ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿Cu2OÔÚÍ¿ÁÏ¡¢ÓÐÉ«²£Á§ºÍ´ß»¯¼ÁµÈÁìÓòÓÐ׏㷺µÄÓÃ;¡£Cu2OΪºìÉ«·ÛÄ©£¬²»ÈÜÓÚË®£¬Ò×ÈÜÓÚÑÎËáºÍÏ¡ÁòËá¡£¿ÉÓõç½â·¨ÖƱ¸Ñõ»¯ÑÇÍ£ºÓÃÍ×÷Ñô¼«£¬îÑƬ×÷Òõ¼«£¬µç½âҺΪһ¶¨Å¨¶ÈµÄNaClºÍNaOHµÄ»ìºÏÈÜÒº£»Ñô¼«¼°ÆäÈÜÒºÖÐÓйØת»¯ÈçͼËùʾ¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ
A.µç½âÒºÖеÄNaOH²»ÄÜÓÃÑÎËá´úÌæ
B.¹ý³Ì¢ÙÖÐCu±»Ñõ»¯Éú³ÉCuCl£
C.¹ý³Ì¢ÜµÄÀë×Ó·½³ÌʽΪ2Cu(OH)Cl£=Cu2O+2Cl£+H2O
D.µ±µç·ÖÐÓÐ0.05mol e£Í¨¹ýʱ£¬ÏûºÄ0.32g Cu
¡¾´ð°¸¡¿BD
¡¾½âÎö¡¿
A.²úÎïCu2OÒ×ÈÜÓÚÑÎËáºÍÏ¡ÁòËᣬ¹ÊNaOH²»ÄÜÓÃÑÎËá´úÌ棬AÕýÈ·£»
B.ÓÉͼʾ¿ÉÖª£¬¹ý³Ì¢ÙÖÐΪÎü¸½¹ý³Ì£¬ÍÔªËØ»¯ºÏ¼Ûδ±ä£¬Î´·¢ÉúÑõ»¯»¹Ô·´Ó¦£¬B´íÎó£»
C.¸ù¾Ýͼʾ¹ý³Ì¢ÜÖÐCu(OH)Cl£·´Ó¦Éú³ÉCu2OºÍCl££¬ÔòÀë×Ó·½³ÌʽΪ2Cu(OH)Cl£=Cu2O+2Cl£+H2O£¬CÕýÈ·£»
D.Õû¸ö¹ý³ÌÖе¥ÖÊÍ·´Ó¦Éú³ÉÁËCu2O£¬ÍÔªËØ»¯ºÏ¼ÛÓÉ0¼Û±äΪ+1¼Û£¬¹Êµ±µç·ÖÐÓÐ0.05mol e£Í¨¹ýʱ£¬ÏûºÄCuµÄÖÊÁ¿Îª64g/mol0.05mol=3.2g£¬D´íÎó£»
´ð°¸Ñ¡BD¡£
¡¾ÌâÄ¿¡¿ÎªÓ¦¶ÔʯÓͶÌȱ£¬Ò»Ì¼»¯Ñ§Ñо¿±¸ÊܹØ×¢¡£Ò»Ì¼»¯Ñ§ÊÇÖ¸ÒÔ·Ö×ÓÖÐÖ»º¬Ò»¸ö̼Ô×ӵĻ¯ºÏÎïÈç¼×´¼¡¢Ò»Ñõ»¯Ì¼µÈΪÔÁÏ£¬ÖÆÔì²úÆ·µÄ»¯Ñ§ÌåϵµÄ×ܳơ£
(1)CH3OH(g)ºÍNH3(g)ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦¿ÉÖƵüװ·CH3NH2(g)¡£
¢ÙÒÑÖª¸Ã·´Ó¦ÖÐÏà¹ØµÄ»¯Ñ§¼ü¼üÄÜÊý¾ÝÈçÏ£º
¹²¼Û¼ü | C¡ªO | N¡ªH | C¡ªN | C¡ªH |
E£¯(kJ£¯mol) | a | b | c | d |
ÔòH¡ªO¼üµÄ¼üÄÜΪ_________________kJ£¯mol(Óú¬ÓÐ×ÖĸµÄ´úÊýʽ±íʾ)
¢ÚÔÚijºãÈÝÃܱÕÈÝÆ÷ÖнøÐи÷´Ó¦£¬ÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬·Ö±ð²âµÃÆðʼʱCH3OH(g)µÄÎïÖʵÄÁ¿ºÍζȶÔƽºâʱCH3NH2(g)µÄÌå»ý·ÖÊýµÄÓ°Ï죬ÈçͼËùʾ£º(ͼÖÐT1¡¢T2±íʾζÈ)
ÔòT1_________T2(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±)£»____________(Ìî¡°a¡±¡¢¡°b¡±»ò¡°c¡±)µã¶ÔÓ¦µÄƽºâ״̬Öз´Ó¦ÎïNH3(g)µÄת»¯ÂÊ×î´ó¡£b¡¢dÁ½µãµÄƽºâ³£Êý´óС¹ØϵΪKb________Kd(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±)¡£
(2)¼×´¼ÔÚ¹¤ÒµÉÏ¿ÉÀûÓÃˮúÆøÀ´ºÏ³É£º¡£½«1mol COºÍ2mol H2ͨÈëÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£¬ÔÚÒ»¶¨Î¶ȺÍѹǿÏ´ﵽƽºâʱ¼×´¼µÄÌå»ý·ÖÊý (CH3OH)±ä»¯Ç÷ÊÆÈçͼËùʾ£º
ͼÖÐYÖá±íʾµÄÍâ½çÌõ¼þΪ________________£¬ÅжϵÄÀíÓÉÊÇ______________________¡£
ÒÑÖªv(Õý)=k(Õý)¡¤p(CO)¡¤p(H2)2£¬v(Äæ)=k(Äæ)¡¤p(CH3OH)£¬ÆäÖÐk(Õý)¡¢k(Äæ)·Ö±ðΪÕý¡¢Äæ·´Ó¦ËÙÂʳ£Êý£¬pΪ¸÷×é·ÖµÄ·Öѹ¡£ÔÚMµãËù´¦µÄζÈ(T3¡æ)ºÍѹǿ(p0kPa)Ï£¬·´Ó¦ÔÚ20·ÖÖӴﵽƽºâʱ (CH3OH)=10£¥£¬¸ÃζÈÏ·´Ó¦µÄƽ³£ÊýKP=____________kPa£2(ÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È¼ÆË㣬·Öѹ=×Üѹ¡ÁÎïÖʵÄÁ¿·ÖÊý)¡£ÈôÔÚ15·ÖÖÓʱ£¬´Ëʱ______________(¼ÆËã½á¹û±£ÁôÁ½Î»Ð¡Êý)