ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿(1)ǦÐîµç³ØÊdz£ÓõĻ¯Ñ§µçÔ´£¬Æäµç¼«²ÄÁÏÊÇPbºÍPbO2£¬µç½âҺΪϡÁòËá¡£¹¤×÷ʱ¸Ãµç³Ø×Ü·´Ó¦Ê½Îª£ºPb+PbO2 +2H2SO4=2PbSO4+2H2O£¬¾Ý´ËÅжϣº
¢ÙǦÐîµç³ØµÄ¸º¼«²ÄÁÏÊÇ________£¨Ð´»¯Ñ§Ê½£©£»
¢Ú¹¤×÷ʱÕý¼«·´Ó¦Îª__________________________________________£»
¢Û¹¤×÷ʱ£¬µç½âÖÊÈÜÒºÖÐÒõÀë×ÓÒÆÏò________¼«(Ìî¡°Õý¡±»ò¡°¸º¡±)¡£
(2)ÈçͼA Ö±Á÷µçÔ´£¬BΪ½þ͸±¥ºÍÂÈ»¯ÄÆÈÜÒººÍ·Ó̪ÊÔÒºµÄÂËÖ½£¬CΪµç¶Æ²Û£¬½Óͨµç·ºó£¬·¢ÏÖBÉϵÄcµãÏÔºìÉ«£¬ÇëÌî¿Õ£º
¢ÙµçÔ´AÉϵÄaΪ________¼«(Ìî¡°Õý¡±»ò¡°¸º¡±)£»
¢ÚÂËÖ½BÉÏ·¢ÉúµÄ×Ü»¯Ñ§·½³ÌʽΪ___________________£»
¢ÛÓûÔÚµç²ÛÖÐʵÏÖÌúÉ϶Æп£¬½ÓͨKµã£¬Ê¹c¡¢dÁ½µã¶Ì·£¬Ôòµç¼«fÉÏ·¢ÉúµÄ·´Ó¦Îª________________________¡£µç²ÛÖзŵĶÆÒº¿ÉÒÔÊÇ___________(Ö»Ðèд³öÒ»ÖÖ¼´¿É)¡£
¡¾´ð°¸¡¿Pb PbO2 + 4H+ +SO42- +2e-==PbSO4+2H2O ¸º Õý 2NaCl+2H2OH2¡ü+Cl2¡ü+2NaOH Zn2++2e-===Zn ZnSO4ÈÜÒº¡¢ZnCl2ÈÜÒºµÈ
¡¾½âÎö¡¿
(1)¸º¼«Ê§È¥µç×Ó£¬·¢ÉúÑõ»¯·´Ó¦£¬Õý¼«ÉÏÎïÖʵõ½µç×Ó£¬·¢Éú»¹Ô·´Ó¦£¬ÈÜÒºÖÐÑôÀë×ÓÏòÕý¼«Òƶ¯£¬ÒõÀë×ÓÏò¸º¼«Òƶ¯·ÖÎö£»
(2)¢ÙÓÉBÉϵÄcµãÏÔºìÉ«ÅжÏcµÄµç¼«£¬¸ù¾ÝcµÄµç¼«·´Ó¦ÏÖÏóÅжÏa¡¢bµÄµç¼«£»
¢Ú¸ù¾ÝÈÜÒºÖÐÀë×ӵķŵç˳ÐòÅжÏÉú³ÉÎÓÉ·´Ó¦Îï¡¢Éú³ÉÎïд³öÏàÓ¦µÄ·½³Ìʽ£»
¢ÛÏÈÅжÏe¡¢fµÄµç¼«£¬ÔÙ¸ù¾ÝÒõÑô¼«ÉÏ·¢ÉúµÄ·´Ó¦Ð´³öÏàÓ¦µÄµç¼«·´Ó¦Ê½£»¸ù¾Ýµç¶ÆÔÀíÑ¡È¡µç½âÖÊ¡£
(1)¢Ù¸ù¾Ýµç³Ø·´Ó¦Ê½Öª£¬PbÔªËØ»¯ºÏ¼ÛÓÉ0¼Û¡¢+4¼Û±äΪ+2¼Û£¬ÔªËØ»¯ºÏ¼ÛÉý¸ßµÄ½ðÊôΪ¸º¼«£¬ÔòPbΪ¸º¼«£»
¢ÚPbO2ΪÕý¼«£¬µÃµ½µç×Ó£¬·¢Éú»¹Ô·´Ó¦£¬Õý¼«µÄµç¼«·´Ó¦Ê½ÊÇPbO2+ 4H+ +SO42- +2e-=PbSO4+2H2O£»
¢Û·Åµçʱ£¬µç½âÖÊÈÜÒºÖÐÑôÀë×ÓÏòÕý¼«Òƶ¯¡¢ÒõÀë×ÓÏò¸º¼«Òƶ¯£¬´ð°¸Îª¸º¼«£»
(2)¢ÙBÍâ½ÓµçÔ´£¬BÊǵç½âÂÈ»¯ÄÆÈÜÒºµÄµç½â³Ø£»BÉϵÄcµãÏÔºìÉ«£¬ËµÃ÷cµãÓÐÇâÑõ¸ùÀë×ÓÉú³É£¬c(OH-)Ôö´ó£¬¸ù¾ÝÀë×ӵķŵç˳ÐòÖª£¬¸Ãµç¼«ÉÏÊÇH+µÃµç×Ó±äΪH2£¬ËùÒԵ缫cÊÇÒõ¼«£¬ÍâµçÔ´bÊǸº¼«£¬aÊÇÕý¼«£»
¢Úµç½âÂÈ»¯ÄÆÈÜҺʱ£¬ÈÜÒºÖеÄÀë×ӷŵç˳ÐòΪ£ºH+>Na+£¬Cl->OH-£¬ËùÒÔµç½âÂÈ»¯ÄÆʱÉú³ÉÎïÊÇÂÈÆø¡¢ÇâÆø¡¢ÇâÑõ»¯ÄÆ£¬·´Ó¦·½³ÌʽΪ 2NaCl+2H2O2NaOH+ H2¡ü+ Cl2¡ü£»
¢ÛÓÉÓÚÍâµçÔ´ÉÏ£¬aÊÇÕý¼«£¬bÊǸº¼«£¬ËùÒÔµç¶Æʱ£¬eÊÇÑô¼«£¬fÊÇÒõ¼«£¬¶Æ²ã½ðÊôп×÷Ñô¼«£¬¶Æ¼þÌú×÷Òõ¼«£¬Ñô¼«ÉÏZnʧµç×Ó±ä³ÉZn2+Àë×Ó½øÈëÈÜÒº£¬Òõ¼«ÉÏпÀë×ӵõç×ÓÉú³Éпµ¥ÖÊ£¬Òõ¼«fÉϵĵ缫·´Ó¦Ê½ÎªZn2++2e-=Zn£»µç¶ÆÒºµÄÑ¡È¡£ºÓú¬ÓжƲã½ðÊôZnÀë×ÓµÄÈÜÒº×÷µç¶ÆÒº£¬ËùÒÔ¿ÉÑ¡ZnSO4¡¢Zn(NO3)2¡¢ZnCl2ÈÜÒº×÷µç¶ÆÒº¡£
¡¾ÌâÄ¿¡¿ÔÚζȡ¢ÈÝ»ýÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î±£³ÖºãΡ¢ºãÈÝ£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ£¨ÒÑÖª kJ¡¤mol£©ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
ÈÝÆ÷ | ¼× | ÒÒ | ±û |
·´Ó¦ÎïͶÈëÁ¿ | 1mol N2¡¢3mol H2 | 2mol NH3 | 4mol NH3 |
NH3µÄŨ¶È£¨mol¡¤L£© | c1 | c2 | c3 |
·´Ó¦µÄÄÜÁ¿±ä»¯ | ·Å³öakJ | ÎüÊÕbkJ | ÎüÊÕckJ |
Ìåϵѹǿ£¨Pa£© | p1 | p2 | p3 |
·´Ó¦Îïת»¯ÂÊ |
A. B. C. D.
¡¾ÌâÄ¿¡¿¼×´¼ÊÇÓ¦Óù㷺µÄ»¯¹¤ÔÁϺÍÇ°¾°ÀÖ¹ÛµÄÎÞÉ«ÒºÌåȼÁÏ¡£Çë°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌâ¡£
(1)ÒÑÖª25¡æ¡¢101 kpaʱһЩÎïÖʵÄȼÉÕÈÈÈçÏÂ±í£º
ÎïÖÊ | CH3OH(l) | CO(g) | H2(g) |
ȼÉÕÈÈ/(kJ/mol) | 726.8 | 283.0 | 285.8 |
д³öÓÉCOºÍH2·´Ó¦Éú³ÉCH3OH(l)µÄÈÈ»¯Ñ§·½³Ìʽ£º _________________________¡£
(2)Ò»¶¨Î¶ÈÏ£¬ÔÚÈÝ»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£ºCO(g)+2H2(g)=CH3OH(g)£¬ÆäÏà¹ØÊý¾ÝÈçͼËùʾ¡£
¢Ù´Ó·´Ó¦¿ªÊ¼ÖÁ5minʱ£¬ÓÃCH3OH±íʾµÄ·´Ó¦Æ½¾ùËÙ¶ÈΪ____________¡£
¢ÚͼÖз´Ó¦´ïƽºâʱ£¬K=_______(mol/L)-2£»COµÄƽºâת»¯ÂÊΪ________________¡£
(3)ÈËÃÇÀûÓü״¼ÖƵÃÄÜÁ¿×ª»¯Âʸߡ¢¶Ô»·¾³ÎÞÎÛȾµÄȼÁϵç³Ø£¬Æ乤×÷ÔÀíÈçͼËùʾ£¬¸Ã×°Öù¤×÷ʱ£¬a¼«·´Ó¦Ê½Îª____________________¡£
ÈôÓøõç³Ø¼°¶èÐԵ缫µç½â2L±¥ºÍºÍʳÑÎË®²úÉú224mL(±ê×¼Îó²î2)Cl2ʱ(¼ÙÉèÈ«²¿¾äÒݳö²¢ÊÕ¼¯£¬ºöÂÔÈÜÒºÌå»ýµÄ±ä»¯)£¬³£ÎÂÏÂËùµÃÈÜÒºµÄpHΪ________¡£
(4)¼×´¼ÔÚÒ»¶¨Ìõ¼þÏ¿Éת»¯Îª¼×Ëá¡£³£ÎÂÏ£¬Ïò0.1mol/L HCOOHÈÜÒºÖеμÓ0.1mol/LNaOHÈÜÒºÖÁpH=7[ÒÑÖª£¬K(HCOOH)=1.8¡Á10-4]¡£´Ëʱ»ìºÏÓÎѧÖÐÁ½ÈÜÖʵÄÎïÖʵÄÁ¿Ö®±Èn(HCOOH)£ºn(HCOONa)____________¡£