ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÑÇÏõõ£ÂÈ(ClNO)³£ÓÃ×÷´ß»¯¼ÁºÍºÏ³ÉÏ´µÓ¼Á£¬Æä·ÐµãΪ-5.5¡æ£¬Ò×Ë®½âÉú³ÉHNO2ºÍHCl¡£Ä³Ñ§Ï°Ð¡×éÔÚʵÑéÊÒÖÐÓÃÏÂͼËùʾװÖÃÖƱ¸ClNO¡£ÒÑÖª: HNO2¼ÈÓÐÑõ»¯ÐÔÓÖÓл¹Ô­ÐÔ£»AgNO2΢ÈÜÓÚË®£¬ÄÜÈÜÓÚÏõËᣬAgNO2+HNO3=AgNO3+HNO2¡£

»Ø´ðÏÂÁÐÎÊÌâ

£¨1£©ÒÇÆ÷aµÄÃû³ÆΪ_____________£¬ aÔÚ×°ÖÃCÖеÄ×÷ÓÃÊÇ____________¡£

£¨2£©×°ÖÃBµÄ×÷ÓÃÊÇ____________¡£

£¨3£©ÊµÑ鿪ʼʱ£¬ÏÈ´ò¿ªK1¡¢K2£¬¹Ø±ÕK3£¬ÔÙ´ò¿ª·ÖҺ©¶·»îÈûµÎÈëÊÊÁ¿Ï¡ÏõËᣬµ±¹Û²ìµ½CÖÐ______ʱ¹Ø±ÕK1¡¢K2¡£Ïò×°ÖÃDÈý¾±Æ¿ÖÐͨÈë¸ÉÔï´¿¾»Cl2£¬µ±Æ¿ÖгäÂú»ÆÂÌÉ«ÆøÌåʱ£¬ÔÙ´ò¿ªK1¡¢K3£¬ÖƱ¸ClNO¡£

£¨4£©×°ÖÃDÖиÉÔï¹ÜµÄ×÷ÓÃÊÇ__________________¡£

£¨5£©ÊµÑé¹ý³ÌÖУ¬ÈôѧϰС×éͬѧÓþƾ«µÆ´ó»ð¼ÓÈÈÖÆÈ¡NO£¬¶Ô±¾ÊµÑéÔì³ÉµÄ²»ÀûÓ°Ïì³ýÁËÓз´Ó¦ËÙÂʹý¿ì£¬Ê¹NOÀ´²»¼°·´Ó¦¼´´óÁ¿ÒݳöÍ⣬»¹¿ÉÄÜÓУº___________¡£

£¨6£©¢ÙÒªÑéÖ¤ClNOÓëH2O·´Ó¦ºóµÄÈÜÒºÖдæÔÚCl-ºÍHNO2£¬ºÏÀíµÄ²Ù×÷²½Öè¼°ÕýÈ·µÄ˳ÐòÊÇ_____(ÌîÐòºÅ)¡£

a.ÏòÉÕ±­ÖеμӹýÁ¿KIµí·ÛÈÜÒº£¬ÈÜÒº±äÀ¶É«

b.È¡1.0mL Èý¾±Æ¿ÖвúÆ·ÓÚÉÕ±­ÖУ¬¼ÓÈë10.0mLH2O³ä·Ö·´Ó¦

c.ÏòÉÕ±­ÖеμÓËáÐÔKMnO4ÈÜÒº£¬ÈÜÒº×ÏÉ«ÍÊÈ¥

d.ÏòÉÕ±­ÖеμÓ×ãÁ¿AgNO3ÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬¼ÓÈëÏ¡ÏõËᣬ½Á°è£¬ÈÔÓа×É«³Áµí

¢ÚÅäÖÆÍõË®µÄ¹ý³ÌÖлᷴӦ²úÉúClNO£¬Çëд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________¡£

¡¾´ð°¸¡¿ ³¤¾±Â©¶· ƽºâÆøѹ£¬±ÜÃâCÖÐѹǿ¹ý´ó ³ýÈ¥NOÖеÄHNO3¡¢NO2ÆøÌå ºì×ØÉ«ÍêÈ«Ïûʧ ·ÀֹˮÕôÆø½øÈëÈý¾±Æ¿ÖÐÓëClNO·´Ó¦ ζȹý¸ßÔì³ÉHNO3·Ö½â£¨»ò»Ó·¢£© bdc HNO3+ 3HCl== 2H2O + Cl2+ ClNO

¡¾½âÎö¡¿ÊµÑ鿪ʼʱ£¬ÏÈ´ò¿ªK1¡¢K2£¬¹Ø±ÕK3£¬´ò¿ª·ÖҺ©¶·»îÈûµÎÈëÊÊÁ¿Ï¡ÏõËᣬÖÁCÖкì×ØÉ«ÍêÈ«Ïûʧºó£¬¹Ø±ÕK1¡¢K2£¬×°ÖÃAÊÇÏ¡ÏõËáºÍÍ­·´Ó¦Éú³ÉNOÆøÌ壬ͨ¹ý×°ÖÃBÖÐË®³ýÈ¥»Ó·¢µÄÏõËá¼°NOºÍ¿ÕÆøÖÐÑõÆø·´Ó¦Éú³ÉµÄ¶þÑõ»¯µª£¬Ê¹´¿NO½øÈë×°ÖÃC£¬´ËʱװÖÃCµÄ×÷ÓÃΪ´¢´æAÖвúÉúµÄNOÆøÌ壬ÏòD×°ÖÃÖÐͨÈë¸ÉÔï´¿¾»µÄCl2£¬µ±DÖÐÖгäÂú»ÆÂÌÉ«ÆøÌåʱ£¬´ò¿ªK1¡¢K2£¬ÔÚDÖÐÖƱ¸ClNO£¬×°ÓÐÎÞË®CaCl2µÄ¸ÉÔï¹ÜÆä×÷ÓÃΪ·ÀֹˮÕôÆø½øÈëDÖУ¬Ê¹NOC1Ë®½â£¬·´Ó¦½áËÙºó´ò¿ªK3£¬½øÐÐβÆø´¦Àí£¬·ÀÖ¹ÎÛȾ¿ÕÆø¡£

(1)ÒÇÆ÷aΪ³¤¾±Â©¶·£»Èô¹Ø±ÕK1£¬Ëæ×Å·´Ó¦µÄ½øÐУ¬CÖÐѹǿÔö¼Ó£¬³¤¾±Â©¶·µÄ×÷ÓÃΪƽºâϵͳÄÚÍâѹǿ£¬±ÜÃâCÖÐѹǿ¹ý´ó£¬¹Ê´ð°¸Îª£º³¤¾±Â©¶·£»Æ½ºâÆøѹ£¬±ÜÃâCÖÐѹǿ¹ý´ó£»

(2)NO²»ÈÜÓÚË®£¬¶ø¿ÉÄÜ´æÔÚµÄÔÓÖÊHNO3¡¢NO2¾ùÒ×ÈÜÓÚË®£¬¹ÊBµÄ×÷ÓÃΪ³ýÈ¥NOÖеÄHNO3¡¢NO2µÈÔÓÖÊ£¬¹Ê´ð°¸Îª£º³ýÈ¥NOÖеÄHNO3¡¢NO2ÆøÌ壻

(3)ΪÁ˵õ½±È½Ï´¿¾»µÄNO£¬µ±CÖкì×ØÉ«ÍêÈ«Ïûʧʱ£¬²»ÔÙ´æÔÚNO2ÆøÌ壬¹Ê´ð°¸Îª£ººì×ØÉ«ÍêÈ«Ïûʧ£»

(4)ÒòΪÑÇÏõõ£ÂÈ(ClNO)Ò×ÓëË®·´Ó¦Ë®½â£¬ËùÒÔ±ØÐë·ÀÖ¹ÓÐË®ÕôÆø½øÈëÈý¾±ÉÕÆ¿ÖÐÓëClNO·´Ó¦£¬×°ÖÃDÖиÉÔï¹Ü¿ÉÒÔ·ÀֹˮÕôÆø½øÈëÈý¾±Æ¿ÖÐÓëClNO·´Ó¦£¬¹Ê´ð°¸Îª£º·ÀֹˮÕôÆø½øÈëÈý¾±Æ¿ÖÐÓëClNO·´Ó¦£»

(5)¶ÔÕôÁóÉÕÆ¿´ó»ð¼ÓÈÈ£¬»áʹµÃ·´Ó¦Ñ¸ËÙ£¬´óÁ¿Éú²úNOÆøÌ壬»áʹ½Ï¶àµÄNOÆøÌåδ²Î¼Ó·´Ó¦±ãÒݳöÖÁ¿ÕÆøÖУ»Í¬Ê±£¬½Ï¸ßµÄζȿÉÄÜÊÇÏõËá·Ö½â»ò»Ó·¢£¬²úÉú½Ï¶àÔÓÖÊÆøÌ壬¹Ê´ð°¸Îª£ºÎ¶ȹý¸ßÔì³ÉHNO3·Ö½â(»ò»Ó·¢)£»

(6)¢ÙÊ×ÏÈҪʹClNOÓëH2O·´Ó¦£¬Ñ¡Ôñb£»ÒòΪµâ»¯ÒøΪ³Áµí£¬¶øËáÐÔ¸ßÃÌËá¼ØÄܽ«ÂÈÀë×ÓÑõ»¯£¬ËùÒÔÐèÒªÏÈÑéÖ¤ÂÈÀë×ӵĴæÔÚ£¬Ñ¡Ôñd£»ÓÉÓÚÒýÈëÁË×ãÁ¿ÒøÀë×Ó£¬ËùÒÔ½öÄÜʹÓÃËáÐÔ¸ßÃÌËá¼ØÑéÖ¤ÑÇÏõËᣬѡÔñc£»¹Ê´ð°¸Îª£ºbdc£»

¢ÚÅäÖÆÍõË®µÄ¹ý³ÌÖлᷴӦ²úÉúClNO£¬ÊÇÏõËáÓëÑÎËá·´Ó¦Éú³ÉÁËClNO£¬ClNOÖÐClÔªËØΪ-1¼Û£¬NÔªËØΪ+3¼Û£¬¸ù¾Ý»¯ºÏ¼ÛÉý½µÊغ㣬ÓÐÂÈÆøÉú³É£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪHNO3+ 3HCl== 2H2O + Cl2+ ClNO£¬¹Ê´ð°¸Îª£ºHNO3+ 3HCl== 2H2O + Cl2+ ClNO¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿SO2µÄº¬Á¿ÊǺâ×î´óÆøÎÛȾµÄÒ»¸öÖØÒªÖ¸±ê£¬¹¤ÒµÉϳ£²ÉÓô߻¯»¹Ô­·¨»òÎüÊÕ·¨´¦ÀíSO2¡£ÀûÓô߻¯»¹Ô­SO2²»½ö¿ÉÏû³ýSO2ÎÛȾ£¬¶øÇҿɵõ½Óо­¼Ã¼ÛÖµµÄµ¥ÖÊS¡£

£¨1£©ÔÚ¸´ºÏ×é·Ö´ß»¯¼Á×÷ÓÃÏ£¬CH4¿ÉʹSO2ת»¯ÎªS£¬Í¬Ê±Éú³ÉCO2ºÍH2O¡£

¼ºÖªCH4ºÍSµÄȼÉÕÈÈ(¡÷H)·Ö±ðΪ-890.3k/molºÍ-297.2kJ/mol£¬ÔòCH4ºÍSO2·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_____________¡£

£¨2£©ÓÃH2»¹Ô­SO2Éú³ÉSµÄ·´Ó¦·ÖÁ½²½Íê³É£¬Èçͼ1Ëùʾ£¬¸Ã¹ý³ÌÖÐÏà¹ØÎïÖʵÄÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼äµÄ±ä»¯¹ØϵÈçͼ2Ëùʾ:

¢Ù·ÖÎö¿ÉÖªXΪ______(д»¯Ñ§Ê½)£¬0¡«t1ʱ¼ä¶ÎµÄζÈΪ_____£¬0¡«t1ʱ¼ä¶ÎÓÃSOz±íʾµÄ»¯Ñ§·´Ó¦ËÙÂÊΪ________¡£

¢Ú×Ü·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________¡£

£¨3£©½¹Ì¿´ß»¯»¹Ô­SO2Éú³ÉS2,»¯Ñ§·½³ÌʽΪ:2C(s)+2SO2(g)S2(g)+2CO2(g)£¬ºãÈÝÈÝÆ÷ÖУ¬1mol/LSO2Óë×ãÁ¿µÄ½¹Ì¿·´Ó¦£¬SO2µÄת»¯ÂÊËæζȵı仯Èçͼ3Ëùʾ¡£

¢Ù¸Ã·´Ó¦µÄ¡÷H____0(Ìî¡°>¡±»ò¡°<¡±)

¢ÚËãaµãµÄƽºâ³£ÊýΪ_________¡£

£¨4£©¹¤ÒµÉÏ¿ÉÓÃNa2SO3ÈÜÒºÎüÊÕ·¨´¦ÀíSO2,25¡æʱÓÃ1mo/LµÄNa2SO3ÈÜÒºÎüÊÕSO2¡£µ±ÈÜÒºpH=7ʱ£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈµÄ´óС¹ØϵΪ________¡£ÒÑÖª:H2SO3µÄµçÀë³£ÊýK1=1.3¡Á10-2£¬K2=6.2¡Á10-8

¡¾ÌâÄ¿¡¿¡°¸ÖÊÇ»¢£¬·°ÊÇÒí£¬¸Öº¬·°ÓÌÈ绢ÌíÒí¡±£¬½ðÊô·°±»ÓþΪ¡°ºÏ½ðµÄάÉúËØ¡±¡£´Ó·Ï·°(Ö÷Òª³É·ÖΪV2O5¡¢Fe2O3¡¢SiO2µÈ)ÖлØÊÕV2O5µÄÒ»ÖÖ¹¤ÒÕÁ÷³ÌÈçÏÂͼËùʾ:

ÒÑÖª:²½Öè¢Ú¡¢¢ÛÖеı仯¹ý³Ì¿É¼ò»¯Îª:Rn+(Ë®²ã)+nHA(Óлú²ã)RAn(Óлú²ã)+nH+(Ë®²ã)(ʽÖÐRn+±íʾVO2+»òFe3+£¬HA±íʾÓлúÝÍÈ¡¼Á)¡£

»Ø´ðÏÂÁÐÎÊÌâ:

£¨1£©²½Öè¢ÙËá½þ¹ý³ÌÖз¢ÉúÑõ»¯»¹Ô­·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________¡£

£¨2£©ÝÍȡʱӦ¼ÓÈëÊÊÁ¿¼îµÄ×÷ÓÃÊÇ___________________¡£

£¨3£©²½Öè¢ÜÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________¡£

£¨4£©²½Öè¢Ý25¡ãCʱ£¬È¡Ñù½øÐÐÊÔÑé·ÖÎö£¬µÃµ½·°³ÁµíÂʺÍÈÜÒºpHÖ®¼ä¹ØϵÈçÏÂ±í£º

pH

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

2.1

·°³ÁµíÂÊ%

88.1

94.8

96.5

98.0

98.8

98.8

96.4

93.1

89.3

ͨ¹ý±íÖÐÊý¾Ý·ÖÎö£¬ÔÚʵ¼ÊÉú²úÖУ¬¢ÝÖмÓÈ백ˮ£¬µ÷½ÚÈÜÒºµÄ×î¼ÑpH·¶Î§Îª______£»

Èô¼ÓÈ백ˮµ÷½ÚÈÜÒºpH=2£¬·°³ÁµíÂÊ´ïµ½93%ÇÒ²»²úÉúFe(OH)3³Áµí£¬Ôò´ËʱÈÜÒºÖÐc(Fe3+)<_____mol/L(°´25¡æ¼ÆË㣬25¡æʱKsp[Fe(OH)3]=2.6¡Á10-39)¡£

£¨5£©V2O5ÊÇÁ½ÐÔÑõ»¯ÎÔÚÇ¿ËáÐÔÈÜÒºÖÐÒÔVO2+ÐÎʽ´æÔÚ£¬VO2+¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܽ«I-Ñõ»¯ÎªI2£¬±¾Éí±»»¹Ô­ÎªVO+£¬ÔòV2O5ÓëÇâµâËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________¡£

£¨6£©ÎªÌá¸ß·°µÄ»ØÊÕÂÊ£¬²½Öè¢ÚºÍ¢ÛÐè¶à´Î½øÐУ¬¼ÙÉèËá½þËùµÃ¡°Ç¿ËáÐÔ½þ³öÒº¡±ÖÐc(VO2+)=amol/L£¬²½Öè¢ÚºÍ¢Ûÿ½øÐÐÒ»´Î£¬VO2+ÝÍÈ¡ÂÊΪ80%£¬4´Î²Ù×÷ºó£¬¡°Ç¿ËáÐÔ½þ³öÒºÖС±c(VO2+)=_______mol/L(ÝÍÈ¡ÂÊ=)

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

¾«Ó¢¼Ò½ÌÍø