ÌâÄ¿ÄÚÈÝ

X¡¢Y¡¢ZÈýÖÖÖ÷×åÔªËصĵ¥ÖÊÔÚ³£ÎÂ϶¼Êdz£¼ûµÄÎÞÉ«ÆøÌ壬ÔÚÊʵ±Ìõ¼þÏ£¬ÈýÕßÖ®¼ä¿ÉÒÔÁ½Á½·¢Éú·´Ó¦Éú³É·Ö±ðÊÇË«ºË¡¢ÈýºËºÍËĺ˵ļס¢ÒÒ¡¢±ûÈýÖÖ·Ö×Ó£¬ÇÒÒÒ¡¢±û·Ö×ÓÖк¬ÓÐXÔªËصÄÔ­×Ó¸öÊý±ÈΪ2¡Ã3¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÔªËØXµÄÃû³ÆÊÇ________£¬±û·Ö×ӵĵç×ÓʽΪ________¡£
(2)Èô¼×ÓëYµ¥ÖÊÔÚ³£ÎÂÏ»ìºÏ¾ÍÓÐÃ÷ÏÔÏÖÏó£¬Ôò¼×µÄ»¯Ñ§Ê½Îª________¡£±ûÔÚÒ»¶¨Ìõ¼þÏÂת»¯Îª¼×ºÍÒҵķ´Ó¦·½³ÌʽΪ___________________________¡£
(3)»¯ºÏÎﶡº¬X¡¢Y¡¢ZÈýÖÖÔªËØ£¬¶¡ÊÇÒ»ÖÖ³£¼ûµÄÇ¿Ëᣬ½«¶¡Óë±û°´ÎïÖʵÄÁ¿Ö®±È1¡Ã1»ìºÏºóËùµÃÎïÖÊÎìµÄ¾§Ìå½á¹¹Öк¬ÓеĻ¯Ñ§¼üΪ________(Ñ¡ÌîÐòºÅ)¡£
a£®Ö»º¬¹²¼Û¼ü    b£®Ö»º¬Àë×Ó¼ü   c£®¼Èº¬Àë×Ó¼ü£¬ÓÖº¬¹²¼Û¼ü
(1)Çâ¡¡ 
(2)NO¡¡4NH3£«5O24NO£«6H2O
(3)c
´Ó·Ö×ÓµÄÌصãºÍµ¥ÖÊΪÎÞÉ«ÆøÌåÈ¥·ÖÎö¿ÉÈ·¶¨£ºX¡¢Y¡¢Z·Ö±ðΪH¡¢O¡¢N£»¼×¡¢ÒÒ¡¢±û·Ö±ðΪNO¡¢H2O¡¢NH3£»¶¡ÎªHNO3£»ÎìΪNH4NO3¡£ÔÚNH4NO3¾§ÌåÖмȺ¬ÓÐÀë×Ó¼ü£¬ÓÖº¬Óм«ÐÔ¹²¼Û¼ü¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÖÆÀä¼ÁÊÇÒ»ÖÖÒ×±»Ñ¹Ëõ¡¢Òº»¯µÄÆøÌ壬Һ»¯ºóÔÚ¹ÜÄÚÑ­»·£¬Õô·¢Ê±ÎüÊÕÈÈÁ¿£¬Ê¹»·¾³Î¶ȽµµÍ£¬´ïµ½ÖÆÀäµÄÄ¿µÄ¡£ÈËÃÇÔø²ÉÓÃÒÒÃÑ¡¢NH3¡¢CH3ClµÈ×÷ÖÆÀäÖÆ£¬µ«ËüÃDz»ÊÇÓж¾¾ÍÊÇÒ×ȼ£¬ÓÚÊÇ¿Æѧ¼Ò¸ù¾ÝÔªËØÐÔÖʵĵݱä¹æÂÉÀ´¿ª·¢ÐµÄÖÆÀä¼Á¡£¸ù¾ÝÒÑÓÐ֪ʶ£¬Ä³Ð©ÔªËØ»¯ºÏÎïµÄÒ×ȼÐÔ¡¢¶¾ÐԱ仯Ç÷ÊÆÈçÏ£º
(1)Ç⻯ÎïµÄÒ×ȼÐÔ£º
SiH4£¾PH3£¾H2S£¾HCl£¬Ôò________£¾________£¾H2O£¾HF(ÌîÎïÖʵĻ¯Ñ§Ê½)¡£
(2)»¯ºÏÎïµÄ¶¾ÐÔ£º
PH3£¾NH3£¬CCl4£¾CF4£¬ÔòH2S________H2O£¬CS2________CO2(Ìî¡°£¾¡±¡°£½¡±»ò¡°£¼¡±)¡£
ÓÚÊÇ¿Æѧ¼ÒÃÇ¿ªÊ¼°Ñ×¢ÒâÁ¦¼¯ÖÐÔÚº¬F¡¢ClµÄ»¯ºÏÎïÉÏ¡£
(3)ÒÑÖªCCl4µÄ·ÐµãΪ76.8 ¡æ£¬CF4µÄ·ÐµãΪ£­128 ¡æ£¬ÐµÄÖÆÀä¼ÁµÄ·Ðµã·¶Î§Ó¦½éÓÚ¶þÕßÖ®¼ä£¬¾­¹ý½Ï³¤Ê±¼äµÄ·´¸´ÊµÑ飬·¢ÏÖÁËÖÆÀä¼ÁCF2Cl2(·úÀû°º)£¬ÆäËûÀàËƵÄÖÆÀä¼Á¿ÉÒÔÊÇ________¡£
(4)È»¶øÕâÖÖÖÆÀä¼ÁÔì³ÉµÄµ±½ñijһ»·¾³ÎÊÌâÊÇ_____________________________¡£
µ«ÇóÖúÓÚÖÜÆÚ±íÖÐÔªËؼ°Æ仯ºÏÎïµÄ______(ÌîдÏÂÁÐÑ¡ÏîµÄ±àºÅ)±ä»¯Ç÷ÊÆ¿ª·¢ÖÆÀä¼ÁµÄ¿Æѧ˼ά·½·¨ÊÇÖµµÃ½è¼øµÄ¡£
¢Ù¶¾ÐÔ£»¢Ú·Ðµã£»¢ÛÒ×ȼÐÔ£»¢ÜË®ÈÜÐÔ£»¢ÝÑÕÉ«
A£®¢Ù¢Ú¢ÛB£®¢Ú¢Ü¢ÝC£®¢Ú¢Û¢ÜD£®¢Ù¢Ú¢Ü¢Ý
ÈýÖÖ¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z£¬ËüÃǵÄÔ­×ÓÐòÊýÖ®ºÍΪ16£¬X¡¢Y¡¢ZÈýÖÖÔªËصij£¼ûµ¥ÖÊÔÚ³£ÎÂ϶¼ÊÇÎÞÉ«ÆøÌå¡£ÒÑÖªXÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇÄÚ²ãµç×ÓÊýµÄ3±¶£¬XºÍYµÄµ¥ÖÊÖ±½Ó»¯ºÏÐγÉÆøÌåA£¬XºÍZµÄµ¥ÖÊÖ±½Ó»¯ºÏÐγÉҺ̬»¯ºÏÎïB£¬YºÍZµÄµ¥ÖÊÖ±½Ó»¯ºÏÐγɵĻ¯ºÏÎïCÊÇÒ»ÖÖÎÞÉ«Óд̼¤ÐÔÆøζµÄÆøÌå¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©YÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ       ¡£
£¨2£©C¿ÉÔÚXµÄµ¥ÖÊÖÐȼÉյõ½YµÄµ¥Öʺͻ¯ºÏÎïB£¬ÀûÓô˷´Ó¦¿ÉÖƳÉÐÂÐ͵Ļ¯Ñ§µçÔ´(KOHÈÜÒº×öµç½âÖÊÈÜÒº)£¬Á½¸öµç¼«¾ùÓɶà¿×̼ÖƳɣ¬Í¨È˵ÄÆøÌåÓÉ¿×϶ÖÐÒݳö£¬²¢Ôڵ缫±íÃæ·Åµç£¬ÔòÕý¼«Í¨ÈëµÄÎïÖÊÊÇ        (ÌîÎïÖÊÃû³Æ)£»¸º¼«µÄµç¼«·´Ó¦Ê½Îª                       ¡£
£¨3£©CÓëXµÄµ¥ÖÊ·´Ó¦Éú³ÉAµÄ»¯Ñ§·½³ÌʽΪ          ¡£
£¨4£©³£ÎÂÏ£¬CµÄË®ÈÜÒºµÄpH=12£¬Ôò¸ÃÈÜÒºÖÐÓÉË®µçÀëµÄC(OH-)=      ¡£ÈôÏòCÈÜÒºÖмÓÈëµÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÑÎËᣬËùµÃÈÜÒºÖÐË®µÄµçÀë³Ì¶È      (Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)ÏàͬÌõ¼þÏÂCÈÜÒºÖÐË®µÄµçÀë³Ì¶È¡£
£¨5£©ÔÚ2LÃܱÕÈÝÆ÷ÖзÅÈë1molCÆøÌ壬ÔÚÒ»¶¨Î¶ȽøÐÐÈçÏ·´Ó¦£º
2C(g) Y2(g)+3Z2(g)£¬·´Ó¦Ê±¼ä(t)ÓëÈÝÆ÷ÄÚÆøÌå×Üѹǿ(p)µÄÊý¾Ý¼ûϱí
ʱ¼ät£¯min
0
1
2
3
4
5
×ÜѹǿP l00 kPa
4
4.6
5.4
5.8
6
6
 
¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽÊÇ             (ÓþßÌåÎïÖʵĻ¯Ñ§Ê½±íʾ)£»Æ½ºâʱCµÄת»¯ÂÊΪ             ¡£
£¨6£©ÒÑÖª£º¢ÙY2(g)+2X2(g)=2YX2(g) H=+67£®7 kJ¡¤mol-1¡£
¢ÚY2Z4(g)+X2(g)=Y2(g)+2Z2X(g)   H="-534" kJ¡¤mol-1¡£
Ôò2Y2Z4(g)+2YX2(g)=3Y2(g)+4Z2X(g) H=    kJ¡¤mol-1

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

¾«Ó¢¼Ò½ÌÍø