ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿×îÐÂÑÐÖƵÄÈËÔìÄÉÃס°Öñ×Ó¡±¿É³ä·ÖÀûÓÃÌ«ÑôÄÜ£¬²¢½«ÆäÓÐЧת»¯ÎªÇâÄÜÔ´¡£¡°Öñ×Ó¡±µÄÖñ½ÚºÍÖñ¾¥£¬·Ö±ðÓÉÁò»¯ïÓºÍÁò»¯Ð¿Á½ÖÖ²»Í¬µÄ°ëµ¼Ìå²ÄÁÏ×é³É¡£

£¨1£©ÒÑÖªCdÓëZnλÓÚͬһ¸±×壬ÇÒÔÚZnµÄÏÂÒ»ÖÜÆÚ£¬ÔòCdµÄ¼Ûµç×ÓÅŲ¼Í¼Îª___¡£ÒÑÖªZnÓëCuµÄÄÜÁ¿±ä»¯ÈçͼËùʾ£¬ÊÔ½âÊͲ½Öè¢ÚÎüÊÕÄÜÁ¿´óÓÚ²½Öè¢ÜµÄÔ­Òò___¡£

£¨2£©CdSȼÉÕ¿ÉÉú³ÉCdOºÍSO2£¬SO2ÔÚ¿ÕÆøÖÐÓöµ½Î¢³¾»á»ºÂýת»¯ÎªSO3¡£SO2ÖÐSµÄÔÓ»¯·½Ê½Îª___£¬SO3·Ö×ӵĿռ乹ÐÍΪ___¡£

£¨3£©O¡¢S¡¢SeΪͬÖ÷×åÔªËØ£¬ÒÑÖªÆä¶ÔÓ¦Ç⻯ÎïµÄÏà¹ØÊý¾ÝÈçÏÂ±í£º

¢ÙH2Se·Ðµã¸ßÓÚH2SµÄÔ­ÒòΪ___¡£

¢ÚH2OµÄ·Ö½âζȸßÓÚH2SµÄÔ­ÒòΪ___¡£

£¨4£©ÒÑÖªZnSÈÛµãΪ2830¡æ£»CdSÈÛµãΪ1750¡æ£¬Ôò¶þÕßÊôÓÚ___¾§Ì壬ZnSÈÛµã¸ü¸ßµÄÔ­ÒòΪ___¡£

£¨5£©ÈçͼΪZnSµÄ¾§°ûͼ¡£ÈôÒÑÖª×î½üµÄпÀë×ÓÖ®¼ä¾àÀëΪapm£¬ÔòZnS¾§ÌåµÄÃܶÈΪ___g¡¤cm-3£¨Áгö¼ÆËãʽ¼´¿É£©¡£

¡¾´ð°¸¡¿ Cu+µÄ3d¹ìµÀΪȫÂú½á¹¹£¬ÄÜÁ¿µÍ£¬¸üÎȶ¨ sp2 ƽÃæÈý½ÇÐÎ H2SeºÍH2S¾ùΪ·Ö×Ó¾§Ì壬H2SÏà¶ÔÔ­×ÓÖÊÁ¿Ð¡£¬·Ö×Ó¼ä×÷ÓÃÁ¦Ð¡£¬·ÐµãµÍ Ô­×Ӱ뾶£ºO£¼S£¬¹Ê¹²¼Û¼üO-H¼üµÄ¼ü³¤Ð¡ÓÚS-H¼ü£¬¼üÄÜ£ºO-H£¾S-H£¬Òò´ËH2OµÄÎȶ¨ÐÔ¸ßÓÚH2S£¬·Ö½âζȸü¸ß£¨»òO-H¼ü±ÈS-H¼üÎȶ¨£¬H2OµÄ·Ö½âζȱÈH2S¸ü¸ß£¬»òÆäËûºÏÀí´ð°¸£© Àë×Ó ÓÉÓÚÀë×Ӱ뾶Cd2+´óÓÚZn2+£¬¹Ê¾§¸ñÄÜZnS£¾CdS£¬ÔòÈÛµãZnS£¾CdS

¡¾½âÎö¡¿

(1)ZnµÄ¼Û²ãµç×ÓÅŲ¼Ê½Îª3d104s2£¬¸ù¾ÝZnÓëCdÔÚÖÜÆÚ±íµÄλÖùØϵ·ÖÎö£»È«Âú״̬ʱ½á¹¹Îȶ¨£¬ÔÙʧȥһ¸öµç×ÓʱÐèÒªµÄÄÜÁ¿¸ü¶à£»

(2)ÀûÓÃÖÐÐÄÔ­×Ó¼Ûµç×Ó¶Ô»¥³âÀíÂÛ·ÖÎöÔÓ»¯ÀàÐͺͿռ乹ÐÍ£»

(3)¢Ù·Ö×Ó¾§ÌåµÄÏà¶Ô·Ö×ÓÖÊÁ¿Ô½´óÈ۷еãÔ½¸ß£»

¢Ú°ë¾¶Ô½Ð¡£¬¼ü³¤Ô½¶Ì£¬¼üÄÜÔ½´ó·ÖÎö£»

(4)¸ù¾ÝÈ۷еãÅжϾ§ÌåÀàÐÍ£»Àë×Ӱ뾶ԽС£¬¾§¸ñÄÜÔ½´ó£¬È۷еãÔ½¸ß£»

(5)¸ù¾Ý¾§°ûµÄ½á¹¹£¬ÀûÓÃÃܶȹ«Ê½¼ÆËã¡£

(1)ÓÉÓÚZnµÄ¼Û²ãµç×ÓÅŲ¼Ê½Îª3d104s2£¬¿É¼ûÓëZnͬһ¸±×åÇÒÔÚZnÏÂÒ»ÖÜÆÚµÄCdµÄ¼Ûµç×ÓÅŲ¼Îª4d105s2£»¼Ûµç×ÓÅŲ¼Í¼Îª£º£»ÓÉÄÜÁ¿±ä»¯Í¼¿ÉÖªI1(Cu)£¼I1(Zn)£¬I2(Cu)£¾I2(Zn)£¬ÓÉÓÚCu+µÄ3d¹ìµÀΪȫÂú½á¹¹£¬ÄÜÁ¿µÍ£¬½ÏÎȶ¨£¬¶øZn+µÄ4s¹ìµÀÓÐÒ»¸öµç×Ó£¬Cu+ÔÙʧȥһ¸öµç×ÓʱÐèÒªµÄÄÜÁ¿¸ü¶à£»

(2)SO2ÖÐSµÄµÄ¼Û²ãµç×Ó¶ÔÊý=2+(6-2¡Á2)=3£¬ËùÒÔÁòÔ­×Ó²ÉÓÃsp2ÔÓ»¯£¬SO3ÖÐSµÄµÄ¼Û²ãµç×Ó¶ÔÊý=3+(6-3¡Á2)=3£¬ËùÒÔÁòÔ­×Ó²ÉÓÃsp2ÔÓ»¯£¬¿Õ¼ä¹¹ÐÍΪƽÃæÈý½ÇÐΣ»

(3)¢ÙH2SeºÍH2S¾ùΪ·Ö×Ó¾§Ì壬ÓÉÓÚ·Ö×Ó¾§ÌåÏà¶ÔÔ­×ÓÖÊÁ¿H2Se£¾H2S£¬¹Ê·Ö×Ó¼ä×÷ÓÃÁ¦H2Se£¾H2S£¬Òò´Ë·ÐµãH2Se£¾H2S£»

¢ÚÓÉÓÚÔ­×Ӱ뾶£ºO£¼S£¬¹Ê¹²¼Û¼üO-H¼üµÄ¼ü³¤Ð¡ÓÚS-H¼ü£¬¼üÄÜ£ºO-H£¾S-H£¬Òò´ËH2OµÄÎȶ¨ÐÔ¸ßÓÚH2S£¬Ôò·Ö½âζȸßÓÚH2S£»

(4)ÓÉÓÚZnSÈÛµãΪ2830¡æ£»CdSÈÛµãΪ1750¡æ£¬¶¼±È½Ï¸ß£¬¹Ê¶¼ÊôÓÚÀë×Ó¾§Ì壻ÓÉÓÚÀë×Ӱ뾶Cd2+´óÓÚZn2+£¬¹Ê¾§¸ñÄÜZnS£¾CdS£¬ÔòÈÛµãZnS£¾CdS£»

(5)ÓÉÓÚ×î½üµÄZnÔ­×Ó¼äµÄ¾àÀ루¼´¶¥µãÓëÃæÐÄÖ®¼äµÄ¾àÀ룩Ϊapm£¬¹ÊÃæ¶Ô½ÇÏßΪ2apm£¬Ôò¾§°ûµÄ±ß³¤Îªapm=a¡Á10-10cm£¬¹Ê¾§°ûµÄÌå»ýΪ(a¡Á10-10)3cm3£¬¶ø¾§°ûÖÐZnÔ­×Ó¶¼ÔÚ¾§°ûµÄ¶¥µãºÍÃæÉÏ£¬ÊýÄ¿ÓÐ8¡Á+6¡Á=4¸ö£¬SÔ­×ÓÔÚ¾§°ûµÄÄÚ²¿£¬¹²ÓÐ4¸ö£¬¼´Ò»¸ö¾§°ûÖÐÓÐ4¸öZnS£¬ÔòÖÊÁ¿Îª4¡Ág£¬¹ÊÃܶÈΪg¡¤cm-3¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÑÇÏõõ£ÂÈ(ClNO)ÊÇÓлúÎïºÏ³ÉÖеÄÖØÒªÊÔ¼Á£¬Æä·ÐµãΪ£­5.5¡æ£¬Ò×Ë®½â¡£ÒÑÖª£ºAgNO2΢ÈÜÓÚË®£¬ÄÜÈÜÓÚÏõËᣬAgNO2+HNO3=AgNO3 +HNO2£¬Ä³Ñ§Ï°Ð¡×éÔÚʵÑéÊÒÓÃCl2ºÍNOÖƱ¸ClNO²¢²â¶¨Æä´¿¶È£¬Ïà¹ØʵÑé×°ÖÃÈçͼËùʾ¡£

(1)ÖƱ¸Cl2µÄ·¢Éú×°ÖÿÉÒÔÑ¡ÓÃ___________(Ìî×Öĸ´úºÅ)×°Ö㬷¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________¡£

(2)ÓûÊÕ¼¯Ò»Æ¿¸ÉÔïµÄÂÈÆø£¬Ñ¡ÔñºÏÊʵÄ×°Öã¬ÆäÁ¬½Ó˳ÐòΪ a¡ú________________(°´ÆøÁ÷·½Ïò£¬ÓÃСд×Öĸ±íʾ)¡£

(3)ʵÑéÊÒ¿ÉÓÃͼʾװÖÃÖƱ¸ÑÇÏõõ£ÂÈ£º

¢ÙʵÑéÊÒÒ²¿ÉÓà B ×°ÖÃÖƱ¸ NO £¬ X×°ÖõÄÓŵãΪ___________________________________¡£

¢Ú¼ìÑé×°ÖÃÆøÃÜÐÔ²¢×°ÈëÒ©Æ·£¬´ò¿ªK2£¬È»ºóÔÙ´ò¿ªK3£¬Í¨ÈëÒ»¶Îʱ¼äÆøÌ壬ÆäÄ¿µÄÊÇ____________£¬È»ºó½øÐÐÆäËû²Ù×÷£¬µ±ZÖÐÓÐÒ»¶¨Á¿ÒºÌåÉú³Éʱ£¬Í£Ö¹ÊµÑé¡£

(4)ÒÑÖª£ºClNO ÓëH2O·´Ó¦Éú³ÉHNO2ºÍ HCl¡£

¢ÙÉè¼ÆʵÑéÖ¤Ã÷ HNO2ÊÇÈõË᣺_____________¡£(½öÌṩµÄÊÔ¼Á£º1 molL-1ÑÎËá¡¢ 1 molL-1HNO2ÈÜÒº¡¢ NaNO2ÈÜÒº¡¢ºìɫʯÈïÊÔÖ½¡¢À¶É«Ê¯ÈïÊÔÖ½)¡£

¢Úͨ¹ýÒÔÏÂʵÑé²â¶¨ClNOÑùÆ·µÄ´¿¶È¡£È¡ZÖÐËùµÃÒºÌåm g ÈÜÓÚË®£¬ÅäÖƳÉ250 mL ÈÜÒº£»È¡³ö25.00 mLÑùÆ·ÈÜÓÚ׶ÐÎÆ¿ÖУ¬ÒÔK2CrO4ÈÜҺΪָʾ¼Á£¬ÓÃc molL-1 AgNO3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ±ê×¼ÈÜÒºµÄÌå»ýΪ20.00mL¡£µÎ¶¨ÖÕµãµÄÏÖÏóÊÇ__________£¬ÑÇÏõõ£ÂÈ(ClNO)µÄÖÊÁ¿·ÖÊýΪ_________¡£(ÒÑÖª£º Ag2CrO4ΪשºìÉ«¹ÌÌ壻 Ksp(AgCl)£½1.56¡Á10-10£¬Ksp(Ag2CrO4)£½1¡Á10-12)

¡¾ÌâÄ¿¡¿µªÑõ»¯ÎNOx£©Ôì³ÉËáÓê¡¢¹â»¯Ñ§ÑÌÎí¡¢³ôÑõ²ãÆÆ»µµÈΣº¦£¬²»½öÆÆ»µ×ÔÈ»Éú̬»·¾³£¬¶øÇÒÑÏÖØΣº¦ÈËÀཡ¿µ¡£²ÉÓúÏÊʵĻ¹Ô­¼ÁÄܹ»ÊµÏÖÑÌÆøµÄ¸ßЧÍÑÏõ¡£

£¨1£©»îÐÔÌ¿»¹Ô­ÍÑÏõ¿É·ÀÖ¹µªÑõ»¯ÎïÎÛȾ£¬ÒÑÖª£º

¢ÙN2£¨g£©+O2£¨g£©¨T2NO£¨g£© ¡÷H1=+180.5kJmol-1

¢Ú2C£¨s£©+O2£¨g£©¨T2CO£¨g£© ¡÷H2=-221.0kJmol-1

¢Û2CO£¨g£©+O2£¨g£©¨T2CO2£¨g£© ¡÷H3=-556.0kJmol-1

Ôò·´Ó¦C£¨s£©+2NO£¨g£©N2£¨g£©+CO2£¨g£©¡÷H=__kJmol-1

£¨2£©ÓûîÐÔÌ¿¶ÔNO½øÐл¹Ô­£¬²ÉÓÃÏàͬÖÊÁ¿²»Í¬Á£¾¶µÄͬÖÖ´ß»¯¼ÁMºÍN£¬²âÁ¿Ïàͬʱ¼äÄÚÑÌÆøµÄÍѵªÂÊ£¬½á¹ûÈçͼËùʾ¡£

¢ÙÔÚM¡¢NÁ½ÖÖ²»Í¬Á£¾¶´ß»¯¼Á×÷ÓÃÏ£¬³öÏÖMºÍNÁ½Ìõ²»Í¬µÄÍѵªÂÊÓëζȵı仯ÇúÏßµÄÔ­ÒòÊÇ___¡£

¢ÚÅжÏMÇúÏßÉÏ×î¸ßµãAµã¶ÔÓ¦µÄÍѵªÂÊ__£¨Ìî¡°ÊÇ¡°»ò¡°²»ÊÇ¡±£©¸ÃζÈϵÄƽºâÍѵªÂÊ¡£

¢Û25¡æÏ£¬ÓÃNaOHÈÜÒº×÷²¶Ìá¼ÁÎüÊÕ²úÉúµÄCO2£¬²»½ö¿ÉÒÔ½µµÍ̼ÅÅ·Å£¬¶øÇҿɵõ½ÖØÒªµÄ»¯¹¤²úÆ·¡£Ä³´Î²¶×½ºóµÃµ½pH=12µÄÈÜÒº£¬ÒÑÖª£º25¡æÏÂKa2£¨H2CO3£©=5.6¡Á10-11£¬ÊÔͨ¹ý¼ÆËãÈÜÒºÖÐc£¨CO32-£©£ºc£¨HCO3-£©=__¡£

£¨3£©ÔÚÒ»ºãÈÝÃܱÕÈÝÆ÷ÖУ¬Ê¹ÓÃijÖÖ´ß»¯¼Á¶Ô·´Ó¦NO2£¨g£©+SO2£¨g£©SO3£¨g£©+NO£¨g£©¡÷H£¼0½øÐÐÏà¹ØʵÑé̽¾¿¡£¸Ä±äͶÁϱȣ¨n0£¨SO2£©£ºn0£¨NO2£©]½øÐжà×éʵÑ飨¸÷´ÎʵÑéµÄζȿÉÄÜÏàͬ£¬Ò²¿ÉÄܲ»Í¬£©£¬²â¶¨SO2µÄƽºâת»¯ÂÊ[¦Á£¨SO2£©]¡£ÒÑÖª£ºKR=16£¬KZ=1£¬²¿·ÖʵÑé½á¹ûÈçͼËùʾ¡£

¢ÙÈç¹ûÒª½«Í¼ÖÐRµãµÄƽºâ״̬¸Ä±äΪXµãµÄƽºâ״̬£¬Ó¦²ÉÈ¡µÄ´ëÊ©ÊÇ__¡£

¢ÚͼÖÐR¡¢X¡¢Y¡¢ZËĵã¶ÔÓ¦µÄʵÑéζȷֱðΪTR¡¢TX¡¢TY£¬TZͨ¹ý¼ÆËãÑ¡ÔñÏÂÁÐÑ¡ÏîÖÐÁ½¸öζÈÏàµÈÊÇ__£¨Ìî±êºÅ£©¡£

A£®TRºÍTY B£®TRºÍTZ C£®TXºÍTZ D£®TYºÍTZ

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

¾«Ó¢¼Ò½ÌÍø