ÌâÄ¿ÄÚÈÝ
9£®X¡¢Y¡¢Z¡¢WÊÇÔªËØÖÜÆÚ±íÇ°ËÄÖÜÆÚ³£¼ûµÄËÄÖÖÔªËØ£¬Ô×ÓÐòÊýÒÀ´ÎÔö´ó£®XµÄµ¥ÖʼÈÓЦҼüÓÖÓЦмü£¬ÇÒXµÄµÚÒ»µçÀëÄܱÈÆäÏàÁÚÔªËض¼´ó£¬YµÄµç¸ºÐÔ±ÈXµÄС£¬Æä»ù̬Ô×Ó×îÍâ²ãµÄpµç×ÓÊǸòãsµç×ÓµÄÁ½±¶£¬ZÊÇÇ°36ºÅÔªËØÖÐÔ×Ӱ뾶×î´óµÄÖ÷×åÔªËØ£¬WλÓÚÖÜÆÚ±íµÄµÚ8ÁУ®£¨1£©ZλÓÚÔªËØÖÜÆÚ±íµÚÈýÖÜÆÚµÚ¢ñA×壬»ù̬W2+µÄºËÍâµç×ÓÅŲ¼Ê½ÊÇ1s22s22p63s23p63d6£®
£¨2£©XµÄÑõ»¯ÎïXO2ÓëH2O±È½Ï£¬ÈÛµã½Ï¸ßµÄÊÇH2O£¨Ìѧʽ£©£¬ZµÄÇ⻯ÎïÊôÓÚÀë×Ó¾§Ì壬Z2O2µÄ»¯Ñ§¼üÀàÐÍÊÇAC£¨Ìî±àºÅ£©£®
A£®Àë×Ó¼ü¡¡¡¡¡¡¡¡B£®¼«ÐÔ¹²¼Û¼ü C£®·Ç¼«ÐÔ¹²¼Û¼üD£®½ðÊô¼ü
£¨3£©¼ÓÈÈÌõ¼þÏ£¬YµÄµ¥ÖÊÓë×ãÁ¿µÄXµÄ×î¸ß¼ÛÑõ»¯ÎïËù¶ÔÓ¦µÄË®»¯ÎïµÄŨÈÜÒº·´Ó¦£¬Éú³ÉYµÄ×î¸ß¼Ûº¬ÑõËᣬд³ö´Ë·´Ó¦µÄ»¯Ñ§·½³ÌʽS+6HNO3£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$H2SO4+6NO2¡ü+2H2O£®
£¨4£©ÔÚ25¡æ£¬101kPaÏ£¬WµÄµ¥ÖÊÔÚÑõÆøÖÐȼÉÕºó»Ö¸´ÖÁÔζȺÍѹǿ£¬Æ½¾ùÿתÒÆ1molµç×ӷųöQkJµÄÈÈÁ¿£¬ÔòWµ¥ÖÊȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽÊÇ3Fe£¨s£©+2O2£¨g£©=Fe3O4£¨s£©¡÷H=-8Q KJ/mol£®
·ÖÎö X¡¢Y¡¢Z¡¢WÊÇÔªËØÖÜÆÚ±íÇ°ËÄÖÜÆÚ³£¼ûµÄËÄÖÖÔªËØ£¬Ô×ÓÐòÊýÒÀ´ÎÔö´ó£®XµÄµ¥ÖʼÈÓЦҼüÓÖÓЦмü£¬Æäµ¥Öʺ¬ÓÐË«¼ü»òÈý¼ü£¬Îª·Ç½ðÊôÔªËØ£¬ÇÒXµÄµÚÒ»µçÀëÄܱÈÆäÏàÁÚÔªËض¼´ó£¬ÍâΧµç×ÓÅŲ¼Îªns2np3£¬¹ÊXΪµªÔªËØ£»ZÊǶÌÖÜÆÚÖÐÔ×Ӱ뾶×î´óµÄÖ÷×åÔªËØ£¬ÔòZΪNaÔªËØ£»WλÓÚÖÜÆÚ±íµÄµÚ8ÁУ¬ÔòWΪFeÔªËØ£»Y»ù̬Ô×Ó×îÍâ²ãµÄpµç×ÓÊǸòãsµç×ÓµÄÁ½±¶£¬ÍâΧµç×ÓÅŲ¼Îªns2np4£¬YµÄµç¸ºÐÔ±ÈXµÄС£¬Y²»¿ÉÄÜÓëX´¦ÓÚͬһÖÜÆÚ£¬Ö»ÄÜ´¦ÓÚµÚÈýÖÜÆÚ£¬¹ÊYΪSÔªËØ£¬¾Ý´Ë½â´ð£®
½â´ð ½â£ºX¡¢Y¡¢Z¡¢WÊÇÔªËØÖÜÆÚ±íÇ°ËÄÖÜÆÚ³£¼ûµÄËÄÖÖÔªËØ£¬Ô×ÓÐòÊýÒÀ´ÎÔö´ó£®XµÄµ¥ÖʼÈÓЦҼüÓÖÓЦмü£¬Æäµ¥Öʺ¬ÓÐË«¼ü»òÈý¼ü£¬Îª·Ç½ðÊôÔªËØ£¬ÇÒXµÄµÚÒ»µçÀëÄܱÈÆäÏàÁÚÔªËض¼´ó£¬ÍâΧµç×ÓÅŲ¼Îªns2np3£¬¹ÊXΪµªÔªËØ£»ZÊǶÌÖÜÆÚÖÐÔ×Ӱ뾶×î´óµÄÖ÷×åÔªËØ£¬ÔòZΪNaÔªËØ£»WλÓÚÖÜÆÚ±íµÄµÚ8ÁУ¬ÔòWΪFeÔªËØ£»Y»ù̬Ô×Ó×îÍâ²ãµÄpµç×ÓÊǸòãsµç×ÓµÄÁ½±¶£¬ÍâΧµç×ÓÅŲ¼Îªns2np4£¬YµÄµç¸ºÐÔ±ÈXµÄС£¬Y²»¿ÉÄÜÓëX´¦ÓÚͬһÖÜÆÚ£¬Ö»ÄÜ´¦ÓÚµÚÈýÖÜÆÚ£¬¹ÊYΪSÔªËØ£¬
£¨1£©ZÊÇNaÔªËØλÓÚÔªËØÖÜÆÚ±íµÚÈýÖÜÆÚµÚ¢ñA×壬»ù̬W2+ÊÇÑÇÌúÀë×Ó£¬ÆäºËÍâµç×ÓÅŲ¼Ê½ÊÇ1s22s22p63s23p63d6£¬
¹Ê´ð°¸Îª£ºÈý£»¢ñA£»1s22s22p63s23p63d6£»
£¨2£©NµÄÑõ»¯ÎïNO2ÓëH2O±È½Ï£¬ÒòΪˮ·Ö×Ó¼ä´æÔÚÇâ¼ü£¬ËùÒÔË®µÄÈÛµã¸ß£¬ZΪNaÔªËØ£¬ÆäÇ⻯ÎïÊÇNaHÊÇÀë×Ó¾§Ì壬¹ýÑõ»¯ÄÆÖдæÔÚÀë×Ó¼üºÍ·Ç¼«ÐÔ¹²¼Û¼ü£¬
¹Ê´ð°¸Îª£ºH2O£»Àë×Ó£»AC£»
£¨3£©¼ÓÈÈÌõ¼þÏ£¬Sµ¥ÖÊÓë×ãÁ¿HNO3ŨÈÜÒº·´Ó¦£¬Éú³ÉH2SO4£¬HNO3±»»¹ÔΪNO2£¬·´Ó¦·½³ÌʽΪ£ºS+6HNO3£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$H2SO4+6NO2¡ü+2H2O£¬
¹Ê´ð°¸Îª£ºS+6HNO3£¨Å¨£©$\frac{\underline{\;\;¡÷\;\;}}{\;}$H2SO4+6NO2¡ü+2H2O£»
£¨4£©ÔÚ25¡ãC£¬1O1kPaÏ£¬FeµÄµ¥ÖÊÔÚÑõÆøÖÐȼÉÕÉú³ÉFe3O4£¬»Ö¸´ÖÁÔζȺÍѹǿ£¬Æ½¾ùÿתÒÆ1molµç×ӷųöQkJµÄÈÈÁ¿£¬Ôò3moFe·´Ó¦·Å³öµÄÈÈÁ¿Îª$\frac{3mol¡Á\frac{8}{3}}{1mol}$¡ÁQkJ=8QkJ£¬Feµ¥ÖÊȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽÊÇ3Fe£¨s£©+2O2£¨g£©=Fe3O4£¨s£©¡÷H=-8Q KJ/mol£¬
¹Ê´ð°¸Îª£º3Fe£¨s£©+2O2£¨g£©=Fe3O4£¨s£©¡÷H=-8Q KJ/mol£®
µãÆÀ ±¾Ì⿼²é½á¹¹Î»ÖÃÐÔÖʹØϵ¡¢ºËÍâµç×ÓÅŲ¼¹æÂÉ¡¢Ñõ»¯»¹Ô·´Ó¦¡¢ÈÈ»¯Ñ§·½³ÌʽÊéдµÈ£¬ÄѶÈÖеȣ¬ÍƶÏÔªËØÊǽâÌâµÄ¹Ø¼ü£¬×¢Òâ¸ù¾ÝXµ¥ÖÊ»¯Ñ§¼üÓëÔªËصçÀëÄÜÈ·¶¨XΪµªÔªËØ£®
A£® | H2O | B£® | CH3COOC2H5 | C£® | CH3CH2OH | D£® | CH3COOH |
A£® | ÒÒ´¼Óë½ðÊôÄÆ·´Ó¦ | B£® | ±½ºÍÒºäå·´Ó¦Éú³Éäå±½ | ||
C£® | ÒÒÏ©ºÍäåµÄËÄÂÈ»¯Ì¼ÈÜÒº·´Ó¦ | D£® | ÒÒËáºÍÒÒ´¼·´Ó¦Éú³ÉÒÒËáÒÒõ¥ |
£¨1£©ÔÚʵÑéÊÒÖУ¬ÓûÓÃ98%µÄŨÁòËᣨÃܶÈΪ1.84g•mL-1£©ÅäÖÆ500mL1.0mol•L-1µÄÁòËᣬÐèÒªµÄÒÇÆ÷³ýÁ¿Í²¡¢ÉÕ±¡¢²£Á§°ôÍ⣬»¹ÓнºÍ·µÎ¹Ü¡¢500mLÈÝÁ¿Æ¿£®
£¨2£©¸ÃС×éͬѧÉè¼ÆÈçÏÂ×°ÖÃÄ£Äâ·ÏÔüÔÚ¹ýÁ¿ÑõÆøÖбºÉÕ£¬²¢ÑéÖ¤·ÏÔüÖк¬ÁòÔªËØ£®
¢Ù×°ÖÃAÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Na2O2+2H2O=4NaOH+O2¡ü£»Îª¿ØÖÆ·´Ó¦²»¹ýÓÚ¼¤ÁÒ²¢²úÉúƽÎÈÆøÁ÷£¬²ÉÈ¡µÄ²Ù×÷¼°ÏÖÏóÊÇ´ò¿ª·ÖҺ©¶·ÉÏ¿Ú»îÈû£¬¿ØÖÆ·ÖҺ©¶·ÐýÈû£¬Ê¹Ë®ÔÈËÙÖðµÎµÎÏ£»B´¦Ó¦Á¬½ÓÊ¢Óмîʯ»ÒµÄ¸ÉÔï¹Ü£¨»òUÐιܣ©»òŨÁòËáµÄÏ´ÆøÆ¿£¨ÌîдÊÔ¼Á¼°ÒÇÆ÷Ãû³Æ£©£®
¢ÚE×°ÖÃÖмÓÈëÆ·ºìÈÜÒºµÄÄ¿µÄÊǼìÑéÆøÌåaÖеÄSO2£»µ±F×°ÖÃÖгöÏÖ°×É«³Áµíʱ£¬·´Ó¦Àë×Ó·½³ÌʽΪ2SO2+O2+2H2O+2Ba2+=2BaSO4¡ý+4H+£®
£¨3£©ÏÂÁвÙ×÷ÖУ¬²»ÊôÓÚ²½Öè¢ÝÖнøÐеIJÙ×÷µÄÊÇad£¨ÌîÏÂÁи÷ÏîÖÐÐòºÅ£©£®
²½Öè¢ÞÖÐÓÉ´Ö͵õ½´¿ÍµÄ·½·¨Îªµç½â¾«Á¶£¨ÌîдÃû³Æ£©£®
£¨4£©Îª²â¶¨²úÆ·ÖÐÂÌ·¯µÄÖÊÁ¿·ÖÊý£¬³ÆÈ¡30.000gÑùÆ·ÈÜÓÚË®Åä³É250mLÈÜÒº£¬È¡25.00mLÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬ÓÃ0.1000mol•L-1ËáÐÔKMnO4ÈÜÒº½øÐе樣¬·´Ó¦Îª£º10FeSO4+8H2SO4+2KMnO4=2MnSO4+5Fe2£¨SO4£©3+K2SO4+8H2O£®ÊµÑéËùµÃÊý¾ÝÈçϱíËùʾ£º
µÎ¶¨´ÎÊý | 1 | 2 | 3 | 4 |
KMnO4ÈÜÒºÌå»ý/mL | 20.90 | 20.02 | 19.98 | 20.00 |
a£®ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºóδÓñê×¼ÒºÈóÏ´
b£®×¶ÐÎÆ¿Ï´¾»ºóδ¸ÉÔï
c£®µÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý
d£®µÎ¶¨ÖÕµãʱÑöÊÓ¶ÁÊý
¢Ú¸ù¾Ý±íÖÐÊý¾Ý£¬¼ÆËãËùµÃ²úÆ·ÖÐÂÌ·¯µÄÖÊÁ¿·ÖÊýΪ92.7%£®
A£® | µªÑõ»¯ÎïÊÇÐγɹ⻯ѧÑÌÎíºÍËáÓêµÄÒ»¸öÖØÒªÔÒò | |
B£® | ²£Á§¸ÖÊǸÖÓë²£Á§ÏËάÐγɵĸ´ºÏ²ÄÁÏ | |
C£® | ʯÓÍ»¯¹¤·ÏÆúÎïÒ×Ôì³ÉË®Ì帻ӪÑø»¯ | |
D£® | ÁòËáï§Êdz£ÓõÄÒ»ÖÖÏõ̬µª·Ê |
A£® | ÓÃʳ´×ÇåÏ´ÈÈˮƿÖеÄË®¹¸ | |
B£® | ¹¤³§Öг£Óõľ²µç³ý³¾×°ÖÃÊǸù¾Ý½ºÌå´øµçÕâ¸öÐÔÖʶøÉè¼ÆµÄ | |
C£® | ÇâÑõ»¯ÂÁ¿ÉÓÃÓÚÖÎÁÆθËá¹ý¶à | |
D£® | ΪÁËÑÓ³¤¹ûʵ»ò»¨¶äµÄ³ÉÊìÆÚ£¬¿ÉÓýþÅݹý¸ßÃÌËá¼ØÈÜÒºµÄ¹èÍÁÎüÊÕË®¹û»ò»¨¶ä²úÉúµÄÒÒÏ© |
A£® | ¸ÃÈÜÒºÖÐÇâÀë×ÓµÄŨ¶È£ºc£¨H+£©=1¡Á10-11mol•L | |
B£® | pH=7µÄNH4ClºÍNH3•H2OµÄ»ìºÏÈÜÒº£ºc£¨Cl-£©£¾c£¨NH4+£©£¾c£¨H+£©£¾c£¨H+£©=c£¨OH-£© | |
C£® | Ũ¶È¾ùΪ0.1mol•L-1µÄNH3•H2OºÍNH4ClÈÜÒºµÈÌå»ý»ìºÏºóµÄ¼îÐÔÈÜÒºÖУºc£¨NH4+£©£¾c£¨Cl-£©£¾c£¨NH3•H2O£©£¾c£¨OH-£©£¾c£¨H+£© | |
D£® | 0.1mol•L-1µÄ°±Ë®Óë0.05molµÄH2SO4ÈÜÒºµÈÌå»ý»ìºÏºóËùµÃÈÜÒºÖУº2c£¨NH4+£©+2c£¨NH3•H2O£©=c£¨SO42-£© |
A£® | 1 mol Cl2²Î¼Ó·´Ó¦×ªÒƵç×ÓÊýÒ»¶¨Îª2NA | |
B£® | ÔÚ·´Ó¦KIO3+6HI¨TKI+3I2+3H2OÖУ¬Ã¿Éú³É3 mol I2תÒƵĵç×ÓÊýΪ6NA | |
C£® | Na2O2ÓëCO2·´Ó¦Éú³É±ê¿öÏÂ11.2L O2£¬·´Ó¦¹ý³ÌÖÐתÒƵç×ÓÊý1 NA | |
D£® | ·´Ó¦14CuSO4+5FeS2+12H2O=7Cu2S+5FeSO4+12H2SO4¸Ã·´Ó¦FeS2ÖеÄÁòÔªËØÈ«²¿±»Ñõ»¯ |