ÌâÄ¿ÄÚÈÝ

ijÑо¿ÐÔѧϰС×éÉè¼ÆÏÂͼװÖÃÖÆÈ¡ÂÈÆø²¢ÒÔÂÈÆøÎªÔ­ÁϽøÐÐʵÑé¡£

£¨1£©×°ÖÃAÉÕÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                                    ¡£

£¨2£©×°ÖÃBÖб¥ºÍʳÑÎË®µÄ×÷ÓÃÊÇ                                                    £¬×°ÖÃCÖÐŨÁòËáµÄ×÷ÓÃÊÇ                                                         ¡£

£¨3£©ÊµÑéʱ£¬Ïȵãȼ                ´¦µÄ¾Æ¾«µÆ£¬ÔÙµãȼ                  ´¦¾Æ¾«µÆ£¬Ð´³öDÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ                                                                                         ¡£Ð´³öEÖз´Ó¦µÄÀë×Ó·½³Ìʽ                                                                                                ¡£

£¨4£©ÈôʵÑéÖÐÓÃ12mol?L£­1ŨÑÎËá10mLÓë×ãÁ¿µÄMnO2·´Ó¦£¬Éú³ÉCl2µÄÎïÖʵÄÁ¿×ÜÊÇСÓÚ0.03mol£¬ÊÔ·ÖÎö¿ÉÄÜ´æÔÚµÄÔ­ÒòÊÇ¢Ù                                                                   £¬¢Ú                                                                     ¡£Óûʹ·´Ó¦Éú³ÉCl2µÄÎïÖʵÄÁ¿×î´ó³Ì¶ÈµÄ½Ó½ü0.03mol£¬ÔòÔÚ×°ÖÃÆøÃÜÐÔÁ¼ºÃµÄǰÌáÏÂʵÑéÖÐÓ¦²ÉÈ¡µÄ´ëÊ©ÊÇ                                    ¡£

£¨1£©

£¨2£©³ýÈ¥HCl              ÎüÊÕË®ÕôÆø

£¨3£©A          D                 

£¨4£©HCl»Ó·¢       ŨÑÎËá±äÏ¡²»ÔÙ·´Ó¦     ÂýÂýµÎ¼ÓŨÑÎËᣬ»º»º¼ÓÈÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÈËÌåѪҺÀïCa2+µÄŨ¶ÈÒ»°ã²ÉÓÃmg/cm3À´±íʾ£®³éȡһ¶¨Ìå»ýµÄѪÑù£¬¼ÓÊÊÁ¿µÄ²ÝËáï§[£¨NH4£©2C2O4]ÈÜÒº£¬¿ÉÎö³ö²ÝËá¸Æ£¨CaC2O4£©³Áµí£¬½«´Ë²ÝËá¸Æ³ÁµíÏ´µÓºóÈÜÓÚÇ¿Ëá¿ÉµÃ²ÝËᣨH2C2O4£©£¬ÔÙÓÃKMnO4ÈÜÒºµÎ¶¨¼´¿É²â¶¨ÑªÒºÑùÆ·ÖÐCa2+µÄŨ¶È£®Ä³Ñо¿ÐÔѧϰС×éÉè¼ÆÈçÏÂʵÑé²½Öè²â¶¨ÑªÒºÑùÆ·ÖÐCa2+µÄŨ¶È£®
ÅäÖÆKMnO4±ê×¼ÈÜÒºÈçͼËùʾÊÇÅäÖÆ50mL KMnO4 ±ê×¼ÈÜÒºµÄ¹ý³ÌʾÒâͼ£®

£¨1£©ÇëÄã¹Û²ìͼʾÅжϣ¬ÆäÖв»ÕýÈ·µÄ²Ù×÷ÓУ¨ÌîÐòºÅ£©
¢Ú¢Ý
¢Ú¢Ý
£»
£¨2£©ÆäÖÐÈ·¶¨50mLÈÜÒºÌå»ýµÄÈÝÆ÷ÊÇ£¨ÌîÃû³Æ£©
ÈÝÁ¿Æ¿
ÈÝÁ¿Æ¿
£»
£¨3£©Èç¹ûÓÃͼʾµÄ²Ù×÷ÅäÖÆÈÜÒº£¬ËùÅäÖÆµÄÈÜҺŨ¶È½«
ƫС
ƫС
£¨Ìî¡°Æ«´ó¡±»ò¡°Æ«Ð¡¡±£©£®
²â¶¨ÑªÒºÑùÆ·ÖÐCa2+µÄŨ¶È£º³éȡѪÑù20.00mL£¬¾­¹ýÉÏÊö´¦ÀíºóµÃµ½²ÝËᣬÔÙÓÃ0.020mol/L KMnO4 ÈÜÒºµÎ¶¨£¬Ê¹²ÝËáת»¯³ÉCO2Òݳö£¬Õâʱ¹²ÏûºÄ12.00mL KMnO4ÈÜÒº£®
£¨4£©ÒÑÖª²ÝËá¸úKMnO4·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2MnO
 
-
4
+5H2C2O4+6H+¨T2Mnx++10CO2¡ü+8H2O
Ôò·½³ÌʽÖеÄx=
2
2
£®
£¨5£©¾­¹ý¼ÆË㣬ѪҺÑùÆ·ÖÐCa2+µÄŨ¶ÈΪ
1.2
1.2
mg/cm3£®
ÈçͼÊÇijÑо¿ÐÔѧϰС×éÉè¼ÆµÄ¶ÔÒ»ÖַϾɺϽðµÄ¸÷³É·Ö£¨º¬ÓÐCu¡¢Fe¡¢Si ÈýÖֳɷ֣©½øÐзÖÀë¡¢»ØÊÕÔÙÀûÓõĹ¤ÒµÁ÷³Ì£¬Í¨¹ý¸ÃÁ÷³Ì½«¸÷³É·Öת»¯Îª³£Óõĵ¥Öʼ°»¯ºÏÎ
¾«Ó¢¼Ò½ÌÍø
ÒÑÖª£º298Kʱ£¬Ksp[Cu£¨OH£©2]=2.2¡Á10-20£¬Ksp[Fe£¨OH£©3]=4.0¡Á10-38£¬Ksp[Mn£¨OH£©2]=1.9¡Á10-13£¬¸ù¾ÝÉÏÃæÁ÷³Ì»Ø´ðÓйØÎÊÌ⣺
£¨1£©²Ù×÷¢ñ¡¢¢ò¡¢¢óÖ¸µÄÊÇ
 
£®
£¨2£©¼ÓÈë¹ýÁ¿FeCl3ÈÜÒº¹ý³ÌÖпÉÄÜÉæ¼°µÄ»¯Ñ§·½³Ìʽ£º
 
£®
£¨3£©¹ýÁ¿µÄ»¹Ô­¼ÁÓ¦ÊÇ
 
£®
£¨4£©¢ÙÏòÈÜÒºbÖмÓÈëËáÐÔKMnO4ÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
 
£®
¢ÚÈôÓÃX mol/L KMnO4ÈÜÒº´¦ÀíÈÜÒºb£¬µ±Ç¡ºÃ½«ÈÜÒºÖеÄÑôÀë×ÓÍêÈ«Ñõ»¯Ê±ÏûºÄKMnO4ÈÜÒºY mL£¬Ôò×îºóËùµÃºìרɫ¹ÌÌåCµÄÖÊÁ¿Îª
 
g£¨Óú¬X¡¢YµÄ´úÊýʽ±íʾ£©£®
£¨5£©³£ÎÂÏ£¬ÈôÈÜÒºcÖÐËùº¬µÄ½ðÊôÑôÀë×ÓŨ¶ÈÏàµÈ£¬ÏòÈÜÒºcÖÐÖðµÎ¼ÓÈëKOHÈÜÒº£¬ÔòÈýÖÖ½ðÊôÑôÀë×Ó³ÁµíµÄÏȺó˳ÐòΪ£º
 
£®£¨Ìî½ðÊôÑôÀë×Ó£©
£¨6£©×îºóÒ»²½µç½âÈôÓöèÐԵ缫µç½âÒ»¶Îʱ¼äºó£¬Îö³ö¹ÌÌåBµÄÖÊÁ¿ÎªZ g£¬Í¬Ê±²âµÃÒõÑôÁ½¼«ÊÕ¼¯µ½µÄÆøÌåÌå»ýÏàµÈ£¬Ôò±ê¿öÏÂÑô¼«Éú³ÉµÄ×îºóÒ»ÖÖÆøÌåÌå»ýΪ
 
L£¨Óú¬ZµÄ´úÊýʽ±íʾ£©£»¸Ãµç¼«µÄ·´Ó¦Ê½Îª
 
£®

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

¾«Ó¢¼Ò½ÌÍø