÓëÒÒ´¼¡¢ÒÒÃÑ»ìÈÜ

ÎÞÉ«»òµ­»ÆÉ«ÓÍ×´ÒºÌå

78

²»ÈÜÓÚË®£¬ÓëÒÒ´¼¡¢ÒÒÃÑ»ìÈÜ

ÎÞÉ«ÓÍ×´ÒºÌå

118

ÓëË®¡¢ÒÒ´¼»ìÈÜ£¬²»ÈÜÓÚÒÒÃÑ

Çë»Ø´ð£º

£¨1£©ÒÇÆ÷AµÄÃû³ÆΪ_____________.

£¨2£©²½Öè1ÖзÖÀë³öÑÇÏõËᶡõ¥µÄ²Ù×÷Ãû³ÆΪ_____________£»²½Öè1ÖÐÓÃNaClºÍNaHCO3µÄ»ìºÏÈÜҺϴµÓµÄÄ¿µÄÊÇ__________________________.

£¨3£©²½Öè2ÖбùÔ¡ÀäÈ´µÄÄ¿µÄÊÇ__________________________£»²½Öè2ÖиÉÔï²úÆ·µÄζȿØÖÆÔÚ55~60¡æ£¬Ô­ÒòÊÇ__________________________

£¨4£©ÈçÐèÌá¸ß²úÆ·µÄ´¿¶È£¬¿ÉÔÚ_____________£¨Ìî±àºÅ£©ÖнøÐÐÖؽᾧ¡£

A.ÎÞË®ÒÒ´¼ B.ÎÞË®ÒÒÃÑ C.Ë® D.ÒÒ´¼µÄË®ÈÜÒº

¢ò.·Ö¹â¹â¶È·¨²â¶¨²úÆ·µÄ´¿¶È

Ô­Àí£ºÓë·´Ó¦·Ç³£ÁéÃô£¬Éú³ÉºìÉ«ÂçºÏÎÔÚÒ»¶¨²¨³¤Ï²âÁ¿ºìÉ«ÈÜÒºµÄÎü¹â¶È£¬ÀûÓá°Îü¹â¶È¡±ÇúÏßÈ·¶¨ÑùÆ·ÈÜÒºÖеġ£²â¶¨²½ÖèÈçÏ£º

¢ÙÓÃÆ·ÌåÅäÖƱê×¼ÈÜÒº£»

¢ÚÅäÖÆÒ»×éÏàͬÌå»ý£¨£©²»Í¬Å¨¶ÈµÄ±ê×¼ÈÜÒº£¬·Ö±ð¼ÓÈ루×ãÁ¿£©±ê×¼ÈÜÒº£¬Ò¡ÔÈ£¬²âÁ¿Îü¹â¶È£¬»æÖƱê×¼ÈÜÒºµÄÓëÎü¹â¶ÈµÄ¹ØϵÇúÏߣ¬Èçͼ£»

¢Û²úÆ·²â¶¨£º³ÆÈ¡0.360g²úÆ·£¬Åä³ÉÈÜÒº£¬È¡³öÓÚ±ê×¼¹ÜÖУ¬¼ÓÈ루×ãÁ¿£©±ê×¼ÈÜÒº£¬Ò¡ÔÈ£¬²âµÃÎü¹â¶ÈΪ0.6¡£

£¨5£©ÊµÑéÊÒÓþ§ÌåÅäÖƱê×¼ÈÜÒºµÄ·½·¨Îª_________________.

£¨6£©²úÆ·µÄ´¿¶ÈΪ_________________£»Èô¢ÛÖмÓÈëµÄ±ê×¼ÈÜÒº²»×ãÒÔ½«²úÆ·ÍêÈ«·´Ó¦£¬Ôò²âµÃµÄ²úÆ·´¿¶È________________£¨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø