ÌâÄ¿ÄÚÈÝ

13£®Ä³Í¬Ñ§ÓÃ18mol/LµÄŨÁòËá ÅäÖÆ240mL 0.9mol/LµÄÏ¡ÁòËᣬ²¢½øÐÐÓйØʵÑ飮Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÐèÒªÁ¿È¡Å¨ÁòËá12.5mL£®
£¨2£©´ÓÏÂÁÐÓÃÆ·ÖÐÑ¡³öʵÑéËùÐèÒªµÄÒÇÆ÷ΪBCFI£¨ÌîÐòºÅ£©£®
A£®500mLÉÕ±­£»    B£®100mLÉÕ±­£»    C£®25mLÁ¿Í²£» D£®100mLÁ¿Í²£»
E£®500mLÈÝÁ¿Æ¿£»  F£®250mLÈÝÁ¿Æ¿£»   G£®¹ã¿ÚÆ¿£»     H£®ÍÐÅÌÌìƽ£»
I£®²£Á§°ô
³ýÑ¡ÓÃÉÏÊöÒÇÆ÷Í⣬ÉÐȱÉٵıØÒªµÄÒÇÆ÷»òÓÃÆ·ÊǽºÍ·µÎ¹Ü£®
£¨3£©ÔÚÅäÖƹý³ÌÖУ¬Èô¶¨ÈÝʱ¸©Êӿ̶ÈÏߣ¬ÔòËùµÃÈÜҺŨ¶È´óÓÚ0.9mol•L-10.9mol•L-1£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©£®
£¨4£©È¡ËùÅäÖƵÄÏ¡ÁòËá100mL£¬ÓëÒ»¶¨ÖÊÁ¿µÄп³ä·Ö·´Ó¦£¬Ð¿È«²¿Èܽâºó£¬Éú³ÉµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ0.448L£¬Ôò²Î¼Ó·´Ó¦µÄпµÄÖÊÁ¿Îª1.3g£»Éè·´Ó¦ºóÈÜÒºµÄÌå»ýÈÔΪ100mL£¬Ôò·´Ó¦ºóÈÜÒºÖÐH+µÄÎïÖʵÄÁ¿Å¨¶ÈΪ1.4mol/L£®

·ÖÎö £¨1£©ÒÀ¾ÝÅäÖÆÈÜÒºµÄÌå»ýÑ¡ÔñºÏÊʵÄÈÝÁ¿Æ¿£¬ÒÀ¾ÝÈÜҺϡÊ͹ý³ÌÖÐËùº¬ÈÜÖʵÄÎïÖʵÄÁ¿²»±ä¼ÆËãÐèҪŨÁòËáÌå»ý£»
£¨2£©¸ù¾ÝÓÃŨÈÜÒºÅäÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÏ¡ÈÜÒºµÄ²Ù×÷²½ÖèѡȡʵÑéÒÇÆ÷£»
£¨3£©ÔÚÅäÖƹý³ÌÖУ¬Èô¶¨ÈÝʱ¸©Êӿ̶ÈÏßµ¼ÖÂÈÜÒºÌå»ýƫС£¬ÒÀ¾ÝC=$\frac{n}{V}$½øÐзÖÎö£»
£¨4£©¸ù¾Ý·½³Ìʽ¼ÆËã²Î¼Ó·´Ó¦µÄп¡¢ÁòËáµÄÎïÖʵÄÁ¿£»¸ù¾Ým=nM¼ÆËãпµÄÖÊÁ¿£¬ÈÜÒºÖÐn£¨H+£©=n£¨H2SO4£©£¬¼ÆËã³öÁòËáµÄÎïÖʵÄÁ¿Å¨¶È±ä»¯Á¿£¬¼´ÇâÀë×ÓŨ¶È±ä»¯Á¿£¬ÇâÀë×Ó¿ªÊ¼Å¨¶È¼õÈ¥ÇâÀë×ÓŨ¶È±ä»¯Á¿µÈÓÚ·´Ó¦ºóÈÜÒºÖеÄÇâÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶È£®

½â´ð ½â£º£¨1£©ÅäÖÆ240mL 0.9mol/LµÄÏ¡ÁòËᣬÐèҪѡÔñ250mLÈÝÁ¿Æ¿£¬ÉèÐèҪŨÁòËáÌå»ýΪV£¬ÔòÒÀ¾ÝÈÜҺϡÊ͹ý³ÌÖÐËùº¬ÈÜÖʵÄÎïÖʵÄÁ¿²»±äµÃ£ºV¡Á18mol/L=0.9mol/L¡Á0.25L£¬½âµÃV=0.0125L£»
¹Ê´ð°¸Îª£º12.5£»
£¨2£©ÓÃÓÃŨÈÜÒºÅäÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÏ¡ÈÜÒºµÄ²Ù×÷²½ÖèΪ£º¼ÆËã¡¢Á¿È¡¡¢Ï¡ÊÍ¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔȵȲÙ×÷£¬Óõ½µÄÒÇÆ÷£ºB£®100mLÉÕ±­£»C£®25mLÁ¿Í²£»F£®250mLÈÝÁ¿Æ¿£»I£®²£Á§°ô£»
»¹È±ÉÙµÄÒÇÆ÷Ϊ£º½ºÍ·µÎ¹Ü£»
¹Ê´ð°¸Îª£ºBCFI£»½ºÍ·µÎ¹Ü£»
£¨3£©ÔÚÅäÖƹý³ÌÖУ¬Èô¶¨ÈÝʱ¸©Êӿ̶ÈÏßµ¼ÖÂÈÜÒºÌå»ýƫС£¬ÒÀ¾ÝC=$\frac{n}{V}$¿ÉÖª£¬ÈÜҺŨ¶ÈÆ«¸ß£»
¹Ê´ð°¸Îª£º´óÓÚ0.9mol•L-1£»
£¨4£©ÇâÆøµÄÎïµÄÁ¿Îª£º$\frac{0.448L}{22.4L/mol}$=0.02mol£¬
Zn+H2SO4=ZnSO4+H2¡ü
 1mol 1mol     1mol
  x    y     0.02mol
ËùÒÔx=0.02mol£»
y=0.02mol£»
ËùÒԲμӷ´Ó¦Ð¿µÄÖÊÁ¿Îª0.02mol¡Á65g/mol=1.3g£»
ÇâÀë×ÓŨ¶È±ä»¯Á¿Îª¡÷c£¨H+£©=$\frac{0.02mol¡Á2}{0.1L}$=0.4mol/L£®
ËùÒÔ·´Ó¦ºóÈÜÒºÖеÄÇâÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈΪ2¡Á0.9mol/L-0.4mol/L=1.4mol/L£¬
¹Ê´ð°¸Îª£º1.3g£»1.4mol/L£®

µãÆÀ ±¾Ì⿼²éÁËÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÅäÖÆ£¬Óйط½³ÌʽµÄ¼ÆË㣬Ã÷È·ÅäÖÆÔ­Àí¡¢²½ÖèºÍÎó²î·ÖÎöµÄ·½·¨ÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø