¡¾ÌâÄ¿¡¿ÒÑÖª£ºÇâÆøºÍÑõÆøµÄ»ìºÏÆøÌåÓö»ðÐǼ´·¢Éú±¬Õ¨£¬Éú³ÉË®£»Ë®ÔÚ1 000 ¡æÒÔÉϳÖÐø¼ÓÈÈ·Ö½âΪÇâÆøºÍÑõÆø£»Ë®µç½âÉú³ÉÇâÆøºÍÑõÆø¡£ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©ÇâÆøºÍÑõÆø·´Ó¦Éú³ÉË®µÄ·´Ó¦ÊÇ__________(Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±)·´Ó¦£¬2molÇâÆøºÍ1molÑõÆøµÄ×ÜÄÜÁ¿__________(Ìî¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±) 2molË®µÄ×ÜÄÜÁ¿£¬´Ë·´Ó¦Öл¯Ñ§ÄÜת»¯Îª_________ÄÜ¡£

£¨2£©Ë®ÔÚ¸ßÎÂÏ·ֽâµÄ·´Ó¦Îª______(Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±)·´Ó¦£¬·´Ó¦ÖÐÈÈÄÜת»¯Îª__________ÄÜ¡£µç½âË®µÄ¹ý³ÌÊÇ__________ÄÜת»¯Îª__________ÄܵĹý³Ì¡£

£¨3£©ÏÂÁÐÕýÈ·µÄÊÇ£º__________

¢ÙÐèÒª¼ÓÈȲÅÄÜ·¢ÉúµÄ·´Ó¦Ò»¶¨ÊÇÎüÈÈ·´Ó¦£»

¢Ú·ÅÈÈ·´Ó¦ÔÚ³£ÎÂÏÂÒ»¶¨ºÜÈÝÒ×·¢Éú£»

¢ÛNa2O2ÖмȺ¬ÓÐÀë×Ó¼üÒ²º¬ÓзǼ«ÐÔ¼ü

¢ÜÓеÄÎüÈÈ·´Ó¦ÔÚ³£ÎÂÏÂÒ²ÄÜ·¢Éú

¢ÝŨÁòËáÈÜÓÚË®ÊÇ·ÅÈÈ·´Ó¦

¢Þº¬ÓÐÀë×Ó¼üµÄ»¯ºÏÎïÒ»¶¨ÊÇÀë×Ó»¯ºÏÎº¬Óй²¼Û¼üµÄÒ»¶¨Êǹ²¼Û»¯ºÏÎï¡£

£¨4£©ÏÂÁÐÎïÖÊÖУº¢ÙNa2O2 ¢ÚHe ¢ÛNaOH ¢ÜN2 ¢ÝMgCl2 ¢ÞNH3 ¢ßH2O ¢àCl2 ¢áNaF

ÊÇÀë×Ó»¯ºÏÎïµÄÊÇ_____________________£¬Êǹ²¼Û»¯ºÏÎïµÄÊÇ______________________¡£

¡¾´ð°¸¡¿ ·ÅÈÈ ´óÓÚ ÈÈÄÜ ÎüÈÈ »¯Ñ§ÄÜ µçÄÜ »¯Ñ§ÄÜ ¢Û¢Ü ¢Ù¢Û¢Ý¢á ¢Þ¢ß

¡¾½âÎö¡¿£¨1£©ÇâÆøºÍÑõÆø·´Ó¦Éú³ÉË®µÄ·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬2molÇâÆøºÍ1molÑõÆøµÄ×ÜÄÜÁ¿´óÓÚ2molË®µÄ×ÜÄÜÁ¿£¬´Ë·´Ó¦Öл¯Ñ§ÄÜת»¯ÎªÈÈÄÜ¡££¨2£©Ë®ÔÚ¸ßÎÂÏ·ֽâµÄ·´Ó¦ÎªÎüÈÈ·´Ó¦£¬·´Ó¦ÖÐÈÈÄÜת»¯Îª»¯Ñ§ÄÜ¡£µç½âË®µÄ¹ý³ÌÊǵçÄÜÄÜת»¯Îª»¯Ñ§ÄܵĹý³Ì¡££¨3£©¢ÙÐèÒª¼ÓÈȲÅÄÜ·¢ÉúµÄ·´Ó¦²»Ò»¶¨ÊÇÎüÈÈ·´Ó¦£¬ÀýÈç̼ȼÉյȣ¬´íÎ󣻢ڷÅÈÈ·´Ó¦ÔÚ³£ÎÂϲ»Ò»¶¨ºÜÈÝÒ×·¢Éú£¬ÀýÈçÇâÆø£¬´íÎ󣻢ÛNa2O2ÖмȺ¬ÓÐÀë×Ó¼üÒ²º¬ÓзǼ«ÐÔ¼ü£¬ÕýÈ·£»¢ÜÓеÄÎüÈÈ·´Ó¦ÔÚ³£ÎÂÏÂÒ²ÄÜ·¢Éú£¬ÀýÈçÇâÑõ»¯±µ¾§ÌåºÍÂÈ»¯ï§·´Ó¦£¬ÕýÈ·£»¢ÝŨÁòËáÈÜÓÚË®ÊÇ·ÅÈÈ£¬ÊôÓÚÎïÀí±ä»¯£¬²»ÊÇ·ÅÈÈ·´Ó¦£¬´íÎ󣻢޺¬ÓÐÀë×Ó¼üµÄ»¯ºÏÎïÒ»¶¨ÊÇÀë×Ó»¯ºÏÎº¬Óй²¼Û¼üµÄ²»Ò»¶¨Êǹ²¼Û»¯ºÏÎÀýÈçÇâÑõ»¯ÄÆ£¬´íÎó¡£´ð°¸Ñ¡¢Û¢Ü£»£¨4£©º¬ÓÐÀë×Ó¼üµÄ»¯ºÏÎïÊÇÀë×Ó»¯ºÏÎÒò´ËÊôÓÚÀë×Ó»¯ºÏÎïµÄÊǹýÑõ»¯ÄÆ¡¢ÇâÑõ»¯ÄÆ¡¢ÂÈ»¯Ã¾¡¢·ú»¯ÄƵȣ»È«²¿Óɹ²¼Û¼üÐγɵϝºÏÎïÊǹ²¼Û»¯ºÏÎÊôÓÚ¹²¼Û»¯ºÏÎïµÄÊǰ±ÆøºÍË®¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¢ñ.Áª°±(ÓÖ³Æë£¬N2H4£¬ÎÞɫҺÌå)ÊÇÒ»ÖÖÓ¦Óù㷺µÄ»¯¹¤Ô­ÁÏ£¬¿ÉÓÃ×÷»ð¼ýȼÁÏ¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ʵÑéÊÒÖпÉÓôÎÂÈËáÄÆÈÜÒºÓë°±·´Ó¦ÖƱ¸Áª°±£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________¡£

(2)¢Ù2O2(g)£«N2(g)===N2O4(l)¡¡¦¤H1

¢ÚN2(g)£«2H2(g)===N2H4(l)¡¡¦¤H2

¢ÛO2(g)£«2H2(g)===2H2O(g)¡¡¦¤H3

¢Ü2N2H4(l)£«N2O4(l)===3N2(g)£«4H2O(g)¡¡¦¤H4£½£­1048.9 kJ¡¤mol£­1

ÉÏÊö·´Ó¦ÈÈЧӦ֮¼äµÄ¹ØÏµÊ½Îª¦¤H4£½__________________£¬Áª°±ºÍN2O4¿É×÷Ϊ»ð¼ýÍÆ½ø¼ÁµÄÖ÷ÒªÔ­ÒòΪ________________¡£

¢ò.úȼÉÕÅŷŵÄÑÌÆøº¬SO2ºÍNO2£¬ÐγÉËáÓ꣬ÎÛȾ´óÆø¡£ÏÖÓÃNaClO¡¢Ca(ClO)2´¦Àí£¬µÃµ½½ÏºÃµÄÑÌÆøÍÑÁòЧ¹û¡£

(3)ÒÑÖªÏÂÁз´Ó¦£º

SO2(g)£«2OH£­(aq)===SO32£­ (aq)£«H2O(l)¡¡¦¤H1

ClO£­(aq)£«SO32£­ (aq)===SO42£­ (aq)£«Cl£­(aq)¡¡¦¤H2

CaSO4(s)===Ca2£«(aq)£«SO42£­ (aq)¡¡¦¤H3

Ôò·´Ó¦SO2(g)£«Ca2£«(aq)£«ClO£­(aq)£«2OH£­(aq)===CaSO4(s)£«H2O(l)£«Cl£­(aq)µÄ¦¤H£½________¡£

¢ó.¹¤ÒµÉϳ£ÓÃÁ×¾«¿ó[Ca5(PO4)3F]ºÍÁòËá·´Ó¦ÖÆ±¸Á×Ëá¡£ÒÑÖª25 ¡æ£¬101 kPaʱ£º

CaO(s)£«H2SO4(l)===CaSO4(s)£«H2O(l) ¦¤H£½£­271 kJ/mol

5CaO(s)£«3H3PO4(l)£«HF(g)===Ca5(PO4)3F(s)£«5H2O(l)¡¡¦¤H£½£­937 kJ/mol

ÔòCa5(PO4)3FºÍÁòËá·´Ó¦Éú³ÉÁ×ËáµÄÈÈ»¯Ñ§·½³ÌʽÊÇ__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖ¶ÌÖÜÆÚÔªËØ£¬ËüÃǵĺ˵çºÉÊý°´C¡¢A¡¢B¡¢D¡¢EµÄ˳ÐòÔö´ó¡£C¡¢D¶¼ÄÜ·Ö±ðÓëA°´Ô­×Ó¸öÊý±ÈΪ1¡Ã1»ò2¡Ã1Ðγɻ¯ºÏÎCB¿ÉÓëEA2·´Ó¦Éú³ÉC2AÓëÆøÌ¬ÎïÖÊEB4£»EµÄM²ãµç×ÓÊýÊÇK²ãµç×ÓÊýµÄ2±¶¡£

(1)д³öÏÂÁÐÔªËØµÄÃû³Æ£ºB£º________E£º________¡£

(2)д³öEB4µÄ»¯Ñ§Ê½£º__________¡£

(3)D2A2Öк¬ÓеĻ¯Ñ§¼üÀàÐÍΪ____________¡£

(4)ÓÉA¡¢C¡¢DÈýÖÖÔªËØ×é³ÉµÄ»¯ºÏÎïMÊôÓÚ________(Ìî¡°Àë×Ó¡±»ò¡°¹²¼Û¡±)»¯ºÏÎMÈÜÓÚË®ºó________(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)·¢Éú»¯Ñ§±ä»¯£¬Ô­ÒòÊÇ(´Ó»¯Ñ§¼ü±ä»¯µÄ½Ç¶È½âÊÍ)£º

___________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬ÏÂÁи÷×éÀë×ÓÔÚÖ¸¶¨ÈÜÒºÖÐÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ

A. 0.1 mol¡¤L¡ª1µÄNaOHÈÜÒº£ºK+¡¢Na+¡¢SO42¡ª¡¢CO32¡ª

B. 0.1 mol¡¤L¡ª1µÄNa2CO3ÈÜÒº£ºK+¡¢Ba2+¡¢NO3¡ª¡¢Cl¡ª

C. 0.1 mol¡¤L¡ª1FeCl3ÈÜÒº£ºK+¡¢NH4+¡¢I¡ª¡¢SCN¡ª

D. 1.0molL¡ª1µÄKNO3ÈÜÒº£ºH+¡¢Fe2+¡¢Cl¡ª¡¢SO42¡ª

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªCH4(g)+2O2(g)=2CO2(g)+2H2O(l)£»¡÷H=£­Q1kJ/mol£¬2H2(g)+O2(g)=2H2O(g) £»¡÷H=£­Q2 kJ/mol£¬2H2(g)+O2(g)=2H2O(l) £»¡÷H=£­Q3 kJ/mol£¬³£ÎÂÏÂÈ¡Ìå»ý±ÈΪ4£º1µÄCH4ºÍH2µÄ»ìºÏÆø11.2L(±ê¿ö)¾­ÍêȫȼÉÕºó»Ö¸´ÖÁ³£Î£¬·Å³öµÄÈÈÁ¿ÊÇ £¨ £©

A£®0.4Q1+0.05Q3 B£®0.4Q1+0.05Q2

C£®0.4Q1+0.1Q3 D£®0.4Q1+0.2Q2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÃºÈ¼Éյķ´Ó¦ÈÈ¿Éͨ¹ýÒÔÏÂÁ½¸ö;¾¶À´ÀûÓãºa.ÀûÓÃúÔÚ³ä×ãµÄ¿ÕÆøÖÐÖ±½ÓȼÉÕ²úÉúµÄ·´Ó¦ÈÈ£»b.ÏÈʹúÓëË®ÕôÆø·´Ó¦µÃµ½ÇâÆøºÍÒ»Ñõ»¯Ì¼£¬È»ºóʹµÃµ½µÄÇâÆøºÍÒ»Ñõ»¯Ì¼ÔÚ³ä×ãµÄ¿ÕÆøÖÐȼÉÕ¡£ÕâÁ½¸ö¹ý³ÌµÄÈÈ»¯Ñ§·½³ÌʽΪ£º

a£®C(s)£«O2(g)===CO2(g) ¦¤H£½E1¢Ù

b£®C(s)£«H2O(g)===CO(g)£«H2(g) ¦¤H£½E2¢Ú

H2(g)£«1/2O2(g)===H2O(g) ¦¤H£½E3¢Û

CO(g)£«1/2O2(g)===CO2(g) ¦¤H£½E4¢Ü

ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©Óë;¾¶aÏà±È£¬Í¾¾¶bÓн϶àµÄÓŵ㣬¼´___________________________________¡£

£¨2£©ÉÏÊöËĸöÈÈ»¯Ñ§·½³ÌʽÖЦ¤H>0µÄ·´Ó¦ÓÐ_____________________________________ ___________________________________¡£

£¨3£©µÈÖÊÁ¿µÄú·Ö±ðͨ¹ýÒÔÉÏÁ½Ìõ²»Í¬µÄ;¾¶²úÉúµÄ¿ÉÀûÓõÄ×ÜÄÜÁ¿¹ØÏµÕýÈ·µÄÊÇ________¡£

A£®a±Èb¶à B£®a±ÈbÉÙ

C£®aÓëbÔÚÀíÂÛÉÏÏàͬ D£®Á½ÕßÎÞ·¨±È½Ï

£¨4£©¸ù¾ÝÄÜÁ¿Êغ㶨ÂÉ£¬E1¡¢E2¡¢E3¡¢E4Ö®¼äµÄ¹ØÏµÎª________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª4NH3(g)+5O2(g) =" 4NO(g)" +6H2O(l) ¡÷H=-xkJ/mol¡£Õô·¢1mol H2O£¨l£©ÐèÒªÎüÊÕµÄÄÜÁ¿Îª44kJ£¬ÆäËüÏà¹ØÊý¾ÝÈçÏÂ±í£º


NH3(g)

O2(g)

NO(g)

H2O(g)

1mol·Ö×ÓÖеĻ¯Ñ§¼ü¶ÏÁÑʱÐèÒªÎüÊÕµÄÄÜÁ¿/kJ

a

b

z

d

Ôò±íÖÐz£¨ÓÃx¡¢a¡¢b¡¢d±íʾ£©µÄ´óСΪ

A£®(x+4a+5b-6d-44)/4 B£®(x+12a+5b-12d-264)/4

C£®(x+4a+5b-6d-264)/4 D£®(x+12a+5b-12d-44)/4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿W¡¢X¡¢Y¡¢ZÊÇÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÖ÷×åÔªËØ¡£XµÄµ¥Öʾ­³£×÷Ϊ±£»¤Æø£¬ÕâËÄÖÖÔªËØ¿ÉÒÔ×é³ÉÒõÑôÀë×Ó¸öÊý±ÈÊÇ1£º1µÄÀë×Ó»¯ºÏÎï¼×£¬ÓÉY¡¢ZÐγɵÄÒ»ÖÖ»¯ºÏÎïÒÒºÍW¡¢Z ÐγɵϝºÏÎï±û·´Ó¦Éú³Éµ­»ÆÉ«¹ÌÌå¡£ÏÂÁÐ˵·¨Öв»ÕýÈ·µÄÊÇ

A. »¯ºÏÎï¼×Ò»¶¨ÄÜÓëNaOHÈÜÒº·´Ó¦

B. Ô­×Ó°ë¾¶´óС˳ÐòÊÇZ>X>Y>W

C. W¡¢X×é³ÉµÄ»¯ºÏÎïX2W4Êǹ²¼Û»¯ºÏÎï

D. W¡¢X¡¢Y×é³ÉµÄ»¯ºÏÎïË®ÈÜÒºÒ»¶¨³ÊËáÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚ25¡æÊ±£¬Ä³Ï¡ÈÜÒºÖÐÓÉË®µçÀë²úÉúµÄc£¨OH-£©=10-10molL-1£®ÏÂÁÐÓйظÃÈÜÒºµÄÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©

A. ¸ÃÈÜÒºÒ»¶¨³ÊËáÐÔ

B. ¸ÃÈÜÒºÖÐc£¨H+£©¿ÉÄܵÈÓÚ10-5molL-1

C. ¸ÃÈÜÒºµÄpH¿ÉÄÜΪ4Ò²¿ÉÄÜΪ10

D. ¸ÃÈÜÒºÓпÉÄܳÊÖÐÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸