ϱíÖеÄÊý¾ÝÊÇÆÆ»µ1 molÎïÖÊÖеĻ¯Ñ§¼üËùÏûºÄµÄÄÜÁ¿(kJ)£º

ÎïÖÊ

H2(g)

O2(g)

H2O(g)

ÄÜÁ¿

436

496

926

(1)·´Ó¦2H2(g)£«O2(g)===2H2O(g)ÊÇ________(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)·´Ó¦£¬Õâ˵Ã÷2 mol H2(g)ºÍ1 mol O2(g)¾ßÓеÄÄÜÁ¿±È2 mol H2O(g)¾ßÓеÄÄÜÁ¿________(Ìî¡°¸ß¡±»ò¡°µÍ¡±)¡£

(2)ÇëÓÃͼʾ±íʾ³ö2 mol H2(g)Óë1 mol O2(g)Éú³É2 mol H2O(g)µÄ·´Ó¦¹ý³Ì£º

(3)¸ù¾Ý±íÖÐÊý¾Ý£¬Ð´³öH2(g)ÓëO2(g)ÍêÈ«·´Ó¦Éú³ÉH2O(g)µÄÈÈ»¯Ñ§·½³Ìʽ£º________________________________________________________________________¡£

(4)ÈôÒÑÖª£ºH2O(g)===H2O(l)¡¡¡¡¦¤H£½£­44 kJ·mol£­1£¬Ð´³öH2(g)ÓëO2(g)ÍêÈ«·´Ó¦Éú³ÉH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ______________________¡£


´ð°¸¡¡(1)·ÅÈÈ¡¡¸ß

(2)»ò

(3)2H2(g)£«O2(g)===2H2O(g)

¦¤H£½£­484 kJ·mol£­1

(4)2H2(g)£«O2(g)===2H2O(l)

¦¤H£½£­572 kJ·mol£­1


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ïòº¬ÓÐ0.78 mo lµÄFeCl2ÈÜÒºÖÐͨÈë0.009molCl2£¬ÔÙ¼ÓÈ뺬0.01molX2O72-µÄËáÐÔÈÜÒº£¬Ê¹ÈÜÒºÖеÄFe2+Ç¡ºÃÈ«²¿Ñõ»¯£¬²¢Ê¹X2O72-»¹Ô­ÎªXn+£¬ÔònµÄÖµÊÇ£¨   £©

   A£®2    B£®3     C£®4    D£®5 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÂÁÓÃ;¹ã·º£¬ÓÃÂÁÍÁ¿ó(Ö÷Òª³É·ÖΪAl2O3·nH2O¡¢ÉÙÁ¿SiO2ºÍFe2O3)ÖÆÈ¡AlÓÐÈçÏÂ;¾¶£º

(1)ÂËÒºA¿ÉÓÃÓÚ¾»Ë®£¬Æ侻ˮԭÀíÓÃÀë×Ó·½³Ìʽ±íʾΪ________________________¡£

(2)×ÆÉÕʱʢ·ÅÒ©Æ·µÄÒÇÆ÷Ãû³ÆÊÇ__________¡£

(3)²½Öè¢õÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ____________________________________¡£

(4)²½Öè¢óÖÐÉú³É¹ÌÌåC·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________________¡£

(5)È¡ÂËÒºB 100 mL£¬¼ÓÈë1 mol·L£­1ÑÎËá200 mL£¬³ÁµíÁ¿´ïµ½×î´óÇÒÖÊÁ¿Îª11.7 g¡£ÔòÂËÒºBÖÐc(AlO)£½______£¬c(Na£«)______(Ìî¡°>¡±¡¢¡°£½¡±»ò¡°<¡±)2 mol·L£­1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÊÒÎÂÏ£¬ÒÑ֪ijÈÜÒºÖÐÓÉË®µçÀëÉú³ÉµÄH£«ºÍOH£­µÄÎïÖʵÄÁ¿Å¨¶ÈµÄ³Ë»ýΪ

10£­24 mol2¡¤L£­2£¬ÔòÔÚ¸ÃÈÜÒºÖУ¬Ò»¶¨²»ÄÜ´óÁ¿´æÔÚµÄÀë×ÓÊÇ(   )

A£®SO32£­                B£®NH4£«         C£®NO3£­                 D£®HCO3-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁдëÊ©¶ÔË®µÄµçÀëÎÞÓ°ÏìµÄÊÇ(   )

A.Éý¸ßζȠ             B.¼ÓÈëÏ¡´×Ëá

C.¼ÓÈëÇâÑõ»¯ÄÆ          D.¼ÓÈëʳÑÎ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ñ°ÕÒÇå½àÄÜÔ´Ò»Ö±ÊÇ»¯Ñ§¼ÒŬÁ¦µÄ·½Ïò£¬ÏÂÁйØÓÚÄÜÔ´µÄ˵·¨ÖдíÎóµÄÊÇ(¡¡¡¡)

A£®ÇâÆøȼÉÕÈȸߣ¬ÆäȼÉÕ²úÎïÊÇË®£¬ÊÇÒ»ÖÖÀíÏëµÄÇå½àȼÁÏ

B£®ÀûÓÃÌ«ÑôÄܵÈÇå½àÄÜÔ´´úÌ滯ʯȼÁÏ£¬ÓÐÀûÓÚ½ÚÔ¼×ÊÔ´¡¢±£»¤»·¾³

C£®ÃºµÄÆø»¯¼¼ÊõÔÚÒ»¶¨³Ì¶ÈÉÏʵÏÖÁËúµÄ¸ßЧ¡¢Çå½àÀûÓÃ

D£®Ê¯ÓÍ×÷ΪÖØÒªµÄ¿ÉÔÙÉúÄÜÔ´Ó¦¸Ã±»¾¡Á¿µØÀûÓÃ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


·´Ó¦ A£«B¡ª¡úC(¦¤H£¼0)·ÖÁ½²½½øÐУº¢ÙA£«B¡ª¡úX (¦¤H £¾0)£¬¢ÚX¡ª¡úC(¦¤H£¼0)¡£ÏÂÁÐʾÒâͼÖУ¬ÄÜÕýÈ·±íʾ×Ü·´Ó¦¹ý³ÌÖÐÄÜÁ¿±ä»¯µÄÊÇ(¡¡¡¡)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÑÖª£ºHCN(aq)ÓëNaOH(aq)·´Ó¦µÄ¦¤H£½£­12.1 kJ·mol£­1£»HCl(aq)ÓëNaOH(aq)·´Ó¦µÄ¦¤H£½£­55.6 kJ·mol£­1£¬ÔòHCNÔÚË®ÈÜÒºÖеçÀëµÄ¦¤HµÈÓÚ(¡¡¡¡)

A£®£­67.7 kJ·mol£­1  B£®£­43.5 kJ·mol£­1

C£®£«43.5 kJ·mol£­1  D£®£«67.7 kJ·mol£­1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÄûÃÊÏ©ÊÇÒ»ÖÖʳÓÃÏãÁÏ£¬Æä½á¹¹¼òʽÈçͼ£¬ÓйØÄûÃÊÏ©µÄ·ÖÎö´íÎóµÄÊÇ
 A£®ÔÚÒ»¶¨Ìõ¼þÏ£¬1molÄûÃÊÏ©¿ÉÓë2molH2ÍêÈ«¼Ó³É

B£®ÄûÃÊÏ©µÄÒ»ÂÈ´úÎïÓÐ7ÖÖ

C£®ÔÚÒ»¶¨Ìõ¼þÏ£¬ÄûÃÊÏ©¿É·¢Éú¼Ó³É¡¢È¡´ú¡¢Ñõ»¯¡¢»¹Ô­·´Ó¦

D£®ÄûÃÊÏ©·Ö×ÓÖÐËùÓÐ̼ԭ×Ó²»¿ÉÄܶ¼ÔÚͬһƽÃæ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸