ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ:¢Ù2H2(g)+O2(g)=2H2O(l) ¡÷H=-571.6kJ¡¤mol-1

¢ÚC3H8(g)+5O2(g)=3CO2(g) +4H2O(g) ¡÷H=-2044.0kJ¡¤mol-1

£¨1£©ÇâÆøµÄȼÉÕÈÈÊÇ__________

£¨2£©ÒÑÖª£ºH2O(l)=H2O(g) ¡÷H=+44.0kJ¡¤mol-1£¬Ð´³ö±ûÍé(C3H8)ȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º__________

£¨3£©ÊµÑé²âµÃH2ºÍC3H8µÄ»ìºÏÆøÌå¹²3mol,ÍêȫȼÉÕÉú³ÉҺ̬ˮʱ·ÅÈÈ2791.6kJ£¬¼ÆËã»ìºÏÆøÌåÖÐH2ºÍC3H8µÄÌå»ý±ÈÊÇ_____

£¨4£©ºãκãÈÝÌõ¼þÏÂ,Áò¿ÉÒÔ·¢ÉúÈçÏ·´Ó¦£¬Æä·´Ó¦¹ý³ÌºÍÄÜÁ¿¹ØϵÈçͼËùʾ£¬ÒÑÖª2SO2(g)+O2(g)=2SO3(g) ¡÷H=-196.6kJ¡¤mol-1

¢Ùд³öÄܱíʾÁòµÄȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º__________

¢Ú¡÷H2=__________kJ¡¤mol-1

¡¾´ð°¸¡¿285.8 kJ¡¤mol-1 C3H8(g)+5O2(g)=3CO2(g) +4H2O(l) ¡÷H=-2220.0kJ¡¤mol-1 3:1 S(s)+O2(g)=SO2(g) ¡÷H=-297kJ¡¤mol-1 -78.64kJ¡¤mol-1

¡¾½âÎö¡¿

(1) ÇâÆøµÄȼÉÕÈÈÖ¸µÄÊÇ1molH2ÍêȫȼÉÕÉú³Éˮʱ·Å³öµÄÈÈÁ¿£¬ÓÉÈÈ»¯Ñ§·½³Ìʽ¢ÙÈ·¶¨ÇâÆøµÄȼÉÕÈÈ£»

(2)½«ÈÈ»¯Ñ§·½³Ìʽ¢ÚÓëÒÑÖªH2O(l)=H2O(g) ¡÷H=+44.0kJ¡¤mol-1½áºÏ¿ÉµÃ³ö±ûÍé(C3H8)ȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£»

(3) Éè»ìºÏÆøÌåÖÐH2µÄÎïÖʵÄÁ¿Îªxmol£¬ÄÇô571.6x+2220.0(3-x)=2791.6£¬½âµÃ»ìºÏÆøÌåÖÐH2µÄÎïÖʵÄÁ¿ºÍ±ûÍéµÄÎïÖʵÄÁ¿£¬¸ù¾ÝÎïÖʵÄÁ¿Ö®±ÈµÈÓÚÌå»ýÖ®±ÈµÃ³ö»ìºÏÆøÌåÖÐH2ºÍC3H8µÄÌå»ý±È£»

(4) ¢ÙȼÉÕÈÈÖ¸1mol´¿ÎïÖÊÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯Îïʱ·Å³öµÄÈÈÁ¿£»

¢Ú¸ù¾ÝÈÈ»¯Ñ§·½³Ìʽ2SO2(g)+O2(g)=2SO3(g) ¡÷H=-196.6kJ¡¤mol-1½øÐмÆËã¡£

(1)ÓÉÈÈ»¯Ñ§·½³Ìʽ¢Ù¿ÉÖªÇâÆøµÄȼÉÕÈÈΪ285.8 kJ¡¤mol-1£»

(2) ½«ÈÈ»¯Ñ§·½³Ìʽ¢ÚÓëÒÑÖªH2O(l)=H2O(g) ¡÷H=+44.0kJ¡¤mol-1½áºÏ£¬µÃ³ö±ûÍé(C3H8)ȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ£ºC3H8(g)+5O2(g)=3CO2(g) +4H2O(l) ¡÷H=-2220.0kJ¡¤mol-1£»

(3)Éè»ìºÏÆøÌåÖÐH2µÄÎïÖʵÄÁ¿Îªxmol£¬ÄÇô571.6x+2220.0(3-x)=2791.6£¬½âµÃx=2£¬¼´»ìºÏÆøÌåÖÐH2µÄÎïÖʵÄÁ¿Îª2.3mol£¬C3H8µÄÎïÖʵÄÁ¿Îª0.7mol£¬Òò´Ë»ìºÏÆøÌåÖÐH2ºÍC3H8µÄÌå»ý±ÈԼΪ3:1£»

(4) ¢Ù1molS(s)ÍêȫȼÉÕÉú³ÉSO2(g)ʱËùÊͷųöµÄÈÈÁ¿ÎªÈ¼ÉÕÈÈ£¬Òò´ËÁòµÄȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ£ºS(s)+O2(g)=SO2(g) ¡÷H=-297kJ¡¤mol-1£»

¢ÚÓÉͼ¿ÉÖª²ÎÓë·´Ó¦µÄSO2µÄÎïÖʵÄÁ¿Îª1mol-0.2mol=0.8mol£¬¸ù¾ÝÈÈ»¯Ñ§·½³Ìʽ2SO2(g)+O2(g)=2SO3(g) ¡÷H=-196.6kJ¡¤mol-1¿ÉÖª¡÷H2==0.4¡Á¡÷H=0.4¡Á(-196.6kJ¡¤mol-1)=-78.64kJ¡¤mol-1.

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÈçͼËùʾΪ³£¼ûÆøÌåÖƱ¸¡¢·ÖÀë¡¢¸ÉÔïºÍÐÔÖÊÑéÖ¤µÄ²¿·ÖÒÇÆ÷×°Ö㨼ÓÈÈÉ豸¼°¼Ð³Ö¹Ì¶¨×°ÖþùÂÔÈ¥£©£¬Çë¸ù¾ÝÒªÇóÍê³ÉÏÂÁи÷Ì⣨ÒÇÆ÷×°ÖÿÉÈÎÒâÑ¡Ó㬱ØҪʱ¿ÉÖظ´Ñ¡Ôñ£¬a¡¢bΪ»îÈû£©¡£

£¨1£©ÊµÑéÇ°£¬ÈçºÎ¼ì²éA×°ÖõÄÆøÃÜ£º__¡£

£¨2£©ÈôÆøÌåÈë¿ÚͨÈëCOºÍCO2µÄ»ìºÏÆøÌ壬EÄÚ·ÅÖÃCuO£¬Ñ¡Ôñ×°ÖûñµÃ´¿¾»¸ÉÔïµÄCO£¬²¢ÑéÖ¤Æ仹ԭÐÔ¼°Ñõ»¯²úÎËùѡװÖõÄÁ¬½Ó˳ÐòΪ__£¨Ìî´úºÅ£©¡£ÄÜÑéÖ¤COÑõ»¯²úÎïµÄÏÖÏóÊÇ__¡£

£¨3£©ÈôAÖÐÆøÌåÈë¿Ú¸Äͨ¿ÕÆø£¬·ÖҺ©¶·ÄڸļÓŨ°±Ë®£¬Ô²µ×ÉÕÆ¿ÄڸļÓNaOH¹ÌÌ壬EÄÚ·ÅÖò¬îîºÏ½ðÍø£¬HΪ¿ÕÆ¿£¬DÖиÄΪʯÈïÊÔÒº£¬°´A¡úG¡úE¡úH¡úD×°ÖÃ˳ÐòÖÆÈ¡¸ÉÔïµÄ°±Æø£¬²¢ÑéÖ¤°±µÄijЩÐÔÖÊ¡£

¢ÙʵÑéÖÐÏÈÓþƾ«µÆ¼ÓÈÈ´ß»¯¹Ü£¬ÔÙͨÈë»ìºÏÆøÌ壬¿É¹Û²ìµ½HÄÚÓкì×ØÉ«ÆøÌå³öÏÖ£¬DÖÐʯÈïÊÔÒº±äºì£¬ÔòEÖз¢Éú·´Ó¦µÄ·½³ÌʽΪ__¡£

¢ÚÈô°Ñ¾Æ¾«µÆ³·È¥£¬¼ÌÐøͨÈëÆøÌ壬´ß»¯¼Á¿É¼ÌÐø±£³ÖºìÈÈ״̬£¬·´Ó¦¼ÌÐø½øÐУ®ÄÜʹ´ß»¯¼Á¼ÌÐø±£³ÖºìÈȵÄÔ­ÒòÊÇ£º__¡£

¢ÛijͬѧÔÚ×öÉÏÊöʵÑéʱ³öÏÖÁËÓëÆäËûͬѧ²»Í¬µÄÏÖÏó£¬Ëû·¢ÏÖ´ß»¯¼ÁºìÈÈÏÖÏóÃ÷ÏÔ¶øÕý³££¬µ«Ê¯ÈïÊÔÒº²»±äºì£®Çë·ÖÎö¿ÉÄܵÄÔ­Òò£º£¨´ð1µã¼´¿É£©__¡£

¡¾ÌâÄ¿¡¿Ä¦¶ûÑÎÔÚ¹¤ÒµÉÏÓÐÖØÒªµÄÓÃ;¡£ÒÑÖªÆäÓÉÒ»ÖÖÒõÀë×Ó£¬Á½ÖÖÑôÀë×Ó×é³ÉµÄ¾§Ì壬ijѧϰС×é°´ÈçÏÂʵÑé²â¶¨Ä¦¶ûÑÎÑùÆ·µÄ×é³É¡£²½ÖèÈçÏ£º

¢Ù³ÆÈ¡3.920gĦ¶ûÑÎÑùÆ·ÅäÖÆ250mLÈÜÒº¡£

¢ÚÈ¡ÉÙÁ¿ÅäÖÆÈÜÒº£¬¼ÓÈëKSCNÈÜÒº£¬ÎÞÃ÷ÏÔÏÖÏó¡£

¢ÛÁíÈ¡ÉÙÁ¿ÅäÖÆÈÜÒº£¬¼ÓÈë¹ýÁ¿Å¨ÇâÑõ»¯ÄÆÈÜÒº²¢¼ÓÈÈ£¬²úÉúʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌåºÍºìºÖÉ«³Áµí¡£

¢Ü¶¨Á¿²â¶¨ÈçÏ£º

µÎ¶¨ÊµÑé½á¹û¼Ç¼ÈçÏ£º

ʵÑé´ÎÊý

µÚÒ»´Î

µÚ¶þ´Î

µÚÈý´Î

ÏûºÄ¸ßÃÌËá¼ØÈÜÒºÌå»ý/mL

10.32

10.02

9.98

Íê³ÉÏÂÁÐÌî¿Õ£º

(1)²½Öè¢ÙÖÐÐèÒªµÄ¶¨Á¿ÒÇÆ÷Ϊ________________ ¡¢__________________¡£

(2)²½Öè¢ÚµÄÄ¿µÄÊÇ_____________________________________________________¡£²úÉúºìºÖÉ«³ÁµíµÄÀë×Ó·½³Ìʽ_____________________________________¡£

(3)²½Öè¢ÜÖвÙ×÷XΪ_________________________________£¨°´²Ù×÷˳ÐòÌîд£©¡£

(4)²½Öè¢ÜÖÐËáÐÔ¸ßÃÌËá¼ØÈÜÒºÄÜ·ñÓõâµÄ¾Æ¾«ÈÜÒº´úÌ棬_______£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©£¬Çë˵Ã÷ÀíÓÉ__________________________________________________¡£

(5)²½Öè¢ÜÈôÔڵζ¨¹ý³ÌÖУ¬´ý²âÒº¾ÃÖã¬ÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ý½«__________¡££¨Ñ¡Ìî¡° Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£©¡£

(6)ͨ¹ýÉÏÊöʵÑé²â¶¨½á¹û£¬ÍƶÏĦ¶ûÑλ¯Ñ§Ê½Îª______________________________¡£

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

¾«Ó¢¼Ò½ÌÍø