ÌâÄ¿ÄÚÈÝ

ÏÂͼ5¸ö×°Öö¼ÊÇÖÐѧ»¯Ñ§Öг£¼ûµÄʵÑé×°Öã¬Ä³Ñ§Ï°Ð¡×éµÄͬѧÓûÓÃÕâЩװÖýøÐг£¼ûÎïÖʵÄÖÆÈ¡²¢Ì½¾¿ÆäÐÔÖÊ£¨Í¼ÖÐa¡¢b¡¢c±íʾֹˮ¼Ð£©£¬Çë¶ÔÆä½øÐÐÍêÉÆ»òÆÀ¼Û£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©½«A¡¢C¡¢EÏàÁ¬ºó£¬ÒÔŨÑÎËáºÍ                £¨ÌîдÃû³Æ£©ÎªÔ­ÁÏÖÆÈ¡Cl2£¬ÒÇÆ÷ÒÒµÄÃû³ÆÊÇ                   ¡£

£¨2£©ÀûÓã¨1£©ÖÐ×°ÖúÍÒ©Æ·£¬ÔÚ±ûÖмÓÈëÊÊÁ¿Ë®£¬¼´¿ÉÖÆµÃÂÈË®¡£½«ËùµÃÂÈË®·ÖΪÁ½·Ý£¬½øÐТñ¡¢¢òÁ½¸öʵÑ飬ʵÑé²Ù×÷¡¢ÏÖÏó¡¢½áÂÛÈçÏ£º

ʵÑé

ÐòºÅ

ʵÑé²Ù×÷

ÏÖÏó

½áÂÛ

¢ñ

½«ËùµÃÂÈË®µÎÈëÆ·ºìÈÜÒº

Æ·ºìÈÜÒºÍÊÉ«

ÂÈÆøÓëË®·´Ó¦µÄ²úÎïÓÐÆ¯°×ÐÔ

¢ò

ÏòËùµÃÂÈË®ÖмÓÈë̼ËáÇâÄÆ·ÛÄ©

ÓÐÎÞÉ«ÆøÅݲúÉú

ÂÈÆøÓëË®·´Ó¦ÖÁÉÙ²úÉúÒ»ÖÖËáÐÔÇ¿ÓÚ̼ËáµÄÎïÖÊ

ʵÑé¢ñ»ñµÃ½áÂÛÊÇ·ñºÏÀí£¿         £¨Ìî¡°ºÏÀí¡±»ò¡°²»ºÏÀí¡±£©¡£ÈôÑ¡¡°²»ºÏÀí¡±£¬Çë˵Ã÷ÀíÓÉ£¨ÈôÑ¡¡°ºÏÀí¡±£¬ÔòÎÞÐèÌîдÀíÓÉ£©£º                                   ¡£

ʵÑé¢ò»ñµÃ½áÂÛÊÇ·ñºÏÀí£¿         £¨Ìî¡°ºÏÀí¡±»ò¡°²»ºÏÀí¡±£©¡£ÈôÑ¡¡°²»ºÏÀí¡±£¬Çë˵Ã÷ÀíÓÉ£¨ÈôÑ¡¡°ºÏÀí¡±£¬ÔòÎÞÐèÌîдÀíÓÉ£©£º                                   ¡£

£¨3£©ÀûÓã¨1£©ÖÐ×°Öû¹¿ÉÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑé±È½ÏCl£­ºÍS2£­µÄ»¹Ô­ÐÔÇ¿Èõ¡£ÔòCÖÐÔ¤ÆÚ³öÏÖµÄÏÖÏóÊÇ                                                              ¡£

£¨4£©½«B¡¢D¡¢E×°ÖÃÏàÁ¬½Ó£¨´ò¿ªÖ¹Ë®¼ÐaºÍֹˮ¼Ðb£¬¹Ø±Õֹˮ¼Ðc£©£¬ÔÚBÖÐʢװŨÏõËáºÍͭƬ£¨½«Í­Æ¬·ÅÔÚÓп×ËÜÁϰåÉÏ£©£¬¿ÉÖÆµÃNO2¡£Ò»¶Îʱ¼äºó£¬ÓûÓÃD×°ÖÃ̽¾¿NO2ÓëË®µÄ·´Ó¦£¬Æä²Ù×÷²½ÖèΪ£ºÏÈ                                                £¬ÔÙ                                                                            ÒÔʹÉÕ±­ÖеÄË®½øÈëÊԹܶ¡£¬¹Û²ìÏÖÏó¡£

(1)£¨4·Ö£©¶þÑõ»¯ÃÌ   £¨Ô²µ×£©ÉÕÆ¿

(2)£¨4·Ö£©²»ºÏÀí£»Ã»ÓÐÊÂÏÈÖ¤Ã÷¸ÉÔïµÄÂÈÆøÎÞÆ¯°×ÐÔ¡£

²»ºÏÀí£»ÖÆÈ¡µÄÂÈÆøÖк¬ÓÐHClÆøÌ壬HClÈÜÓÚË®ºóÄÜÓë̼ËáÇâÄÆ·´Ó¦²úÉúÆøÅÝ¡£

(3)£¨2·Ö£©²úÉúµ­»ÆÉ«³Áµí

(4) (2·Ö)¹Ø±Õa¡¢b£¬´ò¿ªc£»Ë«ÊÖ½ôÎÕ(»ò΢ÈÈ)ÊԹܶ¡£¬Ê¹NO2ÒݳöÓëË®½Ó´¥£¨»òÆäËüºÏÀíµÄ´ð°¸£©

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Ëá¼îÖкͷ´Ó¦ÊÇÖÐѧ»¯Ñ§Ò»ÀàÖØÒªµÄ·´Ó¦£¬Æä·´Ó¦µÄ±¾Öʶ¼ÊÇH+ºÍOH£­·´Ó¦Éú³ÉH2O£®Ä³Ñ§ÉúʵÑéС×é¶Ô¿Î±¾ÖеÄÁ½¸öËá¼îÖкÍʵÑé½øÐÐÁËÄ£ÄâÑо¿£®

(1)ÓÃÏÂͼװÖýøÐÐÇ¿Ëá(50 mL¡¡0.5 mol/LµÄHCI)ºÍÇ¿¼î(50 mL¡¡0.55 mol/LµÄNaOH)·´Ó¦µÄÖкÍÈȲⶨ£®

¢Ù´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙµÄÒ»ÖÖ²£Á§ÒÇÆ÷ÊÇ________£»

¢Ú´óСÉÕ±­Ö®¼äÌîÂúֽмµÄ×÷ÓÃÊÇ________£»

¢ÛÏò×é×°ºÃµÄÁ¿ÈÈÆ÷ÖмÓÈë50 mLÑÎËá²¢²â¶¨Î¶ȣ¬È¡³öζȼƲ¢³å

¾»ºó£¬ÔÙ½«ÒѲⶨºÃζȵÄNaOHÈÜÒº50 mL¼ÓÈëÁ¿ÈÈÆ÷£®ÏÂÁмÓÈë50 mL¡¡NaOHÈÜÒºµÄ²Ù×÷ÕýÈ·µÄÊÇ________£®

A£®ÎªÁËʹËá¼î³ä·ÖÖкͣ¬±ØÐë±ßµÎ¼ÓNaOHÈÜÒº±ßѸËÙ½Á°è

B£®°Ñ50 mL¡¡NaOHÈÜÒºÒ»´ÎѸËÙµ¹È룬¸ÇÉÏÖ½°åѸËÙ½Á°è£®

¢ÜÈôÓô×Ëá´úÌæÉÏÊöÑÎËáʵÑ飬ʵÑé²âµÃµÄÖкÍÈÈÖµ________(ÌîÆ«´ó¡¢Æ«Ð¡»òÎÞÓ°Ïì)£®

(1)¡¢ÓÃʵÑéÊÒ׼ȷÅäÖÆµÄ0.100 mol/LµÄNaOHÈÜÒº²â¶¨Ä³Î´ÖªÅ¨¶ÈµÄÏ¡ÑÎËᣮ

Æä¾ßÌåʵÑé²½ÖèÈçÏ£º

¢Ùȡһ֧¼îʽµÎ¶¨¹ÜÏÂͼÖÐÓÐÁ½Ö»µÎ¶¨¹Ü¡«É϶Ëδ»­³ö£¬ÄãÑ¡ÔñÄÄÒ»Ö»£¿________(Ìîд¶ÔÓ¦×Öĸ)£¬ÓÃÉÙÁ¿±ê×¼NaOHÈÜÒºÈóÏ´2¡«3´ÎˮϴºóµÄ¼îʽµÎ¶¨¹Ü£¬ÔÙ¼ÓÈë±ê×¼µÄ0.100 mol/LµÄNaOHÈÜÒº²¢¼ÇÂ¼ÒºÃæ¿Ì¶È¶ÁÊý£»

¢ÚÓÃËáʽµÎ¶¨¹Ü¾«È·µÄ·Å³ö25.00 mL´ý²âÑÎËᣬÖÃÓÚÓÃÕôÁóˮϴ¾»µÄ×¶ÐÎÆ¿ÖУ¬ÔÙ¼ÓÈë·Ó̪(±äÉ«·¶Î§£º8.2¡«10.0)ÊÔÒº2µÎ£»

¢ÛµÎ¶¨Ê±£¬±ßµÎ±ßÕñµ´£¬Í¬½ø×¢ÊÓÑÛ¾¦×¢ÊÓ×¶ÐÎÆ¿ÄÚÈÝÒ²ÑÕÉ«µÄ±ä»¯£¬µ±×¶ÐÎÆ¿ÄÚÈÜÒºÓÉ________(ÌîдÑÕÉ«±ä»¯)ÇÒ°ë·ÖÖÓÄÚ²»ÍÊɫʱ£¬¼´´ïµ½µÎ¶¨Öյ㣻

¢Ü¼ÇÂ¼ÒºÃæ¿Ì¶È¶ÁÊý£®¸ù¾ÝµÎ¶¨¹ÜµÄÁ½´Î¶ÁÊýµÃ³öÏûºÄ±ê×¼ÑÎËáµÄÌå»ý£¬ÔÙÖØ¸´²â¶¨Á½´Î£¬ÊµÑé½á¹û¼Ç¼¼ûÏÂ±í£º

²âµÃδ֪ϡÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ________(±£ÁôСÊýµãºó3λ)£®

¢ÝÈç¹ûµÎ¶¨½áÊøÊ±¸©ÊÓ¼îʽµÎ¶¨¹Ü¿Ì¶È¶ÁÊý(ÆäËü²Ù×÷¾ùÕýÈ·)£¬Ôò¶ÔµÎ¶¨½á¹û¡«Ï¡ÑÎËáŨ¶ÈµÄÓ°ÏìÊÇ________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)£®

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

¾«Ó¢¼Ò½ÌÍø