ÌâÄ¿ÄÚÈÝ

ÅðÔªËØÔÚ»¯Ñ§ÖÐÓкÜÖØÒªµÄµØ룬Åð¼°Æ仯ºÏÎï¹ã·ºÓ¦ÓÃÓÚÓÀ´Å²ÄÁÏ¡¢³¬µ¼²ÄÁÏ¡¢¸»È¼ÁϲÄÁÏ¡¢¸´ºÏ²ÄÁϵȸßвÄÁÏÁìÓò¡£

£¨1£©Èý·ú»¯ÅðÔÚ³£Î³£Ñ¹ÏÂΪ¾ßÓд̱Ƕñ³ôºÍÇ¿´Ì¼¤ÐÔµÄÎÞÉ«Óж¾¸¯Ê´ÐÔÆøÌ壬Æä·Ö×ÓµÄÁ¢Ìå¹¹ÐÍΪ________£¬BÔ­×ÓµÄÔÓ»¯ÀàÐÍΪ________¡£

£¨2£©Á×»¯ÅðÊÇÒ»ÖÖÊܵ½¸ß¶È¹Ø×¢µÄÄÍĥͿÁÏ£¬Ëü¿ÉÓÃ×÷½ðÊôµÄ±íÃæ±£»¤²ã¡£Í¼£¨a£©ÊÇÁ×»¯Åð¾§ÌåµÄ¾§°ûʾÒâͼ£¬ÔòÁ×»¯ÅðµÄ»¯Ñ§Ê½Îª________£¬¸Ã¾§ÌåµÄ¾§ÌåÀàÐÍÊÇ________¡£

£¨3£©ÕýÅðËᣨH3BO3£©ÊÇÒ»ÖÖƬ²ã×´½á¹¹°×É«¾§Ì壬²ãÄÚµÄH3BO3·Ö×Ó¼äͨ¹ýÇâ¼üÏàÁ¬[Èçͼ£¨b£©]¡£

¢ÙÅðËá·Ö×ÓÖÐB×îÍâ²ãÓÐ________¸öµç×Ó£¬1 mol H3BO3µÄ¾§ÌåÖÐÓÐ________molÇâ¼ü¡£

¢ÚÅðËáÈÜÓÚË®Éú³ÉÈõµç½âÖÊһˮºÏÅðËáB£¨OH£©3¡¤H2O£¬ËüµçÀëÉú³ÉÉÙÁ¿[B£¨OH£©4]£­ºÍH£«Àë×Ó¡£ÔòÅðËáΪ________ÔªËᣬ[B£¨OH£©4]£­º¬ÓеĻ¯Ñ§¼üÀàÐÍΪ________¡£

 

 

£¨1£©Æ½ÃæÈý½ÇÐΡ¡sp2

£¨2£©BP¡¡Ô­×Ó¾§Ìå

£¨3£©¢Ù6¡¡3¡¡¢ÚÒ»¡¡¹²¼Û¼ü¡¢Åäλ¼ü

¡¾½âÎö¡¿£¨1£©BF3·Ö×ÓÖУ¬BÔ­×ÓÐγÉÁË3¸ö¦Ò¼ü£¬²»º¬¹Âµç×Ó¶Ô£¬¹ÊÔÓ»¯¹ìµÀÊýΪ3£¬ÔÓ»¯·½Ê½Îªsp2ÔÓ»¯£¬BF3·Ö×ÓÁ¢Ìå¹¹ÐÍΪƽÃæÈý½ÇÐΡ££¨2£©Ò»¸ö¾§°ûÖк¬ÓÐBÔ­×ӵĸöÊýΪ8¡Á£«6¡Á£½4¸ö£¬PÔ­×Ó¸öÊýΪ4¸ö£¬¹Ê»¯Ñ§Ê½ÎªBP£»Á×»¯Åð¾§ÌåÖÐÖ»º¬¹²¼Û¼üÇÒÄÍÄ¥£¬ÎªÔ­×Ó¾§Ìå¡££¨3£©¢ÙÓÉͼ¿ÉÖª£¬BÔ­×ÓÐγÉÁËÈý¸ö¹²¼Ûµ¥¼ü£¬×îÍâ²ãµç×ÓÊýΪ6£»1 mol H3BO3µÄ¾§ÌåÖк¬ÓÐ3 mol HÔ­×Ó£¬Ã¿¸öHÔ­×Ó¶¼ÓëÏàÁÚµÄOÔ­×ÓÐγÉÇâ¼ü£¬¹ÊÇâ¼ü¸öÊýΪ3 mol£»¢Ú1 mol B£¨OH£©3¡¤H2O¿ÉµçÀë³ö1 mol H£«£¬¹ÊÅðËáΪһԪË᣻[B£¨OH£©4]£­Àë×ÓÖÐBÔ­×ÓÓÐÒ»¸ö¿ÕµÄ2p¹ìµÀ£¬¶øOH£­º¬Óйµç×Ó¶Ô£¬Á½ÕßÖ®¼ä¿ÉÒÔÐγÉÅäλ¼ü¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Ñо¿CO2µÄÀûÓöԴٽøµÍ̼Éç»áµÄ¹¹½¨¾ßÓÐÖØÒªµÄÒâÒå¡£

(1)½«CO2Ó뽹̿×÷ÓÃÉú³ÉCO£¬CO¿ÉÓÃÓÚÁ¶ÌúµÈ¡£

¢ÙÒÑÖª£ºFe2O3(s)£«3C(ʯī)=2Fe(s)£«3CO(g)¡¡¦¤H1£½£«489.0 kJ¡¤mol£­1

C(ʯī)£«CO2(g)=2CO(g)¡¡¦¤H2£½£«172.5 kJ¡¤mol£­1

ÔòCO»¹Ô­Fe2O3µÄÈÈ»¯Ñ§·½³ÌʽΪ___________________________

¢ÚÀûÓÃȼÉÕ·´Ó¦¿ÉÉè¼Æ³ÉCO/O2ȼÁϵç³Ø(ÒÔKOHÈÜҺΪµç½âÒº)£¬Ð´³ö¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½___________________________________________

(2)ijʵÑ齫CO2ºÍH2³äÈëÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚÁ½ÖÖ²»Í¬Ìõ¼þÏ·´Ó¦£º

CO2(g)£«3H2(g)CH3OH(g)£«H2O(g)¡¡ ¦¤H£½£­49.0 kJ¡¤mol£­1

 

²âµÃCH3OHµÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçÉÏͼËùʾ£¬»Ø´ðÎÊÌ⣺

¢ÙÏÂÁдëÊ©ÖÐÄÜʹn(CH3OH)/n(CO2)Ôö´óµÄÊÇ________¡£

A£®Éý¸ßÎÂ¶È B£®³äÈëHe(g)ʹÌåϵѹǿÔö´ó

C£®½«H2O(g)´ÓÌåϵÖзÖÀë D£®ÔÙ³äÈë1 mol CO2ºÍ3 mol H2

¢ÚÇúÏߢñ¡¢¢ò¶ÔÓ¦µÄƽºâ³£Êý´óС¹ØϵΪK¢ñ________K¢ò(Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)¡£

¢ÛÒ»¶¨Î¶ÈÏ£¬ÔÚÈÝ»ýÏàͬÇҹ̶¨µÄÁ½¸öÃܱÕÈÝÆ÷ÖУ¬°´ÈçÏ·½Ê½Í¶Èë·´Ó¦Îһ¶Îʱ¼äºó´ïµ½Æ½ºâ¡£

 

ÈÝÆ÷

¼×

ÒÒ

·´Ó¦ÎïͶÈëÁ¿

1 mol CO2¡¢3 mol H2

a mol CO2¡¢b mol H2¡¢c mol CH3OH(g)¡¢c mol H2O(g)

 

Èô¼×ÖÐƽºâºóÆøÌåµÄѹǿΪ¿ªÊ¼Ê±µÄ£¬ÒªÊ¹Æ½ºâºóÒÒÓë¼×ÖÐÏàͬ×é·ÖµÄÌå»ý·ÖÊýÏàµÈ£¬ÇÒÆðʼʱά³Ö·´Ó¦ÄæÏò½øÐУ¬ÔòcµÄÈ¡Öµ·¶Î§Îª________¡£

(3)ÓÃ0.10 mol¡¤L£­1ÑÎËá·Ö±ðµÎ¶¨20.00 mL 0.10 mol¡¤L£­1µÄNaOHÈÜÒººÍ20.00 mL 0.10 mol¡¤L£­1°±Ë®ËùµÃµÄµÎ¶¨ÇúÏßÈçÏ£º

 

ÇëÖ¸³öÑÎËáµÎ¶¨°±Ë®µÄÇúÏßΪ________(Ìî¡°A¡±»ò¡°B¡±)£¬Çëд³öÇúÏßaµãËù¶ÔÓ¦µÄÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄÅÅÁÐ˳Ðò________¡£

 

 

°±»ù¼×Ëá泥¨NH2COONH4£©ÊÇÒ»ÖÖ°×É«¹ÌÌ壬Ò׷ֽ⡢Ò×Ë®½â£¬¿ÉÓÃ×÷·ÊÁÏ¡¢Ãð»ð¼Á¡¢Ï´µÓ¼ÁµÈ¡£Ä³»¯Ñ§ÐËȤС×éÄ£Ä⹤ҵԭÀíÖƱ¸°±»ù¼×Ëá泥¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÈçÏ£º

2NH3£¨g£©£«CO2£¨g£©??NH2COONH4£¨s£©¡¡¦¤H£¼0¡£

£¨1£©ÈçͼËùʾװÖÃÖÆÈ¡°±Æø£¬ÄãËùÑ¡ÔñµÄÊÔ¼ÁÊÇ________________________¡£

£¨2£©ÖƱ¸°±»ù¼×Ëá淋Ä×°ÖÃÈçͼ13£­7Ëùʾ£¬°Ñ°±ÆøºÍ¶þÑõ»¯Ì¼Í¨ÈëËÄÂÈ»¯Ì¼ÖУ¬²»¶Ï½Á°è»ìºÏ£¬Éú³ÉµÄ°±»ù¼×Ëáï§Ð¡¾§ÌåÐü¸¡ÔÚËÄÂÈ»¯Ì¼ÖС£µ±Ðü¸¡Îï½Ï¶àʱ£¬Í£Ö¹ÖƱ¸¡£

×¢£ºËÄÂÈ»¯Ì¼ÓëÒºÌåʯÀ¯¾ùΪ¶èÐÔ½éÖÊ¡£

¢Ù·¢ÉúÆ÷ÓñùË®ÀäÈ´µÄÔ­ÒòÊÇ________________________________________________________________________________________________________________________________________________¡£

ÒºÌåʯÀ¯¹ÄÅÝÆ¿µÄ×÷ÓÃÊÇ________________________________________________________________________¡£

¢Ú´Ó·´Ó¦ºóµÄ»ìºÏÎïÖзÖÀë³ö²úÆ·µÄʵÑé·½·¨ÊÇ________________________________________________________________________

£¨Ìîд²Ù×÷Ãû³Æ£©¡£ÎªÁ˵õ½¸ÉÔï²úÆ·£¬Ó¦²ÉÈ¡µÄ·½·¨ÊÇ________£¨ÌîдѡÏîÐòºÅ£©¡£

a£®³£Ñ¹¼ÓÈȺæ¸É

b£®¸ßѹ¼ÓÈȺæ¸É

c£®Õæ¿Õ40 ¡æÒÔϺæ¸É

¢ÛβÆø´¦Àí×°ÖÃÈçͼËùʾ¡£

˫ͨ²£Á§¹ÜµÄ×÷Óãº____________£»Å¨ÁòËáµÄ×÷Óãº______________________¡¢__________________________________________________________________¡£

£¨3£©È¡Òò²¿·Ö±äÖʶø»ìÓÐ̼ËáÇâ淋ݱ»ù¼×Ëáï§ÑùÆ·1.173 0 g£¬ÓÃ×ãÁ¿Ê¯»ÒË®³ä·Ö´¦Àíºó£¬Ê¹Ì¼ÔªËØÍêȫת»¯ÎªÌ¼Ëá¸Æ£¬¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ²âµÃÖÊÁ¿Îª1.500 g¡£ÔòÑùÆ·Öа±»ù¼×Ëá淋ÄÎïÖʵÄÁ¿·ÖÊýΪ________¡£[Mr£¨NH2COONH4£©£½78£¬Mr£¨NH4HCO3£©£½79£¬Mr£¨CaCO3£©£½100]

 

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

¾«Ó¢¼Ò½ÌÍø