ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ò»¶¨Î¶ÈÏ£¬ÔÚÌå»ýΪ10LµÄÃܱÕÈÝÆ÷ÖУ¬3molXºÍ1molY½øÐÐÓ¦£º2X£¨g£©+Y£¨g£© Z£¨g£©£¬¾­2min´ïµ½Æ½ºâ£¬Éú³É0.6molZ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. ÒÔXŨ¶È±ä»¯±íʾµÄ·´Ó¦ËÙÂÊΪ0.01mol/£¨L¡¤s£©

B. ½«ÈÝÆ÷Ìå»ý±äΪ20L£¬ZµÄƽºâŨ¶ÈΪԭÀ´µÄ1/2

C. ÈôÔö´óѹǿ£¬ÔòÎïÖÊYµÄת»¯ÂʼõС

D. ÈôÉý¸ßζȣ¬XµÄÌå»ý·ÖÊýÔö´ó£¬Ôò¸Ã·´Ó¦µÄ¡÷H<0

¡¾´ð°¸¡¿D

¡¾½âÎö¡¿

A¡¢ÒÀ¾Ý»¯Ñ§·´Ó¦ËÙÂʹ«Ê½¼ÆËãZµÄËÙÂÊ£¬ÔÙÒÀ¾Ý»¯Ñ§·´Ó¦ËÙÂÊÖ®±ÈµÈÓÚ»¯Ñ§ÖÊÁ¿ÊýÖ®±È¼ÆËãXËÙÂÊ£»

B¡¢¸Ã·´Ó¦ÊÇÒ»¸öÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬ÌåϵµÄѹǿ¼õС£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£»

C¡¢¸Ã·´Ó¦ÊÇÒ»¸öÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬Ôö´óѹǿ£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£»

D¡¢Éý¸ßζȣ¬Æ½ºâÏòÎüÈÈ·½ÏòÒƶ¯¡£

AÏî¡¢¾­2min´ïµ½Æ½ºâ£¬Éú³É0.6molZ£¬ZµÄŨ¶È±ä»¯Á¿Îª0.06mol/L£¬ZµÄ·´Ó¦ËÙÂÊv£¨Z£©Îª0.06mol/L¡Â120s£½0.0005mol/£¨Ls£©£¬¸ù¾Ý»¯Ñ§·´Ó¦ËÙÂÊÖ®±ÈµÈÓÚ»¯Ñ§ÖÊÁ¿ÊýÖ®±È£¬ÓÉ·½³Ìʽ¿ÉÖªv£¨X£©=2v£¨Z£©=2¡Á0.0005mol/£¨Ls£©=0.00lmol/£¨Ls£©£¬¹ÊA´íÎó£»

BÏî¡¢½«ÈÝÆ÷Ìå»ý±äΪ20L£¬Èôƽºâ²»Òƶ¯£¬ZµÄŨ¶È±äΪԭÀ´µÄ1/2£¬¸Ã·´Ó¦ÊÇÒ»¸öÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬ÌåϵµÄѹǿ¼õС£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬ZµÄƽºâŨ¶ÈСÓÚÔ­À´µÄ1/2£¬¹ÊB´íÎó£»

CÏî¡¢¸Ã·´Ó¦ÊÇÒ»¸öÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬Ôö´óѹǿ£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬·´Ó¦ÎïYµÄת»¯ÂÊÔö´ó£¬¹ÊC´íÎó£»

DÏî¡¢Éý¸ßζȣ¬XµÄÌå»ý·ÖÊýÔö´ó£¬ËµÃ÷Éý¸ßζÈƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬Éý¸ßζȣ¬Æ½ºâÏòÎüÈÈ·½ÏòÒƶ¯£¬ÔòÕý·´Ó¦µÄ¡÷H£¼0£¬¹ÊDÕýÈ·¡£

¹ÊÑ¡D¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿µªµÄ»¯ºÏÎïºÏ³É¡¢Ó¦Óü°µªµÄ¹Ì¶¨Ò»Ö±ÊÇ¿ÆѧÑо¿µÄÈȵ㡣

(1) ÒÔCO2ÓëNH3ΪԭÁϺϳɻ¯·ÊÄòËصÄÖ÷Òª·´Ó¦ÈçÏ£º

¢Ù 2NH3(g)£«CO2(g)£½NH2CO2NH4(s)£»¦¤H£½£­159.47 kJ¡¤mol1

¢Ú NH2CO2NH4(s)£½CO (NH2)2(s)£«H2O(g)£»¦¤H£½a kJ¡¤mol1

¢Û 2NH3(g)£«CO2(g)£½CO(NH2)2(s)£«H2O(g)£»¦¤H£½£­86.98 kJ¡¤mol1

ÔòaΪ____¡£

(2) ÄòËØ¿ÉÓÃÓÚʪ·¨ÑÌÆøÍѵª¹¤ÒÕ£¬Æä·´Ó¦Ô­ÀíΪ£ºNO£«NO2£«H2O£½2HNO2£»2HNO2£«CO(NH2)2£½2N2¡ü£«CO2¡ü£«3H2O¡£

¢Ù µ±ÑÌÆøÖÐNO¡¢NO2°´ÉÏÊö·´Ó¦ÖÐϵÊý±ÈʱÍѵªÐ§¹û×î¼Ñ¡£ÈôÑÌÆøÖÐV(NO)¡ÃV(NO2)£½5¡Ã1ʱ£¬¿ÉͨÈëÒ»¶¨Á¿µÄ¿ÕÆø£¬Í¬ÎÂͬѹÏ£¬V(¿ÕÆø)¡ÃV(NO)£½____(¿ÕÆøÖÐÑõÆøµÄÌå»ýº¬Á¿´óԼΪ20%)¡£

¢Ú Èçͼ±íʾÄòËغ¬Á¿¶ÔÍѵªÐ§ÂʵÄÓ°Ï죬´Ó¾­¼ÃÒòËØÉÏ¿¼ÂÇ£¬Ò»°ãÑ¡ÔñÄòËØŨ¶ÈԼΪ____%¡£

(3)ÓÃÏ¡ÏõËáÎüÊÕNOx£¬µÃµ½HNO3ºÍHNO2(ÈõËá)µÄ»ìºÏÈÜÒº£¬µç½â¸Ã»ìºÏÈÜÒº¿É»ñµÃ½ÏŨµÄÏõËᡣд³öµç½âʱÑô¼«µÄµç¼«·´Ó¦Ê½£º____¡£

(4)ÀûÓÃÍѵª¾ú¿É¾»»¯µÍŨ¶ÈNO·ÏÆø¡£µ±·ÏÆøÔÚËþÄÚÍ£Áôʱ¼ä¾ùΪ90sµÄÇé¿öÏ£¬²âµÃ²»Í¬Ìõ¼þÏÂNOµÄÍѵªÂÊÈçͼËùʾ¡£

¢Ù ÓÉͼ֪£¬µ±·ÏÆøÖеÄNOº¬Á¿Ôö¼Óʱ£¬ÒËÑ¡ÓÃ____·¨Ìá¸ßÍѵªµÄЧÂÊ¡£

¢Ú ͼ¢òÖУ¬Ñ­»·ÎüÊÕÒº¼ÓÈëFe2+¡¢Mn2+£¬Ìá¸ßÁËÍѵªµÄЧÂÊ£¬Æä¿ÉÄÜÔ­ÒòΪ____¡£

(5)Ñо¿±íÃ÷£ºNaClO/H2O2ËáÐÔ¸´ºÏÎüÊÕ¼Á¿ÉͬʱÓÐЧÍÑÁò¡¢ÍÑÏõ¡£Í¼ËùʾΪ¸´ºÏÎüÊÕ¼Á×é³ÉÒ»¶¨Ê±£¬Î¶ȶÔÍÑÁòÍÑÏõµÄÓ°Ï졣ζȸßÓÚ60¡æºó£¬NOÈ¥³ýÂÊϽµµÄÔ­ÒòΪ____¡£

¡¾ÌâÄ¿¡¿ÔËÓû¯Ñ§·´Ó¦Ô­Àí֪ʶÑо¿ÈçºÎÀûÓÃCO¡¢SO2µÈÎÛȾÎïÓÐÖØÒªÒâÒå¡£

£¨1£©ÓÃCO¿ÉÒԺϳɼ״¼¡£ÒÑÖª£º

¢ÙCH3OH(g)£«3/2O2(g)CO2(g)£«2H2O(l) ¦¤H£½£­764.5 kJ¡¤mol-1

¢ÚCO(g)£«1/2O2(g)CO2(g) ¦¤H£½£­283.0 kJ¡¤mol-1

¢ÛH2(g)£«1/2O2(g)H2O(l) ¦¤H£½£­285.8 kJ¡¤mol-1

ÔòCO(g)£«2H2 (g)CH3OH(g) ¦¤H£½________ kJ¡¤mol-1¡£

£¨2£©ÏÂÁдëÊ©ÖÐÄܹ»Ôö´óÉÏÊöºÏ³É¼×´¼·´Ó¦µÄ·´Ó¦ËÙÂʵÄÊÇ________(ÌîдÐòºÅ)¡£

a. ʹÓøßЧ´ß»¯¼Á b. ½µµÍ·´Ó¦Î¶È

c. Ôö´óÌåϵѹǿ d. ²»¶Ï½«CH3OH´Ó·´Ó¦»ìºÏÎïÖзÖÀë³öÀ´

£¨3£©ÔÚÒ»¶¨Ñ¹Ç¿Ï£¬ÈÝ»ýΪV LµÄÈÝÆ÷ÖгäÈëa mol COÓë2a mol H2£¬ÔÚ´ß»¯¼Á×÷ÓÃÏ·´Ó¦Éú³É¼×´¼£¬Æ½ºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼËùʾ¡£

¢Ùp1________p2(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)£»

¢Ú100 ¡æʱ£¬¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK£½_______£»

¢Û100¡æʱ£¬´ïµ½Æ½ºâºó£¬±£³ÖѹǿP1²»±äµÄÇé¿öÏ£¬ÔÙÏòÈÝÆ÷ÖÐͨÈëCO¡¢H2ºÍCH3OH¸÷0.5a mol£¬Ôòƽºâ_______ (Ìî¡°ÏòÓÒ¡±¡¢¡°Ïò×ó¡±»ò¡°²»¡±Òƶ¯)¡£

¢ÜÔÚÆäËüÌõ¼þ²»±äµÄÇé¿öÏ£¬ÔÙÔö¼Óa mol COºÍ2a molH2£¬´ïµ½ÐÂƽºâʱ£¬COµÄת»¯ÂÊ________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)¡£

£¨4£©ÔÚÈÝ»ý¾ùΪ1LµÄÃܱÕÈÝÆ÷(a¡¢b¡¢c¡¢d¡¢e)ÖУ¬·Ö±ð³äÈë1molCOºÍ2molH2µÈÁ¿»ìºÏÆøÌ壬ÔÚ²»Í¬µÄζÈÏÂ(ζȷֱðΪT1¡¢T2¡¢T3¡¢T4¡¢T5)£¬¾­ÏàͬµÄʱ¼ä£¬ÔÚtʱ¿Ì£¬²âµÃÈÝÆ÷¼×´¼µÄÌå»ý·ÖÊýÈçͼËùʾ¡£ÔÚT1- T2¼°T4- T5Á½¸öζÈÇø¼ä£¬ÈÝÆ÷ÄÚ¼×´¼µÄÌå»ý·ÖÊýµÄ±ä»¯Ç÷ÊÆÈçͼËùʾ£¬ÆäÔ­ÒòÊÇ_______¡£

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

¾«Ó¢¼Ò½ÌÍø