ÌâÄ¿ÄÚÈÝ

ijÑо¿Ð¡×é²â¶¨²¤²ËÖвÝËá¼°²ÝËáÑκ¬Á¿£¨ÒÔC2O42£­¼Æ£©£¬ÊµÑé²½ÖèÈçÏ£º
¢Ù½«²¤²ËÑùÆ·Ô¤´¦Àí£¬µÃµ½º¬ÓвÝËá¼°²ÝËáÑεÄÈÜÒº¡£
¢Úµ÷½ÚÈÜÒºÖÁÈõ¼îÐÔ£¬µÎ¼Ó×ãÁ¿CaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£¬¹ýÂ˵õ½CaC2O4¹ÌÌå¡£
¢ÛÓÃÏ¡HClÈܽâCaC2O4Åä³É100mLÈÜÒº£¬È»ºóÓÃKMnO4±ê×¼ÈÜÒºµÎ¶¨2¡«3´Î¡£
£¨1£©²½Öè¢ÙÖС°¡°ÑùÆ·Ô¤´¦Àí¡±µÄ·½·¨ÊÇ      ¡£
A£®×ÆÉճɻҡ¢½þÅÝ¡¢¹ýÂË¡¡¡¡    B£®ÑÐÄ¥Õ¥Ö­¡¢½þÅÝ¡¢¹ýÂË
£¨2£©²½Öè¢ÚÖС°µ÷½ÚÈÜÒºÖÁÈõ¼îÐÔ¡±µÄÄ¿µÄÊÇ______________________   ¡£
£¨3£©²½Öè¢Ú½øÐÐÖÐÓ¦¼°Ê±ÑéÖ¤CaCl2ÈÜÒºÒÑ¡°×ãÁ¿¡±£¬²Ù×÷·½·¨ºÍÏÖÏóÊÇ£º         _¡£
£¨4£©²½Öè¢ÛÖÐÓõ½µÄ²£Á§ÒÇÆ÷³ýÉÕ±­¡¢×¶ÐÎÆ¿¡¢½ºÍ·µÎ¹Ü¡¢²£Á§°ôÍ⣬»¹ÓС¡  ¡¡¡¡¡£
£¨5£©±¾ÊµÑéÖе樲»ÐèÒªÁí¼Óָʾ¼Á£¬µÎ¶¨ÖÕµãµÄÏÖÏóÊÇ______________________¡£

£¨1£©B¡¡£¨3·Ö£©
£¨2£©½«H2C2O4ת»¯ÎªC2O42£­£¬·ñÔòÎÞ·¨×ª»¯Îª³Áµí £¨3·Ö£©
£¨3£©¾²Ö㬴ýÉϲãÈÜÒº³ÎÇåºó¼ÌÐøµÎ¼ÓCaCl2ÈÜÒº£¬²»ÔÙ³öÏÖ»ë×Ç˵Ã÷CaCl2ÒÑ×ãÁ¿£¨3·Ö£©
£¨4£©ËáʽµÎ¶¨¹Ü¡¢ÈÝÁ¿Æ¿£¨4·Ö£©
£¨5£©ÈÜÒºÓÉÎÞÉ«±ä³Édz×ÏÉ«ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2010?¹ãÖÝһģ£©Ä³Ñо¿Ð¡×é²â¶¨²¤²ËÖвÝËá¼°²ÝËáÑκ¬Á¿£¨ÒÔC2O42-¼Æ£©£¬ÊµÑé²½ÖèÈçÏ£º
¢Ù½«²¤²ËÑùÆ·Ô¤´¦Àíºó£¬ÈÈË®½þÅÝ£¬¹ýÂ˵õ½º¬ÓвÝËá¼°²ÝËáÑεÄÈÜÒº£®
¢Úµ÷½ÚÈÜÒºµÄËá¼îÐÔ£¬µÎ¼Ó×ãÁ¿CaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£»¼ÓÈë×ãÁ¿´×ËᣬʹCaCO3Èܽ⣻¹ýÂ˵õ½CaC2O4¹ÌÌ壮
¢ÛÓÃÏ¡HClÈܽâCaC2O4£¬²¢¼ÓË®ÅäÖƳÉ100mLÈÜÒº£®Ã¿´Î׼ȷÒÆÈ¡È¡25.00mL¸ÃÈÜÒº£¬ÓÃ0.0100mol?L-1KMnO4±ê×¼ÈÜÒºµÎ¶¨£¬Æ½¾ùÏûºÄ±ê×¼ÈÜÒºVmL£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²½Öè¢ÙÖС°ÑùÆ·Ô¤´¦Àí¡±µÄ·½·¨ÊÇ
B
B
£®£¨A£®×ÆÉճɻҠB£®ÑÐÄ¥³ÉÖ­£©
£¨2£©²½Öè¢ÚÖС°µ÷½ÚÈÜÒºµÄËá¼îÐÔ¡±ÖÁ
A£¨»òC£©
A£¨»òC£©
£®£¨A£®Èõ¼îÐÔ B£®ÈõËáÐÔ C£®ÖÐÐÔ£©ÑéÖ¤CaCl2ÈÜÒºÒÑ¡°×ãÁ¿¡±µÄ²Ù×÷ºÍÏÖÏóÊÇ£º
¾²Ö㬴ýÉϲãÈÜÒº³ÎÇåºóµÎÈëCaCl2ÈÜÒº£¬²»³öÏÖ»ë×Ç
¾²Ö㬴ýÉϲãÈÜÒº³ÎÇåºóµÎÈëCaCl2ÈÜÒº£¬²»³öÏÖ»ë×Ç

£¨3£©²½Öè¢ÛÖÐÓõ½µÄ²£Á§ÒÇÆ÷³ýÉÕ±­¡¢×¶ÐÎÆ¿¡¢½ºÍ·µÎ¹Ü¡¢²£Á§°ôÍ⣬»¹ÓÐ
ËáʽµÎ¶¨¹Ü¡¢100mLÈÝÁ¿Æ¿
ËáʽµÎ¶¨¹Ü¡¢100mLÈÝÁ¿Æ¿
£®
£¨4£©¾­¼ÆË㣬¸Ã²¤²ËÑùÆ·ÖÐC2O42-µÄ×ÜÎïÖʵÄÁ¿Îª
V¡Á10-4mol
V¡Á10-4mol
£¬ÈôÑùÆ·µÄÖÊÁ¿Îªmg£¬Ôò²¤²ËÖвÝËá¼°²ÝËáÑΣ¨ÒÔC2O42-¼Æ£©µÄÖÊÁ¿·ÖÊýΪ
8.8¡Á10-3V
m
¡Á100%
8.8¡Á10-3V
m
¡Á100%
£¬£¨ÒÑÖª£º2MnO4-+5C2O42-+16H+=2Mn2++10CO2¡ü+8H2O£¬C2O42-µÄĦ¶ûÖÊÁ¿Îª88g?mol-1£©
ijÑо¿Ð¡×é²â¶¨²¤²ËÖвÝËá¼°²ÝËáÑκ¬Á¿£¨ÒÔC2O42-¼Æ£©£¬ÊµÑé²½ÖèÈçÏ£º
¢Ù½«²¤²ËÑùÆ·Ô¤´¦Àíºó£¬ÈÈË®½þÅÝ£¬¹ýÂ˵õ½º¬ÓвÝËá¼°²ÝËáÑεÄÈÜÒº¡£
¢Úµ÷½ÚÈÜÒºµÄËá¼îÐÔ£¬µÎ¼Ó×ãÁ¿CaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£»¼ÓÈë×ãÁ¿´×ËᣬʹCaCO3Èܽ⣻¹ýÂ˵õ½CaC2O4¹ÌÌå¡£
¢ÛÓÃÏ¡HClÈܽâCaC2O4£¬²¢¼ÓË®ÅäÖƳÉl00mLÈÜÒº¡£Ã¿´Î׼ȷÒÆÈ¡25.00mL ¸ÃÈÜÒº£¬ÓÃ0.0100 mol¡¤L-1 KMnO4±ê×¼ÈÜÒºµÎ¶¨£¬Æ½¾ùÏûºÄ±ê×¼ÈÜÒºV mL¡£ »Ø´ðÏÂÁÐÎÊÌ⣺
(1)²½Öè¢ÙÖС°ÑùÆ·Ô¤´¦Àí¡±µÄ·½·¨ÊÇ___________¡££¨A£®×ÆÉճɻÒB£®ÑÐÄ¥³ÉÖ­£©
(2)²½Öè¢ÚÖС°µ÷½ÚÈÜÒºµÄËá¼îÐÔ¡±ÖÁ___________¡££¨A£®Èõ¼îÐÔB£®ÈõËáÐÔC£®ÖÐÐÔ£© ÑéÖ¤CaCl2ÈÜÒºÒÑ¡°×ãÁ¿¡±µÄ²Ù×÷ºÍÏÖÏóÊÇ£º____________¡£
(3)²½Öè¢ÛÖÐÓõ½µÄ²£Á§ÒÇÆ÷³ýÉÕ±­¡¢×¶ÐÎÆ¿¡¢½ºÍ·µÎ¹Ü¡¢²£Á§°ôÍ⣬»¹ÓÐ__________ ¡£
(4)¾­¼ÆË㣬¸Ã²¤²ËÑùÆ·ÖÐC2O42-µÄ×ÜÎïÖʵÄÁ¿Îª__________£¬ÈôÑùÆ·µÄÖÊÁ¿Îªmg£¬Ôò ²¤²ËÖвÝËá¼°²ÝËáÑÎ(ÒÔC2O42-¼Æ£©µÄÖÊÁ¿·ÖÊýΪ___________ ¡£
£¨ÒÑÖª£º2MnO4-+ 5C2O42-+16H+=2Mn2+ +10CO2¡ü+8H2O£¬C2O42-µÄĦ¶ûÖÊÁ¿Îª88 g¡¤mol-1£©

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

¾«Ó¢¼Ò½ÌÍø