( 14·Ö)

(1)ÏÂÁÐÀ¨ºÅÄÚµÄÎïÖÊΪÔÓÖÊ£¬½«³ýÈ¥ÏÂÁи÷×é»ìºÏÎïÖÐÔÓÖÊËùÐèµÄÊÔ¼ÁÌîдÔÚºáÏßÉÏ£º

±½(¼×±½)        £»±½(ÒÒ´¼)        £»¼×±½(äå)        ¡£

£¨2£©ÊµÑéÊÒÓɵçʯÖеÄ̼»¯¸ÆºÍË®·´Ó¦ÖÆÈ¡ÒÒȲ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ      £¬ÊµÑéÖÐΪÁ˼õ»º·´Ó¦ËÙÂÊ£¬³£Óà      ´úÌæË®£¬ÈôʵÑéÖвúÉúµÄÆøÌå´æÔÚÄÑÎŵÄÆø棬¿ÉÒÔÓÃ_________________£¨ÌîÊÔ¼Á£©¼ÓÒÔ³ýÈ¥¡£

£¨3£©Ä³ÐËȤС×éµÄͬѧÔÚʵÑéÊÒÀïÖÆÈ¡µÄÒÒÏ©Öг£»ìÓÐÉÙÁ¿¶þÑõ»¯Áò£¬ËûÃÇÉè¼ÆÁËÏÂÁÐʵÑéͼÒÔÈ·ÈÏÉÏÊö»ìºÏÆøÌåÖÐÓÐC2H4ºÍSO2¡£»Ø´ðÏÂÁÐÎÊÌ⣺

1£©I¡¢II¡¢III¡¢IV×°ÖÿÉÊ¢·ÅµÄÊÔ¼ÁÒÀ´ÎΪ       £¨Ìî×Öĸ£©

¢ÙÆ·ºìÈÜÒº   ¢ÚNaOHÈÜÒº  ¢ÛŨÁòËá    ¢ÜËáÐÔKMnO4ÈÜÒº

A. ¢Ü¢Ú¢Ù¢Û   B.¢Ù¢Ú¢Ù¢Û C. ¢Ù¢Ú¢Ù¢Ü D. ¢Ü¢Ú¢Ù¢Ü

2£©ÄÜ˵Ã÷SO2ÆøÌå´æÔÚµÄÏÖÏóÊÇ     ¡£

3£©Ê¹ÓÃ×°ÖÃIIIµÄÄ¿µÄÊÇ           ¡£

4£©È·¶¨º¬ÓÐÒÒÏ©µÄÏÖÏóÊÇ          ¡£

 

( 14·Ö)

£¨1£©¸ßÃÌËá¼ØËáÐÔÈÜÒº£» Ë® £»NaOHÈÜÒº ¡£

£¨2£©CaC2 + 2 H2O¡ú HCºCH¡ü+ Ca(OH)2 £¬±¥ºÍʳÑÎË®£¬   ÁòËáÍ­ÈÜÒº£¨»òNaOHÈÜÒº£©

£¨3£©  1£©C    2£©IÖÐÆ·ºìÍÊÉ«£»  3£©¼ìÑéSO2ÊÇ·ñ³ý¾¡£» 4£©IVÖÐKMnO4ÈÜÒºÍÊÉ«

½âÎö:

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(14·Ö)

(1)ÏÂͼÖÐA¡«K·Ö±ð´ú±íÓйط´Ó¦µÄÒ»ÖÖ·´Ó¦Îï»òÉú³ÉÎÆäÖÐA¡¢C¡¢F¡¢KÊǹÌÌå¡£ÒÑÖªAÊÇÒ»ÖÖ²»º¬½ðÊôÔªËصÄÑΣ¬ÊÜÈÈ·Ö½âÄܵõ½ÎïÖʵÄÁ¿ÏàµÈµÄÈýÖÖ²úÎA¼ÓÈȺóÉú³ÉµÄ»ìºÏÆøÌåÈôͨ¹ý¼îʯ»Ò£¬Ö»Ê£ÓàÆøÌåB(BΪÎÞÉ«Óд̼¤ÐÔÆøζµÄÆøÌå)£¬Èôͨ¹ýŨÁòËáÔòֻʣÓàÆøÌåD(DΪÎÞÉ«ÎÞζÆøÌå)£»CΪµ­»ÆÉ«¹ÌÌå¡£¸÷ÎïÖʼäµÄת»¯¹ØϵÈçÏÂͼËùʾ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

    ¢Ùд³öCµÄµç×Óʽ£º                     ¡£

¢Úд³öAµÄÈÜÒºÓë×ãÁ¿NaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º

                                                                    ¡£

¢Ûд³öʵÑéÊÒÖÆÈ¡BµÄ»¯Ñ§·½³Ìʽ£º

                                                                     ¡£

¢Üд³öJÓëK·´Ó¦µÄ»¯Ñ§·½³Ì£º

                                                                    ¡£

 (2)»ðÐÇ̽²â³µÔÚ»ðÐÇ´óÆøÖмì²âµ½ÁËÆøÌåM¡£×ÊÁÏÏÔʾ£¬M·Ö×ÓÊÇÈýÔ­×Ó·Ö×Ó£¬ÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª60£¬ÔÚµØÇò»·¾³ÏÂMÒ׷ֽ⡣·Ûĩ״µÄKSCNÓë95%µÄÁòËáÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦¿ÉµÃµ½ÆøÌåMºÍÁ½ÖÖÁòËáÇâÑΣ¬Éú³ÉÎïµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ1©s1©s1¡£ÔòÁ½ÖÖÁòËáÇâÑεĻ¯Ñ§Ê½Îª          ¡¢          £¬ÆøÌåMµÄ½á¹¹Ê½ÊÇ                      ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê¹þ¶û±õÊеÚÁùÖÐѧ¸ßÈýµÚ¶þ´ÎÄ£Ä⿼ÊÔ ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)³£ÎÂÏ£¬½«Ä³Ò»ÔªËáºÍÈÜÒºµÈÌå»ý»ìºÏ£¬»ìºÏÇ°Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºµÄÈçϱí(Éè»ìºÏºóÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ)£º

ʵÑé±àºÅ
ÎïÖʵÄÁ¿Å¨¶È
£¨£©
ÎïÖʵÄÁ¿Å¨¶È
£¨£©
»ìºÏÈÜÒºµÄ
¼×



ÒÒ



±û



¶¡



Çë»Ø´ð£»
(1)½öÒÀ¾Ý¼××éÊý¾Ý·ÖÎö£¬µÄÈ¡Öµ·¶Î§Îª         ¡£
(2)´ÓÒÒ×éÇé¿ö·ÖÎö£¬ÊÇ·ñÒ»¶¨µÈÓÚ      (Ñ¡Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)¡£Èô ÎªÈõËᣬ¼ÆËã³ö¸Ã»ìºÏÈÜÒº´ïƽºâʱµçÀë³£ÊýµÄ½üËÆÖµÊÇ   (Óú¬´úÊýʽ±íʾ)¡£
(3)±û×é»ìºÏÈÜÒºµÄÔ­ÒòÊÇ   £¬¸Ã×é»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС¹ØϵÊÇ   ¡£
(4)Èô°ÑijÀë×ÓË®½âµÄÎïÖʵÄÁ¿Õ¼Ä³Àë×Ó×ÜÎïÖʵÄÁ¿µÄ°Ù·ÖÊý½Ð×ö¡°Ë®½â¶È¡±£¬ÒÀ¾Ý¶¡×éÊý¾Ý£¬ÍÆÖªÔÚ¸ÃÌõ¼þϵÄË®½â¶ÈÊÇ    £¬¸ÃÈÜÒºÖУ¬Ë®µçÀë³öµÄÊÇ    
¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìÉÂÎ÷Ê¡áªÉ½ÏظßÒ»ÏÂѧÆÚÆÚÄ©ÖÊÁ¿¼ì²â»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÍƶÏÌâ

(14·Ö£©Ï±íÊÇÔªËØÖÜÆÚ±íµÄÇ°ÈýÖÜÆÚ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¿ÉÓÃÓÚ°ëµ¼Ìå²ÄÁϵÄÔªËØÊÇ       (ÌîÔªËØ·ûºÅ£©£¬ËüÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃΪ        £»

£¨2£©A¡¢GÁ½ÖÖÔªËØ·Ö±ðÓëEÔªËض¼ÄÜÐγÉÁ½ÖÖ»¯ºÏÎÆäÖÐÊôÓÚÀë×Ó»¯ºÏÎïµÄÊÇ         £¨Ð´»¯Ñ§Ê½£¬ÏÂͬ£©£¬ÊôÓÚ¹²¼Û»¯ºÏÎïµÄÊÇ         £»ÕâËÄÖÖ»¯ºÏÎïÖк¬ÓзǼ«ÐÔ¼üµÄ»¯ºÏÎïÊÇ                  ¡£

£¨3£©±í¸ñÖÐÊ®ÖÖÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÖУ¬¼îÐÔ×îÇ¿µÄÊÇ        £¬ÊôÓÚÁ½ÐÔÇâÑõ»¯ÎïµÄÊÇ          £¨Óû¯Ñ§Ê½±íʾ£©£»

£¨4£©Ö»º¬ÓÐA¡¢CÁ½ÖÖÔªËصĻ¯ºÏÎï³Æ×ö        £¬ÕâЩ»¯ºÏÎïÖÐÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄÊÇ         £¬¸Ã»¯ºÏÎï·Ö×ӵĿռ乹ÐÍÊÇ         £»

£¨5£©»­³öÔªËØHµÄÔ­×ӽṹʾÒâͼ        £¬ÔªËØHÓëÔªËØJ×é³ÉµÄ»¯ºÏÎïµÄË®ÈÜÒºÖмÓÈëÉÙÁ¿ÉÕ¼îÈÜÒº£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ                ¡£

£¨6£©ÓÉÔªËØA¡¢D¡¢J×é³ÉµÄ¼Èº¬ÓÐÀë×Ó¼üÓÖº¬Óй²¼Û¼üµÄ»¯ºÏÎïµÄ»¯Ñ§Ê½Îª             ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÄÚÃɹŸ߶þÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)(1)ÔÚ·´Ó¦2KMnO4£«16HBr===5Br2£«2MnBr2£«2KBr£«8H2OÖУ¬»¹Ô­¼ÁÊÇ      ¡£

(2)ÒÑÖªBrFxÓëH2O°´ÎïÖʵÄÁ¿Ö®±È3¡Ã5·´Ó¦µÄ²úÎïÊÇHF¡¢HBrO3¡¢Br2¡¢O2£¬¸Ã·´Ó¦ÖеÄÑõ»¯¼ÁÊÇ           £¬»¹Ô­¼ÁÊÇ          _£¬BrFxÖеÄx£½____________¡£

(3)ŨÑÎËáÔÚ·´Ó¦KClO3£«HCl¨D¡úKCl£«ClO2£«Cl2£«      (²¹³äÍêÕû)ÖÐÏÔʾ³öÀ´µÄÐÔÖÊÊÇ

                    _¡£

(4)ÔÚÒ»¶¨Ìõ¼þÏ£¬PbO2ÓëCr3£«·´Ó¦£¬²úÎïÊÇCr2OºÍPb2£«£¬ÔòÓë1 mol Cr3£«·´Ó¦ËùÐèPbO2µÄÎïÖʵÄÁ¿Îª                              ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÌì½òÊи߶þÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)  ³£ÎÂÏ£¬½«Ä³Ò»ÔªËáHAºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºµÄpHÈçÏÂ±í£º

ʵÑé±àºÅ

HAÎïÖʵÄÁ¿Å¨¶È£¨mol/L£©

NaOHÎïÖʵÄÁ¿Å¨¶È£¨mol/L£©

»ìºÏºóÈÜÒºµÄpH

¼×

0.2

0.2

pH = a

ÒÒ

C1

0.2

pH = 7

±û

0.2

0.1

pH >8

¶¡

0.1

0.1

pH = 9

Çë»Ø´ð

£¨1£©²»¿¼ÂÇÆäËû×éµÄʵÑé½á¹û£¬µ¥´Ó¼××éÇé¿ö·ÖÎö£¬ÈçºÎÓÃaÀ´ËµÃ÷HAÊÇÇ¿ËỹÊÇÈõËá¡£

    _______________________________________________________________________________¡£

£¨2£©²»¿¼ÂÇÆäËû×éµÄʵÑé½á¹û£¬µ¥´ÓÒÒ×éÇé¿ö·ÖÎö£¬C1ÊÇ·ñÒ»¶¨µÈÓÚ0.2 mol/L________ £¬ »ìºÏºóÈÜÒºÖÐÀë×ÓŨ¶Èc(A-)ºÍc(Na+)µÄ´óС¹ØϵÊÇc(A-)_______ c(Na+) (Ìî>¡¢<¡¢=)¡£ 

(3)´Ó±û×éʵÑé½á¹û·ÖÎö£¬HAÊÇ_______Ëá(Ç¿¡¢Èõ)£¬¸Ã»ìºÏÒºÖеÄÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ______________________________________

£¨4£©¶¡×é»ìºÏÒºÖУ¬ÓÉË®µçÀëµÄc(H£«) =        mol¡¤L-1£¬c (Na+)£­c (A-)=        mol¡¤L-1£»

£¨5£©ÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶϣ¨NH4£©2CO3ÈÜÒºµÄpH         7 (Ìî>¡¢<¡¢= )£»

£¨6£©½«ÏàͬζÈÏÂÏàͬŨ¶ÈµÄËÄÖÖÑÎÈÜÒº£º

A£®NH4HCO3     B£®NH4A       C£®(NH4)2SO4       D£®NH4Cl

°´c(NH4+)ÓÉ´óµ½Ð¡µÄ˳ÐòÅÅÁР                      £¨ÌîÐòºÅ£©

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸