ÌâÄ¿ÄÚÈÝ

(12·Ö) ijʵÑéС×éÓÃÏÂͼװÖýøÐÐÒÒ´¼´ß»¯Ñõ»¯µÄʵÑé¡£

(1)ʵÑé¹ý³ÌÖÐÍ­Íø³öÏÖºìÉ«ºÍºÚÉ«½»ÌæµÄÏÖÏó£¬Çëд³öÏàÓ¦µÄ»¯Ñ§·´Ó¦·½³Ìʽ                                   ¡£
ÔÚ²»¶Ï¹ÄÈë¿ÕÆøµÄÇé¿öÏ£¬Ï¨Ãð¾Æ¾«µÆ£¬·´Ó¦ÈÔÄܼÌÐø½øÐУ¬ËµÃ÷¸ÃÒÒ´¼Ñõ»¯·´Ó¦ÊÇ        ·´Ó¦£¨·ÅÈÈ»òÎüÈÈ£©¡£
(2)¼×ºÍÒÒÁ½¸öˮԡ×÷Óò»Ïàͬ¡£¼×µÄ×÷ÓÃÊÇ            £»
ÒÒµÄ×÷ÓÃÊÇ           ¡£
(3)·´Ó¦½øÐÐÒ»¶Îʱ¼äºó£¬¸ÉÔïÊÔ¹ÜaÖÐÄÜÊÕ¼¯µ½²»Í¬µÄÎïÖÊ£¬ËüÃÇÊÇ           ,
¼¯ÆøÆ¿ÖÐÊÕ¼¯µ½µÄÆøÌåµÄÖ÷Òª³É·ÖÊÇ         ¡£
(4)ÈôÊÔ¹ÜaÖÐÊÕ¼¯µ½µÄÒºÌåÓÃ×ÏɫʯÈïÊÔÖ½¼ìÑ飬ÊÔÖ½ÏÔºìÉ«£¬ËµÃ÷ÒºÌåÖл¹º¬ÓР              ¡£Òª³ýÈ¥¸ÃÎïÖÊ£¬¿ÉÏÈÔÚ»ìºÏÒºÖмÓÈë          (Ìîд×Öĸ)¡£È»ºó£¬ÔÙͨ¹ý      (ÌîʵÑé²Ù×÷Ãû³Æ)¼´¿É³ýÈ¥¡£
a£®ÂÈ»¯ÄÆÈÜÒº    b£®±½ c£®Ì¼ËáÇâÄÆÈÜÒº    d£®ËÄÂÈ»¯Ì¼
£¨12·Ö£©(1)2Cu+O22CuO  CH3CH2OH+CuOCH3CHO+Cu+H2
2CH3CH2OH + O2   2CH3CHO  +  2H2O        ·ÅÈÈ 
(2)¼ÓÈÈÀäÈ´ (3)ÒÒÈ© ÒÒ´¼ Ë® µªÆø (4) ÒÒËá  c ÕôÁó
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ij¹¤³§µÄ·ÏË®Öк¬ÓÐFeSO4¡¢H2SO4¡¢Ag2SO4¡¢Al2£¨SO4£©3¼°Ò»Ð©ÎÛÄࡣijÑо¿ÐÔѧϰ¿ÎÌâ×é²â¶¨ÁË·ÏË®Öи÷ÎïÖʵĺ¬Á¿²¢²éÕÒÁËÈܽâ¶ÈÊý¾Ý£¬ÏÖÁбíÈçÏÂ
±íÒ»·ÏË®Öи÷ÎïÖʵĺ¬Á¿
ÎïÖÊ
FeSO4
H2SO4
Ag2SO4
Al2(SO4)2
ÎÛÄà
ÖÊÁ¿·ÖÊý£¯£¨£¥£©
15.0
7.0
0.40
0.34
5.0
±í¶þ  FeSO4ºÍAl2£¨SO4£©3ÔÚË®ÖеÄÈܽâ¶È
ζÈ/¡æ
0
10
20
30
40
50
FeSO4ÈÜÒº¶È£¨g£©
15.6
20.5
26.5
32.9
40.2
48.6
Al2(SO4)3Èܽâ¶È£¨g£©
31.2
33.5
36.4
40.4
45.7
52.2
   ¸Ã¿ÎÌâ×é¸ù¾Ý±íÖÐÊý¾Ý£¬Éè¼ÆÁËÎÛË®´¦Àí·½°¸£¬ÄâÀûÓøó§µÄ·ÏÌúм£¨ÓÐÉÙÁ¿Ðâ°ß£©¡¢ÉÕ¼îÈÜÒººÍÁòËá´¦Àí´ËÎÛË®£¬»ØÊÕFeSO4¡¤7H2OºÍAg¡£
£¨1£©ÇëÌîдÏÂÁпհף¬Íê³ÉµÃµ½AgµÄʵÑé·½°¸£º
¢Ù½«´øÓÐÐâ°ßµÄ·ÏÌúмÏȺóÓÃÈȵÄÉÕ¼îÈÜÒººÍÈÈË®½øÐÐÏ´µÓ£¬Ä¿µÄÊÇ          ¡£
¢Ú½«¹¤³§·ÏË®¹ýÂË£¬ÓÃÉÙÁ¿Ë®Ï´µÓÂËÔü£¬Ï´µÓÒº²¢ÈëÂËÒººó±£Áô´ýÓã»
¢Û                  £¬Ä¿µÄÊÇʹAg+È«²¿»¹Ô­Îª½ðÊôAg£»
¢Ü               £¬Ä¿µÄÊÇ·ÖÀë³öAg£»
£¨2£©Çëд³öºóÐøµÄ²½Ö裬³ýÈ¥Al3+£¬µÃµ½Ö÷Òª³É·ÖΪFeSO4¡¤7H2O¾§Ìå¡£
¢Ý½«µÚ           ²½ÓëµÚ¢Ü²½ËùµÃÂËÒº»ìºÏºó£¬¼ÓÈëÉÙÁ¿ÁòËáÖÁ»ìºÏÒºµÄpHΪ3-4ºó£¬                  £¬Â˳öFeSO4¡¤7H2O¾§Ìå
£¨3£©Ð´³ö²½Öè¢ÛÖÐËùÓл¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ              ¡£
£¨4£©ÔÚ²½Öè¢ÝÖУ¬¼ÓÈëÉÙÁ¿ÁòËáµ÷½ÚpHµÄÄ¿µÄÊÇ                ¡£
ÒÑ֪ij°×É«»ìºÏÎï·ÛÄ©Öк¬ÓÐCuSO4¡¢K2SO4¡¢NH4HCO3¡¢NH4Cl¡¢NaClÎåÖÖÎïÖÊÖеÄÁ½ÖÖ£¬ÇÒÎïÖʵÄÁ¿Ö®±ÈΪ1:1¡£ÇëÍê³ÉÏÂÊö̽¾¿»ìºÏÎï×é³ÉµÄʵÑé¡£
ÏÞÑ¡µÄÒÇÆ÷¡¢ÓÃÆ·ºÍÊÔ¼Á£ºÉÕ±­¡¢ÊԹܡ¢²£Á§°ô¡¢Á¿Í²¡¢½ºÍ·µÎ¹Ü¡¢Ò©³×¡¢¾Æ¾«µÆ¡¢»ð²ñ¡¢ÊԹܼС¢Ä÷×Ó£»ºìɫʯÈïÊÔÖ½¡¢µí·Ûµâ»¯¼ØÊÔÖ½£»1mol/LÁòËá¡¢1mol/LÏõËá¡¢1mol/LÑÎËá¡¢1mol/L NaOHÈÜÒº¡¢Ba(NO3)2ÈÜÒº¡¢AgNO3ÈÜÒº¡¢ÕôÁóË®¡£
£¨Ò»£©³õ²½Ì½¾¿
È¡ÊÊÁ¿¹ÌÌå»ìºÏÎïÓÚÉÕ±­ÖУ¬¼ÓÈëÕôÁóË®½Á°è£¬»ìºÏÎïÍêÈ«Èܽ⣬µÃµ½ÎÞɫ͸Ã÷ÈÜÒºA£¬Í¬Ê±Îŵ½ÓÐÇá΢µÄ´Ì¼¤ÐÔÆøζ¡£ÓýºÍ·µÎ¹ÜÈ¡ÉÙÁ¿ÈÜÒºAÓÚÊÔ¹ÜÖУ¬ÔٵμÓÏ¡ÏõËᣬÈÜÒºÖÐÓÐÎÞÉ«ÆøÅݲúÉú£»¼ÌÐøµÎ¼Ó¹ýÁ¿Ï¡ÏõËáÖÁÈÜÒºÖв»ÔÙ²úÉúÆøÅÝ£¬µÃµ½ÎÞɫ͸Ã÷ÈÜÒºB¡£
£¨1£©ÉÏÊöʵÑé¿ÉÒԵõ½µÄ³õ²½½áÂÛÊÇ
                                                            ¡£                           
£¨¶þ£©½øÒ»²½Ì½¾¿
£¨2£©ÇëÉè¼ÆʵÑé·½°¸½øÒ»²½È·¶¨¸Ã¹ÌÌå»ìºÏÎïµÄ×é³É¡£ÐðÊöʵÑé²Ù×÷¡¢Ô¤ÆÚÏÖÏóºÍ½áÂÛ¡£
ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
 
 
 
 
 
¼×¡¢ÒÒÁ½Î»Í¬Ñ§Éè¼ÆÓÃʵÑéÈ·¶¨Ä³ËáHAÊÇÈõµç½âÖÊ£¬´æÔÚµçÀëƽºâ£¬ÇҸıäÌõ¼þƽºâ·¢ÉúÒƶ¯¡£ÊµÑé·½°¸ÈçÏ£º
¼×£ºÈ¡´¿¶ÈÏàͬ£¬ÖÊÁ¿¡¢´óСÏàµÈµÄпÁ£ÓÚÁ½Ö»ÊÔ¹ÜÖУ¬Í¬Ê±¼ÓÈë0.1 mol¡¤L-1µÄ HA¡¢HClÈÜÒº¸÷10 mL£¬°´Í¼×°ºÃ£¬¹Û²ìÏÖÏó£»

ÒÒ£º¢Ù ÓÃpH¼Æ²â¶¨ÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1 mol¡¤L-1 HAºÍHClÈÜÒºµÄpH£»
¢Ú ÔÙÈ¡0.1 mol¡¤L-1µÄHAºÍHClÈÜÒº¸÷2µÎ£¨1µÎԼΪ1/25 mL£©·Ö±ðÏ¡ÊÍÖÁ100 mL£¬ÔÙÓÃpH¼Æ²âÆäpH±ä»¯
£¨1£©ÒÒ·½°¸ÖÐ˵Ã÷HAÊÇÈõµç½âÖʵÄÀíÓÉÊÇ£º²âµÃ0.1 mol¡¤L-1µÄHAÈÜÒºµÄpH         1£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£© £»¼×·½°¸ÖУ¬ËµÃ÷HAÊÇÈõµç½âÖʵÄʵÑéÏÖÏóÊÇ£º          
A£®¼ÓÈëHClÈÜÒººó£¬ÊÔ¹ÜÉÏ·½µÄÆøÇò¹ÄÆð¿ì
B£®¼ÓÈëHAÈÜÒººó£¬ ÊÔ¹ÜÉÏ·½µÄÆøÇò¹ÄÆðÂý
C£®¼ÓÈëÁ½ÖÖÏ¡Ëáºó£¬Á½¸öÊÔ¹ÜÉÏ·½µÄÆøÇòͬʱ¹ÄÆð£¬ÇÒÒ»Ñù´ó
£¨2£©ÒÒͬѧÉè¼ÆµÄʵÑéµÚ______²½£¬ÄÜÖ¤Ã÷¸Ä±äÌõ¼þÈõµç½âÖÊƽºâ·¢ÉúÒƶ¯¡£¼×ͬѧΪÁ˽øÒ»²½Ö¤Ã÷Èõµç½âÖʵçÀëƽºâÒƶ¯µÄÇé¿ö£¬Éè¼ÆÈçÏÂʵÑ飺¢ÙʹHAµÄµçÀë³Ì¶ÈºÍc(H+)¶¼¼õС£¬ c(A-)Ôö´ó£¬¿ÉÔÚ0.1 mol¡¤L-1µÄHAÈÜÒºÖУ¬Ñ¡Ôñ¼ÓÈë_________ÊÔ¼Á£¨Ñ¡Ìî¡°A¡±¡°B¡±¡°C¡±¡°D¡±,ÏÂͬ£©£»¢ÚʹHAµÄµçÀë³Ì¶È¼õС£¬c(H+)ºÍc(A-)¶¼Ôö´ó£¬¿ÉÔÚ0.1 mol¡¤L-1µÄHAÈÜÒºÖУ¬Ñ¡Ôñ¼ÓÈë_____ÊÔ¼Á¡£
A£® NaA¹ÌÌ壨¿ÉÍêÈ«ÈÜÓÚË®£©    
B£®1 mol¡¤L-1 NaOHÈÜÒº
C£® 1 mol¡¤L-1 H2SO4       D£®2 mol¡¤L-1 HA
£¨3£©pH£½1µÄÁ½ÖÖËáÈÜÒºA¡¢B¸÷1 mL£¬·Ö±ð¼ÓˮϡÊ͵½1000 mL£¬ÆäpHÓëÈÜÒºÌå»ýVµÄ¹ØϵÈçͼËùʾ£¬ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÓÐ        
 
A£®Á½ÖÖËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÒ»¶¨ÏàµÈ
B£®Ï¡Êͺó£¬AËáÈÜÒºµÄËáÐÔ±ÈBËáÈÜÒºÈõ
C£®Èôa£½4£¬ÔòAÊÇÇ¿ËᣬBÊÇÈõËá
D£®Èô1£¼a£¼4£¬ÔòA¡¢B¶¼ÊÇÈõËá   
E£®Ï¡ÊͺóAÈÜÒºÖÐË®µÄµçÀë³Ì¶È±ÈBÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈС
ijÑо¿Ð¡×éÔÚʵÑéÊÒ̽¾¿°±»ù¼×Ëá泥¨NH2COONH4£©·Ö½â·´Ó¦Æ½ºâ³£ÊýºÍË®½â·´Ó¦ËÙÂʵIJⶨ¡£
£¨1£©½«Ò»¶¨Á¿´¿¾»µÄ°±»ù¼×Ëá粒ÌÌåÖÃÓÚÌØÖƵÄÃܱÕÕæ¿ÕÈÝÆ÷ÖУ¨¼ÙÉèÈÝÆ÷Ìå»ý²»±ä£¬¹ÌÌåÊÔÑùÌå»ýºöÂÔ²»¼Æ£©£¬Ôں㶨ζÈÏÂʹÆä´ïµ½·Ö½âƽºâ£º
NH2COONH4£¨s£©2NH3(g)+CO2(g)
ʵÑé²âµÃ²»Í¬Î¶ÈϵÄƽºâÊý¾ÝÁÐÓÚÏÂ±í£º
ζÈ/¡æ
15.0
20.0
25.0
30.0
35.0
ƽºâ×Üѹǿ/kPa
5.7
8.3
12.0
17.1
24.0
ƽºâÆøÌå×ÜŨ¶È/mol?L-1
2.4¡Á10-3
3.4¡Á10-3
4.8¡Á10-3
6.8¡Á10-3
9.4¡Á10-3
¢Ù¿ÉÒÔÅжϸ÷ֽⷴӦÒѾ­´ïµ½Æ½ºâµÄÊÇ       ¡£
A£®2v(NH2)=v(CO2)B£®ÃܱÕÈÝÆ÷ÖÐ×Üѹǿ²»±ä
C£®ÃܱÕÈÝÆ÷ÖлìºÏÆøÌåµÄÃܶȲ»±äD£®ÃܱÕÈÝÆ÷Öа±ÆøµÄÌå»ý·ÖÊý²»±ä
¢Ú¸ù¾Ý±íÖÐÊý¾Ý£¬ÁÐʽ¼ÆËã25.0¡æʱµÄ·Ö½âƽºâ³£Êý£º
¢ÛÈ¡Ò»¶¨Á¿µÄ°±»ù¼×Ëá粒ÌÌå·ÅÔÚÒ»¸ö´ø»îÈûµÄÃܱÕÕæ¿ÕÈÝÆ÷ÖУ¬ÔÚ25.0¡æÏ´ﵽ·Ö½âƽºâ¡£ÈôÔÚºãÎÂÏÂѹËõÈÝÆ÷Ìå»ý£¬°±»ù¼×Ëá粒ÌÌåµÄÖÊÁ¿ÊÇ         (Ìî¡°Ôö¼Ó¡±£¬¡°¼õÉÙ¡±»ò¡°²»±ä¡±£©¡£
¢Ü°±»ù¼×Ëá立ֽⷴӦµÄìʱä¡÷H       O(Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±),ìرä¡÷S        O
(Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±)¡£
(2)ÒÑÖª£ºNH2COONH4+2H2ONH4HCO2+NH3?H2O
¸ÃÑо¿Ð¡×é·Ö±ðÓÃÈý·Ý²»Í¬³õʼŨ¶ÈµÄ°±»ù¼×Ëáï§ÈÜÒº²â¶¨Ë®½â·´Ó¦ËÙÂÊ£¬µÃµ½c(NH2COO2£©Ê±¼äµÄ±ä»¯Ç÷ÊÆÈçͼËùʾ¡£

¢Ý¼ÆËãʱ£¬0-6min °±»ù¼×Ëáï§Ë®½â·´Ó¦µÄƽÒÖËÙÂÊ ______¡£
¢Þ¸ù¾ÝͼÖÐÐÅÏ¢£¬ÈçºÎ˵Ã÷¸ÃË®½â·´Ó¦ËÙÂÊËæζÈÉý¸ß¶øÔö´ó£º_____¡£
ijʵÑé̽¾¿Ð¡×é¸ù¾ÝÒÒÈ©»¹Ô­ÐÂÖÆCu£¨OH£©2µÄʵÑé²Ù×÷ºÍʵÑéÏÖÏ󡣶Է´Ó¦·½³Ìʽ¡°CH3CHO+2Cu£¨OH£©2CH3COOH+Cu2O¡ý+2H2O¡±Ìá³öÁËÖÊÒÉ£¬²¢½øÐÐÈçÏÂ̽¾¿£º

£¨Ò»£©²éÔÄ×ÊÁÏ
£¨1£©ÖÊÁ¿·ÖÊýΪ10%µÄNaOHÈÜÒºÃܶÈΪ1.1g¡¤cm-3£¬¸ÃÈÜÒºµÄc(NaOH)=     mol/L£»ÖÊÁ¿·ÖÊýΪ2%µÄCuSO4ÈÜÒºÃܶÈΪ1.0g¡¤cm¡ª3£¬Æäc(CuSO4)=0.125mol/L¡£
£¨2£©Cu£¨OH£©2¿ÉÈÜÓÚŨµÄÇ¿¼îÈÜÒºÉú³ÉÉîÀ¶É«
µÄ[Cu£¨OH£©4]2-ÈÜÒº£¬CuSO4ÈÜÒºÓëNaOH
ÈÜÒº·´Ó¦¹ý³ÌÖÐCu2+Óë[Cu£¨OH£©4]2¡ªÅ¨¶È
±ä»¯ÈçÓÒͼËùʾ£º
£¨3£©Cu2O¡¢CuO¾ù¿ÉÈÜÓÚCH3COOH£»ÔÚÈÜÒºÖÐ
¿É·¢Éú2Cu+==Cu+Cu2+·´Ó¦¡£
£¨¶þ£©ÊµÑé²Ù×÷¼°ÊµÑéÏÖÏó
£¨1£©ÔÚ2mL 10%µÄNaOHÈÜÒºÖеμÓ4~6µÎ£¨Ô¼0.2mL£©2%CuSO4ÈÜÒº£¬¾­²â¶¨»ìºÏÈÜÒºÖÐc(OH-)ԼΪ2.5mol/L¡£
£¨2£©Èô¶ÔÉÏÊö»ìºÏÎï½øÐйýÂË£¬¿ÉµÃµ½Ç³À¶É«µÄ           £¨Ìѧʽ£©¹ÌÌ壬ÂËÒº³ÊÉîÀ¶É«£¬ÔòÏÔÉîÀ¶É«µÄÀë×ÓÊÇ            £¨ÌîÀë×Ó·ûºÅ£©¡£
£¨3£©ÔÚµÚ£¨1£©²½ËùµÃµÄ»ìºÏÎïÖмÓÈë0.5mL ÒÒÈ©£¬¼ÓÈÈÖÁ·ÐÌÚ£¬²úÉúשºìÉ«³Áµí¡£
£¨Èý£©¶Ô¡°CH3CHO+2Cu£¨OH£©2CH3COOH+Cu2O¡ý+2H2O¡±ÖÊÒɵÄÀíÓÉ£º
£¨1£©                                  £»£¨2£©                         ¡£
£¨ËÄ£©½áÂÛ£º¸ù¾ÝÉÏÊö̽¾¿£¬ÒÒÈ©ÓëÐÂÖÆCu£¨OH£©2·´Ó¦µÄÀë×Ó·½³ÌʽΪ
                                                            ¡£

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

¾«Ó¢¼Ò½ÌÍø