(10·Ö)Ô­×ÓÐòÊýÖ®ºÍΪ16µÄÈýÖÖ¶ÌÖÜÆÚÔªËصĵ¥ÖÊX¡¢Y¡¢Z£¬³£Î³£Ñ¹Ï¾ùΪÎÞÉ«ÆøÌ壬ÔÚÊʵ±Ìõ¼þÏÂX¡¢Y¡¢ZÖ®¼ä¿ÉÒÔ·¢ÉúÈçͼËùʾµÄ±ä»¯¡£

ÒÑÖªB·Ö×Ó×é³ÉÖÐZÔ­×Ó¸öÊý±ÈC·Ö×ÓÖÐÉÙÒ»¸ö¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1) ÔªËØXλÓÚ          ÖÜÆÚ           ×å
(2) ÔªËØYµÄÔ­×ӽṹʾÒâͼ                         
(3) Óõç×Óʽ±íʾBµÄÐγɹý³Ì£º                                        
(4) BÓëCµÄÎȶ¨ÐÔ´óС˳ÐòΪ                                   £¨Óû¯Ñ§Ê½±íʾ£©
(5) CÓëXÔÚÒ»¶¨Ìõ¼þÏÂÉú³É»¯ºÏÎïAµÄ»¯Ñ§·½³Ìʽ                              

(1)   µÚ¶þÖÜÆÚ¢öA×å   £¨2)
(3)
(4) H2O£¾ NH3             (5) 4NH3+5O24NO+6H2O

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?ÀÖƽÊÐÄ£Ä⣩2008Äê9ÔÂ25ÈÕ21ʱ10·Ö£¬¡°ÉñÆß¡±Ë³ÀûÉý¿Õ£®·É´¬µÄÌ«ÑôÄܵç³Ø°åÓС°·É´¬ÑªÒº¡±Ö®³Æ£¬Ëü¿É½«Ì«ÑôÄÜÖ±½Óת»¯ÎªµçÄÜ£¬ÎÒ¹úÔÚÉ黯ïØÌ«ÑôÄܵç³ØÑо¿·½Ãæ¹ú¼ÊÁìÏÈ£®É飨As£©ºÍïØ£¨Ga£©¶¼ÊǵÚËÄÖÜÆÚÔªËØ£¬·Ö±ðÊôÓÚ¢õAºÍ¢óA×壮ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄê¸ÊËàÊ¡¸ßÈý10ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨°Â°à£©£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(10·Ö) Öйú¹Å´úËÄ´ó·¢Ã÷Ö®Ò»¡ª¡ªºÚ»ðÒ©£¬ËüµÄ±¬Õ¨·´Ó¦Îª£º

2KNO3£«3C£«SA£«N2¡ü£«3CO2¡ü(ÒÑÅäƽ)

£¨1£©¢Ù³ýSÍ⣬ÉÏÁÐÔªËصĵ縺ÐÔ´Ó´óµ½Ð¡ÒÀ´ÎΪ______________________¡£

¢ÚÔÚÉú³ÉÎïÖУ¬AµÄ¾§ÌåÀàÐÍΪ___£¬º¬¼«ÐÔ¹²¼Û¼üµÄ·Ö×ÓµÄÖÐÐÄÔ­×Ó¹ìµÀÔÓ»¯ÀàÐÍΪ____¡£

¢ÛÒÑÖªCN£­ÓëN2ΪµÈµç×ÓÌ壬ÍÆËãHCN·Ö×ÓÖЦҼüÓë¦Ð¼üÊýÄ¿Ö®±ÈΪ________________¡£

£¨2£©Ô­×ÓÐòÊýСÓÚ36µÄÔªËØQºÍT£¬ÔÚÖÜÆÚ±íÖмȴ¦ÓÚͬһÖÜÆÚÓÖλÓÚͬһ×壬ÇÒÔ­×ÓÐòÊýT±ÈQ¶à2¡£TµÄ»ù̬ԭ×ÓÍâΧµç×Ó(¼Ûµç×Ó)ÅŲ¼Îª______£¬Q2£«µÄδ³É¶Ôµç×ÓÊýÊÇ_______¡£

£¨3£©Èôij½ðÊôµ¥Öʾ§ÌåÖÐÔ­×ӵĶѻý·½Ê½ÈçÏÂͼ¼×Ëùʾ£¬Æ侧°ûÌØÕ÷ÈçÏÂͼÒÒËùʾ£¬Ô­×ÓÖ®¼äÏ໥λÖùØϵµÄƽÃæͼÈçÏÂͼ±ûËùʾ¡£Ôò¾§°ûÖиÃÔ­×ÓµÄÅäλÊýΪ           £¬¸Ãµ¥Öʾ§ÌåÖÐÔ­×ӵĶѻý·½Ê½ÎªËÄÖÖ»ù±¾¶Ñ»ý·½Ê½ÖеĠ        £®

£¨4£©ÔÚCrCl3µÄË®ÈÜÒºÖУ¬Ò»¶¨Ìõ¼þÏ´æÔÚ×é³ÉΪ[CrCln(H2O)6£­n]x£«(nºÍx¾ùΪÕýÕûÊý)µÄÅäÀë×Ó£¬½«Æäͨ¹ýÇâÀë×Ó½»»»Ê÷Ö¬(R¡ªH)£¬¿É·¢ÉúÀë×Ó½»»»·´Ó¦£º[CrCln(H2O)6£­n]x£«£«xR¡ªH¨D¡úRx[CrCln(H2O)6£­n]x£«£«xH£«½»»»³öÀ´µÄH£«¾­Öк͵樣¬¼´¿ÉÇó³öxºÍn£¬È·¶¨ÅäÀë×ÓµÄ×é³É¡£½«º¬0.0015 mol [CrCln(H2O)6£­n]x£«µÄÈÜÒº£¬ÓëR¡ªHÍêÈ«½»»»ºó£¬ÖкÍÉú³ÉµÄH£«ÐèŨ¶ÈΪ0.1200mol¡¤L£­1NaOHÈÜÒº25.00 mL£¬¿ÉÖª¸ÃÅäÀë×ӵĻ¯Ñ§Ê½Îª______£¬ÖÐÐÄÀë×ÓµÄÅäλÊýΪ    ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêºÓ±±Ê¡¸ß¶þµÚ¶þѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

(10·Ö)Ô­×ÓÐòÊýÒÀ´ÎÔö´óµÄÎåÖÖ¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z¡¢Q¡¢R£¬ÒÑÖªXÓëQͬÖ÷×壬Y¡¢ZÔ­×ÓÐòÊýÖ®±ÈΪ3£º4£¬ZµÄ×îÍâ²ãµç×ÓÊýÊÇQµÄ×îÍâ²ãµç×ÓÊýµÄ6±¶£¬RΪͬÖÜÆÚÖÐÔ­×Ӱ뾶×îСµÄÔªËØ(Ï¡ÓÐÆøÌåÔªËسýÍâ)£¬XÓëRÐγɵĻ¯ºÏÎï³£ÎÂÏÂΪÆø̬¡£

£¨1£©±íʾYÔªËØÖÐÖÊ×ÓÊýÓëÖÐ×ÓÊýÏàµÈµÄͬλËØ·ûºÅÊÇ          ¡£

£¨2£©QÓëZÐγɵÄÔ­×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïµÄµç×ÓʽÊÇ                    ¡£

£¨3£©A+B¡úC+D+H2OΪÖÐѧ»¯Ñ§Öг£¼ûµÄ·´Ó¦ÐÎʽ£¬A¡¢B¡¢C¡¢D¾ùΪÓÉÉÏÊöÎåÖÖÔªËØ×é³ÉµÄµ¥ÖÊ»ò»¯ºÏÎï¡£

¢ÙÈçAΪNaOH¡¢BΪCO2ʱ£¬¿Éд³ÉaNaOH + bCO2 = cNa2CO3 + d NaHCO3 + nH2O£¬aÓëbµÄÎïÖʵÄÁ¿Ö®±ÈÓ¦Âú×ãµÄ¹ØϵÊÇ         ¡£ÏÖÏòl00 mL¡¢3 mol£¯LµÄNaOHÈÜÒºÖлºÂýͨÈë±ê×¼×´¿öÏÂ4£®48 LµÄCO2ÆøÌ壬ÓÃl¸ö»¯Ñ§·½³Ìʽ±íʾÒÔÉÏ·´Ó¦

                                     (»¯Ñ§¼ÆÁ¿ÊýΪ×î¼òÕûÊý)£¬´ËʱÈÜÒºÖи÷ÖÖÀë×ÓÎïÖʵÄÁ¿Å¨¶ÈÓɸߵ½µÍµÄÅÅÁÐ˳ÐòÊÇ_                                     _¡£

¢ÚÇëÈÎдһ¸ö·ûºÏA+B¡úC+D+H2OÐÎʽµÄÑõ»¯»¹Ô­·´Ó¦·½³Ìʽ£¬ÆäÖÐC¡¢D¾ùΪÑÎ

                                                        ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(10·Ö)Ô­×ÓÐòÊýÒÀ´ÎÔö´óµÄÎåÖÖ¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z¡¢Q¡¢R£¬ÒÑÖªXÓëQͬÖ÷×壬Y¡¢ZÔ­×ÓÐòÊýÖ®±ÈΪ3£º4£¬ZµÄ×îÍâ²ãµç×ÓÊýÊÇQµÄ×îÍâ²ãµç×ÓÊýµÄ6±¶£¬RΪͬÖÜÆÚÖÐÔ­×Ӱ뾶×îСµÄÔªËØ(Ï¡ÓÐÆøÌåÔªËسýÍâ)£¬XÓëRÐγɵĻ¯ºÏÎï³£ÎÂÏÂΪÆø̬¡£

£¨1£©±íʾYÔªËØÖÐÖÊ×ÓÊýÓëÖÐ×ÓÊýÏàµÈµÄͬλËØ·ûºÅÊÇ         ¡£

£¨2£©QÓëZÐγɵÄÔ­×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïµÄµç×ÓʽÊÇ                   ¡£

£¨3£©A+B¡úC+D+H2OΪÖÐѧ»¯Ñ§Öг£¼ûµÄ·´Ó¦ÐÎʽ£¬A¡¢B¡¢C¡¢D¾ùΪÓÉÉÏÊöÎåÖÖÔªËØ×é³ÉµÄµ¥ÖÊ»ò»¯ºÏÎï¡£

¢ÙÈçAΪNaOH¡¢BΪCO2ʱ£¬¿Éд³ÉaNaOH + bCO2 =cNa2CO3 + d NaHCO3 + nH2O£¬aÓëbµÄÎïÖʵÄÁ¿Ö®±ÈÓ¦Âú×ãµÄ¹ØϵÊÇ         ¡£ÏÖÏòl00 mL¡¢3 mol£¯LµÄNaOHÈÜÒºÖлºÂýͨÈë±ê×¼×´¿öÏÂ4£®48 LµÄCO2ÆøÌ壬ÓÃl¸ö»¯Ñ§·½³Ìʽ±íʾÒÔÉÏ·´Ó¦

                                    (»¯Ñ§¼ÆÁ¿ÊýΪ×î¼òÕûÊý)£¬´ËʱÈÜÒºÖи÷ÖÖÀë×ÓÎïÖʵÄÁ¿Å¨¶ÈÓɸߵ½µÍµÄÅÅÁÐ˳ÐòÊÇ_                                    _¡£

¢ÚÇëÈÎдһ¸ö·ûºÏA+B¡úC+D+H2OÐÎʽµÄÑõ»¯»¹Ô­·´Ó¦·½³Ìʽ£¬ÆäÖÐC¡¢D¾ùΪÑÎ

                                                       ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêºÓ±±Ê¡Õý¶¨ÖÐѧ¸ß¶þµÚ¶þѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

(10·Ö)Ô­×ÓÐòÊýÒÀ´ÎÔö´óµÄÎåÖÖ¶ÌÖÜÆÚÔªËØX¡¢Y¡¢Z¡¢Q¡¢R£¬ÒÑÖªXÓëQͬÖ÷×壬Y¡¢ZÔ­×ÓÐòÊýÖ®±ÈΪ3£º4£¬ZµÄ×îÍâ²ãµç×ÓÊýÊÇQµÄ×îÍâ²ãµç×ÓÊýµÄ6±¶£¬RΪͬÖÜÆÚÖÐÔ­×Ӱ뾶×îСµÄÔªËØ(Ï¡ÓÐÆøÌåÔªËسýÍâ)£¬XÓëRÐγɵĻ¯ºÏÎï³£ÎÂÏÂΪÆø̬¡£
£¨1£©±íʾYÔªËØÖÐÖÊ×ÓÊýÓëÖÐ×ÓÊýÏàµÈµÄͬλËØ·ûºÅÊÇ         ¡£
£¨2£©QÓëZÐγɵÄÔ­×Ó¸öÊý±ÈΪ1£º1µÄ»¯ºÏÎïµÄµç×ÓʽÊÇ                   ¡£
£¨3£©A+B¡úC+D+H2OΪÖÐѧ»¯Ñ§Öг£¼ûµÄ·´Ó¦ÐÎʽ£¬A¡¢B¡¢C¡¢D¾ùΪÓÉÉÏÊöÎåÖÖÔªËØ×é³ÉµÄµ¥ÖÊ»ò»¯ºÏÎï¡£
¢ÙÈçAΪNaOH¡¢BΪCO2ʱ£¬¿Éд³ÉaNaOH + bCO2 = cNa2CO3 + d NaHCO3 + nH2O£¬aÓëbµÄÎïÖʵÄÁ¿Ö®±ÈÓ¦Âú×ãµÄ¹ØϵÊÇ        ¡£ÏÖÏòl00 mL¡¢3 mol£¯LµÄNaOHÈÜÒºÖлºÂýͨÈë±ê×¼×´¿öÏÂ4£®48 LµÄCO2ÆøÌ壬ÓÃl¸ö»¯Ñ§·½³Ìʽ±íʾÒÔÉÏ·´Ó¦
                                    (»¯Ñ§¼ÆÁ¿ÊýΪ×î¼òÕûÊý)£¬´ËʱÈÜÒºÖи÷ÖÖÀë×ÓÎïÖʵÄÁ¿Å¨¶ÈÓɸߵ½µÍµÄÅÅÁÐ˳ÐòÊÇ_                                    _¡£
¢ÚÇëÈÎдһ¸ö·ûºÏA+B¡úC+D+H2OÐÎʽµÄÑõ»¯»¹Ô­·´Ó¦·½³Ìʽ£¬ÆäÖÐC¡¢D¾ùΪÑÎ
                                                       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸