ÏÂÁйØÓÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄ˵·¨£¬ÕýÈ·µÄ×éºÏÊÇ
[     ]
¢ÙÍÐÅÌÌìƽ¿É¶ÁÈ¡µ½Ð¡Êýµãºóһλ(ÒÔ¡°g¡±Îªµ¥Î»)
¢ÚÍÐÅÌÌìƽֻÄÜ´ÖÂԵسÆÁ¿ÎïÖʵÄÖÊÁ¿£¬Á¿Í²Ö»ÄÜ´ÖÂÔµØÁ¿È¡ÒºÌåµÄÌå»ý£¬ÑϸñµØ˵£¬ËüÃǶ¼²»ÄÜÓëÈÝÁ¿Æ¿¡ª¡ª¾«È·ÒÇÆ÷ÅäÌ×ʹÓÃ
¢ÛÁ¿Í²ÄڵIJÐÒº±ØÐë³åÏ´ÏÂÀ´£¬·ÅÈë ÈÝÁ¿Æ¿ÖÐ
¢Ü³ÆÁ¿µÄ¹ÌÌå(»òÁ¿È¡µÄÒºÌå)¿ÉÖ±½Ó·ÅÈëÈÝÁ¿Æ¿ÖÐÈܽâ(»òÏ¡ÊÍ)
¢ÝÒýÁ÷ʱ£¬²£Á§°ô²»ÄÜ¿¿ÔÚÆ¿¿ÚÉÏ
¢Þ¶¨ÈÝÒ¡ÔȺó£¬ÈôÒºÃæµÍÓڿ̶ÈÏߣ¬¿ÉÔٴμÓË®²¹Æë 
A£®¢Ù¢Ú¢Ý  
B£®¢Ù¢Û¢Ü
C£®¢Ú¢Ü¢Þ  
D£®¢Ü¢Ý¢Þ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÑ§Ï°¸ßÊÖ±ØÐÞÒ»»¯Ñ§È˽̰æ È˽̰æ ÌâÐÍ£º021

ÏÂÁйØÓÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄ˵·¨£¬ÕýÈ·µÄ×éºÏÊÇ

¢ÙÍÐÅÌÌìƽ¿É¶ÁÈ¡µ½Ð¡Êýµãºóһλ(ÒÔ¿ËΪµ¥Î»)£¬ÈÝÁ¿Æ¿¿É¾«È·µ½Ð¡ÊýµãºóÁ½Î»(ÒÔºÁÉýΪµ¥Î»)

¢ÚÍÐÅÌÌìƽֻÄÜ´ÖÂԵسÆÁ¿ÎïÖʵÄÖÊÁ¿£¬Á¿Í²Ö»ÄÜ´ÖÂÔµØÁ¿È¡ÒºÌåµÄÌå»ý£¬ÑϸñµØ˵£¬ËüÃǶ¼²»ÄÜÓëÈÝÁ¿Æ¿¡ª¡ª¾«È·ÒÇÆ÷ÅäÌ×ʹÓÃ

¢ÛÁ¿Í²ÄڵIJÐÒº±ØÐë³åÏ´ÏÂÀ´£¬·ÅÈëÈÝÁ¿Æ¿ÖÐ

¢Ü³ÆÁ¿µÄ¹ÌÌå(»òÁ¿È¡µÄÒºÌå)¿ÉÖ±½Ó·ÅÈëÈÝÁ¿Æ¿ÖÐÈܽâ(»òÏ¡ÊÍ)

¢ÝÒýÁ÷ʱ£¬²£Á§°ô²»ÄÜ¿¿ÔÚÆ¿¿ÚÉÏ

¢Þ¶¨ÈÝÒ¡ÔȺó£¬ÈôÒºÃæµÍÓڿ̶ÈÏߣ¬¿ÉÔٴμÓË®²¹Æë

A£®

¢Ù¢Ú¢Ý

B£®

¢Ù¢Û¢Ü

C£®

¢Ú¢Ü¢Þ

D£®

¢Ü¢Ý¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁйØÓÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄ˵·¨£¬ÕýÈ·µÄ×éºÏÊÇ£¨    £©

¢ÙÍÐÅÌÌìƽ¿É¶ÁÈ¡µ½Ð¡Êýµãºóһ루ÒÔ¿ËΪµ¥Î»£©£¬ÈÝÁ¿Æ¿¿É¾«È·µ½Ð¡ÊýµãºóÁ½Î»£¨ÒÔºÁÉýΪµ¥Î»£©  ¢ÚÍÐÅÌÌìƽֻÄÜ´ÖÂԵسÆÁ¿ÎïÖʵÄÖÊÁ¿£¬Á¿Í²Ö»ÄÜ´ÖÂÔµØÁ¿È¡ÒºÌåµÄÌå»ý£¬ÑϸñµØ˵£¬ËüÃǶ¼²»ÄÜÓëÈÝÁ¿Æ¿¡ª¡ª¾«È·ÒÇÆ÷ÅäÌ×ʹÓà ¢ÛÁ¿Í²ÄڵIJÐÒº±ØÐë³åÏ´ÏÂÀ´£¬·ÅÈëÈÝÁ¿Æ¿ÖР ¢Ü³ÆÁ¿µÄ¹ÌÌ壨»òÁ¿È¡µÄÒºÌ壩¿ÉÖ±½Ó·ÅÈëÈÝÁ¿Æ¿ÖÐÈܽ⣨»òÏ¡ÊÍ£©  ¢ÝÒýÁ÷ʱ£¬²£Á§°ô²»ÄÜ¿¿ÔÚÆ¿¿ÚÉÏ  ¢Þ¶¨ÈÝÒ¡ÔȺó£¬ÈôÒºÃæµÍÓڿ̶ÈÏߣ¬¿ÉÔٴμÓË®²¹Æë

A.¢Ù¢Ú¢Ý               B.¢Ù¢Û¢Ü                    C.¢Ú¢Ü¢Þ                D.¢Ü¢Ý¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ÄêÖØÇìµÚÊ®°ËÖÐѧ¸ßÒ»ÉÏѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁйØÓÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜҺ˵·¨ÕýÈ·µÄÊÇ£¨    £©

A£®°Ñ40gNaOHÈܽâÓÚ1LË®Öм´µÃµ½1mol/LµÄµÄNaOHÈÜÒº
B£®½«4.48LNH3ÈܽâÓÚË®Åä³É1LÈÜÒº¼´µÃµ½0.2mol/LµÄ°±Ë®ÈÜÒº
C£®ÅäÖÆ0.2mol/LµÄµÄCuSO4ÈÜÒº480mL£¬ÐèÁòËáÍ­¾§ÌåµÄÖÊÁ¿Îª25g
D£®ÅäÖÆ1L0.2mol/LµÄH2SO4ÈÜÒº£¬¼ÓË®¶¨ÈÝʱÑöÊӿ̶ÈÏß²Ù×÷£¬ÔòÅä³öŨ¶ÈÆ«¸ß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015½ìÖØÇìµÚÊ®°ËÖÐѧ¸ßÒ»ÉÏѧÆÚµÚÒ»´ÎÔ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁйØÓÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜҺ˵·¨ÕýÈ·µÄÊÇ£¨    £©

A£®°Ñ40gNaOHÈܽâÓÚ1LË®Öм´µÃµ½1mol/LµÄµÄNaOHÈÜÒº

B£®½«4.48LNH3ÈܽâÓÚË®Åä³É1LÈÜÒº¼´µÃµ½0.2mol/LµÄ°±Ë®ÈÜÒº

C£®ÅäÖÆ0.2mol/LµÄµÄCuSO4ÈÜÒº480mL£¬ÐèÁòËáÍ­¾§ÌåµÄÖÊÁ¿Îª25g

D£®ÅäÖÆ1L0.2mol/LµÄH2SO4ÈÜÒº£¬¼ÓË®¶¨ÈÝʱÑöÊӿ̶ÈÏß²Ù×÷£¬ÔòÅä³öŨ¶ÈÆ«¸ß

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁйØÓÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜҺ˵·¨ÕýÈ·µÄÊÇ


  1. A.
    °Ñ40gNaOHÈܽâÓÚ1LË®Öм´µÃµ½1mol/LµÄµÄNaOHÈÜÒº
  2. B.
    ½«4.48LNH3ÈܽâÓÚË®Åä³É1LÈÜÒº¼´µÃµ½0.2mol/LµÄ°±Ë®ÈÜÒº
  3. C.
    ÅäÖÆ0.2mol/LµÄµÄCuSO4ÈÜÒº480mL£¬ÐèÁòËáÍ­¾§ÌåµÄÖÊÁ¿Îª25g
  4. D.
    ÅäÖÆ1L0.2mol/LµÄH2SO4ÈÜÒº£¬¼ÓË®¶¨ÈÝʱÑöÊӿ̶ÈÏß²Ù×÷£¬ÔòÅä³öŨ¶ÈÆ«¸ß

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸