ÌâÄ¿ÄÚÈÝ
A£®ÓÃÁ¿Í²Á¿È¡5.4mLŨÁòËᣬ»ºÂýµ¹Èë×°ÓÐÔ¼50mLÕôÁóË®µÄÉÕ±À²¢Óò£Á§°ô½Á°è
B£®ÓÃÔ¼30mLÕôÁóË®£¬·Ö³ÉÈý´ÎÏ´µÓÉÕ±ºÍ²£Á§°ô£¬½«Ï´µÓÒº¶¼µ¹ÈëÈÝÁ¿Æ¿ÖÐ
C£®½«Ï¡ÊͺóµÄÁòËáСÐĵص¹ÈëÈÝÁ¿Æ¿ÖÐ
D£®¼ì²é100mLÈÝÁ¿Æ¿¿ÚÊÇ·ñÓЩҺÏÖÏó
E£®½«ÕôÁóˮֱ½Ó¼ÓÈëÈÝÁ¿Æ¿£¬ÖÁÒºÃæ¾à¿Ì¶ÈÏß1¡«2cm´¦
F£®¸Ç½ôÆ¿Èû£¬·´¸´µßµ¹Õñµ´£¬Ò¡ÔÈÈÜÒº
G£®ÓýºÍ·µÎ¹ÜÏòÈÝÁ¿Æ¿ÀïÖðµÎ¼ÓÈëÕôÁóË®£¬ÖÁÒºÃæ×îµÍµãÓë¿Ì¶ÈÏßÏàÇÐ
£¨1£©ÕýÈ·µÄ²Ù×÷˳ÐòÓ¦¸ÃÊÇ
£¨2£©½øÐÐA²½²Ù×÷µÄʱºò£¬Ó¦¸ÃÑ¡ÓõÄÊÇ
¢Ù10mLÁ¿Í²£»¢Ú50mLÁ¿Í²£»¢Û500mLÁ¿Í²£»¢Ü1000mLÁ¿Í²
Èç¹û¶Ô×°ÓÐŨÁòËáµÄÁ¿Í²¶ÁÊýÈçͼËùʾ£¬ÅäÖÆµÄÏ¡ÁòËáµÄŨ¶È½«
£¨3£©½øÐÐA²½²Ù×÷ºó£¬±ØÐë
¿¼µã£ºÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº
רÌ⣺ʵÑéÌâ
·ÖÎö£º£¨1£©¸ù¾ÝÈÝÁ¿Æ¿Ê¹ÓÃǰ¼ì©ºÍʵÑé²Ù×÷µÄ²½Ö裨¼ÆËã¡¢Á¿È¡¡¢Ï¡ÊÍ¡¢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔȵȲÙ×÷£©À´È·¶¨Ë³Ðò£»
£¨2£©Ïȸù¾ÝÏ¡ÊÍǰºóÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËã³öŨÑÎËáµÄÌå»ý£¬¼´C£¨Å¨£©¡ÁV£¨Å¨£©=C£¨Ï¡£©¡ÁV£¨Ï¡£©£¬¸ù¾Ý¼ÆËãËùµÃÌå»ýÀ´Ñ¡ÔñÁ¿Í²µÄ¹æ¸ñ£»
¸ù¾Ýc=
¼ÆËã²»µ±²Ù×÷¶Ôn»òVµÄÓ°Ï죬Èç¹ûnÆ«´ó»òVƫС£¬ÔòËùÅäÖÆÈÜҺŨ¶ÈÆ«¸ß£»
£¨3£©¸ù¾ÝŨÁòËáÏ¡ÊÍ·ÅÈÈ£¬ÈÈÈÜÒºµÄÌå»ý´ó£¬Ó°ÏìÈÜÒºµÄŨ¶È£»
£¨2£©Ïȸù¾ÝÏ¡ÊÍǰºóÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËã³öŨÑÎËáµÄÌå»ý£¬¼´C£¨Å¨£©¡ÁV£¨Å¨£©=C£¨Ï¡£©¡ÁV£¨Ï¡£©£¬¸ù¾Ý¼ÆËãËùµÃÌå»ýÀ´Ñ¡ÔñÁ¿Í²µÄ¹æ¸ñ£»
¸ù¾Ýc=
| n |
| V |
£¨3£©¸ù¾ÝŨÁòËáÏ¡ÊÍ·ÅÈÈ£¬ÈÈÈÜÒºµÄÌå»ý´ó£¬Ó°ÏìÈÜÒºµÄŨ¶È£»
½â´ð£º
½â£º£¨1£©ÒòÈÝÁ¿Æ¿Ê¹ÓÃǰ¼ì©£¬ÅäÖÆË³ÐòÊÇ£º¼ÆËã¡úÁ¿È¡¡úÏ¡ÊÍ¡¢ÀäÈ´¡úÒÆÒº¡ú¶¨ÈÝ¡úÒ¡ÔÈ¡ú×°Æ¿¡úÌùÇ©£¬ËùÒÔÕýÈ·µÄ²Ù×÷˳ÐòÊÇΪDACBEGF£¬¹Ê´ð°¸Îª£ºDACBEGF£»
£¨2£©ÉèŨÁòËáµÄÌå»ýΪVL£¬Ôò18.4mol/L¡ÁVL=1mol/L¡Á0.1L£¬½âµÃV=0.0054L£¬¼´5.4mL£»
ÒòÐèÒªµÄŨÁòËáµÄÌå»ýΪ5.4mL£¬ÓÃ10mLµÄÁ¿Í²À´Á¿È¡£¬¹ÊÑ¡£º¢Ù£®
Èç¹û¶Ô×°ÓÐŨÁòËáµÄÁ¿Í²ÑöÊÓ¶ÁÊý£¬Á¿È¡µÄŨÁòËáµÄÌå»ýÆ«´ó£¬ÈÜÖʵÄÖÊÁ¿Æ«¶à£¬ËùÅäÖÆµÄÈÜÒºµÄŨ¶ÈÆ«´ó£¬¹Ê´ð°¸Îª£ºÆ«¸ß£»
£¨3£©ÒòŨÁòËáÏ¡ÊÍ·ÅÈÈ£¬ÈÈÈÜÒºµÄÌå»ý´ó£¬Ò»µ©ÀäÈ´ÏÂÀ´Ìå»ýƫС£¬Ó°ÏìÈÜÒºµÄŨ¶È£¬ËùÒÔ±ØÐëÀäÈ´ÖÁÊÒβÅÄÜ×ªÒÆÖÁÈÝÁ¿Æ¿£¬¹Ê´ð°¸Îª£º´ýÈÜÒºÀäÈ´µ½ÊÒΣ®
£¨2£©ÉèŨÁòËáµÄÌå»ýΪVL£¬Ôò18.4mol/L¡ÁVL=1mol/L¡Á0.1L£¬½âµÃV=0.0054L£¬¼´5.4mL£»
ÒòÐèÒªµÄŨÁòËáµÄÌå»ýΪ5.4mL£¬ÓÃ10mLµÄÁ¿Í²À´Á¿È¡£¬¹ÊÑ¡£º¢Ù£®
Èç¹û¶Ô×°ÓÐŨÁòËáµÄÁ¿Í²ÑöÊÓ¶ÁÊý£¬Á¿È¡µÄŨÁòËáµÄÌå»ýÆ«´ó£¬ÈÜÖʵÄÖÊÁ¿Æ«¶à£¬ËùÅäÖÆµÄÈÜÒºµÄŨ¶ÈÆ«´ó£¬¹Ê´ð°¸Îª£ºÆ«¸ß£»
£¨3£©ÒòŨÁòËáÏ¡ÊÍ·ÅÈÈ£¬ÈÈÈÜÒºµÄÌå»ý´ó£¬Ò»µ©ÀäÈ´ÏÂÀ´Ìå»ýƫС£¬Ó°ÏìÈÜÒºµÄŨ¶È£¬ËùÒÔ±ØÐëÀäÈ´ÖÁÊÒβÅÄÜ×ªÒÆÖÁÈÝÁ¿Æ¿£¬¹Ê´ð°¸Îª£º´ýÈÜÒºÀäÈ´µ½ÊÒΣ®
µãÆÀ£º±¾Ì⿼²éÁËÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÅäÖÆµÄ¼ÆËã¡¢²½ÖèÒÔ¼°Îó²î·ÖÎö£¬ÄѶȲ»´ó£¬×¢ÒâʵÑéµÄ»ù±¾²Ù×÷·½·¨ºÍ×¢ÒâÊÂÏ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÂÁÐÀë×Ó·½³Ìʽ´íÎóµÄÊÇ£¨¡¡¡¡£©
| A¡¢ÇâÑõ»¯±µÈÜÒººÍÁòËáÍÈÜÒº·´Ó¦£ºBa2++2OH-+Cu2++SO42-=BaSO4¡ý+Cu£¨OH£©2¡ý |
| B¡¢ÂÈ»¯ÂÁÈÜÒº¼ÓÈë¹ýÁ¿°±Ë®£ºAl3++3OH-=Al£¨OH£©3¡ý |
| C¡¢Ï¡ÏõËáºÍÍÆ¬£º3Cu+8H++2NO3-=3Cu2++2NO¡ü+4H2O |
| D¡¢ÂÈÆøºÍË®·´Ó¦£ºCl2+2H2O=2H++ClO-+Cl-¡ü |
ÉèNAΪ°¢·ð¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢Ù±ê×¼×´¿öÏ£¬11.2LÒÔÈÎÒâ±ÈÀý»ìºÏµÄµªÆøºÍÑõÆøËùº¬µÄÔ×ÓÊýΪNA
¢Ú56g FeÓëÈκÎÎïÖÊÍêÈ«·´Ó¦Ê±×ªÒƵĵç×ÓÊýÒ»¶¨Îª2NA
¢Û³£Î³£Ñ¹Ï£¬18gH2Oº¬ÓеÄÔ×Ó×ÜÊýΪ3NA
¢Ü±ê×¼×´¿öÏÂ22.4L SO3ÖзÖ×ÓÊýΪNA
¢ÝͬÎÂͬѹÏ£¬Ìå»ýÏàͬµÄÇâÆøºÍë²ÆøËùº¬µÄ·Ö×ÓÊýÏàµÈ£®
¢Ù±ê×¼×´¿öÏ£¬11.2LÒÔÈÎÒâ±ÈÀý»ìºÏµÄµªÆøºÍÑõÆøËùº¬µÄÔ×ÓÊýΪNA
¢Ú56g FeÓëÈκÎÎïÖÊÍêÈ«·´Ó¦Ê±×ªÒƵĵç×ÓÊýÒ»¶¨Îª2NA
¢Û³£Î³£Ñ¹Ï£¬18gH2Oº¬ÓеÄÔ×Ó×ÜÊýΪ3NA
¢Ü±ê×¼×´¿öÏÂ22.4L SO3ÖзÖ×ÓÊýΪNA
¢ÝͬÎÂͬѹÏ£¬Ìå»ýÏàͬµÄÇâÆøºÍë²ÆøËùº¬µÄ·Ö×ÓÊýÏàµÈ£®
| A¡¢¢Ù¢Ú¢Û | B¡¢¢Ù¢Û¢Ü |
| C¡¢¢Ù¢Û¢Ý | D¡¢¢Û¢Ü¢Ý |
ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A¡¢³£ÎÂÏ£¬3.2 g O2ÓëO3µÈÎïÖʵÄÁ¿»ìºÏºóµÄ»ìºÏÎïÖÐËùº¬OÔ×ÓÊýΪ0.1NA |
| B¡¢1 L 1 mol/L FeCl3ÈÜÒºÖÐCl-ÊýΪNA |
| C¡¢±ê×¼×´¿öÏ£¬2.24 L H2OËùº¬Ô×ÓÊýΪ0.3NA |
| D¡¢22 g¶þÑõ»¯Ì¼Óë±ê×¼×´¿öÏÂ11.2 L HClº¬ÓеķÖ×ÓÊý¾ùΪ0.5NA |
ÏÂÁÐÎïÖÊÊôÓÚ´¿¾»ÎïµÄÊÇ£¨¡¡¡¡£©
| A¡¢±ùË®»ìºÏÎï | B¡¢½à¾»µÄ¿ÕÆø |
| C¡¢Ê³´× | D¡¢Ç峺µÄº£Ë® |