ÌâÄ¿ÄÚÈÝ

£¨21·Ö£©£¨1£©ÓÃ50mL 0.55mol/L NaOHÈÜÒººÍ50mL 0.25mol/LH2SO4ÈÜÒºÀ´×öÖкÍÈȲⶨµÄʵÑ飬²âµÃÈÜÒºÔÚ·´Ó¦Ç°ºóµÄζȱ仯Ϊt1¡æ¡«t2¡æ £¨t2>t1£©,»ìºÏºóÈÜÒºµÄ±ÈÈÈÈÝΪc = 4.18J£¯£¨g¡¤¡æ£©,ÈÜÒºµÄÃܶȶ¼½üËÆΪ1g/mL£¬ÖкÍÈÈ¡÷H=__________(Ìî±í´ïʽ£¬²»Óû¯¼ò)£»Èô½«H2SO4ÈÜÒº»»³ÉÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄCH3COOHÈÜÒº£¬²âµÃµÄ¡÷H______Ìî(¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°Ïàͬ¡±£©£»Èô½«Ï¡ÁòËá»»³ÉŨÁòËáÀ´×ö¸ÃʵÑ飬²âµÃµÄ¡÷H_______£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°Ïàͬ¡±£©¡£

      £¨2£©Ä³¸ßÄÜ»¯Ñ§ÎïÖÊN2H2ÄÚ£¬µªÔªËصÄÔÓ»¯ÀàÐÍΪ____£¬µç×ÓʽΪ____£¬Ò»¸ö·Ö×ÓÖÐÓЦҼü    ¸ö  £¬¦Ð ¼ü     ¸ö¡£

£¨3£©£®ÔÚÃܱÕÈÝÆ÷ÖнøÐпÉÄæ·´Ó¦£º CO(g)£«NO2(g)CO2(g)£«NO(g)£¬(Õý·´Ó¦·ÅÈÈ)£¬´ïµ½Æ½ºâºó£¬Ö»¸Ä±äÆäÖÐÒ»¸öÌõ¼þ£¬¶ÔƽºâµÄÓ°ÏìÊÇ£º

¢ÙÔö´óÈÝÆ÷µÄÌå»ý£¬Æ½ºâ¡¡ ¡¡ Òƶ¯(Ìî¡°ÕýÏò¡±¡°ÄæÏò¡±¡°²»¡±)£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡¡¡¡¡¡£(Ìî¡°±äÉ¡°±ädz¡±¡°²»±ä¡±)

¢ÚÈÝÆ÷Ìå»ý²»±ä£ºÈôͨÈËCO2ÆøÌ壬ƽºâ¡¡  ¡¡Òƶ¯£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡ ¡¡ ¡£ÈôͨÈËN2ÆøÌ壬ƽºâ¡¡  ¡¡Òƶ¯£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡ ¡¡ ¡£¢Û¼ÓÈë´ß»¯¼Á£¬Æ½ºâ¡¡¡¡¡¡Òƶ¯¡£

   (4)ϱíÊǼ¸ÖÖ³£ÓÃȼÁÏ(1 mol)ÍêȫȼÉÕʱ·Å³öµÄÈÈÁ¿£º

ÎïÖÊ

Ì¿·Û(C)

Ò»Ñõ»¯Ì¼(CO)

ÇâÆø(H2)

¼×Íé(CH4)

ÒÒ´¼(C2H5OH)

״̬

¹ÌÌå

ÆøÌå

ÆøÌå

ÆøÌå

ÒºÌå

ÈÈÁ¿(kJ)

392.8

282.6

285.8

890.3

1 367

¢Ù´ÓÈÈÁ¿½Ç¶È·ÖÎö£¬Ä¿Ç°×îÊʺϼÒͥʹÓõÄÓÅÖÊÆøÌåȼÁÏÊÇ________¡£

¢Úд³ö¹ÜµÀúÆøÖеÄÒ»Ñõ»¯Ì¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ_______________________

¢Û³ä·ÖȼÉÕ1mol±íÖи÷ÖÖȼÁÏ£¬Åŷųö¶þÑõ»¯Ì¼µÄÁ¿×î¶àµÄÊÇ________¡£

¢Ü¿óÎïȼÁÏ´¢Á¿ÓÐÏÞ£¬¶øÇÒÔÚȼÉÕ¹ý³ÌÖлá²úÉúÎÛȾ¡£¸ù¾ÝÄÜÔ´¶àÑù»¯µÄ·¢Õ¹Õ½ÂÔ£¬ÎÒ¹ú¿ª·¢ÀûÓõÄÂÌÉ«ÄÜÔ´ÓÐ________µÈ¡£

 

£¨21·Ö£©£¨1£©       Æ«´ó       ƫС£¨¸÷2·Ö£©

   £¨2£©sp2ÔÓ»¯             3       1          £¨¸÷1·Ö£©

   £¨3£©²»£¬±ädz   ÄæÏò   ±äÉî   ²»    ²»±ä    £¨¸÷1·Ö£©

   £¨4£© (1)¼×Íé £¨1·Ö£©  (2)CO(g)£«1/2O2(g)===CO2(g)  ¦¤H£½£­282.6 kJ¡¤mol£­1

 £¨2·Ö£©

(3)ÒÒ´¼£¨1·Ö£©    (4)ÇâÄÜ¡¢Ì«ÑôÄÜ(»ò·çÄÜ¡¢µØÈÈÄÜ)£¨1·Ö£©

½âÎö:

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨21·Ö£©£¨1£©ÓÃ50mL 0.55mol/L NaOHÈÜÒººÍ50mL 0.25mol/L H2SO4ÈÜÒºÀ´×öÖкÍÈȲⶨµÄʵÑ飬²âµÃÈÜÒºÔÚ·´Ó¦Ç°ºóµÄζȱ仯Ϊt1¡æ¡«t2¡æ£¨t2>t1£©,»ìºÏºóÈÜÒºµÄ±ÈÈÈÈÝΪc = 4.18J£¯£¨g¡¤¡æ£©,ÈÜÒºµÄÃܶȶ¼½üËÆΪ1g/mL£¬ÖкÍÈÈ¡÷H=__________(Ìî±í´ïʽ£¬²»Óû¯¼ò)£»Èô½«H2SO4ÈÜÒº»»³ÉÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄCH3COOHÈÜÒº£¬²âµÃµÄ¡÷H______Ìî(¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°Ïàͬ¡±£©£»Èô½«Ï¡ÁòËá»»³ÉŨÁòËáÀ´×ö¸ÃʵÑ飬²âµÃµÄ¡÷H_______£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°Ïàͬ¡±£©¡£
£¨2£©Ä³¸ßÄÜ»¯Ñ§ÎïÖÊN2H2ÄÚ£¬µªÔªËصÄÔÓ»¯ÀàÐÍΪ____£¬µç×ÓʽΪ____£¬Ò»¸ö·Ö×ÓÖÐÓЦҼü    ¸ö £¬¦Ð ¼ü    ¸ö¡£
£¨3£©£®ÔÚÃܱÕÈÝÆ÷ÖнøÐпÉÄæ·´Ó¦£º CO(g)£«NO2(g)CO2(g)£«NO(g)£¬(Õý·´Ó¦·ÅÈÈ)£¬´ïµ½Æ½ºâºó£¬Ö»¸Ä±äÆäÖÐÒ»¸öÌõ¼þ£¬¶ÔƽºâµÄÓ°ÏìÊÇ£º

¢ÙÔö´óÈÝÆ÷µÄÌå»ý£¬Æ½ºâ¡¡¡¡Òƶ¯(Ìî¡°ÕýÏò¡±¡°ÄæÏò¡±¡°²»¡±)£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡¡¡¡¡¡£(Ìî¡°±äÉ¡°±ädz¡±¡°²»±ä¡±)
¢ÚÈÝÆ÷Ìå»ý²»±ä£ºÈôͨÈËCO2ÆøÌ壬ƽºâ¡¡ ¡¡Òƶ¯£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡¡¡¡£ÈôͨÈËN2ÆøÌ壬ƽºâ¡¡ ¡¡Òƶ¯£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡¡¡¡£¢Û¼ÓÈë´ß»¯¼Á£¬Æ½ºâ¡¡¡¡¡¡Òƶ¯¡£
(4)ϱíÊǼ¸ÖÖ³£ÓÃȼÁÏ(1 mol)ÍêȫȼÉÕʱ·Å³öµÄÈÈÁ¿£º

ÎïÖÊ
Ì¿·Û(C)
Ò»Ñõ»¯Ì¼(CO)
ÇâÆø(H2)
¼×Íé(CH4)
ÒÒ´¼(C2H5OH)
״̬
¹ÌÌå
ÆøÌå
ÆøÌå
ÆøÌå
ÒºÌå
ÈÈÁ¿(kJ)
392.8
282.6
285.8
890.3
1 367
¢Ù´ÓÈÈÁ¿½Ç¶È·ÖÎö£¬Ä¿Ç°×îÊʺϼÒͥʹÓõÄÓÅÖÊÆøÌåȼÁÏÊÇ________¡£
¢Úд³ö¹ÜµÀúÆøÖеÄÒ»Ñõ»¯Ì¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ_______________________
¢Û³ä·ÖȼÉÕ1 mol±íÖи÷ÖÖȼÁÏ£¬Åŷųö¶þÑõ»¯Ì¼µÄÁ¿×î¶àµÄÊÇ________¡£
¢Ü¿óÎïȼÁÏ´¢Á¿ÓÐÏÞ£¬¶øÇÒÔÚȼÉÕ¹ý³ÌÖлá²úÉúÎÛȾ¡£¸ù¾ÝÄÜÔ´¶àÑù»¯µÄ·¢Õ¹Õ½ÂÔ£¬ÎÒ¹ú¿ª·¢ÀûÓõÄÂÌÉ«ÄÜÔ´ÓÐ________µÈ¡£

£¨16·Ö£©£¨1£©ÔÚÖк͵樲Ù×÷¹ý³ÌÖУ¬ÓÐÒÔϸ÷ÏîÒò²Ù×÷²»µ±ÒýÆðµÄʵÑéÎó²î£¬Óá°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족Ìî¿Õ£º
¢ÙµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºó£¬Î´ÓÃÒÑ֪Ũ¶ÈµÄ±ê×¼ÈÜÒºÈóÏ´£¬Ê¹µÎ¶¨½á¹û        £»
¢Ú׶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºó£¬ÓÖÓôý²âÈÜÒºÈóÏ´£¬Ê¹µÎ¶¨½á¹û          £»
¢ÛµÎ¶¨¹Ü(×°±ê×¼ÈÜÒº)Ôڵζ¨Ç°¼â×ì´¦ÓÐÆøÅÝ£¬µÎ¶¨ÖÕÁËÎÞÆøÅÝ£¬Ê¹µÎ¶¨½á¹û     £»
¢ÜµÎ¶¨Ç°Æ½ÊÓ£¬µÎ¶¨ÖÕÁ˸©ÊÓ£¬Ê¹µÎ¶¨½á¹û          £»
¢ÝÓú¬Na2OÔÓÖʵÄNaOH¹ÌÌåÀ´ÅäÖÆÒÑ֪Ũ¶ÈµÄ±ê×¼ÈÜÒº£¬ÓÃÓڵζ¨Î´ÖªÅ¨¶ÈµÄÑÎËᣬʹ²âµÃÑÎËáµÄŨ¶È           £»
¢ÞÏ´µÓ׶ÐÎƿʱ£¬Îó°ÑϡʳÑÎË®µ±×öÕôÁóË®£¬È»ºóÓÃ׶ÐÎƿװ´ý²âµÄÑÎËᣬÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨Ê±£¬¶Ô²âµÃµÄ½á¹û           ¡£
£¨2£©ÒÑÖªH+(aq)+OH-(aq) = H2O(l)  ¦¤H= £­57.3kJ¡¤mol£­1¡£ÓÃ50mL 0.50mol/LÑÎËáÓë50mL 0.55mol/L NaOHÈÜÒºÔÚÈçÏÂͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙÒ»ÖÖ²£Á§ÒÇÆ÷ÊÇ                ¡£
¢Ú´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýÖµ        (Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£
¢ÛÈôͨ¹ýʵÑé²â¶¨ÖкÍÈȵĦ¤H£¬Æä½á¹û³£³£´óÓÚ£­57.3kJ¡¤mol£­1£¬ÆäÔ­Òò¿ÉÄÜÊÇ£º
                                                   ¡£
£¨3£©·Ö±ðÉè¼Æ»¯Ñ§ÊµÑ飬ÓÃ×î¼Ñ·½·¨Ö¤Ã÷Ã÷·¯ÈÜÓÚˮʱ·¢ÉúµÄÏÂÁб仯(¹©Ñ¡ÓõÄÒ©Æ·ºÍÒÇÆ÷£ºÃ÷·¯ÈÜÒº¡¢Ê¯ÈïÊÔÒº¡¢·Ó̪ÊÔÒº¡¢pHÊÔÖ½¡¢¾Æ¾«µÆ)£º
¢ÙÖ¤Ã÷Ã÷·¯·¢ÉúÁËË®½â·´Ó¦                                              ¡£
¢ÚÖ¤Ã÷ÆäË®½â·´Ó¦ÊÇÒ»¸öÎüÈÈ·´Ó¦                                        ¡£

£¨16·Ö£©£¨1£©ÔÚÖк͵樲Ù×÷¹ý³ÌÖУ¬ÓÐÒÔϸ÷ÏîÒò²Ù×÷²»µ±ÒýÆðµÄʵÑéÎó²î£¬Óá°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족Ìî¿Õ£º

¢ÙµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºó£¬Î´ÓÃÒÑ֪Ũ¶ÈµÄ±ê×¼ÈÜÒºÈóÏ´£¬Ê¹µÎ¶¨½á¹û        £»

¢Ú׶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºó£¬ÓÖÓôý²âÈÜÒºÈóÏ´£¬Ê¹µÎ¶¨½á¹û          £»

¢ÛµÎ¶¨¹Ü(×°±ê×¼ÈÜÒº)Ôڵζ¨Ç°¼â×ì´¦ÓÐÆøÅÝ£¬µÎ¶¨ÖÕÁËÎÞÆøÅÝ£¬Ê¹µÎ¶¨½á¹û     £»

¢ÜµÎ¶¨Ç°Æ½ÊÓ£¬µÎ¶¨ÖÕÁ˸©ÊÓ£¬Ê¹µÎ¶¨½á¹û          £»

¢ÝÓú¬Na2OÔÓÖʵÄNaOH¹ÌÌåÀ´ÅäÖÆÒÑ֪Ũ¶ÈµÄ±ê×¼ÈÜÒº£¬ÓÃÓڵζ¨Î´ÖªÅ¨¶ÈµÄÑÎËᣬʹ²âµÃÑÎËáµÄŨ¶È           £»

¢ÞÏ´µÓ׶ÐÎƿʱ£¬Îó°ÑϡʳÑÎË®µ±×öÕôÁóË®£¬È»ºóÓÃ׶ÐÎƿװ´ý²âµÄÑÎËᣬÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨Ê±£¬¶Ô²âµÃµÄ½á¹û           ¡£

£¨2£©ÒÑÖªH+(aq)+OH-(aq) = H2O(l)  ¦¤H= £­57.3kJ¡¤mol£­1¡£ÓÃ50mL 0.50mol/LÑÎËáÓë50mL 0.55mol/L NaOHÈÜÒºÔÚÈçÏÂͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ù´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙÒ»ÖÖ²£Á§ÒÇÆ÷ÊÇ                ¡£

¢Ú´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýÖµ        (Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

¢ÛÈôͨ¹ýʵÑé²â¶¨ÖкÍÈȵĦ¤H£¬Æä½á¹û³£³£´óÓÚ£­57.3kJ¡¤mol£­1£¬ÆäÔ­Òò¿ÉÄÜÊÇ£º

                                                   ¡£

£¨3£©·Ö±ðÉè¼Æ»¯Ñ§ÊµÑ飬ÓÃ×î¼Ñ·½·¨Ö¤Ã÷Ã÷·¯ÈÜÓÚˮʱ·¢ÉúµÄÏÂÁб仯(¹©Ñ¡ÓõÄÒ©Æ·ºÍÒÇÆ÷£ºÃ÷·¯ÈÜÒº¡¢Ê¯ÈïÊÔÒº¡¢·Ó̪ÊÔÒº¡¢pHÊÔÖ½¡¢¾Æ¾«µÆ)£º

¢ÙÖ¤Ã÷Ã÷·¯·¢ÉúÁËË®½â·´Ó¦                                              ¡£

¢ÚÖ¤Ã÷ÆäË®½â·´Ó¦ÊÇÒ»¸öÎüÈÈ·´Ó¦                                        ¡£

 

£¨21·Ö£©£¨1£©ÓÃ50mL 0.55mol/L NaOHÈÜÒººÍ50mL 0.25mol/L H2SO4ÈÜÒºÀ´×öÖкÍÈȲⶨµÄʵÑ飬²âµÃÈÜÒºÔÚ·´Ó¦Ç°ºóµÄζȱ仯Ϊt1¡æ¡«t2¡æ £¨t2>t1£©,»ìºÏºóÈÜÒºµÄ±ÈÈÈÈÝΪc = 4.18J£¯£¨g¡¤¡æ£©,ÈÜÒºµÄÃܶȶ¼½üËÆΪ1g/mL£¬ÖкÍÈÈ¡÷H=__________(Ìî±í´ïʽ£¬²»Óû¯¼ò)£»Èô½«H2SO4ÈÜÒº»»³ÉÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄCH3COOHÈÜÒº£¬²âµÃµÄ¡÷H______Ìî(¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°Ïàͬ¡±£©£»Èô½«Ï¡ÁòËá»»³ÉŨÁòËáÀ´×ö¸ÃʵÑ飬²âµÃµÄ¡÷H_______£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°Ïàͬ¡±£©¡£

      £¨2£©Ä³¸ßÄÜ»¯Ñ§ÎïÖÊN2H2ÄÚ£¬µªÔªËصÄÔÓ»¯ÀàÐÍΪ____£¬µç×ÓʽΪ____£¬Ò»¸ö·Ö×ÓÖÐÓЦҼü     ¸ö  £¬¦Ð ¼ü     ¸ö¡£

£¨3£©£®ÔÚÃܱÕÈÝÆ÷ÖнøÐпÉÄæ·´Ó¦£º CO(g)£«NO2(g)CO2(g)£«NO(g)£¬(Õý·´Ó¦·ÅÈÈ)£¬´ïµ½Æ½ºâºó£¬Ö»¸Ä±äÆäÖÐÒ»¸öÌõ¼þ£¬¶ÔƽºâµÄÓ°ÏìÊÇ£º

¢ÙÔö´óÈÝÆ÷µÄÌå»ý£¬Æ½ºâ¡¡ ¡¡ Òƶ¯(Ìî¡°ÕýÏò¡±¡°ÄæÏò¡±¡°²»¡±)£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡¡¡¡¡¡£(Ìî¡°±äÉ¡°±ädz¡±¡°²»±ä¡±)

¢ÚÈÝÆ÷Ìå»ý²»±ä£ºÈôͨÈËCO2ÆøÌ壬ƽºâ¡¡  ¡¡Òƶ¯£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡ ¡¡ ¡£ÈôͨÈËN2ÆøÌ壬ƽºâ¡¡  ¡¡Òƶ¯£¬·´Ó¦»ìºÏÎïµÄÑÕÉ«¡¡ ¡¡ ¡£¢Û¼ÓÈë´ß»¯¼Á£¬Æ½ºâ¡¡¡¡¡¡Òƶ¯¡£

   (4)ϱíÊǼ¸ÖÖ³£ÓÃȼÁÏ(1 mol)ÍêȫȼÉÕʱ·Å³öµÄÈÈÁ¿£º

ÎïÖÊ

Ì¿·Û(C)

Ò»Ñõ»¯Ì¼(CO)

ÇâÆø(H2)

¼×Íé(CH4)

ÒÒ´¼(C2H5OH)

״̬

¹ÌÌå

ÆøÌå

ÆøÌå

ÆøÌå

ÒºÌå

ÈÈÁ¿(kJ)

392.8

282.6

285.8

890.3

1 367

¢Ù´ÓÈÈÁ¿½Ç¶È·ÖÎö£¬Ä¿Ç°×îÊʺϼÒͥʹÓõÄÓÅÖÊÆøÌåȼÁÏÊÇ________¡£

¢Úд³ö¹ÜµÀúÆøÖеÄÒ»Ñõ»¯Ì¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ_______________________

¢Û³ä·ÖȼÉÕ1 mol±íÖи÷ÖÖȼÁÏ£¬Åŷųö¶þÑõ»¯Ì¼µÄÁ¿×î¶àµÄÊÇ________¡£

¢Ü¿óÎïȼÁÏ´¢Á¿ÓÐÏÞ£¬¶øÇÒÔÚȼÉÕ¹ý³ÌÖлá²úÉúÎÛȾ¡£¸ù¾ÝÄÜÔ´¶àÑù»¯µÄ·¢Õ¹Õ½ÂÔ£¬ÎÒ¹ú¿ª·¢ÀûÓõÄÂÌÉ«ÄÜÔ´ÓÐ________µÈ¡£

 

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

¾«Ó¢¼Ò½ÌÍø