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

2KNO3£«3C£«S£½A£«N2£«3CO2¡ü(ÒÑÅäƽ)

¢Ù³ýSÍ⣬ÉÏÁÐÔªËصĵ縺ÐÔ´Ó´óµ½Ð¡ÒÀ´ÎΪ________£®

¢ÚÔÚÉú³ÉÎïÖУ¬AµÄ»¯Ñ§¼üÀàÐÍΪ________£¬º¬¼«ÐÔ¹²¼Û¼üµÄ·Ö×ÓµÄÖÐÐÄÔ­×Ó¹ìµÀÔÓ»¯ÀàÐÍΪ________ÊôÓÚ________·Ö×Ó£®

¢ÛÒÑÖªCN£­ÓëN2½á¹¹ÏàËÆ£¬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)ÔÚCrCl3µÄË®ÈÜÒºÖУ¬Ò»¶¨Ìõ¼þÏ´æÔÚ×é³ÉΪ[CrCln(H2O)6£­n]x+(nºÍx¾ùΪÕýÕûÊý)µÄÅäÀë×Ó£¬½«Æäͨ¹ýÇâÀë×Ó½»»»Ê÷Ö¬(R£­H)£¬¿É·¢ÉúÀë×Ó½»»»·´Ó¦£º

[CrCln(H2O)6£­n]x+£«xR£­HRx[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£­HÍêÈ«½»»»ºó£¬ÖкÍÉú³ÉµÄH+ÐèŨ¶ÈΪ0.1200 mol¡¤L£­1¡¡NaOHÈÜÒº25.00 ml£¬¸ÃÅäÀë×ӵĻ¯Ñ§Ê½Îª________£®

(4)ÔÚÌúºÍÑõ»¯ÌúµÄ»ìºÏÎï15 gÖмÓÈëÏ¡ÁòËá150 mL£¬±ê×¼×´¿öÏ·ųöÇâÆø1.68 L£¬Í¬Ê±ÌúºÍÑõ»¯Ìú¾ùÎÞÊ£Ó࣮ÏòÈÜÒºÖеÎÈëKSCNδ¼ûÑÕÉ«±ä»¯£®ÎªÁËÖк͹ýÁ¿µÄÁòËᣬÇÒʹFe2+Íêȫת»¯ÎªÇâÑõ»¯ÑÇÌú£¬¹²ÏûºÄ3 mol/LµÄÇâÑõ»¯ÄÆÈÜÒº200 mL£¬ÔòÔ­ÁòËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ________mol/L£®

´ð°¸£º
½âÎö£º

¡¡¡¡(1)¢ÙO£¾N£¾C£¾K

¡¡¡¡¢ÚÀë×Ó¼ü¡¡SPÔÓ»¯ÊôÓÚ¡¡·Ç¼«ÐÔ·Ö×Ó£®

¡¡¡¡¢Û1¡Ã1

¡¡¡¡(2)3d84s2¡¡4

¡¡¡¡(3)»¯Ñ§Ê½Îª[CrCl(H2O)5]2+

¡¡¡¡(4)2 mol/L£®


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêËÄ´¨Ê¡³É¶¼ÆßÖи߶þ10Ô½׶ÎÐÔ¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(7·Ö)(1)Öйú¹Å´úËÄ´ó·¢Ã÷Ö®Ò»¡ª¡ªºÚ»ðÒ©£¬ËüµÄ±¬Õ¨·´Ó¦Îª£º
2KNO3+ 3C+SA+N2¡ü+3CO2¡ü£¨ÒÑÅäƽ£©
¢Ù³ý S Í⣬ÉÏÁÐÔªËصĵ縺ÐÔ´Ó´óµ½Ð¡ÒÀ´ÎΪ_________________________;
¢ÚÉú³ÉÎïÖУ¬A µÄµç×ÓʽΪ__________________£¬º¬¼«ÐÔ¹²¼Û¼üµÄ·Ö×ÓµÄÖÐÐÄÔ­×Ó¹ìµÀÔÓ»¯ÀàÐÍ____________;
¢ÛÒÑÖªCN£­ÓëN2½á¹¹ÏàËÆ£¬ÍÆËãHCN·Ö×ÓÖЦҼüÓë¦Ð¼üÊýÄ¿Ö®±ÈΪ____________¡£
(2)Ô­×ÓÐòÊýСÓÚ36µÄÔªËØQºÍT£¬ÔÚÖÜÆÚ±íÖмȴ¦ÓÚͬһÖÜÆÚÓÖλÓÚͬһ×壬ÇÒÔ­×ÓÐòÊýT±ÈQ¶à2¡£TµÄ»ù̬ԭ×ÓÍâΧµç×Ó(¼Ûµç×Ó)ÅŲ¼Ê½Îª_______________________________£¬Q2£«µÄδ³É¶Ôµç×ÓÊýÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêËÄ´¨Ê¡ÕѾõÖÐѧ¸ß¶þ10ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(1)Öйú¹Å´úËÄ´ó·¢Ã÷Ö®Ò»¡ª¡ªºÚ»ðÒ©£¬ËüµÄ±¬Õ¨·´Ó¦Îª£º
2KNO3+ 3C+SA+N2¡ü+3CO2¡ü£¨ÒÑÅäƽ£©
¢Ù    ³ý S Í⣬ÉÏÁÐÔªËصĵ縺ÐÔ´Ó´óµ½Ð¡ÒÀ´ÎΪ_________________________;
¢ÚÉú³ÉÎïÖУ¬A µÄµç×ÓʽΪ__________________ £¬º¬¼«ÐÔ¹²¼Û¼üµÄ·Ö×ÓµÄÖÐÐÄÔ­×Ó¹ìµÀÔÓ»¯ÀàÐÍ____________;
¢ÛÒÑÖªCN£­ÓëN2½á¹¹ÏàËÆ£¬ÍÆËãHCN·Ö×ÓÖЦҼüÓë¦Ð¼üÊýÄ¿Ö®±ÈΪ____________¡£
(2)Ô­×ÓÐòÊýСÓÚ36µÄÔªËØQºÍT£¬ÔÚÖÜÆÚ±íÖмȴ¦ÓÚͬһÖÜÆÚÓÖλÓÚͬһ×壬ÇÒÔ­×ÓÐòÊýT±ÈQ¶à2¡£TµÄ»ù̬ԭ×ÓÍâΧµç×Ó(¼Ûµç×Ó)ÅŲ¼Ê½Îª_______________________________£¬Q2£«µÄδ³É¶Ôµç×ÓÊýÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìËÄ´¨Ê¡¸ß¶þ10ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

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

2KNO3+ 3C+SA+N2¡ü+3CO2¡ü£¨ÒÑÅäƽ£©

¢Ù    ³ý S Í⣬ÉÏÁÐÔªËصĵ縺ÐÔ´Ó´óµ½Ð¡ÒÀ´ÎΪ_________________________;

¢ÚÉú³ÉÎïÖУ¬A µÄµç×ÓʽΪ__________________ £¬º¬¼«ÐÔ¹²¼Û¼üµÄ·Ö×ÓµÄÖÐÐÄÔ­×Ó¹ìµÀÔÓ»¯ÀàÐÍ____________;

¢ÛÒÑÖªCN£­ÓëN2½á¹¹ÏàËÆ£¬ÍÆËãHCN·Ö×ÓÖЦҼüÓë¦Ð¼üÊýÄ¿Ö®±ÈΪ____________¡£

(2)Ô­×ÓÐòÊýСÓÚ36µÄÔªËØQºÍT£¬ÔÚÖÜÆÚ±íÖмȴ¦ÓÚͬһÖÜÆÚÓÖλÓÚͬһ×壬ÇÒÔ­×ÓÐòÊýT±ÈQ¶à2¡£TµÄ»ù̬ԭ×ÓÍâΧµç×Ó(¼Ûµç×Ó)ÅŲ¼Ê½Îª_______________________________£¬Q2£«µÄδ³É¶Ôµç×ÓÊýÊÇ________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìËÄ´¨Ê¡¸ß¶þ10Ô½׶ÎÐÔ¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

 (7·Ö)(1)Öйú¹Å´úËÄ´ó·¢Ã÷Ö®Ò»¡ª¡ªºÚ»ðÒ©£¬ËüµÄ±¬Õ¨·´Ó¦Îª£º

2KNO3+ 3C+SA+N2¡ü+3CO2¡ü£¨ÒÑÅäƽ£©

¢Ù³ý S Í⣬ÉÏÁÐÔªËصĵ縺ÐÔ´Ó´óµ½Ð¡ÒÀ´ÎΪ_________________________;

¢ÚÉú³ÉÎïÖУ¬A µÄµç×ÓʽΪ__________________£¬º¬¼«ÐÔ¹²¼Û¼üµÄ·Ö×ÓµÄÖÐÐÄÔ­×Ó¹ìµÀÔÓ»¯ÀàÐÍ____________;

¢ÛÒÑÖªCN£­ÓëN2½á¹¹ÏàËÆ£¬ÍÆËãHCN·Ö×ÓÖЦҼüÓë¦Ð¼üÊýÄ¿Ö®±ÈΪ____________¡£

(2)Ô­×ÓÐòÊýСÓÚ36µÄÔªËØQºÍT£¬ÔÚÖÜÆÚ±íÖмȴ¦ÓÚͬһÖÜÆÚÓÖλÓÚͬһ×壬ÇÒÔ­×ÓÐòÊýT±ÈQ¶à2¡£TµÄ»ù̬ԭ×ÓÍâΧµç×Ó(¼Ûµç×Ó)ÅŲ¼Ê½Îª_______________________________£¬Q2£«µÄδ³É¶Ôµç×ÓÊýÊÇ________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÍ¬²½Ìâ ÌâÐÍ£ºÌî¿ÕÌâ

(1)Öйú¹Å´úËÄ´ó·¢Ã÷Ö®Ò»--ºÚ»ðÒ©£¬ËüµÄ±¬Õ¨·´Ó¦Îª£º2KNO3+3C+SA+ N2¡ü +3CO2¡ü
¢Ù³ýSÍ⣬ÉÏÁÐÔªËصĵ縺ÐÔ´Ó´óµ½Ð¡ÒÀ´ÎΪ_______________¡£
¢ÚÔÚÉú³ÉÎïÖУ¬º¬¼«ÐÔ¹²¼Û¼üµÄ·Ö×ÓµÄÖÐÐÄÔ­ÓÚ¹ìµÀÔÓ»¯ÀàÐÍΪ____________¡£
¢ÛÒÑÖªCN-ÓëN2½á¹¹ÏàËÆ£¬ÍÆËãHCN·Ö×ÓÖЦҼüÓë¦Ð¼üÊýÄ¿Ö®±ÈΪ____________
(2)Ô­×ÓÐòÊýСÓÚ36µÄÔªËØQºÍT£¬ÔÚÖÜÆÚ±íÖмȴ¦ÓÚͬһÖÜÆÚÓÖλÓÚͬһ×壬ÇÒÔ­×ÓÐòÊýT±ÈQ¶à2¡£TµÄ»ù̬ԭ×ÓÍâΧµç×Ó£¨¼Ûµç×Ó£©ÅŲ¼Ê½Îª_____________£¬Q2+µÄδ³É¶Ôµç×ÓÊýÊÇ____________
(3)ÔÚCrCl3µÄË®ÈÜÒºÖУ¬Ò»¶¨Ìõ¼þÏ´æÔÚ×é³ÉΪ[CrCln(H2O)6-n]x+£¨nºÍx¾ùΪÕýÕûÊý£©µÄÅäÀë×Ó£¬½«Æäͨ¹ýÇâÀë×Ó½»»»Ê÷Ö¬(R-H)£¬¿É·¢ÉúÀë×Ó½»»»·´Ó¦£º[CrCln(H2O)6-n]x++xR-H¡úRx[CrCln(H2O)6-n]+xH+ ½»»»³öÀ´µÄH+¾­Öк͵樣¬¼´¿ÉÇó³öxºÍn£¬È·¶¨ÅäÀë×ÓµÄ×é³É¡£½«º¬0.001 5 mol[CrCln(H2O)6-n]x+µÄÈÜÒº£¬Óë
R-HÍêÈ«½»»»ºó£¬ÖкÍÉú³ÉµÄH+ÐèŨ¶ÈΪ0.120 0 mol/L NaOHÈÜÒº25. 00 mL£¬¿ÉÖª¸ÃÅäÀë×ӵĻ¯Ñ§Ê½Îª
___________________

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸