£¨06Äê½­ËÕ¾í£©(10·Ö)Ï㶹ËØÊǹ㷺´æÔÚÓÚÖ²ÎïÖеÄÒ»Àà·¼Ïã×廯ºÏÎ´ó¶à¾ßÓйâÃôÐÔ£¬ÓеĻ¹¾ßÓп¹¾úºÍÏûÑ××÷Óá£ËüµÄºËÐĽṹÊÇ·¼ÏãÄÚõ¥A£¬Æä·Ö×ÓʽΪC9H6O2¡£¸Ã·¼ÏãÄÚõ¥A¾­ÏÂÁв½Öèת±äΪˮÑîËáºÍÒÒ¶þËá¡£

Ìáʾ£º

¢ÙCH3CH===CHCH2CH3¢ÙKMnO4¡¢OH-CH3COOH£«CH3CH2COOH

¢ÚR£­CH=CH2HBrR£­CH2£­CH2£­Br

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢Åд³ö»¯ºÏÎïCµÄ½á¹¹¼òʽ_______________¡£

¢Æ»¯ºÏÎïDÓжàÖÖͬ·ÖÒì¹¹Ì壬ÆäÖÐÒ»Ààͬ·ÖÒì¹¹ÌåÊDZ½µÄ¶þÈ¡´úÎÇÒË®½âºóÉú³ÉµÄ²úÎïÖ®Ò»ÄÜ·¢ÉúÒø¾µ·´Ó¦¡£ÕâÀàͬ·ÖÒì¹¹Ìå¹²ÓÐ_______________ÖÖ¡£

¢ÇÔÚÉÏÊöת»¯¹ý³ÌÖУ¬·´Ó¦²½ÖèB¡úCµÄÄ¿µÄÊÇ_______________¡£

¢ÈÇëÉè¼ÆºÏÀí·½°¸´ÓºÏ³É£¨Ó÷´Ó¦Á÷³Ìͼ±íʾ£¬²¢×¢Ã÷·´Ó¦Ìõ¼þ£©¡£

Àý£ºÓÉÒÒ´¼ºÏ³É¾ÛÒÒÏ©µÄ·´Ó¦Á÷³Ìͼ¿É±íʾΪ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨06Äê½­ËÕ¾í£©Ä³»¯Ñ§ÐËȤС×é°´ÕÕÏÂÁз½°¸½øÐС°Óɺ¬Ìú·ÏÂÁÖƱ¸ÁòËáÂÁ¾§Ì塱µÄʵÑ飺

²½Öè1£ºÈ¡Ò»¶¨Á¿º¬Ìú·ÏÂÁ£¬¼Ó×ãÁ¿µÄNaOHÈÜÒº£¬·´Ó¦ÍêÈ«ºó¹ýÂË¡£

²½Öè2£º±ß½Á°è±ßÏòÈÜÒºÖеμÓÏ¡ÁòËáÖÁÈÜÒºpH£½8¡«9£¬¾²ÖᢹýÂË¡¢Ï´µÓ¡£

²½Öè3£º½«²½Öè2Öеõ½µÄ¹ÌÌåÈÜÓÚ×ãÁ¿µÄÏ¡ÁòËá¡£

²½Öè4£º½«µÃµ½µÄÈÜÒº·¶·¢Å¨Ëõ¡¢ÀäÈ´¡¢½á¾§¡¢¹ýÂË¡¢¸ÉÔï¡£

Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©ÉÏÊöʵÑéÖеĹýÂ˲Ù×÷ÐèÒª²£Á§°ô¡¢          ¡¢          µÈ²£Á§ÒÇÆ÷¡£

£¨2£©²½Öè1¹ýÂ˵ÄÄ¿µÄÊÇ                                                ¡£

£¨3£©µ±²½Öè2ÖÐÈÜÒºµÄpH£½8¡«9ʱ£¬¼ìÑé³ÁµíÊÇ·ñÍêÈ«µÄ·½·¨ÊÇ             ¡£

£¨4£©²½Öè2ÖÐÈÜÒºµÄpH½ÏÄÑ¿ØÖÆ£¬¿É¸ÄÓà                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨06Äê½­ËÕ¾í£©(8·Ö)ÂÈ»¯ÑÇÍ­£¨CuCl£©ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£¹ú¼Ò±ê×¼¹æ¶¨ºÏ¸ñµÄCuCl²úÆ·µÄÖ÷ÒªÖÊÁ¿Ö¸±êΪCuClµÄÖÊÁ¿·ÖÊý´óÓÚ96.50%¡£¹¤ÒµÉϳ£Í¨¹ýÏÂÁз´Ó¦ÖƱ¸CuCl

2CuSO4£«Na2SO3£«2NaCl£«Na2CO3===2CuCl¡ý£«3Na2SO4£«CO2¡ü

¢ÅCuClÖƱ¸¹ý³ÌÖÐÐèÒªÅäÖÃÖÊÁ¿·ÖÊýΪ20.0%µÄCuSO4ÈÜÒº£¬ÊÔ¼ÆËãÅäÖøÃÈÜÒºËùÐèµÄCuSO4?5H2OÓëH2OµÄÖÊÁ¿Ö®±È¡£

¢Æ׼ȷ³ÆÈ¡ËùÅäÖõÄ0.2500g CuClÑùÆ·ÖÃÓÚÒ»¶¨Á¿µÄ0.5mol?L£­1 FeCl3ÈÜÒºÖУ¬´ýÑùÆ·ÍêÈ«Èܽâºó£¬¼ÓË®20mL£¬ÓÃ0.1000mol?L£­1µÄCe(SO4)2ÈÜÒºµÎ¶¨µ½ÖÕµã,ÏûºÄ24.60mLCe(SO4)2ÈÜÒº¡£Óйط´»¯Ñ§·´Ó¦Îª

Fe3+£«CuCl===Fe2+£«Cu2+£«Cl-

Ce4+£«Fe2+===Fe3+£«Ce3+

ͨ¹ý¼ÆËã˵Ã÷ÉÏÊöÑùÆ·ÖÐCuClµÄÖÊÁ¿·ÖÊýÊÇ·ñ·ûºÏ±ê×¼¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨06Äê½­ËÕ¾í£©ÂÈ»¯ÑÇÍ­£¨CuCl£©ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£¹ú¼Ò±ê×¼¹æ¶¨ºÏ¸ñµÄCuCl²úÆ·µÄÖ÷ÒªÖÊÁ¿Ö¸±êΪCuClµÄÖÊÁ¿·ÖÊý´óÓÚ96.50%¡£¹¤ÒµÉϳ£Í¨¹ýÏÂÁз´Ó¦ÖƱ¸CuCl¡£

2CuSO4£«Na2SO3£«2NaCl£«Na2CO3===2CuCl¡ý£«3Na2SO4£«CO2¡ü

¢ÅCuClÖƱ¸¹ý³ÌÖÐÐèÒªÅäÖÃÖÊÁ¿·ÖÊýΪ20.0%µÄCuSO4ÈÜÒº£¬ÊÔ¼ÆËãÅäÖøÃÈÜÒºËùÐèµÄCuSO4?5H2OÓëH2OµÄÖÊÁ¿Ö®±È¡£

¢Æ׼ȷ³ÆÈ¡ËùÅäÖõÄ0.2500g CuClÑùÆ·ÖÃÓÚÒ»¶¨Á¿µÄ0.5mol?L£­1 FeCl3ÈÜÒºÖУ¬´ýÑùÆ·ÍêÈ«Èܽâºó£¬¼ÓË®20mL£¬ÓÃ0.1000mol?L£­1µÄCe(SO4)2ÈÜÒºµÎ¶¨µ½ÖÕµã,ÏûºÄ24.60mLCe(SO4)2ÈÜÒº¡£Óйط´»¯Ñ§·´Ó¦Îª

Fe3+£«CuCl===Fe2+£«Cu2+£«Cl-

Ce4+£«Fe2+===Fe3+£«Ce3+

ͨ¹ý¼ÆËã˵Ã÷ÉÏÊöÑùÆ·ÖÐCuClµÄÖÊÁ¿·ÖÊýÊÇ·ñ·ûºÏ±ê×¼¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨06Äê½­ËÕ¾í£©X¡¢Y¡¢ZÊÇ3ÖÖ¶ÌÖÜÆÚÔªËØ£¬ÆäÖÐX¡¢YλÓÚͬһ×壬Y¡¢Z´¦ÓÚͬһÖÜÆÚ¡£XÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇÆäµç×Ó²ãÊýµÄ3±¶¡£ZÔ­×ӵĺËÍâµç×ÓÊý±ÈYÔ­×ÓÉÙ1¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A  ÔªËطǽðÊôÐÔÓÉÈõµ½Ç¿µÄ˳ÐòΪZ£¼Y£¼X

B  YÔªËØ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½¿É±íʾΪH3YO4

C  3ÖÖÔªËصÄÆø̬Ç⻯ÎïÖÐZµÄÆø̬Ç⻯Îï×îÎȶ¨

D  Ô­×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪZ£¾Y£¾X

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨06Äê½­ËÕ¾í£©ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨£©

A  0.1mol?L¨D1°±Ë®ÖУ¬c(OH¨D)£½c(NH4£«)

B  10mL 0.02mol?L¨D1 HClÈÜÒºÓë10mL 0.02mol?L¨D1 Ba(OH)2ÈÜÒº³ä·Ö»ìºÏ£¬Èô»ìºÏºóÈÜÒºµÄÌå»ýΪ20mL£¬ÔòÈÜÒºµÄpH£½12

C  ÔÚ0.1mol?L¨D1 CH3COONaÈÜÒºÖУ¬c(OH¨D)£½c(CH3COOH)£«c(H£«)

D  0.1mol?L¨D1ij¶þÔªÈõËá¼îÐÔNaHAÈÜÒºÖУ¬c(Na£«)£½2c(A2¨D)£«c(HA¨D)£«c(H2A)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸