ÌâÄ¿ÄÚÈÝ


 ÇâÄÜÊÇÖØÒªµÄÐÂÄÜÔ´¡£´¢Çâ×÷ΪÇâÄÜÀûÓõĹؼü¼¼Êõ£¬Êǵ±Ç°¹Ø×¢µÄÈȵãÖ®Ò»¡£

(1)ÇâÆøÊÇÇå½àȼÁÏ£¬ÆäȼÉÕ²úÎïΪ________¡£

(2)NaBH4ÊÇÒ»ÖÖÖØÒªµÄ´¢ÇâÔØÌ壬ÄÜÓëË®·´Ó¦µÃµ½NaBO2£¬ÇÒ·´Ó¦Ç°ºóBµÄ»¯ºÏ¼Û²»±ä£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________________£¬·´Ó¦ÏûºÄ1 mol NaBH4Ê±×ªÒÆµÄµç×ÓÊýĿΪ________¡£

(3)´¢Ç⻹¿É½èÖúÓлúÎÈçÀûÓû·¼ºÍéºÍ±½Ö®¼äµÄ¿ÉÄæ·´Ó¦À´ÊµÏÖÍÑÇâºÍ¼ÓÇ⣺                      

 (g)       (g)£«3H2(g)¡£

ÔÚijζÈÏ£¬ÏòºãÈÝÃܱÕÈÝÆ÷ÖмÓÈë»·¼ºÍ飬ÆäÆðʼŨ¶ÈΪa mol¡¤L£­1£¬Æ½ºâʱ±½µÄŨ¶ÈΪb mol¡¤L£­1£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK£½________¡£

(4)Ò»¶¨Ìõ¼þÏ£¬ÈçͼËùʾװÖÿÉʵÏÖÓлúÎïµÄµç»¯Ñ§´¢Çâ(ºöÂÔÆäËûÓлúÎï)¡£

¢Ùµ¼ÏßÖеç×ÓÒÆ¶¯·½ÏòΪ________¡£(ÓÃA¡¢D±íʾ)

¢ÚÉú³ÉÄ¿±ê²úÎïµÄµç¼«·´Ó¦Ê½Îª__________________¡£

¢Û¸Ã´¢Çâ×°ÖõĵçÁ÷ЧÂʦǣ½____________________¡£(¦Ç£½¡Á100%£¬¼ÆËã½á¹û±£ÁôСÊýµãºó1λ)


´ð°¸]¡¡(1)H2O

(2)NaBH4£«2H2O===NaBO2£«4H2¡ü¡¡

4NA»ò2.408¡Á1024

(3) mol3¡¤L£­3

(4)¢ÙA¡úD¡¡¢ÚC6H6£«6H£«£«6e£­===C6H12¡¡¢Û64.3%[½âÎö] (2)BÔªËØ·´Ó¦Ç°ºó»¯ºÏ¼Û²»±ä£¬ÔòÒÀ·´Ó¦Éú³ÉÎïÊÇNaBO2¿ÉÖª£¬·´Ó¦Ç°NaBH4ÖÐBµÄ»¯ºÏ¼ÛÒ²ÊÇ£«3¼Û£¬ÔòHΪ£­1¼Û£¬Ë®ÖÐÇâÔªËØ»¯ºÏ¼ÛΪ£«1¼Û£¬·¢Éú¹éÖз´Ó¦Éú³ÉÇâÆø£»¸ù¾ÝÇâÔªËØ»¯ºÏ¼Û¿ÉÈ·¶¨Ê§È¥µç×ÓÊý¡£(3)¸ù¾Ý»¯Ñ§·½³Ìʽ¿ÉÖª£¬Éú³ÉH2Ϊ3b mol/L£¬ÏûºÄ»·ÎìÍéb mol/L£¬Ôòƽºâʱ»·ÎìÍéΪ(a£­b) mol/L£¬½ø¶øÁÐʽȷ¶¨Æ½ºâ³£Êý¡£(4)¢Ù±½Éú³É»·ÎìÍéÊǵÃÇâ·´Ó¦£¬Îª»¹Ô­·´Ó¦£¬¼´µç¼«DÊÇÒõ¼«£¬µç¼«EÊÇÑô¼«£¬µ¼ÏßÖеç×ÓÒÆ¶¯·½ÏòA¡úD£»¢ÚÄ¿±ê²úÎïÊÇ»·ÎìÍ飬ÇâÀë×Óͨ¹ýÖÊ×Ó½»»»Ä¤ÏòÒõ¼«Òƶ¯£¬Ôòµç¼«·´Ó¦Ê½ÎªC6H6£«6H£«£«6e£­===C6H12£»¢ÛÑô¼«Éú³É2.8 molÆøÌ壬¸ÃÆøÌåÊÇOH£­ÔÚÑô¼«·Åµç²úÉúµÄÑõÆø£¬×ªÒƵç×Ó11.2 mol£»ÉèÒõ¼«ÏûºÄ±½µÄÎïÖʵÄÁ¿ÊÇx mol£¬Í¬Ê±Éú³Éx mol»·ÎìÍ飬ÒÀµç¼«·´Ó¦Ê½¿ÉÖª×ªÒÆµÄµç×ÓΪ6x mol£¬Ôò¸ù¾Ýµç×ÓÊØºãÖª£¬Í¬Ê±Éú³ÉÇâÆøÊÇ(11.2 mol£­6x mol)¡Â2£½5.6 mol£­3x mol£»ÒÀ¾Ý»ìºÏÆøÌå³É·Ö¿ÉÁÐʽ £½10%£¬½âµÃx£½1.2£¬Òò´Ë´¢Çâ×°ÖõĵçÁ÷ЧÂÊΪ£½64.3%¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

H2O2ÊÇÒ»ÖÖÂÌÉ«Ñõ»¯»¹Ô­ÊÔ¼Á£¬ÔÚ»¯Ñ§Ñо¿ÖÐÓ¦Óù㷺¡£

(1)ijС×éÄâÔÚͬŨ¶ÈFe3£«µÄ´ß»¯Ï£¬Ì½¾¿H2O2Ũ¶È¶ÔH2O2·Ö½â·´Ó¦ËÙÂʵÄÓ°Ïì¡£ÏÞÑ¡ÊÔ¼ÁÓëÒÇÆ÷£º30%H2O2ÈÜÒº¡¢0.1 mol¡¤L£­1Fe2(SO4)3ÈÜÒº¡¢ÕôÁóË®¡¢×¶ÐÎÆ¿¡¢Ë«¿×Èû¡¢Ë®²Û¡¢½º¹Ü¡¢²£Á§µ¼¹Ü¡¢Á¿Í²¡¢Ãë±í¡¢ºãÎÂˮԡ²Û¡¢×¢ÉäÆ÷¡£

¢Ùд³ö±¾ÊµÑéH2O2·Ö½â·´Ó¦·½³Ìʽ²¢±êÃ÷µç×Ó×ªÒÆµÄ·½ÏòºÍÊýÄ¿£º______________________________¡£

¢ÚÉè¼ÆÊµÑé·½°¸£ºÔÚ²»Í¬H2O2Ũ¶ÈÏ£¬²â¶¨________(ÒªÇóËù²âµÃµÄÊý¾ÝÄÜÖ±½ÓÌåÏÖ·´Ó¦ËÙÂÊ´óС)¡£

¢ÛÉè¼ÆÊµÑé×°Öã¬Íê³ÉͼÖеÄ×°ÖÃʾÒâͼ¡£

¢Ü²ÎÕÕϱí¸ñʽ£¬ÄⶨʵÑé±í¸ñ£¬ÍêÕûÌåÏÖʵÑé·½°¸(ÁгöËùÑ¡ÊÔ¼ÁÌå»ý¡¢Ðè¼Ç¼µÄ´ý²âÎïÀíÁ¿ºÍËùÄⶨµÄÊý¾Ý£»Êý¾ÝÓÃ×Öĸ±íʾ)¡£

¡¡¡¡ÎïÀíÁ¿

ʵÑéÐòºÅ ¡¡¡¡

V[0.1 mol¡¤L£­1

Fe2(SO4)3]/mL

¡­¡­

1

a

¡­¡­

2

a

¡­¡­

(2)ÀûÓÃͼ(a)ºÍ(b)ÖеÄÐÅÏ¢£¬°´Í¼(c)×°ÖÃ(Á¬Í¨µÄA¡¢BÆ¿ÖÐÒѳäÓÐNO2ÆøÌå)½øÐÐʵÑé¡£¿É¹Û²ìµ½BÆ¿ÖÐÆøÌåÑÕÉ«±ÈAÆ¿ÖеÄ__________(Ìî¡°É»ò¡°Ç³¡±)£¬ÆäÔ­ÒòÊÇ____________________________¡£

¡¡¡¡¡¡¡¡¡¡(a)¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡(b)

(c)

ͼ21

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

¾«Ó¢¼Ò½ÌÍø