ÌâÄ¿ÄÚÈÝ

ϱíΪ²¿·Ö¶ÌÖÜÆÚÔªËØ»¯ºÏ¼Û¼°ÏàÓ¦Ç⻯Îï·ÐµãµÄÊý¾Ý£º

ÔªËØÐÔÖÊ

ÔªËØ±àºÅ

A

B

C

D

E

F

G

H

Ô­×Ó°ë¾¶

0.102

0.075

0.117

0.074

0.110

0.071

0.099

0.077

×î¸ß»¯ºÏ¼Û

+6

+5

£«4

+5

+7

£«4

×îµÍ»¯ºÏ¼Û

£­2

£­3

£­4

£­2

£­3

£­1

£­1

£­4

ÒÑÖª£º¢ÙAÓëD¿ÉÐγɻ¯ºÏÎïAD2¡¢AD3£¬¢ÚBÓëD¿ÉÐγɶàÖÖ»¯ºÏÎï,ÆäÖÐBD¡¢BD2Êdz£¼ûµÄ»¯ºÏÎï,C¿ÉÓÃÓÚÖÆ¹âµç³Ø¡£
Çë»Ø´ð£º¡¡
 £¨1£©EÔÚÖÜÆÚ±íÖÐλÖÃÊÇ                          £º

£¨2£©CºÍHµÄÆøÌ¬Ç⻯ÎïµÄÎȶ¨ÐÔÇ¿Èõ¹ØÏµÎª:               (Ó÷Ö×Óʽ±íʾ)

£¨3£©32g AD2ÆøÌåºÍD2ÆøÌåÇ¡ºÃÍêÈ«·´Ó¦Éú³ÉAD3ÆøÌ壬·Å³ö49.15kJµÄÈÈÁ¿£¬ÔòÆä·´Ó¦ÈÈ»¯Ñ§·½³ÌʽΪ£º                                      

£¨4£©·Ö×Ó×é³ÉΪADG2µÄÎïÖÊÔÚË®ÖлáÇ¿ÁÒË®½â£¬²úÉúʹƷºìÈÜÒºÍÊÉ«µÄÎÞÉ«ÆøÌåºÍÒ»ÖÖÇ¿Ëá¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º                                              ¡£

£¨5£©¹¤ÒµÉÏ¿ÉÓô¿¼îÈÜÒº´¦ÀíBDºÍBD2£¬¸Ã·´Ó¦ÈçÏ£º
     BD+BD2+Na2CO3=2      ¡¡¡¡¡¡+CO2ºáÏßÉÏijÑεĻ¯Ñ§Ê½Ó¦Îª¡¡¡¡¡¡¡¡¡¡¡¡

(1).µÚÈýÖÜÆÚ,VA×å

(2).CH4>SiH4

(3).2SO2(g) + O2(g) = 2SO3(g) ¦¤H£½-196.6 kJ/mol

(4).SOCl2 + H2O = SO2 + 2HCl

(5)NaNO2(ÿСÌâ3·Ö)

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ϱíΪ²¿·Ö¶ÌÖÜÆÚÔªËØ»¯ºÏ¼Û¼°ÏàÓ¦Ç⻯Îï·ÐµãµÄÊý¾Ý£º
ÔªËØÐÔÖÊ ÔªËØ±àºÅ
A B C D E F G H
Ç⻯ÎïµÄ·Ðµã£¨¡æ£© -60.7 -33.4 -111.5 100 -87.7 19.54 -84.9 -161.5
×î¸ß»¯ºÏ¼Û +6 +5 +4 +5 +7 +4
×îµÍ»¯ºÏ¼Û -2 -3 -4 -2 -3 -1 -1 -4
ÒÑÖª£º¢ÙAÓëD¿ÉÐγɻ¯ºÏÎïAD2¡¢AD3£¬¿ÉÓÃÓÚÖÆ±¸Ç¿Ëá¼×£»¢ÚBÓëD¿ÉÐγɻ¯ºÏÎïBD¡¢BD2£¬¿ÉÓÃÓÚÖÆ±¸Ç¿ËáÒÒ£®
Çë»Ø´ð£º
£¨1£©±íÖÐÊôÓÚµÚÈýÖÜÆÚÔªËØµÄÊÇ
ACEG
ACEG
£¨ÓñíÖÐÔªËØ±àºÅÌîд£©£®
£¨2£©Ð´³öHµÄ×î¸ß¼ÛÑõ»¯ÎïµÄµç×Óʽ£º
£¬±È½ÏA¡¢D¡¢GÈýÖÖ¼òµ¥ÒõÀë×ӵİ뾶´óС£º
r£¨
S2-
S2-
£©£¾r£¨
Cl-
Cl-
£©£¾r£¨
O2-
O2-
£© £¨ÓÃÔªËØ·ûºÅ±íʾ£©
£¨3£©ÓɱíÖÐDÔªËØºÍÇâÔªËØµÄÔ­×Ó°´1£º1×é³ÉµÄ³£¼ûҺ̬»¯ºÏÎïµÄÏ¡ÈÜÒºÒ×±»´ß»¯·Ö½â£¬¿ÉʹÓõĴ߻¯¼ÁΪ£¨ÌîÐòºÅ£©
ab
ab
£®
a£®MnO2b£®FeCl3¡¡¡¡      c£®Na2SO3¡¡¡¡    d£®KMnO4
£¨4£©·Ö×Ó×é³ÉΪADG2µÄÎïÖÊÔÚË®ÖлáÇ¿ÁÒË®½â£¬²úÉúʹƷºìÈÜÒºÍÊÉ«µÄÎÞÉ«ÆøÌåºÍÒ»ÖÖÇ¿Ëᣮ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º
SOCl2+H2O=SO2+2HCl
SOCl2+H2O=SO2+2HCl
£®
£¨5£©¹¤ÒµÉú²úÇ¿ËáÒÒʱ¿ÉÓô¿¼îÈÜÒº´¦ÀíÎ²Æø£¬¸Ã·´Ó¦ÈçÏ£ºBD+BD2+Na2CO3=2
NaNO2
NaNO2
+CO2
¢ÙºáÏßÉÏijÑεĻ¯Ñ§Ê½Ó¦Îª
NaNO2
NaNO2
£»¢Úÿ²úÉú44.8L£¨±ê×¼×´¿ö£©CO2£¬±»ÎüÊÕÎ²ÆøµÄÖÊÁ¿ÊÇ
152
152
g£®
£¨6£©ÇëÉè¼ÆÒ»¸öʵÑé·½°¸£¬Ê¹µÃÍ­ºÍÏ¡µÄÇ¿Ëá¼×·´Ó¦£¬µÃµ½À¶É«ÈÜÒººÍÇâÆø£¬ÔÚ´ðÌ⿨ָ¶¨Î»Öûæ³öʵÑé×°ÖÃͼ£¬±ê³ö±ØÒªµÄ˵Ã÷¼´¿É
£®
ϱíΪ²¿·Ö¶ÌÖÜÆÚÔªËØ»¯ºÏ¼Û¼°ÏàÓ¦Ô­×Ó°ë¾¶µÄÊý¾Ý£º
ÔªËØÐÔÖÊ ÔªËØ±àºÅ
A B C D E F G H
Ô­×Ӱ루nm£© 0.102 0.110 0.117 0.074 0.075 0.071 0.099 0.077
×î¸ß»¯ºÏ¼Û +6 +5 +4 +5 +7 +4
×îµÍ»¯ºÏ¼Û -2 -3 -4 -2 -3 -1 -1 -4
¼ºÖª£º¢ÙAÓëD¿ÉÐγɻ¯ºÏÎïAD2¡¢AD3£¬¢ÚEÓëD¿ÉÐγɶàÖÖ»¯ºÏÎÆäÖÐED¡¢ED2 Êdz£¼ûµÄ»¯ºÏÎC¿ÉÓÃÓÚÖÆ¹âµç³Ø
£¨1£©EÔÚÖÜÆÚ±íÖÐλÖÃÊÇ
 
£»
£¨2£©CºÍHµÄÆøÌ¬Ç⻯ÎïµÄÎȶ¨ÐÔÇ¿Èõ¹ØÏµÎª£º
 
 £¨Ó÷Ö×Óʽ±íʾ£©£»
£¨3£©·Ö×Ó×é³ÉΪADG2µÄÎïÖÊÔÚË®ÖлáÇ¿ÁÒË®½â£¬²úÉúʹƷºìÈÜÒºÍÊÉ«µÄÎÞÉ«ÆøÌåºÍ Ò»ÖÖÇ¿Ëᣮ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º
 
£®
£¨4£©¹¤ÒµÉÏ¿ÉÓô¿¼îÈÜÒº´¦ÀíEDºÍED2£¬¸Ã·´Ó¦ÈçÏ£ºED+ED2+Na2C03=2
 
+CO2ºáÏßÉÏijÑεĻ¯Ñ§Ê½Ó¦Îª
 
£®
£¨5£©ÔÚÒ»ÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦2AD2+D2
´ß»¯¼Á
¼ÓÈÈ
2AD3¡÷H=-47KJ/molÔÚÉÏÊöƽºâÌåϵÖмÓÈë18D2£¬µ±Æ½ºâ·¢ÉúÒÆ¶¯ºó£¬AD2ÖÐ18DµÄ°Ù·Öº¬Á¿
 
 £¨Ìî¡°Ôö¼Ó¡±¡¢¡°¼õÉÙ¡±»ò¡°²»±ä¡±£©ÆäÔ­ÒòΪ
 
£®
£¨6£©ÇëÉè¼ÆÒ»¸öʵÑé·½°¸£¬Ê¹Í­ºÍÏ¡µÄH2AD4ÈÜÒº·´Ó¦£¬µÃµ½À¶É«ÈÜÒººÍÇâÆø£®ÔÚ´ðÌ⿨ָ¶¨Î»ÖûæÉ½¸ÃʵÑé·½°¸×°ÖÃͼ£®

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

¾«Ó¢¼Ò½ÌÍø