ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ba(OH)2ÊÇÒ»ÖÖÇ¿¼î£¬¿ÉÓÃÓڲⶨÌìÈ»ÆøÖÐCO2µÄº¬Á¿¡£

£¨1£©Çëд³ö×ãÁ¿ÇâÑõ»¯±µÈÜÒºÎüÊÕCO2ÆøÌåµÄÀë×Ó·½³Ìʽ£º__________¡£

£¨2£©Ä³¿ÎÍâС×éͨ¹ýÒÔÏÂʵÑé²Ù×÷²â¶¨Ba(OH)2¡¤nH2OÖÐnµÄÖµ¡£

¢Ù³ÆÈ¡5.25gÊÔÑù£¨º¬ÓÐÔÓÖÊ£©Åä³É100mLÈÜÒº¡£ÅäÖÃÈÜÒºÖÐÓõ½µÄÒÇÆ÷ÓÐÌìƽ¡¢__________¡¢__________¡¢__________ºÍ½ºÍ·µÎ¹Ü¡£ÈôÅäÖƹý³ÌÖж¨Èݺó¾­Õñµ´¡¢Ò¡ÔÈ¡¢¾²Ö㬷¢ÏÖÒºÃæϽµ£¬ÔÙ¼ÓÊÊÁ¿ÕôÁóË®£¬ÔòËùµÃÈÜÒºµÄŨ¶È½«__________£¨Ìî¡°Æ«´ó¡±¡¢¡°²»±ä¡±»ò¡°Æ«Ð¡¡±£©¡£

¢ÚÓÃ30.00mL 1mol¡¤L-1ÑÎËáÓëÉÏÊöBa(OH)2ÈÜÒº·´Ó¦£¬ÏûºÄ¸ÃBa(OH)2ÈÜÒº100.00mL£¨ÔÓÖʲ»ÓëËá·´Ó¦£©£¬Ôò¸ÃBa(OH)2ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________¡£

¢ÛÁíÈ¡5.25gÊÔÑù¼ÓÈÈÖÁʧȥȫ²¿½á¾§Ë®£¨ÔÓÖʲ»·Ö½â£©£¬³ÆµÃÊ£Óà¹ÌÌåÖÊÁ¿Îª3.09g£¬ÔòBa(OH)2¡¤nH2OÖÐn=__________¡£

¡¾´ð°¸¡¿CO2+Ba2++2OH-===BaCO3¡ý+H2O 100mLÈÝÁ¿Æ¿ ²£Á§°ô ÉÕ±­ ƫС 0.15mol/L 8

¡¾½âÎö¡¿

(1)×ãÁ¿ÇâÑõ»¯±µÈÜÒºÎüÊÕCO2ÆøÌåÉú³ÉBaCO3£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪCO2+Ba2++2OH-===BaCO3¡ý+H2O£»

(2)¢Ù³ÆÈ¡5.25gÊÔÑù(º¬ÓÐÔÓÖÊ)Åä³É100mLÈÜÒº£¬ÐèÒª¾­¹ýÈܽ⡢תÒƲ¢¶¨ÈÝ£¬ÆäÖÐÈܽâʱÐèÒª²£Á§°ôºÍÉÕ±­£¬×ªÒÆʱÐèÒª²£Á§°ôÒýÁ÷£¬¶¨ÈÝʱÐèÒªÈÝÁ¿Æ¿ºÍ½ºÍ·µÎ¹Ü£¬ÔòÅäÖÃÈÜÒºÖÐÓõ½µÄÒÇÆ÷ÓÐÌìƽ¡¢100mLÈÝÁ¿Æ¿¡¢²£Á§°ô¡¢ÉÕ±­ºÍ½ºÍ·µÎ¹Ü£»ÈôÅäÖƹý³ÌÖж¨Èݺó¾­Õñµ´¡¢Ò¡ÔÈ¡¢¾²Ö㬷¢ÏÖÒºÃæϽµ£¬ÔÙ¼ÓÊÊÁ¿ÕôÁóË®£¬µ¼ÖÂËùÅäÈÜÒºµÄÌå»ýÆ«´ó£¬ÔòËùµÃÈÜÒºµÄŨ¶È½«Æ«Ð¡£»

¢ÚÉèBa(OH)2ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪcmol/L£¬Ôò1mol¡¤L-1¡Á0.03L=cmol/L¡Á0.1L¡Á2£¬½âµÃ£ºc=0.15£¬¸ÃBa(OH)2ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.15mol/L£»

¢Û5.25gÊÔÑùÖк¬ÓÐBa(OH)2µÄÎïÖʵÄÁ¿Îª0.15mol/L¡Á0.1L=0.015mol£¬º¬ÓÐË®µÄÎïÖʵÄÁ¿Îª=0.12mol£¬ËùÒÔ1£ºn=0.015mol£º0.12mol£¬½âµÃn=8¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÔªËØÖÜÆÚ±íÖеÄ28ºÅÔªËØNiÓÐÖØÒªµÄÓÃ;£¬ËüÓÐÁ¼ºÃµÄÄ͸ßΡ¢Ä͸¯Ê´¡¢·ÀÐ⹦ÄÜ£¬ÔÚµç³Ø¡¢´ß»¯¼Á·½ÃæÒ²Óй㷺ӦÓ᣹¤ÒµÉÏÒÔÁò»¯Äø¿ó(º¬ÉÙÁ¿ÔÓÖÊÁò»¯Í­¡¢Áò»¯ÑÇÌú)ΪԭÁÏÖƱ¸²¢¾«ÖÆÄøµÄ»ù±¾Á÷³ÌÈçÏ£º

ÒÑÖª£ºµç¼«µçλ(E)ÄÜÌåÏÖ΢Á£µÄÑõ»¯»¹Ô­ÄÜÁ¦Ç¿Èõ£¬È磺

H2-2e-=2H+ E=0.00V Cu-2e-=Cu2+ E=0.34V

Fe-2e-=Fe2+ E=-0.44V Ni-2e-=Ni2+ E=-0.25V

(1)ÄøÔÚÖÜÆÚ±íÖеÄλÖÃΪ_______________________________¡£

(2)¸ßÄø¿óÆÆËéϸĥµÄ×÷ÓÃ______________________________________¡£

(3)ÑæÉ«·´Ó¦ÊµÑé¿ÉÒÔÓùâ½àÎÞÐâµÄÄøË¿´úÌ沬˿պȡ»¯Ñ§ÊÔ¼Á×ÆÉÕ£¬Ô­ÒòÊÇ______________________¡£

(4)ÔìÔü³ýÌúʱ·¢ÉúµÄ»¯Ñ§·´Ó¦·½³Ìʽ___________________________________£¨²úÎïÒÔÑõ»¯ÐÎʽ±íʾ)¡£

(5)µç½âÖÆ´ÖÄøʱÑô¼«·¢ÉúµÄÖ÷Òªµç¼«·´Ó¦Ê½_____________________________________¡£

(6)¹¤ÒµÉÏÓÉNiSO4ÈÜÒºÖƵÃNi(OH)2ºó£¬ÔٵμÓNaC1OÈÜÒº£¬µÎ¼Ó¹ý³ÌÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________________________________¡£

(7)µç½â¾«Á¶ÄøµÄ¹ý³ÌÐè¿ØÖÆpHΪ2¡«5£¬ÊÔ·ÖÎöÔ­Òò______________________________£¬Ñô¼«ÄàµÄ³É·ÖΪ________________(дÃû³Æ)¡£

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

¾«Ó¢¼Ò½ÌÍø