ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿²ÝËáï§[£¨NH4)2C2O4] ΪÎÞÉ«Öù×´¾§Ì壬²»Îȶ¨ £¬ÊÜÈÈÒ׷ֽ⣬¿ÉÓÃÓڲⶨ Ca2+¡¢Mg2+µÄº¬Á¿¡£

I.ijͬѧÀûÓÃÈçͼËùʾʵÑé×°ÖüìÑé²ÝËá淋ķֽâ²úÎï¡£

£¨1£©ÊµÑé¹ý³ÌÖУ¬¹Û²ìµ½½þÓзÓ̪ÈÜÒºµÄÂËÖ½±äºì£¬×°Öà B ÖгÎÇåʯ»ÒË®±ä»ë×Ç£¬ËµÃ÷·Ö½â²úÎïÖк¬ÓÐ__________________£¨Ìѧʽ£©£»Èô¹Û²ìµ½__________________£¬ËµÃ÷·Ö½â²úÎïÖк¬ÓÐ CO¡£²ÝËá立ֽâµÄ»¯Ñ§·½³ÌʽΪ______________________¡£

£¨2£©·´Ó¦¿ªÊ¼Ç° £¬Í¨È뵪ÆøµÄÄ¿µÄÊÇ________________________¡£

£¨3£©×°Öà C µÄ×÷ÓÃÊÇ_______________________¡£

£¨4£©»¹ÓÐÒ»ÖÖ·Ö½â²úÎïÔÚÒ»¶¨Ìõ¼þÏÂÒ²ÄÜ»¹Ô­CuO , ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________¡£

II.¸ÃͬѧÀûÓòÝËá鱗ⶨѪҺÖиÆÔªËصĺ¬Á¿ ¡£

£¨5£©È¡ 20.00 mL ѪҺÑùÆ· £¬¶¨ÈÝÖÁ l00m L, ·Ö±ðÈ¡Èý·ÝÌå»ý¾ùΪ25.00 mL Ï¡ÊͺóµÄѪҺÑùÆ·£¬¼ÓÈë²ÝËá泥¬Éú³É²ÝËá¸Æ³Áµí£¬¹ýÂË£¬½«¸Ã³ÁµíÈÜÓÚ¹ýÁ¿Ï¡ÁòËáÖУ¬È»ºóÓà 0.0l00mol/L KMnO4 ÈÜÒº½øÐеζ¨¡£µÎ¶¨ÖÁÖÕµãʱµÄʵÑéÏÖÏóΪ___________¡£Èý´ÎµÎ¶¨ÊµÑéÏûºÄ KMnO4 ÈÜÒºµÄÌå»ý·Ö±ðΪ0.43mL , 0.41 m L , 0.52mL, Ôò¸ÃѪҺÑùÆ·ÖиÆÔªËصĺ¬Á¿Îª________m mol/L¡£

¡¾´ð°¸¡¿¡¢ ×°ÖÃEÖÐÑõ»¯Í­ÓɺÚÉ«±äΪºìÉ«£¬×°ÖÃFÖгÎÇåʯ»ÒË®±ä»ë×Ç Åž¡×°ÖÃÄڵĿÕÆø£¬±ÜÃâCOÓë¿ÕÆø»ìºÏ¼ÓÈÈ·¢Éú±¬Õ¨£¬²¢·ÀÖ¹¿ÕÆøÖеĸÉÈÅʵÑé ÎüÊÕ£¬±ÜÃâ¶ÔCOµÄ¼ìÑé²úÉú¸ÉÈÅ µ±µÎÈë×îºóÒ»µÎÈÜҺʱ£¬ÈÜÒºÓÉÎÞÉ«±äΪdzºìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ« 2.10

¡¾½âÎö¡¿

I£®²ÝËáï§ÔÚAÖÐÊÜÈȷֽ⣬Èô²úÎïÖÐÓа±Æø£¬°±ÆøÓëË®·´Ó¦Éú³É°±Ë®£¬°±Ë®ÏÔ¼îÐÔ£¬Ôò½þÓзÓ̪ÈÜÒºµÄÂËÖ½±äºì£¬Í¨Èë³ÎÇåʯ»ÒË®£¬Èô³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ôò²úÎïÓжþÑõ»¯Ì¼£¬ÓÃNaOHÈÜÒº³ýÈ¥¶þÑõ»¯Ì¼£¬ÓÃŨÁòËá³ýȥˮÕôÆø£¬½ÓÏÂÀ´Èô²£Á§¹ÜÄÚ±äºì£¬ÇÒFÖгÎÇåʯ»ÒË®±ä»ë×ÇÔòÖ¤Ã÷²úÎïÖÐÓÐCO£»

II£®¸ÆÀë×ӺͲÝËá¸ùÀë×ÓÉú³É²ÝËá¸Æ³Áµí£¬²ÝËá¸ÆºÍÁòËá·´Ó¦Éú³ÉÁòËá¸ÆºÍ²ÝËᣬÓøßÃÌËá¼ØµÎ¶¨²ÝËá´Ó¶ø¼ä½ÓµÎ¶¨¸ÆÀë×Ó¡£

¢ñ.£¨1£©ÊµÑé¹ý³ÌÖУ¬¹Û²ìµ½½þÓзÓ̪ÈÜÒºµÄÂËÖ½±äΪºìɫ˵Ã÷·Ö½â²úÎïÖк¬Óа±Æø£¬×°ÖÃBÖгÎÇåʯ»ÒË®±ä»ë×Ç£¬ËµÃ÷·Ö½â²úÎïÖк¬ÓжþÑõ»¯Ì¼ÆøÌ壬Èô¹Û²ìµ½×°ÖÃEÖÐÑõ»¯Í­ÓɺÚÉ«±äΪºìÉ«£¬×°ÖÃFÖгÎÇåʯ»ÒË®±ä»ë×Ç£¬ËµÃ÷·Ö½â²úÎïÖк¬ÓÐCO£¬ËùÒÔ²ÝËá立ֽâ²úÉúÁË¡¢¡¢COºÍ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£»

£¨2£©·´Ó¦¿ªÊ¼Ç°£¬Í¨È뵪ÆøµÄÄ¿µÄÊÇÅž¡×°ÖÃÄڵĿÕÆø£¬±ÜÃâCOÓë¿ÕÆø»ìºÏ¼ÓÈÈ·¢Éú±¬Õ¨£¬²¢·ÀÖ¹¿ÕÆøÖеĸÉÈÅʵÑé,¹Ê´ð°¸Îª£ºÅž¡×°ÖÃÄڵĿÕÆø£¬±ÜÃâCOÓë¿ÕÆø»ìºÏ¼ÓÈÈ·¢Éú±¬Õ¨£¬²¢·ÀÖ¹¿ÕÆøÖеĸÉÈÅʵÑ飻

£¨3£©×°ÖÃEºÍFÊÇÑéÖ¤²ÝËá立ֽâ²úÎïÖк¬ÓÐCO£¬ËùÒÔÒª°Ñ·Ö½â²úÉúµÄ³ýÈ¥£¬Òò´Ë×°ÖÃCµÄ×÷ÓÃÊÇ£ºÎüÊÕ£¬±ÜÃâ¶ÔCOµÄ¼ìÑé²úÉú¸ÉÈÅ£¬¹Ê´ð°¸Îª£ºÎüÊÕ£¬±ÜÃâ¶ÔCOµÄ¼ìÑé²úÉú¸ÉÈÅ£»

£¨4£©²ÝËá立ֽâ²úÉúµÄÓл¹Ô­ÐÔ£¬Ò»¶¨Ìõ¼þÏÂÒ²»áÓëCuO·´Ó¦£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£»

¢òÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒºµÎ¶¨²ÝËá¸ùÀë×Ó£¬·¢ÉúµÄÊÇÑõ»¯»¹Ô­·´Ó¦£¬ºìÉ«µÄ¸ßÃÌËá¼ØÈÜÒº»áÍÊÉ«£¬µÎ¶¨ÖÁÖÕµãʱµÄʵÑéÏÖÏóΪ£ºµ±µÎÈë×îºóÒ»µÎÈÜҺʱ£¬ÈÜÒºÓÉÎÞÉ«±äΪdzºìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«£¬¸ù¾ÝÈý´ÎµÎ¶¨ËùÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ý¿ÉÖª£¬µÚÈý´ÎÓëÒ»¡¢¶þ´ÎÌå»ý²î±ðÌ«´ó£¬ÐèÉáÆú£¬ËùÒÔÁ½´ÎµÄƽ¾ùÌå»ýΪ£¬¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦Öеĵç×ÓÊغ㼰ԪËØÊغ㣬Ôò£º£¬½âµÃ£¬ËùÒÔ20mLѪҺÑùÆ·Öк¬ÓеĸÆÔªËصÄÎïÖʵÄÁ¿Îª£¬¼´£¬Ôò¸ÃѪҺÖиÆÔªËصĺ¬Á¿Îª£¬¹Ê´ð°¸Îª£ºµ±µÎÈë×îºóÒ»µÎÈÜҺʱ£¬ÈÜÒºÓÉÎÞÉ«±äΪdzºìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«£»¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿CO¡¢CO2ÊÇ»¯Ê¯È¼ÁÏȼÉÕºóµÄÖ÷Òª²úÎï¡£

£¨1£©Æû³µÅÅÆø¹ÜÄÚ°²×°µÄ´ß»¯×ª»¯Æ÷£¬¿ÉʹÆû³µÎ²ÆøÖеÄÖ÷ÒªÎÛȾÎïת»¯ÎªÎÞ¶¾µÄ´óÆøÑ­»·ÎïÖÊ¡£ÒÑÖª£ºN2(g)£«O2(g)===2NO(g)¦¤H=£«180.5kJ¡¤mol£­1

2C(s)£«O2(g)===2CO(g)¦¤H=£­221.0kJ¡¤mol£­1

C(s)£«O2(g)===CO2(g)¦¤H=£­393.5kJ¡¤mol£­1

Ôò·´Ó¦2NO(g)£«2CO(g)===N2(g)£«2CO2(g)µÄ¦¤H=________kJ¡¤mol£­1¡£

£¨2£©ÒÑÖª£º·´Ó¦CO2(g)CO(g)+O(g)ÔÚÃܱÕÈÝÆ÷ÖÐCO2·Ö½âʵÑéµÄ½á¹ûÈçͼ1£»·´Ó¦2CO2(g)2CO(g)+O2(g)ÖÐ1molCO2ÔÚ²»Í¬Î¶ÈϵÄƽºâ·Ö½âÁ¿Èçͼ2£¬

¢Ù·ÖÎöͼ1£¬Çó2minÄÚv(CO2)=_______£¬5minʱ´ïµ½Æ½ºâ£¬Æ½ºâʱc£¨CO£©=_______¡£

¢Ú·ÖÎöͼ2£¬1500¡æʱ·´Ó¦´ïƽºâ£¬´ËʱÈÝÆ÷Ìå»ýΪ1L£¬Ôò·´Ó¦µÄƽºâ³£ÊýK=______(¼ÆËã½á¹û±£Áô1λСÊý)¡£

£¨3£©ÎªÌ½¾¿²»Í¬´ß»¯¼Á¶ÔCOºÍH2ºÏ³ÉCH3OHµÄÑ¡ÔñÐÔЧ¹û£¬Ä³ÊµÑéÊÒ¿ØÖÆCOºÍH2µÄ³õʼͶÁϱÈΪ1£º3½øÐÐʵÑ飬µÃµ½ÈçÏÂÊý¾Ý£º

Ñ¡Ïî

T/K

ʱ¼ä/min

´ß»¯¼ÁÖÖÀà

¼×´¼µÄº¬Á¿(%)

A

450

10

CuO-ZnO

78

B

450

10

CuO-ZnO-ZrO2

88

C

450

10

ZnO-ZrO2

46

¢ÙÓɱí1¿ÉÖª£¬¸Ã·´Ó¦µÄ×î¼Ñ´ß»¯¼ÁΪ____________£¨Ìî±àºÅ£©£»Í¼3ÖÐa¡¢b¡¢c¡¢dËĵãÊǸÃζÈÏÂCOµÄƽºâת»¯ÂʵÄÊÇ____________¡£

¢ÚÓÐÀûÓÚÌá¸ßCOת»¯ÎªCH3OHµÄƽºâת»¯ÂʵĴëÊ©ÓÐ____________¡£

A£®Ê¹Óô߻¯¼ÁCuO£­ZnO£­ZrO2B£®Êʵ±½µµÍ·´Ó¦Î¶È

C£®Ôö´óCOºÍH2µÄ³õʼͶÁϱÈD£®ºãÈÝÏ£¬ÔÙ³äÈëamolCOºÍ3amolH2

¡¾ÌâÄ¿¡¿

½Ì²Ä²åͼ¾ßÓмò½à¶øÓÖÄÚº­·á¸»µÄÌص㡣Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©µÚÈýÖÜÆÚµÄijÖ÷×åÔªËØ£¬ÆäµÚÒ»ÖÁµÚÎåµçÀëÄÜÊý¾ÝÈçͼ1Ëùʾ£¬Ôò¸ÃÔªËضÔÓ¦µÄÔ­×ÓÓÐ_____ÖÖ²»Í¬Ô˶¯×´Ì¬µÄµç×Ó¡£

£¨2£©Èçͼ2Ëùʾ£¬Ã¿ÌõÕÛÏß±íʾÖÜÆÚ±í¢ôA ~¢÷A ÖеÄijһ×åÔªËØÇ⻯ÎïµÄ·Ðµã±ä»¯¡£Ã¿¸öСºÚµã´ú±íÒ»ÖÖÇ⻯ÎÆäÖÐaµã´ú±íµÄÊÇ___________¡£ÅжÏÒÀ¾ÝÊÇ____________¡£

£¨3£©CO2ÔÚ¸ßθßѹÏÂËùÐγɵľ§ÌåÆ侧°ûÈçͼ3Ëùʾ¡£Ôò¸Ã¾§ÌåµÄÀàÐÍÊôÓÚ_____________¾§Ìå¡£

£¨4£©µÚÒ»µçÀëÄܽéÓÚAl¡¢PÖ®¼äµÄµÚÈýÖÜÆÚÔªËØÓÐ____ÖÖ¡£ GaCl3ÖÐÖÐÐÄÔ­×ÓµÄÔÓ»¯·½Ê½Îª_________£¬Ð´³öÓëGaCl3½á¹¹ÏàͬµÄÒ»Öֵȵç×ÓÌ壨дÀë×Ó£©______________¡£

£¨5£©±ù¡¢¸É±ù¡¢µâ¶¼ÊÇ·Ö×Ó¾§Ì壬±ùµÄ½á¹¹¾ßÓÐÌØÊâÐÔ£¬¶ø¸É±ù¡¢µâµÄ¾§Ìå¾ßÓÐÏàËƵĽṹÌØÕ÷£¬¸É±ù·Ö×ÓÖÐÒ»¸ö·Ö×ÓÖÜΧÓÐ__________¸ö½ôÁÚ·Ö×Ó¡£ DµÄ´×ËáÑξ§Ìå¾Ö²¿½á¹¹Èçͼ£¬¸Ã¾§ÌåÖк¬ÓеĻ¯Ñ§¼üÊÇ_____________£¨Ìî×Öĸ±êºÅ£©¡£

a£®¼«ÐÔ¼ü b£®·Ç¼«ÐÔ¼ü c£®Åäλ¼ü d£®½ðÊô¼ü

£¨6£©FeµÄÒ»ÖÖ¾§ÌåÈç¼×¡¢ÒÒËùʾ£¬Èô°´¼×ÐéÏß·½ÏòÇÐÒҵõ½µÄA~DͼÖÐÕýÈ·µÄÊÇ_____£¨Ìî×Öĸ±êºÅ£©¡£

ÌúÔ­×ÓµÄÅäλÊýÊÇ____________£¬¼ÙÉèÌúÔ­×ӵİ뾶ÊÇr cm£¬¸Ã¾§ÌåµÄÃܶÈÊǦÑg/cm3 £¬ÔòÌúµÄÏà¶ÔÔ­×ÓÖÊÁ¿Îª________________£¨Éè°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪNA£©¡£

¡¾ÌâÄ¿¡¿ÁòËáÄø¹ã·ºÓ¦ÓÃÓÚµç¶Æ¡¢µç³Ø¡¢´ß»¯¼ÁµÈ¹¤Òµ¡£Ä³¿ÆÑÐС×éÒÔ´ÖÁòËáÄø(º¬Cu2+¡¢Fe3+¡¢Ca2+¡¢Mg2+¡¢Zn2+µÈ)ΪԭÁÏ£¬¾­ÈçͼһϵÁгýÔÓ¹ý³ÌÄ£Ä⾫ÖÆÁòËáÄø¹¤ÒÕ¡£»Ø´ðÏÂÁÐÎÊÌâ¡£

£¨1£©ÂËÔü1µÄÖ÷Òª³É·ÖÊÇ__(д»¯Ñ§Ê½)£¬Ð´³ö¡°Áò»¯³ýÍ­¡±¹ý³ÌÖз¢ÉúµÄÑõ»¯»¹Ô­·´Ó¦µÄÀë×Ó·½³Ìʽ___¡£

£¨2£©¡°Ñõ»¯³ýÔÓ¡±Ê±¼ÓÈëCl2ºÍNi(OH)2µÄ×÷Ó÷ֱðÊÇ___¡£

£¨3£©ÒÑÖª25¡æʱ£¬Ksp[CaF2]=3.95¡Á10-11£»Ksp[MgF2]=6.40¡Á10-9¡£Ôò¡°·ú»¯³ýÔÓ¡±¹ýºóÂËÒº3ÖÐ=__¡£(±£ÁôÈýλÓÐЧÊý×Ö)

£¨4£©¡°ÝÍÈ¡¡±Ê±Ê¹ÓÃÝÍÈ¡¼ÁRÔÚÁòËáÑÎÖжÔijЩ½ðÊôÀë×ÓµÄÝÍÈ¡ÂÊÓëÈÜÒºpHµÄ¹ØϵÈçͼ¡£ÔòʵÑéʱÐè¿ØÖƵÄpHÊÊÒË·¶Î§ÊÇ___(Ìî×ÖĸÐòºÅ)¡£

A.1¡«2 B.3¡«4 C.4¡«5 D.5¡«6

£¨5£©½«ÝÍÈ¡ºóËùµÃ¸»º¬ÁòËáÄøµÄÈÜÒº¾­²Ù×÷A¿ÉµÃÁòËáÄø¾§Ì壬Ôò²Ù×÷AΪ___¡¢___¡¢¹ýÂË¡¢Ï´µÓµÈ¡£

£¨6£©³ÆÈ¡2.000gÁòËáÄø¾§Ìå(NiSO4¡¤6H2O)ÑùÆ·Èܽ⣬¶¨ÈÝÖÁ250mL¡£È¡25.00mLÊÔÒº£¬ÓÃ0.0200mol¡¤L-1µÄEDTA(Na2H2Y)±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㡣Öظ´ÊµÑ飬ƽ¾ùÏûºÄEDTA±ê×¼ÈÜÒºÌå»ýΪ36.50mL¡£·´Ó¦ÎªNi2++H2Y2-=NiY2-+2H+¡£¼ÆËãÑùÆ·´¿¶ÈΪ__%¡£(±£ÁôÈýλÓÐЧÊý×Ö,ÇÒ²»¿¼ÂÇÔÓÖÊ·´Ó¦)

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

¾«Ó¢¼Ò½ÌÍø