ÌâÄ¿ÄÚÈÝ

£¨1£©±¾Éú£¨Bunsen£©ÈÈ»¯Ñ§Ñ­»·ÎüÊÕSO2¹¤ÒÕÓÉÏÂÁÐÈý¸ö·´Ó¦×é³É£º

2H2(g)+O2(g)=2H2O(l) ¡÷H1=-572kJ¡¤mol-1

2HI(g)=H2(g)+I2(g) ¡÷H2=+10kJ¡¤mol-1

2H2SO4(l)=2SO2(g)+2H2O(l)+O2(g) ¡÷H3=+462kJ¡¤mol-1

ÔòµÃ

SO2(g)+I2(g)+2H2O(I)=2HI (g)+H2SO4(I) ¡÷H=_______ kJ¡¤mol-1

£¨2£©Ñõ»¯Ð¿ÎüÊÕ·¨¡£ÅäÖÆZnOÐü×ÇÒº£¬ÔÚÎüÊÕËþÖзâ±ÕÑ­»·ÍÑÁò¡£²âµÃpH¡¢ÎüÊÕЧÂʦÇËæʱ¼ätµÄ±ä»¯Èçͼ1Ëùʾ£»ÈÜÒºÖв¿·Ö΢Á£ÓëPHµÄ¹ØϵÈçͼ2Ëùʾ¡£

¢ÙΪÌá¸ßSO2µÄÎüÊÕЧÂʦǣ¬¿É²ÉÈ¡µÄ´ëÊ©ÓУºÔö´óÐü×ÇÒºÖÐZnOµÄÁ¿¡¢________________¡£

¢Úͼ1ÖеÄpH-tÇúÏßab¶Î·¢ÉúµÄÖ÷Òª»¯Ñ§·½³ÌʽΪ___________________

¢ÛpH=7ʱ£¬ÈÜÒºÖÐ =________

£¨3£©Èçͼ3Ëùʾ£¬ÀûÓöèÐÔµç½âµç½âº¬SO2µÄÑÌÆø»ØÊÕS¼°H2SO4£¬ÒÔʵÏÖ·ÏÎïÀûÓá£

¢ÙÒõ¼«µÄµç¼«·´Ó¦Ê½Îª_____________¡£

¢Úÿ´¦Àíº¬19.2g SO2µÄÑÌÆø£¬ÀíÂÛÉÏ»ØÊÕS¡¢H2SO4µÄÎïÖʵÄÁ¿·Ö±ðΪ______¡¢________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÈκÎÒ»¸ö»¯Ñ§·´Ó¦Öж¼»á°éËæÄÜÁ¿±ä»¯ºÍÎïÖʱ仯¡£

£¨1£©ÏÂͼΪÇâÆøºÍÑõÆøÉú³É1molË®ÕôÆøʱµÄÄÜÁ¿±ä»¯Çé¿ö¡£

Éú³É1molË®ÕôÆøʱ£¬ÐγÉл¯Ñ§¼üËùÊͷŵÄ×ÜÄÜÁ¿_____¶Ï¿ª¾É»¯Ñ§¼üËùÎüÊÕµÄ×ÜÄÜÁ¿(Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)£¬Òò´ËË®µÄ·Ö½â·´Ó¦Îª________(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)·´Ó¦¡£

£¨2£©Ëæ×Å¿Æѧ¼¼ÊõµÄ·¢Õ¹ºÍÉç»á½ø²½£¬¸÷ÖÖ¸÷ÑùµÄµçÆ÷²»¶Ï½øÈëÏÖ´úÉç»á£¬»¯Ñ§µçÔ´´ó´ó·á¸»ºÍ·½±ãÁËÎÒÃǵÄÉú»î¡¢Ñ§Ï°ºÍ¹¤×÷¡£Ô­µç³ØÖз¢ÉúµÄ»¯Ñ§·´Ó¦ÊôÓÚ______¡£Ã¾ÂÁÔ­µç³Ø£¬µç½âÖÊÓÃÏ¡ÁòËᣬÔòÆ为¼«µÄµç¼«·´Ó¦Ê½Îª______£»µç½âÖÊÓÃÇâÑõ»¯ÄÆÈÜÒº£¬µç³Ø¹¤×÷ʱ£¬ÈÜÒºÖеÄOH-ÒÆÏò_____¼«(¡°Ã¾¡±»ò¡°ÂÁ¡±)¡£

£¨3£©ÔÚÈý¸öÈÝÆ÷¾ùΪ2LµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£º2HI(g) H2(g)+I2(g)£¬ÒÑÖªH2(g)ºÍI2(g)µÄÆðʼÎïÖʵÄÁ¿¾ùΪ0£¬HI(g)µÄÎïÖʵÄÁ¿(mol)Ë淴Ӧʱ¼ä(min)µÄ±ä»¯Çé¿öÈçϱíËùʾ£¬±íÖÐζȵĵ¥Î»ÎªÉãÊ϶È(¡æ)¡£

¢ÙʵÑé1ºÍʵÑé2ÖУ¬ÓÐÒ»¸öʵÑéʹÓÃÁË´ß»¯¼Á£¬ÔòʹÓÃÁË´ß»¯¼ÁµÄÊÇʵÑé_____(Ìî¡°1¡±»ò¡°2¡±)¡£

¢ÚʵÑé1ÖУ¬20min¡«30min¼äÉú³ÉI2(g)µÄƽ¾ù·´Ó¦ËÙÂÊΪ_____£¬ÊµÑé3µÄ·´Ó¦´ïµ½»¯Ñ§·´Ó¦ÏÞ¶ÈʱH2(g)ÎïÖʵÄÁ¿µÄ°Ù·ÖÊýΪ_______¡£

¢ÛÏÂÁÐÎïÀíÁ¿²»Ôٱ仯ʱ£¬ÄÜÅжÏʵÑé3Öз´Ó¦´ïµ½»¯Ñ§·´Ó¦Ï޶ȵÄÊÇ__________¡£

A.»ìºÏÆøÌåÑÕÉ« B.»ìºÏÆøÌåÃÜ¶È C.»ìºÏÆøÌå×Üѹǿ

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

¾«Ó¢¼Ò½ÌÍø