ÏÖÓÐA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ³£¼û»¯ºÏÎÒÑÖªËüÃÇ£ºÑôÀë×Ó·Ö±ðÓÐK£«¡¢Ag£«¡¢Na£«¡¢Ba2£«¡¢Fe2£«¡¢Al3£«£»ÒõÀë×Ó·Ö±ðÓÐCl£­¡¢OH£­¡¢AlO2£­¡¢NO3£­¡¢SO42£­¡¢CO32£­¡£½«ËüÃÇ·Ö±ðÅä³É0.1mol¡¤L-1µÄÈÜÒº½øÐÐÈçÏÂʵÑé:

¢Ù²âµÃÈÜÒºA¡¢C¡¢E¾ù³Ê¼îÐÔ£¬ÇÒ¼îÐÔA£¾E£¾C£¬EµÄÑæÉ«³Êdz×ÏÉ«£¨Í¸¹ýÀ¶É«îܲ£Á§¹Û²ì£©£»

¢ÚÏòBµÄÈÜÒºÖеμÓÏ¡°±Ë®ÖÁ¹ýÁ¿£¬ÏÈÉú³É³Áµí£¬ºó³ÁµíÈ«²¿Èܽ⣻

¢ÛÏòFµÄÈÜÒºÖеμÓÏ¡ÏõËᣬÈÜÒº±ä³É×Ø»ÆÉ«£¬ÇÒÓÐÎÞÉ«ÆøÌåÉú³É£»

¢ÜÏòDµÄÈÜÒºÖеμÓBa(NO3)2ÈÜÒºÎÞÃ÷ÏÔÏÖÏó¡£

£¨1£©Ð´³öA¡¢D¡¢E¡¢FµÄ»¯Ñ§Ê½

A           D           E            F         

£¨2£©ÓÃÀë×Ó·½³Ìʽ½âÊÍCÈÜÒºµÄ¼îÐÔ

                                                 

£¨3£©Ð´³öʵÑé¢ÛÖз´Ó¦µÄÀë×Ó·½³Ìʽ

                                                 

£¨1£©A   Ba(OH)2     D    AlCl3       KAlO2       F    FeSO4 

£¨2£©½âÊÍCÈÜÒºµÄ¼îÐÔ    CO32£­£«H2O===HCO3£­£«OH£­ 

£¨3£©ÊµÑé¢ÛÀë×Ó·½³Ìʽ  3Fe2£« £« 4H£« £« NO3£­===3Fe3£« £« 2H2O £« NO¡ü 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÏÖÓÐA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬DÓëEµÄÇ⻯Îï·Ö×Ó¹¹ÐͶ¼ÊÇVÐÍ£®A¡¢BµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍÓëCµÄ×îÍâ²ãµç×ÓÊýÏàµÈ£¬AÄÜ·Ö±ðÓëB¡¢C¡¢DÐγɵç×Ó×ÜÊýÏàµÈµÄ·Ö×Ó£¬ÇÒAÓëD¿ÉÐγɵĻ¯ºÏÎ³£ÎÂϾùΪҺ̬£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣨Ìî¿ÕʱÓÃʵ¼Ê·ûºÅ£©£º
£¨1£©CµÄÔªËØ·ûºÅÊÇ
N
N
£»ÔªËØFÔÚÖÜÆÚ±íÖеÄλÖÃ
µÚ3ÖÜÆÚµÚ¢÷A×å
µÚ3ÖÜÆÚµÚ¢÷A×å
£®
£¨2£©BÓëDÒ»°ãÇé¿öÏ¿ÉÐγÉÁ½ÖÖ³£¼ûÆø̬»¯ºÏÎ¼ÙÈôÏÖÔÚ¿Æѧ¼ÒÖƳöÁíÒ»ÖÖÖ±ÏßÐÍÆø̬»¯ºÏÎï B2D2·Ö×Ó£¬ÇÒ¸÷Ô­×Ó×îÍâ²ã¶¼Âú×ã8µç×ӽṹ£¬ÔòB2D2µç×ÓʽΪ
£¬Æä¹ÌÌåʱµÄ¾§ÌåÀàÐÍÊÇ
·Ö×Ó¾§Ìå
·Ö×Ó¾§Ìå
£®
£¨3£©×î½üÒâ´óÀûÂÞÂí´óѧµÄFuNvio CacaceµÈÈË»ñµÃÁ˼«¾ßÀíÂÛÑо¿ÒâÒåµÄC4·Ö×Ó£®C4·Ö×ӽṹÈçͼËùʾ£¬ÒÑÖª¶ÏÁÑ1molC-CÎüÊÕ167kJÈÈÁ¿£¬Éú³É1molC=C·Å³ö942kJÈÈÁ¿£®¸ù¾ÝÒÔÉÏÐÅÏ¢ºÍÊý¾Ý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
¢Ú¢Ü¢ß
¢Ú¢Ü¢ß
£®
¢ÙC4ÊôÓÚÒ»ÖÖÐÂÐ͵Ļ¯ºÏÎï
¢ÚC4·Ðµã±ÈP4£¨°×Á×£©µÍ
¢Ûlmol C4ÆøÌåת±äΪC2ÎüÊÕ882kJÈÈÁ¿
¢ÜC4ÓëC2»¥ÎªÍ¬ËØÒìÐÎÌå
¢ÝC4Îȶ¨ÐÔ±ÈP4£¨°×Á×£©²î
¢ÞC4ÊôÓÚÔ­×Ó¾§Ìå
¢ßC4ºÍP4 £¨°×Á×£©µÄ¾§Ì嶼ÊôÓÚ·Ö×Ó¾§Ìå
¢àC4ÓëC2»¥ÎªÍ¬·ÖÒì¹¹Ìå
£¨4£©CÓëFÁ½ÖÖÔªËØÐγÉÒ»ÖÖ»¯ºÏÎï·Ö×Ó£¬¸÷Ô­×Ó×îÍâ²ã´ï8µç×ӽṹ£¬Ôò¸Ã·Ö×ӵĽṹʽΪ
£¬Æä¿Õ¼ä¹¹ÐÍΪ
Èý½Ç׶ÐÍ
Èý½Ç׶ÐÍ
£®
£¨5£©ÎªÁ˳ýÈ¥»¯ºÏÎïÒÒ£¨A2ED4£©Ï¡ÈÜÒºÖлìÓеÄA2ED3£¬³£²ÉÓÃA2D2ΪÑõ»¯¼Á£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
H2O2+H2SO3=2H++SO42-+H2O
H2O2+H2SO3=2H++SO42-+H2O

£¨6£©EÓëFÐγɵĻ¯ºÏÎïE2F2ÔÚÏ𽺹¤ÒµÉÏÓÐÖØÒªÓÃ;£¬ÓöË®Ò×Ë®½â£¬Æä¿Õ¼ä½á¹¹ÓëA2D2¼«ÎªÏàËÆ£®¶Ô´ËÒÔÏÂ˵·¨ÕýÈ·µÄÊÇ
acd
acd
£®
a£®E2F2µÄ½á¹¹Ê½Îª£ºF-E-E-F
b£®E2F2Ϊº¬Óм«ÐÔ¼ü ºÍ·Ç¼«ÐÔ¼üµÄ·Ç¼«ÐÔ·Ö×Ó
c£®E2Br2ÓëE2F2½á¹¹ÏàËÆ£¬È۷е㣺E2Br2£¾E2F2
d£®E2F2ÓëH2O·´Ó¦µÄ»¯Ñ§·½³Ìʽ¿ÉÄÜΪ£º2E2F2+2H2O=EO2¡ü+3E¡ý+4HF
£¨7£©¾Ù³öÒ»ÖÖÊÂʵ˵Ã÷EÓëFµÄ·Ç½ðÊôÐÔÇ¿Èõ£¨Óû¯Ñ§·½³Ìʽ»òÓÃÓïÑÔÎÄ×Ö±í´ï¾ù¿É£©£º
C12+H2S=S¡ý+2HCl
C12+H2S=S¡ý+2HCl
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÖÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖÔªËØ£¬ËüÃǵÄÖÊ×ÓÊýÒÀ´ÎÔö¶à£®
¢ÙAµÄºËµçºÉÊý´óÓÚ2£¬¿ÉÐγÉÇ⻯ÎïH2A£¬¸ÃÇ⻯ÎïÔÚ³£ÎÂÏÂÊÇÒºÌ壻
¢ÚAºÍBÁ½ÔªËØ¿ÉÐγÉB2A3»¯ºÏÎ¸Ã»¯ºÏÎï¼ÈÄÜÈÜÓÚÇ¿ËᣬÓÖÄÜÈÜÓÚÇ¿¼î£»
¢ÛC+Àë×Ó±ÈB3+Àë×Ó¶à8¸öµç×Ó£»
¢ÜCÓëDÔªËØ¿ÉÒÔÐγɻ¯ºÏÎïCD£»
¢ÝCDµÄÈÜÒºÖÐͨÈëÂÈÆøºó¼Óµí·ÛÈÜÒºÏÔÀ¶É«£»
¢ÞÔÚÖÜÆÚ±íÖÐE´¦ÓÚCµÄÏÂÁ½¸öÖÜÆÚ£¬Eµ¥ÖÊ¿ÉÓëÀäË®·´Ó¦Éú³ÉÇâÆø£¬·´Ó¦Ê±EµÄµ¥ÖÊÓëÉú³ÉµÄÇâÆøµÄÎïÖʵÄÁ¿Ö®±ÈΪ2£º1£®ÊԻشð£º
£¨1£©BÊÇ
ÂÁ
ÂÁ
¡¢EÊÇ
ï¤
ï¤
£®£¨Ð´ÔªËØÃû³Æ£©£º
£¨2£©BµÄÀë×ӽṹʾÒâͼ
£¬DµÄµ¥ÖÊ·Ö×ӵĵç×Óʽ
£¬CÔªËØ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄµç×Óʽ
£®
£¨3£©Óõç×Óʽ±íʾH2AÐγɹý³Ì£º

£¨4£©Ð´³öCDµÄÈÜÒºÖÐͨÈëÂÈÆøµÄÀë×Ó·½³Ìʽ£º
2I-+Cl2¨TI2+2Cl-
2I-+Cl2¨TI2+2Cl-

£¨5£©±È½ÏB¡¢C¡¢EÈýÖÖÔªËØÐγɵļòµ¥Àë×ÓÑõ»¯ÐÔµÄÇ¿Èõ£º£¨B¡¢C¡¢EÀë×ÓÓÃʵ¼ÊÀë×Ó·ûºÅ±íʾ£©Ñõ»¯ÐÔ£º
Al3+£¾K+£¾Cs+
Al3+£¾K+£¾Cs+
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÖÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵÄÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬A¡¢DͬÖ÷×åC¡¢E Í¬Ö÷×壬A¡¢BµÄ×îÍâ²ãµç×ÓÊýÖ®ºÍÓëCµÄ×îÍâ²ãµç×ÓÊýÏàµÈ£¬AÄÜ·Ö±ðÓëB¡¢CÐγɵç×Ó×ÜÊýÏàµÈµÄ·Ö×ÓM¡¢N£¬ÇÒNΪ±È½Ï³£¼ûµÄҺ̬»¯ºÏÎÇë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©BµÄ¼òµ¥ÒõÀë×ӵĽṹʾÒâͼ
£®
£¨2£©A¡¢D¡¢EÈýÖÖÔªËØ×é³ÉµÄ»¯ºÏÎïµç×ÓʽΪ£º

£¨3£©¢ÙC¡¢D¡¢E¶ÔÓ¦µÄ¼òµ¥Àë×ӵİ뾶ÓÉ´óµ½Ð¡Îª
S2-£¾O2-£¾Na+
S2-£¾O2-£¾Na+
£¨ÓÃÀë×Ó·ûºÅ±íʾ£©¢Úд³öÄÜÖ¤Ã÷CºÍE·Ç½ðÊôÐÔÇ¿ÈõµÄÒ»¸ö»¯Ñ§·½³Ìʽ
2H2S+O2¨T2H2O+2S
2H2S+O2¨T2H2O+2S
£®
£¨4£©A¡¢C¡¢EÈýÖÖÔªËØÄÜÐγɶàÖÖ18µç×Ó·Ö×Ó£¬Ð´³öÆäÖÐÁ½ÖÖËùÄÜ·¢ÉúµÄ·´Ó¦·½³Ìʽ
H2O2+H2S=S¡ý+2H2O
H2O2+H2S=S¡ý+2H2O

£¨5£©±ê¿öÏ£¬1gDµ¥ÖÊÔÚCµ¥ÖÊÖÐÍêȫȼÉտɷųöQKJÈÈÁ¿£®Ôò´Ë·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
2Na£¨s£©+O2£¨g£©=Na2O2£¨s£©¡÷H=-46QkJ/mol
2Na£¨s£©+O2£¨g£©=Na2O2£¨s£©¡÷H=-46QkJ/mol
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÖÓÐA¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÔªËØ£¬ÔÚ»¯Ñ§·´Ó¦ÖоùÄÜÐγɼòµ¥µÄÒõÀë×Ó»òÑôÀë×Ó£¬ÇÒA¡¢B¡¢C¡¢DÀë×Ó¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£®ÒÑÖª£º
¢Ù³£ÎÂÏ£¬FµÄµ¥ÖÊÊÇÒ»ÖÖÓÐÉ«ÆøÌ壬³£ÓÃÓÚɱ¾ú¡¢Ïû¶¾£»
¢ÚAµÄÑõ»¯Îï¼ÈÄÜÈÜÓÚNaOHÈÜÒº£¬ÓÖÄÜÈÜÓÚÑÎË᣻
¢ÛCµÄÇ⻯Îï·Ö×ÓGÊǾßÓÐ10µç×ÓµÄ΢Á££¬ÇÒ¿ÉÒÔ·¢ÉúÏÂÁÐת»¯£ºG
DµÄµ¥ÖÊ
P
DµÄµ¥ÖÊ
Q
H2O
M+P
¢ÜEºÍDÊÇͬÖ÷×åÔªËØ£¬Ç⻯ÎH2E£©³£ÎÂÏÂΪÆø̬£®
¢ÝBºÍD¿ÉÐγÉ΢Á£¸öÊý±ÈΪ1£º1ºÍ2£º1µÄÀë×Ó»¯ºÏÎïXºÍY£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AÔªËصÄÔ­×ӽṹʾÒâͼÊÇ
 
£»B2EµÄµç×ÓʽΪ
 
£®
£¨2£©»¯ºÏÎïXËùº¬»¯Ñ§¼üÀàÐÍÓÐ
 
£®
£¨3£©GÆøÌåµÄ¼ìÑé·½·¨Îª
 
£®
£¨4£©Ð´³öFµ¥ÖÊÓëEµÄÇ⻯Îï·´Ó¦µÄ·½³Ìʽ²¢±êÃ÷µç×ÓתÒÆ·½ÏòºÍÊýÄ¿
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÖÓÐA¡¢B¡¢C¡¢DËÄÖÖÎÞÉ«ÈÜÒº£¬ËüÃÇ·Ö±ðÊÇÏ¡ÑÎËá¡¢ÂÈ»¯±µ¡¢ÁòËáÄƺÍ̼ËáÄÆÖеÄijһÖÖ£®ÊÔ¸ù¾ÝÒÔÏÂʵÑéÏÖÏó»Ø´ðÏÂÁÐÎÊÌ⣺
A+B¡ú°×É«³Áµí£»A+C¡úÎÞÏÖÏó£»A+D¡úÎÞÏÖÏó£»B+C¡ú°×É«³Áµí£»B+D¡úÎÞÏÖÏó£»C+D¡úÎÞÉ«ÆøÌ壮
£¨1£©Ð´³ö»¯Ñ§Ê½£ºA
 
£¬B
 
£¬C
 
£¬D
 
£®
£¨2£©Ð´³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ£º
A+B£º
 
£»
C+D£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸