£¨12·Ö£©¹¤ÒµÉÏÒ»°ã¿É²ÉÓÃÈçÏ·´Ó¦À´ºÏ³É¼×´¼£º
CO(g)+2H2(g)CH3OH(g)  ¦¤H£½£­a kJ¡¤mol£­1

£¨1£©ÉÏͼÊǸ÷´Ó¦ÔÚ²»Í¬Î¶ÈÏÂCOµÄת»¯ÂÊËæʱ¼ä±ä»¯µÄÇúÏß¡£
¢Ùa __0£¨Ìî¡°£¾¡± ¡°£¼¡± ¡°£½¡±£©¡£
¢ÚÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_         _£¨ÌîÐòºÅ£©¡£
a£®1mol CO(g)ºÍ2mol H2(g)Ëù¾ßÓеÄÄÜÁ¿Ð¡ÓÚ1mol CH3OH(g)Ëù¾ßÓеÄÄÜÁ¿
b£®½«1mol CO(g)ºÍ2mol H2(g)ÖÃÓÚÒ»ÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦ºó·Å³öa KJµÄÈÈÁ¿
c£®Éý¸ßζȣ¬Æ½ºâÏòÄæ·´Ó¦Òƶ¯£¬ÉÏÊöÈÈ»¯Ñ§·½³ÌʽÖеÄaÖµ½«¼õС
d£®È罫һ¶¨Á¿CO(g) ºÍH2(g)ÖÃÓÚijÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦ºó·ÅÈÈaKJ£¬Ôò´Ë¹ý³ÌÖÐÓÐ1molCO(g)±»»¹Ô­
£¨2£©ÔÚÒ»¶¨Ìõ¼þÏ£¬¿Æѧ¼ÒÀûÓôÓÑ̵ÀÆøÖзÖÀë³öCO2ÓëÌ«ÑôÄܵç³Øµç½âË®²úÉúµÄH2ºÏ³É¼×´¼£¬Æä¹ý³ÌÈçÏÂͼËùʾ£º

¢Ù¸ÃºÏ³É·Ï߶ÔÓÚ»·¾³±£»¤µÄ¼ÛÖµÔÚÓÚ_                                  _¡£
¢Ú15%¡«20%µÄÒÒ´¼°·£¨HOCH2CH2NH2£©Ë®ÈÜÒº¾ßÓÐÈõ¼îÐÔ£¬ÉÏÊöºÏ³ÉÏß·ÖÐÓÃ×÷CO2
ÎüÊÕ¼Á¡£ÓÃÀë×Ó·½³Ìʽ±íʾÒÒ´¼°·Ë®ÈÜÒº³ÊÈõ¼îÐÔµÄÔ­Òò£º
                                                                            ¡£
£¨3£©¼×´¼È¼Áϵç³ØµÄ¹¤×÷Ô­ÀíÈçÏÂ×óͼËùʾ¡£¸Ãµç³Ø¹¤×÷ʱ£¬c¿ÚͨÈëµÄÎïÖÊ·¢ÉúµÄµç¼«
·´Ó¦Ê½Îª£º_                                                                 _¡£

£¨4£©ÒÔÉÏÊöµç³Ø×öµçÔ´£¬ÓÃÉÏÓÒͼËùʾװÖã¬ÔÚʵÑéÊÒÖÐÄ£ÄâÂÁÖÆÆ·±íÃæ¡°¶Û»¯¡±´¦ÀíµÄ¹ý³ÌÖУ¬·¢ÏÖÈÜÒºÖð½¥±ä»ë×Ç£¬Ô­ÒòÊÇ£¨ÓÃÏà¹ØµÄµç¼«·´Ó¦Ê½ºÍÀë×Ó·½³Ìʽ±íʾ£©£º
_                                                                            ¡£
£¨1£©¢Ù >£¨2·Ö£© ¢Úd£¨2·Ö£©
£¨2£©¢ÙÓÐÀûÓÚ·ÀÖ¹ÎÂÊÒЧӦ£¨2·Ö£©
¢ÚHOCH2CH2NH2 £« H2OHOCH2CH2NH3£«£« OH£­£¨2·Ö£©
£¨3£© O2+4e£­+4H+ =2H2O£¨2·Ö£©£» 
£¨4£© Al£­3e£­= Al3+£¨1·Ö£©     Al3++3HCO3£­ = Al(OH)3¡ý+3CO2¡ü£¨2·Ö£©
»òAl£­3e£­+3HCO3£­ = Al(OH)3¡ý+3CO2¡ü£¨2·Ö£©
¿¼²éÍâ½çÌõ¼þ¶ÔƽºâµÄÓ°ÏìÒÔ¼°µç»¯Ñ§µÄÓ¦Óõȡ£
£¨1£©¸ù¾ÝͼÏñ¿ÉÖª£¬Î¶ÈΪT2µÄÇúÏßÏȴﵽƽºâ״̬£¬ËùÒÔT2´óÓÚT1¡£Ëæ×ÅζȵÄÉý¸ß£¬·´Ó¦ÎïµÄת»¯ÂÊÊǽµµÍµÄ£¬ËµÃ÷Éý¸ßζÈƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬ËùÒÔÕý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬a´óÓÚ0£»·´Ó¦·ÅÈÈ£¬ËµÃ÷·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬a²»ÕýÈ·¡£·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬1mol CO(g)ºÍ2mol H2(g)²»¿ÉÄÜÉú³É1mol¼×´¼£¬ËùÒԷųöµÄÈÈÁ¿ÒªÐ¡ÓÚakJ£¬b²»ÕýÈ·¡£·´Ó¦ÈÈÓëÎïÖÊÇ°ÃæµÄ»¯Ñ§¼ÆÁ¿ÊýÓйأ¬ÓëƽºâµÄÒƶ¯·½ÏòÎ޹أ¬c²»ÕýÈ·¡£ËùÒÔÑ¡ÏîdÊÇÕýÈ·µÄ¡£
£¨2£©¸ù¾Ýת»¯¿ÉÖªÉú³ÉµÄCO2ÓÃÀ´ºÏ³É¼×´¼£¬ËùÒÔÓÐÀûÓÚ¿ÉÖªÎÂÊÒЧӦ£»¸ù¾ÝÒÒËá°·µÄ½á¹¹¼òʽ¿ÉÖª£¬·Ö×ÓÖеݱ»ù¿ÉÒÔ½áºÏË®µçÀë³öµÄÇâÀë×Ó£¬´Ó¶øÆÆ»µË®µÄµçÀëƽºâ£¬Ê¹ÈÜÒºÖÐOH£­Å¨¶È´óÓÚÇâÀë×ÓŨ¶È£¬ÈÜÒºÏÔ¼îÐÔ¡£
£¨3£©Ô­µç³ØÖиº¼«ÊÇʧȥµç×ӵģ¬¼×´¼ÔÚ·´Ó¦ÖÐÊÇ»¹Ô­¼Áʧȥµç×Ó±»Ñõ»¯£¬Òò´Ë¼×´¼ÔÚ¸º¼«Í¨Èë¡£¸ù¾ÝȼÁϵç³ØµÄ½á¹¹¿ÉÅжÏÇâÀë×ÓÏòÓÒ²àÒƶ¯£¬ËùÒÔÓÒ²àÊÇÕý¼«£¬×ó²àµç¼«ÊǸº¼«¡£Òò´ËÑõÆøÔÚÓÒ²àͨÈ룬ËùÒÔc´¦µÄµç¼«·´Ó¦Ê½ÎªO2+4e£­+4H+ =2H2O¡£
£¨4£©¸ù¾Ýµç½â³ØµÄ½á¹¹¿ÉÖª£¬ÂÁÊÇÑô¼«£¬Ê§È¥µç×ÓÉú³ÉÂÁÀë×Ó½øÈëÈÜÒºÖС£ÓÉÓÚµç½âÖÊÊÇ̼ËáÇâÄÆÈÜÒº£¬ËùÒÔÉú³ÉµÄÂÁÀë×ÓºÍ̼ËáÇâÄÆË®½âÏ໥´Ù½ø£¬´Ó¶øÉú³ÉÇâÑõ»¯ÂÁ³Áµí¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨3·Ö£©ÔÚÏÂÁÐÎïÖÊÖУº¢Ùµâ¢ÚÉÕ¼î¢ÛNaCl ¢Ü¸É±ù¢ÝÂÈ»¯Çâ¢ÞNa2O2¡£
ÓÃÐòºÅÌîдÏÂÁпհףº
£¨1£©¼Èº¬ÓÐÀë×Ó¼üÓÖº¬ÓзǼ«ÐÔ¼üµÄ»¯ºÏÎïÊÇ                  £»
£¨2£©½öº¬ÓÐÀë×Ó¼üµÄ»¯ºÏÎïÊÇ                                £»
£¨3£©½öº¬Óй²¼Û¼üµÄµ¥ÖÊÊÇ                                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ
A£®½«NaHSO4¾§Ìå¼ÓÈÈÈÛÈÚʱÓÐÁ½ÖÖ»¯Ñ§¼ü¶ÏÁÑ
B£®Ë®·Ö×Ó¼äÒòΪ´æÔÚÇâ¼ü,ËùÒÔ¼ÓÈÈÖÁ½Ï¸ßζÈʱҲÄÑÒÔ·Ö½â
C£®¸ù¾Ý½ðÊô¾§ÌåµÄ¹²ÐÔ¿ÉÖª½ðÊô¼üºÍÀë×Ó¼üÒ»ÑùûÓз½ÏòÐԺͱ¥ºÍÐÔ
D£®ÒÔ¼«ÐÔ¹²¼Û¼üÐγɵķÖ×ÓÒ»¶¨ÊǼ«ÐÔ·Ö×Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

Àë×Ó¼üµÄÇ¿ÈõÖ÷Òª¾ö¶¨ÓÚÀë×Ӱ뾶ºÍÀë×ÓµçºÉÖµ¡£Ò»°ã¹æÂÉÊÇ£ºÀë×Ӱ뾶ԽС£¬µçºÉÖµÔ½´ó£¬ÔòÀë×Ó¼üԽǿ¡£ÊÔ·ÖÎö£º¢ÙK2O  ¢ÚCaO  ¢ÛMgO µÄÀë×Ó¼üÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ   
A£®¢Û¢Ú¢Ù B£®¢Û¢Ù¢ÚC£®¢Ú¢Ù¢ÛD£®¢Ù¢Ú¢Û

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ
A£®ÔÚÈκÎÎïÖÊ·Ö×ÓÖж¼º¬Óл¯Ñ§¼ü
B£®HF¡¢HCl¡¢HBr¡¢HIµÄ·ÐµãÒÀ´ÎÉý¸ß
C£®D2O·Ö×ÓÓëH2O·Ö×ÓÊÇËùº¬µç×Ó×ÜÊý²»ÏàµÈµÄ·Ö×Ó
D£®CO2¡¢PCl3·Ö×ÓÖÐËùÓÐÔ­×Ó¶¼Âú×ã×îÍâ²ã8µç×ÓÎȶ¨½á¹¹

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚ25oC¡¢1.01¡Á105 PaÏ£¬½«2.2 g CO2ͨÈë75 mL 1 mol/L NaOHÈÜÒºÖгä·Ö·´Ó¦£¬²âµÃ·´Ó¦·Å³öx kJ µÄÈÈÁ¿£®ÒÑÖªÔÚ¸ÃÌõ¼þÏ£¬1 mol CO2ͨÈë1 L 2 mol/L NaOHÈÜÒºÖгä·Ö·´Ó¦·Å³öy kJµÄÈÈÁ¿£¬ÔòCO2ÓëNaOHÈÜÒº·´Ó¦Éú³ÉNaHCO3µÄÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ(¡¡)
A£®CO2(g)£«NaOH(aq)==NaHCO3(aq)£»¦¤H£½£­(20y£­x) kJ/mol
B£®CO2(g)£«NaOH(aq)==NaHCO3(aq)£»¦¤H£½£­(4x£­y) kJ/mol
C£®CO2(g)£«NaOH(aq)==NaHCO3(aq)£»¦¤H£½£­(40x£­y) kJ/mol
D£®2CO2(g)£«2NaOH(l)==2NaHCO3(l)£»¦¤H£½£­(80x£­20y) kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚ¢ÙH2O2¡¢¢ÚNaOH¡¢¢ÛCaF2¡¢¢ÜNa2O2ÖУ¬Ö»º¬ÓÐÀë×Ó¼üµÄÊÇ£¨±¾ÌâÌî±àºÅ£©       £¬¼Èº¬Óм«ÐÔ¹²¼Û¼üÓÖº¬ÓÐÀë×Ó¼üµÄÊÇ      £¬¼Èº¬ÓÐÀë×Ó¼üÓÖº¬ÓзǼ«ÐÔ¼üµÄÊÇ      £¬¼Èº¬ÓзǼ«ÐÔ¹²¼Û¼üÓÖº¬Óм«ÐÔ¹²¼Û¼üµÄÊÇ         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨7·Ö£©ÓÐÏÂÁÐÎïÖÊ£º¢ÙH2 ¢ÚNa2S2 ¢ÛKOH ¢ÜHF ¢ÝH2O2 ¢ÞMgCl2 ¢ßNH4Cl£¬°´ÏÂÁÐÒªÇó£¬ÓÃԲȦÊý×Ö£¨È磺¢Ù¡¢¢Ú¡¢¢ÛµÈ£©Ìî¿Õ£º
(1)Ö»ÓÉÀë×Ó¼ü¹¹³ÉµÄÎïÖÊÊÇ________£»
(2)Ö»Óɼ«ÐÔ¼ü¹¹³ÉµÄÎïÖÊÊÇ________£»
(3)Ö»ÓɷǼ«ÐÔ¼ü¹¹³ÉµÄÎïÖÊÊÇ      £»
(4)Ö»ÓɷǽðÊôÔªËØ×é³ÉµÄÀë×Ó»¯ºÏÎïÊÇ        £»
(5)Óɼ«ÐÔ¼üºÍ·Ç¼«ÐÔ¼ü¹¹³ÉµÄÎïÖÊÊÇ___________£»
(6)ÓÉÀë×Ó¼üºÍ¼«ÐÔ¼ü¹¹³ÉµÄÎïÖÊÊÇ_____________£»
(7)ÓÉÀë×Ó¼üºÍ·Ç¼«ÐÔ¼ü¹¹³ÉµÄÎïÖÊÊÇ___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ
A£®ÔÚµ¥Öʵľ§ÌåÖÐÒ»¶¨²»´æÔÚÀë×Ó¼ü
B£®·Ö×Ó¾§ÌåÖÐÖ»´æÔÚ·Ö×Ó¼ä×÷ÓÃÁ¦£¬²»º¬ÓÐÆäËû»¯Ñ§¼ü
C£®Èκξ§ÌåÖУ¬Èôº¬ÓÐÑôÀë×ÓÒ²Ò»¶¨º¬ÓÐÒõÀë×Ó
D£®½ðÊôÔÚ³£ÎÂ϶¼ÒÔ¾§ÌåÐÎʽ´æÔÚ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸