18£®ÏÂÁи÷×éÀë×ÓÔÚÖ¸¶¨ÈÜÒºÖУ¬ÄÜ´óÁ¿¹²´æµÄÊÇ£¨¡¡¡¡£©
¢ÙÎÞÉ«ÈÜÒºÖУºK+  Cl- Mg2+   SO42-  Cr2O72-
¢ÚpH=11µÄÈÜÒºÖУºCO32-  Na+  AlO2-   NO3-  S2-
¢ÛË®µçÀëµÄH+Ũ¶ÈΪ10-2mol/LµÄÈÜÒºÖУºCl-   CO32-   NH4+  SO32- NO3-
¢Ü¼ÓÈëÂÁ·Û·Å³öÇâÆøµÄÈÜÒºÖУºK+  Cl- Mg2+   S2O32-SO42-NH4+
¢Ýʹ¼×»ù³È±äºìµÄÈÜÒºÖУºMnO4-Fe2+  Na+    NO3-SO42-
¢ÞÖÐÐÔÈÜÒºÖУºFe3+   Al3+   NO3-   Cl-
¢ßÄÜÈܽâÍ­·ÛµÄÈÜÒº£ºI- NO3-SO42-Fe3+   Al3+£®
A£®¢Ù¢Ú¢ÝB£®¢Ù¢Û¢ÞC£®¢Ú¢ÜD£®¢Ú

·ÖÎö ¢ÙCr2O72-Ϊ³ÈÉ«£»
¢ÚpH=11µÄÈÜÒº£¬ÏÔ¼îÐÔ£»
¢ÛË®µçÀëµÄH+Ũ¶ÈΪ10-2mol/LµÄÈÜÒº£¬ÈÜÒºÏÔËáÐÔ£»
¢Ü¼ÓÈëÂÁ·Û·Å³öÇâÆøµÄÈÜÒº£¬Îª·ÇÑõ»¯ÐÔËá»òÇ¿¼îÈÜÒº£»
¢Ýʹ¼×»ù³È±äºìµÄÈÜÒº£¬ÏÔËáÐÔ£»
¢ÞÖÐÐÔÈÜÒº²»ÄÜ´óÁ¿´æÔÚFe3+£»
¢ßÄÜÈܽâÍ­·ÛµÄÈÜÒº£¬¾ßÓÐÑõ»¯ÐÔÎïÖÊ£®

½â´ð ½â£º¢ÙCr2O72-Ϊ³ÈÉ«£¬ÓëÎÞÉ«²»·û£¬¹Ê´íÎó£»
¢ÚpH=11µÄÈÜÒº£¬ÏÔ¼îÐÔ£¬¸Ã×éÀë×ÓÖ®¼ä²»·´Ó¦£¬¿É´óÁ¿¹²´æ£¬¹ÊÕýÈ·£»
¢ÛË®µçÀëµÄH+Ũ¶ÈΪ10-2mol/LµÄÈÜÒº£¬ÈÜÒºÏÔËáÐÔ£¬ËáÐÔÈÜÒºÖв»ÄÜ´óÁ¿´æÔÚCO32-¡¢SO32-£¬¹Ê´íÎó£»
¢Ü¼ÓÈëÂÁ·Û·Å³öÇâÆøµÄÈÜÒº£¬Îª·ÇÑõ»¯ÐÔËá»òÇ¿¼îÈÜÒº£¬¼îÐÔÈÜÒºÖв»ÄÜ´æÔÚMg2+¡¢NH4+£¬ËáÐÔÈÜÒºÖÐH+¡¢S2O32-·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬¹Ê´íÎó£»
¢Ýʹ¼×»ù³È±äºìµÄÈÜÒº£¬ÏÔËáÐÔ£¬£¨NO3-£©MnO4-¡¢Fe2+¡¢H+·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬²»ÄÜ´óÁ¿¹²´æ£¬¹Ê´íÎó£»
¢ÞÖÐÐÔÈÜÒº²»ÄÜ´óÁ¿´æÔÚFe3+£¬¹Ê´íÎó£»
¢ßÄÜÈܽâÍ­·ÛµÄÈÜÒº£¬¾ßÓÐÑõ»¯ÐÔÎïÖÊ£¬²»ÄÜ´æÔÚ»¹Ô­ÐÔÀë×ÓI-£¬ÇÒI-¡¢Fe3+·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬²»ÄÜ´óÁ¿¹²´æ£¬¹Ê´íÎó£»
¹ÊÑ¡D£®

µãÆÀ ±¾Ì⿼²éÀë×ӵĹ²´æ£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕÏ°ÌâÖеÄÐÅÏ¢¼°³£¼ûÀë×ÓÖ®¼äµÄ·´Ó¦Îª½â´ðµÄ¹Ø¼ü£¬²àÖظ´·Ö½â·´Ó¦¡¢Ñõ»¯»¹Ô­·´Ó¦µÄÀë×Ó¹²´æ¿¼²é£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®¼º¶þËáÊÇÒ»ÖÖÖØÒªµÄÓлú¶þÔªËᣬÖ÷ÒªÓÃÓÚÖÆÔìÄáÁú¡¢ÅÝÄ­ËÜÁϼ°È󻬼ÁºÍÔöËܼÁµÈ£¬ÓÃ;ʮ·Ö¹ã·º£®¼º¶þËáµÄ´«Í³¡¢ÂÌÉ«ºÏ³É·¨Ô­Àí·Ö±ðÈçͼ1£º

ÂÌÉ«ºÏ³ÉÁ÷³ÌÈçÏ£º
ÔÚ100mLÈý¾±ÉÕÆ¿ÖмÓÈë2.00g´ß»¯¼Á£¨ÎÙËáÄƺͲÝËáºÏ³É£©£¬34.0mL 30%µÄH2O2ºÍ´Å×Ó£®ÊÒÎÂϽÁ°è15-20minºó£¬¼ÓÈë6.00g»·¼ºÏ©£®´îºÃ»ØÁ÷×°Ö㨼ûͼ2£©£¬¼ÌÐø¿ìËÙ¾çÁÒ½Á°è²¢¼ÓÈÈ»ØÁ÷£®·´Ó¦¹ý³ÌÖлØÁ÷ζȽ«ÂýÂýÉý¸ß£¬»ØÁ÷l.5hºó£¬µÃµ½ÈȵĺϳÉÒº£®
×¢£º¢Ù¼º¶þË᣺°×É«¾§Ì壬ÈÛµã153.0-153.1¡æ£¬Ò×ÈÜÓھƾ«£»»·¼ºÏ©£ºÎÞÉ«ÒºÌ壬²»ÈÜÓÚË®£¬ÈÜÓÚÒÒ´¼£®
¢Ú¼¸ÖÖÎïÖÊÔÚË®ÖеÄÈܽâ¶ÈÇúÏßÈçͼ3£º
£¨1£©¶Ô±È´«Í³ÓëÂÌÉ«ÖƱ¸µÄÔ­Àí£¬´«Í³ÖƸ÷·½·¨±»ÌÔÌ­µÄÖ÷ÒªÔ­ÒòÊÇ£ºÅ¨ÏõËá¾ßÓÐÇ¿¸¯Ê´ÐÔ£¬¶ÔÉ豸¸¯Ê´ÑÏÖØ£¬·´Ó¦ÖлáÉú³ÉÓж¾µÄµªÑõ»¯ÎÎÛȾ»·¾³£®
£¨2£©ÂÌÉ«ºÏ³Éʱ£¬Çë½âÊͱØÐë¿ìËÙ¾çÁÒ½Á°èµÄÔ­ÒòÊÇ£º»·¼ºÏ©²»ÈÜÓÚË®£¬¿ìËÙ½Á°èÓÐÀûÓÚ·´Ó¦Îï³ä·Ö½Ó´¥£¬¼Ó¿ì·´Ó¦ËÙÂÊ£®
£¨3£©³éÂËʱ±ØÐëÓõ½µÄÒÇÆ÷£º³éÆø±Ã¡¢°²È«Æ¿¡¢ÎüÂËÆ¿¡¢²¼ÊÏ©¶·¡¢Ïð½ºÈû¡¢ÏðƤÈû¡¢µ¼¹ÜµÈ£¬Ö¸³öÎüÂËÆ¿ÓëÆÕͨ׶ÐÎÆ¿³ý¹æ¸ñÒÔÍâµÄÁ½µã²»Í¬Ö®´¦£ºÎüÂËÆ¿´øÓÐÖ§¹Ü¡¢ÎüÂËÆ¿Æ¿±Ú±È׶ÐÎÆ¿ºñ£®
£¨4£©´ÓÈȺϳÉÒºÖзÖÀëµÃµ½¼º¶þËá¹ÌÌåµÄ²½Ö裺ÀäÈ´½á¾§£¬³éÂË£¬Ï´µÓ£¬¸ÉÔ³ÆÁ¿£®ÇëÄãÑ¡ÔñºÏÊʵÄÏ´µÓ¼Á£ºB£®
A£®ÕôÁóË®    B£®±ùË®»ìºÏÎï    C£®ÒÒ´¼    D£®ÂËÒº
£¨5£©»ØÊÕδ·´Ó¦ÍêµÄ»·¼ºÏ©µÄ·½·¨ÊÇ£º·ÖÒº£»
£¨6£©ÓÃëϸ¹Ü·Ö±ðÈ¡´«Í³·¨ºÍÂÌÉ«·¨ÖƱ¸²úÆ·µÄ¾Æ¾«ÈÜÒº£¬·Ö±ðµãÓÚÀëÂËÖ½ÌõÄ©¶Ë1cm´¦£¬×ö²ãÎöʵÑ飬ÏÔÉ«ºóÈçͼ4Ëùʾ£ºÓÉʵÑé¿ÉÖª£º´«Í³·¨ÖƵõIJúÆ·´¿¶È±ÈÂÌÉ«·¨¸ß£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

9£®AºÍBÊÇÇ°ÈýÖÜÆÚµÄÔªËØ£¬ËüÃǵÄÀë×ÓA2+ºÍB3+¾ßÓÐÏàͬµÄºËÍâµç×Ó²ã½á¹¹£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Ô­×Ӱ뾶£ºA£¾BB£®Ô­×ÓÐòÊý£ºA£¾B
C£®Àë×Ӱ뾶£ºA2+£¼B3+D£®ÖÊÁ¿Êý£ºA£¾B

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

6£®¿ÉÓÃÓÚ·ÖÀë»òÌá´¿ÎïÖʵķ½·¨ÓУº
A£®¹ýÂË¡¡  B£®Éý»ª    C£®ÕôÁó     D£®½á¾§     E£®ÝÍÈ¡      F£®·ÖÒº
ÏÂÁи÷×é»ìºÏÎïµÄ·ÖÀë»òÌᴿӦѡÓÃÉÏÊöÄÄÖÖ·½·¨×îºÏÊÊ£¿£¨Ìî×Öĸ£©
£¨1£©·ÖÀë±¥ºÍʳÑÎË®ºÍɳ×ӵĻìºÏÎÓÃA£®
£¨2£©´ÓÏõËá¼ØºÍÂÈ»¯¼Ø»ìºÏÒºÖлñÈ¡ÏõËá¼Ø£¬ÓÃD£®
£¨3£©³ýÈ¥ÒÒ´¼ÖÐÈܽâµÄ΢Á¿Ê³ÑΣ¬ÓÃC£®
£¨4£©´ÓäåË®ÖÐÌáÈ¡ä壬ÓÃE£®
£¨5£©³ýÈ¥NaCl¾§ÌåÖлìÓеÄÉÙÁ¿µâµ¥ÖÊ£¬ÓÃB£®
£¨6£©·ÖÀëË®ºÍÆûÓ͵ĻìºÏÎÓÃF£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

13£®Ä³Í¬Ñ§°´ÏÂÁв½ÖèÅäÖÆ500mL 0.200mol/L KCl ÈÜÒº£º
¢Ù½«ÉÕ±­ÖеÄÈÜҺתÒÆÖÁ500mL ÈÝÁ¿Æ¿ÖР¢Ú³ÆÁ¿KCl ¹ÌÌå ¢ÛÏòÈÝÁ¿Æ¿ÖмÓÕôÁóË®ÖÁ¿Ì¶ÈÏß ¢Ü¼ÆËãËùÐèKClµÄÖÊÁ¿ ¢Ý½«KCl¼ÓÈë100mL ÉÕ±­ÖУ¬²¢¼ÓÊÊÁ¿Ë®
£¨1£©¢Ü¼ÆËãËùÐèÒªKCl µÄÖÊÁ¿Îª7.5g£»
£¨2£©ÊµÑéµÄÏȺó˳ÐòӦΪ¢Ü¢Ú¢Ý¢Ù¢Û£¨ÌîдÐòºÅ£©£»
£¨3£©ÔÚ ¢ÙʵÑéÖУ¬Îª·ÀÖ¹ÈÜÒº½¦³ö£¬Ó¦²ÉÈ¡µÄ´ëÊ©ÊDz£Á§°ôÒýÁ÷£»
£¨4£©ÔÚ½øÐР¢ÛµÄʵÑé²Ù×÷ʱӦעÒâµÄÎÊÌâÊDZÜÃâ¼ÓË®³¬¹ý¿Ì¶ÈÏߣ»
£¨5£©ÔÚ½øÐР¢Ýʱ£¬ÎªÁ˼ӿìÈܽâËÙÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊDz£Á§°ô½Á°è£®
£¨6£©ÅäÖÆÈÜҺʱ£¬ÏÂÁÐʵÑé²Ù×÷»áʹÅäÖÆÈÜҺŨ¶ÈÆ«¸ßµÄÊÇB
A£®ÈÝÁ¿Æ¿ÄÚÓÐË®£¬Î´¾­¹ý¸ÉÔï´¦Àí
B£®¶¨ÈÝʱ£¬¸©Êӿ̶ÈÏß
C£®ÒÆÒºÍê³Éºó£¬ÈܽâKClµÄÉÕ±­Î´ÓÃÕôÁóˮϴµÓ
D£®¶¨Èݺóµ¹×ªÈÝÁ¿Æ¿¼¸´Î£¬·¢ÏÖÒºÌå×îµÍµãµÍÓڿ̶ÈÏߣ¬ÔÙ²¹¼Ó¼¸µÎË®µ½¿Ì¶ÈÏߣ®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

3£®ÏÂÁÐÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄ²Ù×÷ÖÐʹÎïÖʵÄÁ¿Å¨¶ÈÆ«´óµÄÊÇACD£¬Æ«Ð¡µÄÊÇBF£¬ÎÞÓ°ÏìµÄÊÇE£®£¨Ñ¡Ìî×Öĸ£©
A£®ÅäÖÆÁòËáÍ­ÈÜҺʱ£¬ËùÓõĵ¨·¯²¿·Ö·ç»¯£®
B£®ÅäÖÆÇâÑõ»¯ÄÆÈÜҺʱ£¬³ÆÁ¿¹ÌÌåʱ¼ä¹ý³¤£®
C£®ÅäÖÆÏ¡ÁòËáʱ£¬ÓÃÁ¿Í²Á¿È¡Å¨ÁòËáʱÑöÊÓ¶ÁÊý£®
D£®ÅäÖÆÏ¡ÁòËáʱ£¬ÔÚСÉÕ±­ÖÐÏ¡ÊÍŨÁòËáºóδÀäÈ´Á¢¼´×ªÒƵ½ÈÝÁ¿Æ¿Öв¢¶¨ÈÝ£®
E£®×ªÒÆÈÜҺǰ£¬ÈÝÁ¿Æ¿ÄÚÓÐË®Ö飮
F£®¶¨ÈÝʱ£¬¼ÓË®³¬¹ý¿Ì¶ÈÏߣ¬ÓýºÍ·µÎ¹ÜÎüÈ¡ÒºÌåÖÁ¿Ì¶ÈÏߣ®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

10£®NA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢Ù46g NO2ºÍN2O4µÄ»ìºÏÆøÌåÖк¬ÓеÄÔ­×Ó¸öÊýΪ3NA
¢Ú³£ÎÂÏ£¬4g CH4º¬ÓÐNA¸öC-H¹²¼Û¼ü
¢Û10mLÖÊÁ¿·ÖÊýΪ98%µÄH2SO4£¬¼ÓË®ÖÁ100mL£¬H2SO4µÄÖÊÁ¿·ÖÊýΪ9.8%
¢Ü±ê×¼×´¿öÏ£¬5.6L SO3º¬ÓеķÖ×ÓÊýΪ0.25NA
¢Ý25¡æʱ£¬pH=12µÄ1.0L NaClOÈÜÒºÖÐË®µçÀë³öµÄOH-µÄÊýĿΪ0.01NA
¢Þ0.1mol•L-1Na2CO3ÈÜÒºÖк¬ÓÐCO32-ÊýĿСÓÚ0.1NA
¢ß1mol Na2O2ÓëË®ÍêÈ«·´Ó¦Ê±×ªÒƵç×ÓÊýΪ2NA£®
A£®¢Û¢Þ¢ßB£®¢Ù¢Ú¢ÝC£®¢Ú¢Ü¢ÞD£®¢Ù¢Ú¢Ü¢Ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

7£®£¨1£©µ½Ä¿Ç°ÎªÖ¹£¬ÓÉ»¯Ñ§ÄÜת±äµÄÈÈÄÜ»òµçÄÜÈÔÈ»ÊÇÈËÀàʹÓõÄ×îÖ÷ÒªµÄÄÜÔ´£®ÔÚ25¡æ¡¢101kPaÏ£¬16gµÄ¼×´¼£¨CH3OH£©ÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·Å³ö352kJµÄÈÈÁ¿£¬Ôò±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪCH3OH£¨l£©+$\frac{3}{2}$O2£¨g£©=CO2£¨g+2H2O £¨l£©¡÷H=-704 KJ/mol£®
£¨2£©Í­µ¥Öʼ°Æ仯ºÏÎïÔÚ¹¤ÒµÉú²úºÍ¿ÆÑÐÖÐÓÐÖØÒª×÷Óã®ÏÖÓÃÂÈ»¯Í­¾§Ì壨CuCl2•2H2O£¬º¬ÂÈ»¯ÑÇÌúÔÓÖÊ£©ÖÆÈ¡´¿¾»µÄCuCl2•2H2O£®ÏȽ«ÆäÖƳÉË®ÈÜÒº£¬ºó°´Èçͼ²½Öè½øÐÐÌá´¿£º

ÒÑÖªCu2+¡¢Fe3+ºÍFe2+µÄÇâÑõ»¯Î↑ʼ³ÁµíºÍ³ÁµíÍêȫʱµÄpH¼û±í
½ðÊôÀë×ÓFe3+Fe2+Cu2+
ÇâÑõ»¯Î↑ʼ³ÁµíʱµÄpH1.97.04.7
ÇâÑõ»¯ÎïÍêÈ«³ÁµíʱµÄpH3.29.06.7
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÏÖÓÐÑõ»¯¼ÁNaClO¡¢H2O2¡¢KMnO4£¬X¼ÓÄÄÖֺã¿H2O2ºÃ£¬²»ÒýÈëÔÓÖÊÀë×Ó£»¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪH2O2+2Fe2++2H+¨T2Fe3++2H2O£®
¢ÚÈÜÒºIIÖгýCu2+Í⣬»¹ÓÐFe3+½ðÊôÀë×Ó£¬ÈçºÎ¼ìÑéÆä´æÔÚÈ¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼ÓKSCNÈÜÒº£¬ÈôÈÜÒº£®±äΪѪºìÉ«£¬Ôò˵Ã÷ÓÐFe3+£¬ÈôÈÜÒº²»±äΪѪºìÉ«Ôò˵Ã÷ûÓÐFe3+£®
¢ÛÎïÖÊY²»ÄÜΪÏÂÁеÄef
a£®CuO  b£®Cu£¨OH£©2  c£®CuCO3 d£®Cu2£¨OH£©2CO3 e£®CaO   f£®NaOH
¢ÜÈôÏòÈÜÒº¢òÖмÓÈë̼Ëá¸Æ£¬²úÉúµÄÏÖÏóÊÇ̼Ëá¸ÆÈܽ⣬²úÉúÆøÅݺͺìºÖÉ«³Áµí£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

8£®ÉèNAΪ°¢·üÙ¤µÂÂÞ³£ÊýµÄÊýÖµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®50mL18.4mol£®L-1ŨÁòËáÓë×ãÁ¿Í­Î¢ÈÈ·´Ó¦£¬Éú³ÉSO2·Ö×ÓµÄÊýĿΪ0.46NA
B£®Ä³ÃܱÕÈÝÆ÷Ê¢ÓÐ0.1molNAºÍ0.3molH2£¬³ä·Ö·´Ó¦ºóתÒƵç×ÓµÄÊýĿΪ0.6NA
C£®ÈôÓÉCO2ºÍO2×é³ÉµÄ»ìºÏÎïÖй²ÓÐNA¸ö·Ö×Ó£¬ÔòÆäÖеÄÑõÔ­×ÓÊýΪ2NA
D£®³£ÎÂÏÂ1L0.1mol£®L-1NH4NO3ÈÜÒºÖеÄÇâÔ­×ÓÊýΪ0.4NA

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸