ÌâÄ¿ÄÚÈÝ

£¨1£©ºÏ³É°±¹¤Òµ¶Ô»¯Ñ§¹¤ÒµºÍ¹ú·À¹¤Òµ¾ßÓÐÖØÒªÒâÒå¡£¹¤ÒµºÏ³É°±Éú²úʾÒâͼÈçͼ¼×Ëùʾ¡£

¢ÙXµÄ»¯Ñ§Ê½Îª__________£¬ÊôÓÚ________£¨Ìî¡°¼«ÐÔ¡±»ò¡°·Ç¼«ÐÔ¡±£©·Ö×Ó¡£
¢Úͼ¼×ÖÐÌõ¼þÑ¡¶¨µÄÖ÷ÒªÔ­ÒòÊÇ£¨Ñ¡Ìî×ÖĸÐòºÅ£¬ÏÂͬ£©________¡£
A£®Î¶ȡ¢Ñ¹Ç¿¶Ô»¯Ñ§Æ½ºâµÄÓ°Ïì
B£®Ìú´¥Ã½ÔÚ¸ÃζÈʱ»îÐÔ´ó
C£®¹¤ÒµÉú²úÊܶ¯Á¦¡¢²ÄÁÏ¡¢É豸µÈÌõ¼þµÄÏÞÖÆ
¢Û¸Ä±ä·´Ó¦Ìõ¼þ£¬»áʹƽºâ·¢ÉúÒƶ¯¡£Í¼ÒÒ±íʾËæÌõ¼þ¸Ä±ä£¬°±ÆøµÄ°Ù·Öº¬Á¿µÄ±ä»¯Ç÷ÊÆ¡£µ±ºá×ø±êΪѹǿʱ£¬±ä»¯Ç÷ÊÆÕýÈ·µÄÊÇ________£¬µ±ºá×ø±êΪζÈʱ£¬±ä»¯Ç÷ÊÆÕýÈ·µÄÊÇ__________¡£
£¨2£©³£ÎÂÏ°±Æø¼«Ò×ÈÜÓÚË®£¬ÆäË®ÈÜÒº¿ÉÒÔµ¼µç¡£
¢ÙÓ÷½³Ìʽ±íʾ°±ÆøÈÜÓÚË®µÄ¹ý³ÌÖдæÔڵĿÉÄæ·´Ó¦£º
___________________________________________________________________¡£
¢Ú°±Ë®ÖÐË®µçÀë³öµÄc(H+)___________10£­7 mol/L£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£
¢Û½«°±Ë®ºÍÑÎËá»ìºÏºó£¬Ä³Í¬Ñ§ÍƲâ¸ÃÈÜÒºÖи÷Àë×ÓŨ¶È´óС˳Ðò¿ÉÄÜÓÐÈçÏÂËÄÖÖ¹Øϵ£º
A£®c(Cl£­)£¾c(NH4+)£¾c(H+)£¾c(OH£­)           B£®c(Cl£­)£¾c(NH4+)£¾c(OH£­)£¾c(H+)
C£®c(Cl£­)£¾c(H+)£¾c(NH4+)£¾c(OH£­)           D£®c(NH4+)£¾c(Cl£­)£¾c(OH£­)£¾c(H+)
¢ñ¡¢ÈôÈÜÒºÖÐÖ»ÈܽâÁËÒ»ÖÖÈÜÖÊ£¬¸ÃÈÜÖʵÄÃû³ÆÊÇ             £¬ÉÏÊöÀë×ÓŨ¶È´óС˳Ðò¹ØϵÖÐÕýÈ·µÄÊÇ£¨Ñ¡ÌîÐòºÅ£©                        ¡£
¢ò¡¢ÈôÉÏÊö¹ØϵÖÐCÊÇÕýÈ·µÄ£¬ÔòÈÜÒºÖÐÈÜÖʵĻ¯Ñ§Ê½ÊÇ                   ¡£
¢ó¡¢Èô¸ÃÈÜÒºÖÐÓÉÌå»ýÏàµÈµÄÏ¡ÑÎËáºÍ°±Ë®»ìºÏ¶ø³É£¬ÇÒÇ¡ºÃ³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°
c£¨HCl£©            c£¨NH3¡¤H2O£©£¨Ìî¡°>¡±¡¢¡°<¡±¡¢»ò¡°=¡±£¬ÏÂͬ£©£¬»ìºÏºóÈÜÒºÖÐc£¨NH4+£©Óëc£¨Cl£­£©µÄ¹Øϵc£¨NH4+£©           c£¨Cl£­£©¡£
£¨3£©°±Æø¾ßÓл¹Ô­ÐÔ£¬ÔÚÍ­µÄ´ß»¯×÷ÓÃÏ£¬°±ÆøºÍ·úÆø·´Ó¦Éú³ÉXºÍYÁ½ÖÖÎïÖÊ¡£XΪï§ÑΣ¬YÔÚ±ê×¼×´¿öÏÂΪÆø̬¡£ÔÚ´Ë·´Ó¦ÖУ¬Èôÿ·´Ó¦1Ìå»ý°±Æø£¬Í¬Ê±·´Ó¦0.75Ìå»ý·úÆø£»Èôÿ·´Ó¦8.96 L°±Æø(±ê×¼×´¿ö)£¬Í¬Ê±Éú³É0.3 mol X¡£
¢Ùд³ö°±ÆøºÍ·úÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º ___________________________________¡£
¢ÚÔÚ±ê×¼×´¿öÏ£¬Ã¿Éú³É1 mol Y£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª___________mol¡£
£¨4£©ÒÑ֪Һ̬NH3ÓëH2OÏàËÆ£¬Ò²¿ÉÒÔ·¢Éú΢ÈõµÄµçÀ룬µçÀë³öº¬ÓÐÏàͬµç×ÓÊýµÄ΢Á££¬ÔòҺ̬NH3µÄµçÀë·½³ÌʽΪ£º                                         

(1) ¢ÙNH3 (1·Ö)  ¼«ÐÔ (1·Ö)   ¢ÚB¡¢C (2·Ö)     ¢Ûc (1·Ö)   a (1·Ö)
(2) ¢ÙNH3 + H2O  NH3¡¤H2O  NH4+ + OH-  (2·Ö) £»¢Ú< (1·Ö)  
£¨3£©¢ñ¡¢ÂÈ»¯ï§ £¬A £¨¸÷1·Ö£©£»¢ò¡¢NH4ClºÍHCl£¨1·Ö£©£»¢ó¡¢£¼ £¬£½£¨¸÷1·Ö£©
(4) ¢Ù4NH3 + 3F2NF3 + 3NH4F (2·Ö)     ¢Ú6 (2·Ö)
£¨5£©2NH3 NH4++ NH2-   £¨2·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¹¤ÒµÉ϶Ժ£Ë®×ÊÔ´×ۺϿª·¢ÀûÓõIJ¿·Ö¹¤ÒÕÁ÷³ÌÈçͼ1Ëùʾ£®

£¨1£©´ÖÑÎÖк¬ÓÐCa2+¡¢Mg2+¡¢SO42-µÈÔÓÖÊ£¬´ÖÖƺó¿ÉµÃ±¥ºÍNaClÈÜÒº£¬¾«ÖÆʱͨ³£ÔÚÈÜÒºÖÐÒÀ´ÎÖмÓÈë¹ýÁ¿µÄBaCl2ÈÜÒº¡¢¹ýÁ¿µÄNaOHÈÜÒººÍ¹ýÁ¿µÄNa2CO3ÈÜÒº£¬¹ýÂ˺óÏòÂËÒºÖмÓÈëÑÎËáÖÁÈÜÒº³ÊÖÐÐÔ£®Çëд³ö¼ÓÈëNa2CO3ÈÜÒººóÏà¹Ø»¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ£º
Ba2++CO32-=BaCO3¡ý£»Ca2++CO32-=CaCO3¡ý£»
Ba2++CO32-=BaCO3¡ý£»Ca2++CO32-=CaCO3¡ý£»
£®
£¨2£©±¾¹¤ÒÕÁ÷³ÌÖÐÏȺóÖƵÃBr2¡¢CaSO4¡¢Mg£¨OH£©2£¬ÄÜ·ñ°´Br2¡¢Mg£¨OH£©2¡¢CaSO4µÄ˳ÐòÖƱ¸£¿
·ñ
·ñ
£¨Ìî¡°ÄÜ¡±»ò¡°·ñ¡±£©£¬Ô­ÒòÊÇ
Èç¹ûÏȳÁµíMg£¨OH£©2£¬Ôò³ÁµíÖлá¼ÐÔÓÓÐCaSO4³Áµí£¬²úÆ·²»´¿
Èç¹ûÏȳÁµíMg£¨OH£©2£¬Ôò³ÁµíÖлá¼ÐÔÓÓÐCaSO4³Áµí£¬²úÆ·²»´¿
£®
£¨3£©ÂÈ»¯ÄÆÊÇÖØÒªµÄÂȼҵ»¯¹¤µÄÔ­ÁÏ£®µç½â±¥ºÍʳÑÎË®³£ÓÃÀë×ÓĤµç½â²ÛºÍ¸ôĤµç½â²Û£®Àë×ÓĤºÍ¸ôĤ¾ùÔÊÐíͨ¹ýµÄ·Ö×Ó»òÀë×ÓÊÇ
B
B
£®
A£®Cl-      B£®Na+       C£®OH-        D£®Cl2
£¨4£©±¥ºÍʳÑÎË®µç½âʱÓëµçÔ´Õý¼«ÏàÁ¬µÄµç¼«ÉÏ·¢ÉúµÄ·´Ó¦Îª
Ñõ»¯·´Ó¦£¬2Cl--2e-=Cl2¡ü
Ñõ»¯·´Ó¦£¬2Cl--2e-=Cl2¡ü
£¬ÓëµçÔ´¸º¼«ÏßÁ¬µÄµç¼«¸½½üÈÜÒºpH
±ä´ó
±ä´ó
£¨±ä´ó¡¢²»±ä¡¢±äС£©£®
£¨5£©ÂÈ»¯ÄƵĿÉÓÃÓÚÉú²ú´¿¼î£¬ÎÒ¹ú»¯Ñ§¼ÒºîµÂ°ñ¸Ä¸ï¹úÍâÉú²ú¹¤ÒÕ£¬Éú²úÁ÷³Ì¼òÒª±íʾÈçͼ2£º

¢ÙÉÏÊöÉú²ú´¿¼îµÄ·½·¨³Æ
ÁªºÏÖƼ
ÁªºÏÖƼ
£¬¸±²úÆ·µÄÒ»ÖÖÓÃ;Ϊ
×ö»¯·Ê
×ö»¯·Ê
£®
д³öÉÏÊöÁ÷³ÌÖÐXÎïÖʵķÖ×Óʽ
CO2
CO2
£®Ê¹Ô­ÁÏÂÈ»¯ÄƵÄÀûÓÃÂÊ´Ó70%Ìá¸ßµ½90%ÒÔÉÏ£¬Ö÷ÒªÊÇÉè¼ÆÁË
¢ñ
¢ñ
£¨ÌîÉÏÊöÁ÷³ÌÖеıàºÅ£©µÄÑ­»·£®´Ó³Áµí³ØÖÐÈ¡³ö³ÁµíµÄ²Ù×÷ÊÇ
¹ýÂË
¹ýÂË
£®
¢ÚºÏ³É°±Ô­ÁÏÆøÖеªÆøÖƱ¸µÄ·½·¨Ö®Ò»Îª
Һ̬¿ÕÆøÒÀ¾Ý·Ðµã·ÖÀë
Һ̬¿ÕÆøÒÀ¾Ý·Ðµã·ÖÀë
£¬ÁíÒ»Ô­ÁÏÆøÇâÆøµÄÖÆÈ¡»¯Ñ§·½³ÌʽΪ
C+H2O
 ¸ßΠ
.
 
CO+H2£¬CO+H2O
 ¸ßΠ
.
 
CO2+H2
C+H2O
 ¸ßΠ
.
 
CO+H2£¬CO+H2O
 ¸ßΠ
.
 
CO2+H2
£®
¢Û³Áµí³ØÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ
NH3+CO2+NaCl+H2O=NH4Cl+NaHCO3¡ý
NH3+CO2+NaCl+H2O=NH4Cl+NaHCO3¡ý
£®ÒªÊµÏָ÷´Ó¦£¬ÄãÈÏΪӦ¸ÃÈçºÎ²Ù×÷£º
Ïò°±»¯µÄ±¥ºÍÂÈ»¯ÄÆÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼ÆøÌåµÃµ½Ì¼ËáÇâÄƾ§Ìå
Ïò°±»¯µÄ±¥ºÍÂÈ»¯ÄÆÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼ÆøÌåµÃµ½Ì¼ËáÇâÄƾ§Ìå
£»
¢ÜΪ¼ìÑé²úƷ̼ËáÄÆÖÐÊÇ·ñº¬ÓÐÂÈ»¯ÄÆ£¬¿ÉÈ¡ÉÙÁ¿ÊÔÑùÈÜÓÚË®ºó£¬ÔٵμÓ
ÓÃÏõËáËữµÄÏõËáÒø£¬¹Û²ì²úÉú°×É«³Áµí
ÓÃÏõËáËữµÄÏõËáÒø£¬¹Û²ì²úÉú°×É«³Áµí
£®

µªÊǵØÇòÉϼ«Îª·á¸»µÄÔªËØ£®ÌîдÏÂÁпհף®

(1)NHÖÐNÔ­×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ________£¬

NHµÄ¿Õ¼ä¹¹ÐÍΪ________£®

(2)³£ÎÂÏ£¬ï®¿ÉÓ뵪ÆøÖ±½Ó·´Ó¦Éú³ÉLi3N£¬Li3N¾§ÌåÖеªÒÔN3£­´æÔÚ£¬»ù̬N3£­µÄµç×ÓÅŲ¼Ê½Îª£º________£¬Li3N¾§ÌåÊôÓÚ________¾§Ìå(ÌÌåÀàÐÍ)£®

(3)NH3µÄ·ÐµãΪ234K£¬NF3µÄ·ÐµãΪ154K£¬Á½Õ߽ṹÏàËÆ£¬NH3µÄ·Ðµã¸ßÓÚNF3µÄÔ­ÒòÊÇ£º________£®

(4)COÓëN2ÏàËÆ£¬·Ö×ÓÖж¼´æÔÚÒ»¸ö¹²¼ÛÈþ¼ü£¬COº¬________¸ö¦Ð¼ü£®Ï±íΪCOÓëN2µÄÏà¹ØÐÅÏ¢£®¸ù¾Ý±íÖÐÊý¾Ý£¬ËµÃ÷CO±ÈN2»îÆõÄÔ­Òò£º________£®

(5)¸ù¾ÝϱíÊý¾Ý£¬Ð´³öµªÆøÓëÇâÆø·´Ó¦Éú³É°±ÆøµÄÈÈ»¯Ñ§·½³Ìʽ________£®

(6)COÓëN2»¥ÎªµÈµç×ÓÌ壬Æä̼ԭ×ÓÉÏÓÐÒ»¶Ô¹Â¶Ôµç×Ó£¬Òò´Ë¿É×÷ÅäÌ壬È磺Fe(CO)5¡¢Ni(CO)4¡¢Cr(CO)6µÈ£¬Ôںϳɰ±¹¤ÒµÉÏÖÐÓÃÍ­Ï´ÒºÎüÊÕCO£¬·´Ó¦ÈçÏ£º

¢Ù»ù̬FeÔ­×ÓµÄδ³É¶Ôµç×ÓÊýÓÐ________¸ö£¬

д³öCr¡¢Cu+µÄ¼Û²ãµç×ÓÅŲ¼Ê½£®________¡¢________£®

¢Ú´×Ëá¶þ°±ºÏÍ­(I)ºÍ´×ËáôÊ»ùÈý°±ºÏÍ­(I)¶¼ÊÇÅäºÏÎ[Cu(NH3)3CO]+ÖÐÌṩ¹Â¶Ôµç×ӵķÖ×ÓÊÇ£º________£¬½ÓÊܹ¶Եç×ÓµÄÀë×ÓÊÇ£º________£¬ÓüýºÅ±ê³ö[Cu(NH3)2]+ÐγɵÄÅäλ¼ü£º________£¬[Cu(NH3)2]+ÖÐÁ½¸öÅäλ¼üµÄ¼ü½ÇΪ180¡ã£¬ÔòCu+²ÉÈ¡________ÔÓ»¯ÓëNH3ÐγÉÅäλ¼ü(ÌîÔÓ»¯ÀàÐÍ)£®

¢ÛNi(CO)4ÊÇÎÞÉ«ÒºÌ壬·Ðµã42.1¡æ£¬È۵㣭19.3¡æ£¬ÄÑÈÜÓÚË®£¬Ò×ÈÜÓÚÓлúÈܼÁ£®ÍƲâNi(CO)4ÊÇ________·Ö×Ó(Ìî¡°¼«ÐÔ¡±»ò¡°·Ç¼«ÐÔ¡±)£®

(7)µª»¯¹èÊÇÒ»ÖÖ¸ßÎÂÌմɲÄÁÏ£¬ËüÓ²¶È´ó¡¢ÈÛµã¸ß¡¢»¯Ñ§ÐÔÖÊÎȶ¨£®

¢Ùµª»¯¹è¾§ÌåÊôÓÚ________¾§Ìå(ÌÌåÀàÐÍ)£»

¢ÚÒÑÖªµª»¯¹è¾§Ìå½á¹¹ÖУ¬Ô­×Ӽ䶼ÒÔ¹²¼Û¼üÏàÁ¬£¬ÇÒNÔ­×ÓÓëNÔ­×Ó£¬SiÔ­×ÓÓëSiÔ­×Ó²»Ö±½ÓÏàÁ¬£¬Í¬Ê±Ã¿¸öÔ­×Ó¶¼Âú×ã8µç×ӽṹ£¬Çëд³öµª»¯¹èµÄ»¯Ñ§Ê½£º________£®

(8)¼«´¿µÄµªÆø¿ÉÓɵþµª»¯ÄÆ(NaN3)¼ÓÈÈ·Ö½â¶øµÃµ½£®2NaN3(s)£½2Na(l)£«3N2(g)£¬·´Ó¦¹ý³ÌÖУ¬¶ÏÁѵĻ¯Ñ§¼üÊÇÀë×Ó¼üºÍ¹²¼Û¼ü£¬ÐγɵĻ¯Ñ§¼üÓÐ________£®

£¨12·Ö£©±¾Ìâ°üÀ¨A¡¢BÁ½Ð¡Ì⣬·Ö±ð¶ÔÓ¦ÓÚ¡°ÎïÖʽṹÓëÐÔÖÊ¡±ºÍ¡°ÊµÑ黯ѧ¡±Á½¸öÑ¡ÐÞÄ£¿éµÄÄÚÈÝ¡£ÇëÑ¡¶¨ÆäÖÐÒ»Ì⣬²¢ÔÚÏàÓ¦µÄ´ðÌâÇøÓòÄÚ×÷´ð¡£ÈôÁ½Ìⶼ×ö£¬Ôò°´AÌâÆÀ·Ö¡£
A£®ÓÃÓںϳɰ±µÄ¹¤ÒµÃºÆøÖк¬ÓÐH2S¡¢C2H5SH£¨ÒÒËá´¼£©¡¢COS£¨ôÊ»ùÁò£©¡¢CS2µÈº¬Áò»¯ºÏÎ¹¤ÒµÉÏÎÞ»úÁò³£ÓÃÑõ»¯Ð¿·¨´¦Àí£¬ÓлúÁò¿ÉÓÃîÜîâ´ß»¯¼ÓÇâ´¦Àí¡£
H2S+ZnO=ZnS+H2O£»C2H5SH+ZnO=ZnS+C2H4+H2O
C2H5SH+H2=C2H6+H2S£»COS+H2=CO+H2S;CS2+4H2=CH4+2H2S
£¨1£©îÜÔ­×ÓÔÚ»ù̬ʱºËÍâµç×ÓÅŲ¼Ê½Îª         ¡£
£¨2£©ÏÂÁÐÓйطÖ×ӽṹµÄ˵·¨ÕýÈ·µÄÊÇ           ¡£
A£®C2H4·Ö×ÓÖÐÓÐ5¸ö¼ü´¦1¸ö¼ü
B£®COS·Ö×Ó£¨½á¹¹ÈçÓÒͼ£©ÖмüÄÜC=O>C=S
C£®H2S·Ö×Ó³ÊVÐνṹ
D£®CH4¡¢C2H6·Ö×ÓÖÐ̼ԭ×Ó¾ù²ÉÓÃsp3ÔÓ»¯
£¨3£©ÏÂÁÐÓйØ˵·¨²»ÕýÈ·µÄÊÇ     ¡£
A£®H2O¡¢CO¡¢COS¾ùÊǼ«ÐÔ·Ö×Ó
B£®ÏàͬѹǿÏ·е㣺Cs2>COS>CO2
C£®ÏàͬѹǿÏ·е㣺C2H5SH>C2H5OH
D£®ÏàͬѹǿÏ·е㣺CO>N2
£¨4£©-ZnSµÄ¾§°û½á¹¹ÈçÓÒͼ£¬¾§°ûÖÐS2-ÊýĿΪ£º        ¸ö¡£

£¨5£©¾ßÓÐÏàËƾ§°û½á¹¹µÄZnSºÍZnO£¬ZnSÈÛµãΪ1830¡æ£¬ZnOÈÛµãΪ1975¡æ£¬ºóÕß½ÏÇ°Õ߸ßÊÇÓÉÓÚ            ¡£
£¨6£©îâµÄÒ»ÖÖÅäºÏÎﻯѧʽΪ£ºNa3[Mo(CN)8]¡¤8H2O£¬ÖÐÐÄÔ­×ÓµÄÅäλÊýΪ   ¡£
B£®ÌþÈ©½áºÏ·´Ó¦ÓлúºÏ³ÉÖÐÆÄΪÖØÒª£¬ÂÌÉ«´ß»¯¼ÁµÄ¹ÌÌåîêËáËᱶÊÜÑо¿Õß¹Ø×¢¡£îêËá¾ßÓнϸߵĴ߻¯»îÐÔ¼°Îȶ¨ÐÔ¡£·´Ó¦Ô­ÀíÈçÏ£º

ʵÑé·½·¨ÊÇÔÚ25mLÉÕÆ¿ÖмÓÈëîêËá¡¢10mL¼×´¼ºÍ 0.5mL±½¼×È©£¬ÔÚ»ØÁ÷״̬Ï·´Ó¦2h£¬·´Ó¦µÄ²úÂʺÍת»¯Âʾù·Ç³£¸ß¡£
£¨1£©²ÉÓûØÁ÷·´Ó¦2hµÄÄ¿µÄÊÇ             ¡£
£¨2£©ÔÚ·´Ó¦Öм״¼Ðè¹ýÁ¿£¬ÆäÔ­ÒòÊÇ              ¡£
 £¨3£©²»Í¬îêËáÓÃÁ¿¶Ô²úÂʺÍת»¯ÂÊÓ°Ï죬ÈçÏÂ±í£º

îêËáÓÃÁ¿/mol
0.01
0.02
0.03
0.05
0.1
0.15
0.2
0.6
²úÂÊ%
87.3
88.2
90.3
94.2
92.9
93.1
91.8
92.3
ת»¯ÂÊ%
89.7
92.1
93.9
98.9
94.9
95.7
93.9
94.3
      ÔÚÉÏÊö±½¼×È©Óë¼×´¼ËõºÏ·´Ó¦ÊµÑéÖд߻¯¼ÁîêËáµÄ×î¼ÑÓÃÁ¿Îª        ¡£
£¨4£©´ß»¯¼ÁµÄ»ØÊÕÀûÓÃÐÔÄÜÊÇ¿¼²ì´ß»¯¼ÁµÄÒ»ÏΪÖØÒªµÄÖ¸±ê¡£îêËá´ß»¯¼ÁÑ­»·Ê¹ÓôÎÊý¶Ô²úÂʵÄÓ°ÏìÈçÓÒÏÂͼ£¬Õâ˵Ã÷îêËá´ß»¯¼ÁµÄÓŵãÖ®Ò»ÊÇ         ¡£
£¨5£©ÓÃîêËá×÷´ß»¯¼Áʱ£¬²»Í¬µÄÈ©Óë¼×´¼µÄËõºÏ·´Ó¦µÄת»¯ÂʺͲúÂÊÈçÏÂ±í£º
ÐòºÅ
È©
´¼
ת»¯ÂÊ%
²úÂÊ%
1
ÁÚôÇ»ù±½¼×È©
¼×´¼
94.3
89.6
2
ÁÚôÇ»ù±½¼×È©
¼×´¼
93.6
88.7
3
ÁÚÂȱ½¼×È©
¼×´¼
93.1
87.3
4
¼äÏõ»ù±½¼×È©
¼×´¼
54.2
34.1
5
ÁÚÏõ»ù±½¼×È©
¼×´¼
89.9
79.5
6
¶ÔÏõ»ù±½¼×È©
¼×´¼
65.7
41.9

´Ó±íÖеóöµÄ²»Í¬µÄÈ©Óë¼×´¼ËõºÏ·´Ó¦Ó°Ïìת»¯ÂʺͲúÂʵĹæÂÉÊÇ   ¡£

£¨1£©ºÏ³É°±¹¤Òµ¶Ô»¯Ñ§¹¤ÒµºÍ¹ú·À¹¤Òµ¾ßÓÐÖØÒªÒâÒå¡£

¹¤ÒµºÏ³É°±Éú²úʾÒâͼÈçͼ1Ëùʾ¡£

¢ÙXµÄ»¯Ñ§Ê½Îª              £»

¢ÚͼÖÐÌõ¼þÑ¡¶¨µÄÖ÷ÒªÔ­ÒòÊÇ£¨Ñ¡Ìî×ÖĸÐòºÅ£©         £»

A£®Î¶ȡ¢Ñ¹Ç¿¶Ô»¯Ñ§Æ½ºâÓ°Ïì

B£®Ìú´¥Ã½ÔÚ¸ÃζÈʱ»îÐÔ´ó

C£®¹¤ÒµÉú²úÊܶ¯Á¦¡¢²ÄÁÏ¡¢É豸µÈÌõ¼þµÄÏÞÖÆ

¢Û¸Ä±ä·´Ó¦Ìõ¼þ£¬»áʹƽºâ·¢ÉúÒƶ¯¡£Í¼2±íʾËæÌõ¼þ¸Ä±ä£¬

°±ÆøµÄ°Ù·Öº¬Á¿µÄ±ä»¯Ç÷ÊÆ¡£µ±ºá×ø±êΪѹǿʱ£¬±ä»¯Ç÷ÊÆÕý

È·µÄÊÇ£¨Ñ¡Ìî×Öĸ´úºÅ£©         £¬µ±ºá×ø±êΪζÈʱ£¬±ä        »¯Ç÷ÊÆÕýÈ·µÄÊÇ£¨Ñ¡Ìî×ÖĸÐòºÅ£©          ¡£

£¨2£©³£ÎÂÏ°±Æø¼«Ò×ÈÜÓÚË®£¬ÆäË®ÈÜÒº¿ÉÒÔµ¼µç¡£

¢ÙÓ÷½³Ìʽ±íʾ°±ÆøÈÜÓÚË®µÄ¹ý³ÌÖдæÔڵĿÉÄæ·´Ó¦

                                                                      £»

                                                                      £»

¢Ú°±Ë®ÖÐË®µçÀë³öµÄc(OH£­)        10£­7mol/L£¨Ìîд¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©£»

¢Û½«ÏàͬÌå»ý¡¢ÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄ°±Ë®ºÍÑÎËá»ìºÏºó£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óÒÔСÒÀ´ÎΪ                                       ¡£

£¨3£©°±Æø¾ßÓл¹Ô­ÐÔ£¬ÔÚÍ­µÄ´ß»¯×÷ÓÃÏ£¬°±ÆøºÍ·úÆø·´Ó¦Éú³ÉAºÍBÁ½ÖÖÎïÖÊ¡£AΪï§ÑΣ¬BÔÚ±ê×¼×´¿öÏÂΪÆø̬¡£ÔÚ´Ë·´Ó¦ÖУ¬Èôÿ·´Ó¦1Ìå»ý°±Æø£¬Í¬Ê±·´Ó¦0.75Ìå»ý·úÆø£»Èôÿ·´Ó¦8.96L°±Æø£¨±ê×¼×´¿ö£©£¬Í¬Ê±Éú³É0.3molA¡£

¢Ùд³ö°±ÆøºÍ·úÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                  £»

¢ÚÔÚ±ê×¼×´¿öÏ£¬Ã¿Éú³É1 mol B£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª          mol¡£

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

¾«Ó¢¼Ò½ÌÍø