£¨8·Ö£©ÂÈ»¯¸ÆÊÇÓÃ;¹ã·ºµÄ»¯Ñ§ÊÔ¼Á£¬¿É×÷¸ÉÔï¼Á¡¢À䶳¼Á¡¢·À¶³¼ÁµÈ¡£ÎªÁ˲ⶨijÂÈ»¯¸ÆÑùÆ·ÖиƵĺ¬Á¿£¬½øÐÐÈçÏÂʵÑ飺
¢Ù׼ȷ³ÆÈ¡ÂÈ»¯¸ÆÑùÆ·0.2312g£¬·ÅÈëÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿6mol/LµÄÑÎËáºÍÊÊÁ¿ÕôÁóˮʹÑùÆ·ÍêÈ«Èܽ⣬ÔٵμÓ35mL0.25mol/L (NH4)2C2O4ÈÜÒº£¬Ë®Ô¡¼ÓÈÈ£¬Öð½¥Éú³ÉCaC2O4³Áµí£¬¾­¼ìÑ飬Ca2+ÒѳÁµíÍêÈ«¡£
¢Ú¹ýÂ˲¢Ï´µÓ¢ÙËùµÃ³Áµí¡£
¢Û¼ÓÈë×ãÁ¿µÄ10% H2SO4ºÍÊÊÁ¿µÄÕôÁóË®£¬¢ÚÖгÁµíÍêÈ«Èܽ⣬ÈÜÒº³ÊËáÐÔ£¬¼ÓÈÈÖÁ75¡æ£¬³ÃÈȼÓÈë0.05mol/L KMnO4ÈÜÒº16mL£¬Ç¡ºÃÍêÈ«·´Ó¦¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Åäƽ£º   MnO4£­£«   H2C2O4£«   H£«¨D   Mn2£«£«    CO2¡ü£«   H2O
£¨2£©0.05mol¡¤L£­1KMnO4±ê×¼ÈÜÒºÓ¦ÖÃÓÚ         £¨Ñ¡Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©µÎ¶¨¹ÜÖУ»Åжϵζ¨ÖÕµãµÄÏÖÏóΪ                                                  ¡£
£¨3£©¼ÆËã¸ÃÂÈ»¯¸ÆÑùÆ·ÖиÆÔªËصÄÖÊÁ¿·ÖÊý£¨¾«È·µ½0.01%£©¡£

£¨8·Ö£¬£¨1£©¡¢£¨2£©¸÷2·Ö£¬£¨3£©4·Ö£©
£¨1£©2¡¢5¡¢6¡¢2¡¢10¡¢8
£¨2£©Ëáʽ¡¢µ±¿´µ½µÎÈë1µÎKMnO4ÈÜÒº£¬×¶ÐÎÆ¿ÖÐÈÜÒºÁ¢¼´±ä³É×ϺìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«£¬¼´´ïµ½µÎ¶¨Öյ㡣
£¨3£©34.60%

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÉ½Î÷Ê¡´óͬһÖÐ2008£­2009ѧÄêÉÏѧÆÚ¸ßÈýÆÚÖÐÊÔ¾í(»¯Ñ§) ÌâÐÍ£º038

2008ÄêÎÒ¹úÄÏ·½µØÇøÔâÊÜÁËÀúÊ·ÉϺ±¼ûµÄÌØ´óµÄ±ùÑ©£¬½»Í¨¶ÂÈû£¬µçÍøÖжϣ¬¸øȺÖÚµÄÉú»î´øÀ´ºÜ¶àÀ§ÄÑ£®»¯Ñ§ÊÔ¼ÁÂÈ»¯¸Æ×÷ΪÈÚÑ©¼ÁÔÚ¿¹Ñ©ÔÖÖÐÆðÁËÒ»¶¨µÄ×÷Óã®ÂÈ»¯¸ÆÊÇÓÃ;¹ã·ºµÄ»¯Ñ§ÊÔ¼Á£¬¿É×÷¸ÉÔï¼Á¡¢À䶳¼Á¡¢·À¶³¼ÁµÈ£®ÎªÁ˲ⶨijÂÈ»¯¸ÆÑùÆ·ÖиƵĺ¬Á¿£¬½øÐÐÈçÏÂʵÑ飺

¢Ù׼ȷ³ÆÈ¡ÂÈ»¯¸ÆÑùÆ·0.2312 g£¬·ÅÈëÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿6 mol/LµÄÑÎËáºÍÊÊÁ¿ÕôÁóˮʹÑùÆ·ÍêÈ«Èܽ⣬ÔٵμÓ35.00 mL¡¡0.25 mol/L(NH4)2C2O4ÈÜÒº£¬Ë®Ô¡¼ÓÈÈ£¬Öð½¥Éú³ÉCaC2O4³Áµí£¬¾­¼ìÑ飬Ca2+ÒѳÁµíÍêÈ«£®¢Ú¹ýÂ˲¢Ï´µÓ¢ÙËùµÃ³Áµí£®¢Û¼ÓÈë×ãÁ¿µÄ10£¥H2SO4ºÍÊÊÁ¿µÄÕôÁóË®£¬¢ÜÖгÁµíÍêÈ«Èܽ⣬ÈÜÒº³ÊËáÐÔ£¬¼ÓÈÈÖÁ75¡æ£¬³ÃÈȼÓÈë0.05 mol/L¡¡KMnO4ÈÜÒº16.00 mL£¬Ç¡ºÃÍêÈ«·´Ó¦£®

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)Åäƽ£º________MnO4£­£«________H2C2O4£«________H+¨D________Mn2+£«________CO2¡ü£«________H2O

(2)0.05 mol¡¤L£­1KMnO4±ê×¼ÈÜÒºÓ¦ÖÃÓÚ________(Ñ¡Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±)µÎ¶¨¹ÜÖУ»Åжϵζ¨ÖÕµãµÄÏÖÏóΪ________£®

(3)¼ÆËã¸ÃÂÈ»¯¸ÆÑùÆ·ÖиÆÔªËصÄÖÊÁ¿·ÖÊýΪ________(¾«È·µ½0.01£¥)£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º½­ËÕÊ¡¸ß´¾¸ß¼¶ÖÐѧ2008½ì¸ßÈýÖÊÁ¿¼ì²â»¯Ñ§ÊÔ¾í ÌâÐÍ£º058

ÂÈ»¯¸ÆÊÇÓÃ;¹ã·ºµÄ»¯Ñ§ÊÔ¼Á£¬¿É×÷¸ÉÔï¼Á¡¢À䶳¼Á¡¢·À¶³¼ÁµÈ£®ÎªÁ˲ⶨijÂÈ»¯¸ÆÑùÆ·ÖиƵĺ¬Á¿£¬½øÐÐÈçÏÂʵÑ飺

¢Ù׼ȷ³ÆÈ¡ÂÈ»¯¸ÆÑùÆ·0.2312 g£¬·ÅÈëÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿6 mol/LµÄÑÎËáºÍÊÊÁ¿ÕôÁóˮʹÑùÆ·ÍêÈ«Èܽ⣬ÔٵμÓ35 mL¡¡0.25 mol/L¡¡(NH4)2C2O4ÈÜÒº£¬Ë®Ô¡¼ÓÈÈ£¬Öð½¥Éú³ÉCaC2O4³Áµí£¬¾­¼ìÑ飬Ca2+ÒѳÁµíÍêÈ«£®

¢Ú¹ýÂ˲¢Ï´µÓ¢ÙËùµÃ³Áµí£®

¢Û¼ÓÈë×ãÁ¿µÄ10£¥H2SO4ºÍÊÊÁ¿µÄÕôÁóË®£¬¢ÚÖгÁµíÍêÈ«Èܽ⣬ÈÜÒº³ÊËáÐÔ£¬¼ÓÈÈÖÁ75¡æ£¬³ÃÈȼÓÈë0.05 mol/L¡¡KMnO4ÈÜÒº16 mL£¬Ç¡ºÃÍêÈ«·´Ó¦£®

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)Åäƽ£º________MnO4£­£«________H2C2O4£«________H+¨D________Mn2+£«________CO2¡ü£«________H2O

(2)0.05 mol¡¤L£­1¡¡KMnO4±ê×¼ÈÜÒºÓ¦ÖÃÓÚ________(Ñ¡Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±)µÎ¶¨¹ÜÖУ»Åжϵζ¨ÖÕµãµÄÏÖÏóΪ________________£®

(3)¼ÆËã¸ÃÂÈ»¯¸ÆÑùÆ·ÖиÆÔªËصÄÖÊÁ¿·ÖÊý(¾«È·µ½0.01£¥)£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ì¸£½¨Ê¡¡¢¶þÖи߶þÉÏѧÆÚÆÚÄ©Áª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

Ä¿Ç°ÊÀ½ç¸÷¹ú»ñµÃþµÄÖ÷ÒªÀ´Ô´´Óº£Ë®ÖÐÌáÈ¡£¬ÒÔÏÂÊÇÌáȡþµÄ¹ý³ÌÖÐÉæ¼°µ½µÄ¼¸ÖÖÎïÖʳ£ÎÂϵÄÈܶȻý³£Êý£¬¸ù¾ÝÄãËùѧµÄ֪ʶ»Ø´ðÏÂÃæµÄ¼¸¸öÎÊÌ⣺?

ÎïÖÊ

CaCO3

MgCO3

Ca(OH)2

Mg(OH)2

ÈܶȻý

2.8¡Á10-9

6.8¡Á10-6

5.5¡Á10-6

1.8¡Á10-11

(1)ÔÚ´Óº£Ë®ÖÐÌáȡþʱ£¬ÍùÍùÓõ½±´¿Ç£¨Ö÷Òª³É·ÖÊÇCaCO3£©£¬ÄãÈÏΪ        (Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)ͨ¹ý½«±´¿ÇÑÐÄ¥³É·ÛÄ©Ö±½ÓͶÈ뺣ˮÀïÖƱ¸º¬Ã¾µÄ³Áµí£¬ÀíÓÉÊÇ                                ¡£Èç¹û²»ÄÜÖ±½ÓͶÈ룬Ӧ½«±´¿Ç×÷ºÎ´¦Àí£¬ÊÔд³ö»¯Ñ§·½³Ìʽ£º                          ¡££¨ÈôµÚÒ»¸ö¿Õ¸ñÌî¡°ÄÜ¡±£¬´Ë¿Õ¸ñ²»ÌÈôÌî¡°²»ÄÜ¡±£¬ ´Ë¿Õ¸ñÖ»ÐëµÚÒ»²½´¦ÀíµÄ»¯Ñ§·´Ó¦·½³Ìʽ¡££©

£¨2£©ÒÑ֪ijµØº£Ë®ÖеÄþÀë×ÓµÄŨ¶ÈΪ1.8¡Á10-3 mol¡¤L-1,Ôò³£ÎÂÏÂҪʹþÀë×Ó²úÉú³Áµí£¬ÈÜÒºpH×îµÍӦΪ        ¡£

£¨3£©ÊµÑéÊÒÖг£ÓÃCaCO3ÖÆCO2£¬Æä²úÎïÖ®Ò»µÄÂÈ»¯¸ÆÊÇÓ¦Óù㷺µÄ»¯Ñ§ÊÔ¼Á£¬¿É×÷¸ÉÔï¼Á¡¢À䶳¼ÁµÈ¡£ÎªÁ˲ⶨijÂÈ»¯¸ÆÑùÆ·ÖиÆÔªËصĺ¬Á¿£¬½øÐÐÈçÏÂʵÑ飺

£¨I£©×¼È·³ÆÈ¡ÂÈ»¯¸ÆÑùÆ·0.2000g£¬·ÅÈëÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿6mol/LµÄÑÎËáºÍÊÊÁ¿ÕôÁóˮʹÑùÆ·ÍêÈ«Èܽ⣬ÔÙ¼ÓÈë35mL 0.25mol/L £¨NH4£©2C2O4ÈÜÒº£¬Ë®Ô¡¼ÓÈÈ£¬Öð½¥Éú³ÉCaC2O4³Áµí£¬¾­¼ìÑ飬Ca2+ÒѳÁµíÍêÈ«¡£

£¨II£©¹ýÂ˲¢Ï´µÓ£¨I£©ËùµÃ³Áµí¡£

£¨III£©¼ÓÈë×ãÁ¿µÄ10% H2SO4ÈÜÒººÍÊÊÁ¿µÄÕôÁóË®£¬£¨II£©ÖгÁµíÍêÈ«Èܽ⣬ÈÜÒº³ÊËáÐÔ£¬¼ÓÈÈÖÁ75¡æ£¬³ÃÈÈÖðµÎ¼ÓÈë0.05000 mol/L KMnO4ÈÜÒº16.00mL£¬Ç¡ºÃÍêÈ«·´Ó¦¡£Çë»Ø´ð£º

ÒÑÖªµÎ¶¨¹ý³Ì·¢ÉúµÄ·´Ó¦Îª2MnO4- + 5H2C2O4 + 6H+ ="=2" Mn2+ +10 CO2¡ü+8 H2O£¨ÒÑÅäƽ£©

¢Ù0.05000 mol/L KMnO4ÈÜÒº±ê×¼ÈÜÒºÓ¦ÖÃÓÚ         £¨Ñ¡Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©µÎ¶¨¹ÜÖС£

¢ÚµÎ¶¨ÖÕµãµÄÏÖÏóΪ                                 ¡£

¢Û¸ÃÂÈ»¯¸ÆÑùÆ·ÖиÆÔªËصÄÖÊÁ¿°Ù·ÖÊýΪ            ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÂÈ»¯¸ÆÊÇÓÃ;¹ã·ºµÄ»¯Ñ§ÊÔ¼Á£¬¿É×÷¸ÉÔï¼Á¡¢À䶳¼Á¡¢·À¶³¼ÁµÈ¡£ÎªÁ˲ⶨijÂÈ»¯¸ÆÑùÆ·ÖиƵĺ¬Á¿£¬½øÐÐÈçÏÂʵÑ飺

¢Ù׼ȷ³ÆÈ¡ÂÈ»¯¸ÆÑùÆ·0.2312g£¬·ÅÈëÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿6mol/LµÄÑÎËáºÍÊÊÁ¿ÕôÁóˮʹÑùÆ·ÍêÈ«Èܽ⣬ÔٵμÓ35mL0.25mol/L (NH4)2C2O4ÈÜÒº£¬Ë®Ô¡¼ÓÈÈ£¬Öð½¥Éú³ÉCaC2O4³Áµí£¬¾­¼ìÑ飬Ca2+ÒѳÁµíÍêÈ«¡£

¢Ú¹ýÂ˲¢Ï´µÓ¢ÙËùµÃ³Áµí¡£

¢Û¼ÓÈë×ãÁ¿µÄ10% H2SO4ºÍÊÊÁ¿µÄÕôÁóË®£¬¢ÚÖгÁµíÍêÈ«Èܽ⣬ÈÜÒº³ÊËáÐÔ£¬¼ÓÈÈÖÁ75¡æ£¬³ÃÈȼÓÈë0.05mol/L KMnO4ÈÜÒº16mL£¬Ç¡ºÃÍêÈ«·´Ó¦¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Åäƽ£º    MnO4£­£«    H2C2O4£«    H£«¨D    Mn2£«£«     CO2¡ü£«    H2O

£¨2£©0.05mol¡¤L£­1KMnO4±ê×¼ÈÜÒºÓ¦ÖÃÓÚ          £¨Ñ¡Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©µÎ¶¨¹ÜÖУ»Åжϵζ¨ÖÕµãµÄÏÖÏóΪ                                                   ¡£

£¨3£©¼ÆËã¸ÃÂÈ»¯¸ÆÑùÆ·ÖиÆÔªËصÄÖÊÁ¿·ÖÊý£¨¾«È·µ½0.01%£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸