ij½ðÊôMµÄÇâÑõ»¯ÎïµÄ½á¾§Ë®ºÏÎïM£¨OH£©2¡¤xH2OÓëNa2CO3µÄ»ìºÏÎï¹²36.8g£¬¼ÓÈë×ãÁ¿µÄË®ºó£¬Éú³É°×É«³Áµí£¨³ÁµíÖв»º¬½á¾§Ë®£©£¬½«³Áµí¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É£¬³ÆµÃÆäÖÊÁ¿Îª9.85g£¬½«µÃµ½µÄ³Áµí¸ßÎÂ×ÆÉÕºó£¬ÖÊÁ¿±äΪ7.65g£¬ÂËÒºÓëËá×÷Óò»²úÉúÆøÌ壬ÈôÓÃ×ãÁ¿µÄï§ÑÎÓëÂËÒº¹²ÈÈ£¬Ôò²úÉú4.48LÆøÌ壨±ê×¼×´¿ö£©¡£
£¨1£©ÂËÒºÖÐn£¨OH££©=____________mol
£¨2£©MµÄÏà¶ÔÔ×ÓÖÊÁ¿_____________________
£¨3£©MµÄÇâÑõ»¯ÎïµÄ½á¾§Ë®ºÏÎïµÄ»¯Ñ§Ê½Îª____________________
£¨4£©½«MµÄ´ËÇâÑõ»¯ÎïÓëNH4ClÖÃÓÒͼËùʾװÖ㨲£Á§Æ¬ÖÐÓëÉÕ±µ×²¿Ö®¼äÓÐÒ»±¡²ãË®£©ÖлìºÏ·´Ó¦ºó£¬ÓÃÊÖÄÃÆðÉձʱ£¬²£Á§Æ¬½«___________________£¬Óë¸Ã·´Ó¦¶ÔÓ¦µÄÄÜÁ¿±ä»¯¹ØϵͼÊÇ_________
£¨1£©0.2£¬£¨2£©137 £¨3£© Ba£¨OH£©2¡¤8H2O £¨4£© ËæÉÕ±Ò»Æð±»ÄÃÆðÀ´£¬A
¡¾½âÎö¡¿£ºÓÉÉú³ÉµÄ°±Æø¿ÉÇóµÃn£¨OH££©=0.2mol/L£¬ÒòOH£ÔÚÇ°Ò»½×¶ÎûÓвÎÓë·´Ó¦£¬¹ÊÖªM£¨OH£©2¡¤xH2OΪ0.1mol/L£¬ÓÉÂËÒºÓëËá×÷Óò»²úÉúÆøÌåÖªM£¨OH£©2¡¤xH2OÓëNa2CO3ʱ£¬ºóÕßÈ«²¿·´Ó¦ÍêÁË¡£ÓÖ³ÁµíÊÇ̼ËáÑΣ¬ÓÉMCO3MO+CO2¡ü¿ÉÇóµÃMµÄÏà¶ÔÔ×ÓÖÊÁ¿Îª137£¬¹ÊMÊDZµÔªËØ¡£ÓÉÉú³ÉµÄ9.85gBaCO3¿ÉÇóµÃn£¨Na2CO3£©=0.05mL£¬ÔÙ½áºÏ»ìºÏÎï×ÜÖÊÁ¿¿ÉÇó³öx=8¡£Ba£¨OH£©2¡¤8H2OÓëNH4Cl·´Ó¦Ê±»áÎüÊÕ´óÁ¿µÄÈȶøµ¼ÖÂË®½á±ù£¬½«ÉÕ±Óë²£Á§Æ¬Õ´½áÔÚÒ»Æ𣬸÷´Ó¦ÊÇÎüÈÈ·´Ó¦£¬Éú³ÉÎïÄÜÁ¿¸ßÓÚ·´Ó¦Îï¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(1)MµÄÇâÑõ»¯ÎïÓëNaHCO3·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º?
¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£?
(2)Òª¼ÆËãMµÄÏà¶ÔÔ×ÓÖÊÁ¿£¬ÄãÈÏΪ»¹±ØÐëÌṩÏÂÁÐÄÄÏîÊý¾Ý (Ìî×ÖĸÐòºÅ)¡£
A.MµÄÇâÑõ»¯ÎïÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È(ÉèΪ2 mol¡¤L-1?)?
B.MµÄ̼ËáÑεÄÖÊÁ¿(ÉèΪ39.4 g)?
C.ÓëMµÄ̼ËáÑη´Ó¦µÄÑÎËáµÄÎïÖʵÄÁ¿Å¨¶È(ÉèΪ0.1 mol¡¤L-1?)?
D.ÌâÉèÊý¾Ý³ä×㣬²»ÐèÒª²¹³äÊý¾Ý?
¸ù¾ÝÄãµÄÑ¡Ôñ£¬¼ÆËã½ðÊôMµÄÏà¶ÔÔ×ÓÖÊÁ¿ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£?
(3)Çë¼ÆËãÓëÂËÒº·´Ó¦µÄÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ mol¡¤L-1¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ij½ðÊôMµÄÇâÑõ»¯ÎïµÄ½á¾§Ë®ºÏÎïM(OH)2¡¤xH2OÓëNa2CO3µÄ»ìºÏÎï¹²36.800 g£¬¼ÓÈë×ãÁ¿µÄË®ºó£¬Éú³É°×É«³Áµí(³Áµí²»º¬½á¾§Ë®)£¬½«³ÁµíÂ˳ö£¬Ï´¾»ºæ¸É£¬³ÆµÃÆäÖÊÁ¿Îª9.850 g¡£½«µÃµ½µÄ³Áµí¸ßÎÂ×ÆÉÕºó£¬ÖÊÁ¿±äΪ7.650 g£»ÂËÒºÓëËá×÷Óò»²úÉúÆøÌ壻ÈôÓÃ×ãÁ¿µÄï§ÑÎÓëÂËÒº¹²ÈÈ£¬Ôò²úÉú4.48 LÆøÌå(±ê×¼×´¿ö)¡£Çó£º
¢ÅÂËÒºÖÐOH£µÄÎïÖʵÄÁ¿Îª__________mol¡£
¢ÆÈôMµÄÖÐ×ÓÊýΪ81£¬ÔòMµÄÔªËØ·ûºÅΪ____________¡£
¢ÇMµÄÇâÑõ»¯ÎïµÄ½á¾§Ë®ºÏÎïµÄ»¯Ñ§Ê½Îª______________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ij½ðÊôMµÄÇâÑõ»¯ÎïµÄ½á¾§Ë®ºÏÎïM£¨OH£©2¡¤xH2OÓëNa2CO3µÄ»ìºÏÎï¹²36.8g£¬¼ÓÈë×ãÁ¿µÄË®ºó£¬Éú³É°×É«³Áµí£¨³ÁµíÖв»º¬½á¾§Ë®£©£¬½«³Áµí¹ýÂË¡¢Ï´µÓ¡¢ºæ¸É£¬³ÆµÃÆäÖÊÁ¿Îª9.85g£¬½«µÃµ½µÄ³Áµí¸ßÎÂ×ÆÉÕºó£¬ÖÊÁ¿±äΪ7.65g£¬ÂËÒºÓëËá×÷Óò»²úÉúÆøÌ壬ÈôÓÃ×ãÁ¿µÄï§ÑÎÓëÂËÒº¹²ÈÈ£¬Ôò²úÉú4.48LÆøÌ壨±ê×¼×´¿ö£©¡£
£¨1£©ÂËÒºÖÐn£¨OH££©=____________mol
£¨2£©MµÄÏà¶ÔÔ×ÓÖÊÁ¿_____________________
£¨3£©MµÄÇâÑõ»¯ÎïµÄ½á¾§Ë®ºÏÎïµÄ»¯Ñ§Ê½Îª____________________
£¨4£©½«MµÄ´ËÇâÑõ»¯ÎïÓëNH4ClÖÃÓÒͼËùʾװÖ㨲£Á§Æ¬ÖÐÓëÉÕ±µ×²¿Ö®¼äÓÐÒ»±¡²ãË®£©ÖлìºÏ·´Ó¦ºó£¬ÓÃÊÖÄÃÆðÉձʱ£¬²£Á§Æ¬½«___________________£¬Óë¸Ã·´Ó¦¶ÔÓ¦µÄÄÜÁ¿±ä»¯¹ØϵͼÊÇ_________
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com