ÌâÄ¿ÄÚÈÝ

ʵÑéÊÒ³£ÀûÓü×È©·¨(HCHO)²â¶¨(NH4)2SO4ÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý£¬Æä·´Ó¦Ô­ÀíΪ£º
4NH4£« £«6HCHO =3H£«£«6H2O£«(CH2)6N4H£« £ÛµÎ¶¨Ê±£¬1 mol (CH2)6N4H£«ÏûºÄNaOHÓë l mol H£«Ï൱£Ý£¬È»ºóÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨·´Ó¦Éú³ÉµÄËᡣijÐËȤС×éÓü×È©·¨½øÐÐÁËÈçÏÂʵÑ飺
²½ÖèI ³ÆÈ¡ÑùÆ·1.500 g¡£
²½ÖèII ½«ÑùÆ·Èܽâºó£¬ÍêȫתÒƵ½250 mLÈÝÁ¿Æ¿ÖУ¬¶¨ÈÝ£¬³ä·ÖÒ¡ÔÈ¡£
²½ÖèIII ÒÆÈ¡25.00 mLÑùÆ·ÈÜÒºÓÚ250 mL׶ÐÎÆ¿ÖУ¬¼ÓÈë10 mL 20£¥µÄÖÐÐÔ¼×È©ÈÜÒº£¬Ò¡ÔÈ¡¢¾²ÖÃ5 minºó£¬¼ÓÈë1~2µÎ·Ó̪ÊÔÒº£¬ÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㡣°´ÉÏÊö²Ù×÷·½·¨ÔÙÖظ´2´Î¡£
(1)¸ù¾Ý²½ÖèIIIÌî¿Õ£º
¢Ù¼îʽµÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó£¬Ö±½Ó¼ÓÈëNaOH±ê×¼ÈÜÒº½øÐе樣¬Ôò²âµÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£
¢Ú׶ÐÎÆ¿ÓÃÕôÁóˮϴµÓºó£¬Ë®Î´µ¹¾¡£¬ÔòµÎ¶¨Ê±ÓÃÈ¥NaOH±ê×¼ÈÜÒºµÄÌå»ý________________(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)
¢ÛµÎ¶¨Ê±±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦Ó¦¹Û²ì____________
(A)µÎ¶¨¹ÜÄÚÒºÃæµÄ±ä»¯    (B)׶ÐÎÆ¿ÄÚÈÜÒºÑÕÉ«µÄ±ä»¯
¢ÜµÎ¶¨´ïµ½ÖÕµãʱ£¬·Óָ̪ʾ¼ÁÓÉ_________É«±ä³É_________É«¡£
(2)µÎ¶¨½á¹ûÈçϱíËùʾ£º
µÎ¶¨
´ÎÊý
´ý²âÈÜÒºµÄÌå»ý
/mL
±ê×¼ÈÜÒºµÄÌå»ý/mL
µÎ¶¨Ç°¿Ì¶È
µÎ¶¨ºó¿Ì¶È
1
25.00
1.02
21.03
2
25.00
2.00
21.99
3
25.00
0.20
20.20
µÎ¶¨Ê±ÏûºÄNaOH±ê×¼ÈÜÒºµÄÌå»ýµÄƽ¾ùÖµV="__________mL;" ÈôNaOH±ê×¼ÈÜÒºµÄŨ¶ÈΪ0.1010 mol¡¤L£­1£¬Ôò250 mLÈÜÒºÖеÄn(NH4+)=__________mol,¸ÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊýΪ______¡£
(1)¢ÙÆ«¸ß£¨2·Ö£©¡£  ¢ÚÎÞÓ°Ï죨2·Ö£©    ¢ÛB£¨2·Ö£©¡£   ¢ÜÎÞ£¨1·Ö£©£»               ·Ûºì(»òdzºì) £¨1·Ö£©¡£  (2) 20.00 £¨1·Ö£© 0.0202 £¨1·Ö£© 18.85%¡££¨1·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÓÃÈçÏÂ×°ÖÿÉÒÔÍê³ÉһϵÁÐʵÑ飨ͼÖмгÖ×°ÖÃÒÑÂÔÈ¥£©¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ñ£®ÈôÓÃ×°ÖÃAÑ¡ÓÃŨÁòËáºÍÑÇÁòËáÄƹÌÌåÖÆÈ¡SO2ÆøÌ壨¸ù¾ÝÐèÒª¿ÉÒÔ¼ÓÈÈ£©£¬²¢Í¨¹ý×°ÖÃBÍê³É±íÖÐÉè¼ÆʵÑ飬ÇëÌîд±íÖпհףº
BÖÐÃÞ»¨µÄλÖÃ
¢Ù
¢Ú
¢Û
¢Ü
ËùÕºÊÔ¼Á
ʯÈïÊÔÒº
Æ·ºìÈÜÒº
µí·ÛºÍµâË®»ìºÏÒº
ÇâÁòËá
ÏÖÏó
 
 
ÍÊÉ«
dz»ÆÉ«
ÌåÏÖSO2µÄÐÔÖÊ
 
 
 
 
 
II£®ÈôÓÃ×°ÖÃAÑ¡ÓÃŨÁòËáºÍŨÑÎËá»ìºÏÖÆÈ¡HClÆøÌ壬װÖÃBÖеÄËÄ´¦ÃÞ»¨ÒÀ´Î×öÁËÈçÏ´¦Àí£º¢Ù°üÓÐij¹ÌÌåÎïÖÊ¡¢¢ÚÕºÓÐKIÈÜÒº¡¢¢ÛÕºÓÐʯÈïÈÜÒº¡¢¢ÜÕºÓÐŨNaOHÈÜÒº¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©pÖÐÊ¢×°ÊÔ¼ÁΪ                  ¡£
£¨2£©·´Ó¦¿ªÊ¼ºó£¬¹Û²ìµ½¢Ú´¦ÓÐ×Ø»ÆÉ«ÎïÖʲúÉú£¬Ð´³ö¢Ú´¦·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ       ¡£¢Ù´¦°üÓеÄij¹ÌÌåÎïÖÊ¿ÉÄÜÊÇ              ¡£
a£®MnO2              b£®KMnO4         c£®KCl           d£®Cu
£¨3£©ÔÚÕû¸öʵÑé¹ý³ÌÖУ¬ÔÚ¢Û´¦Äܹ۲쵽                         ¡£
£¨4£©·´Ó¦½Ï³¤Ê±¼äºó£¬¢Ú´¦ÓÐ×Ø»ÆÉ«ÍÊÈ¥£¬Éú³ÉÎÞÉ«µÄIO3£­£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                          ¡£
À¬»øÊÇ·Å´íµØ·½µÄ×ÊÔ´£¬¹¤Òµ·ÏÁÏÒ²¿ÉÒÔÔÙÀûÓá£Ä³»¯Ñ§ÐËȤС×éÔÚʵÑéÊÒÖÐÓ÷ÏÆúº¬ÓÐÂÁ¡¢Ìú¡¢Í­µÄºÏ½ðÖÆÈ¡ÁòËáÂÁÈÜÒº¡¢ÏõËáÍ­¾§ÌåºÍÌúºì(Fe2O3)¡£ÆäʵÑé·½°¸ÈçÏÂ:

£¨1£©Çëд³öÔںϽðÖмÓÈëKOHÈÜÒººóËù·¢ÉúµÄÀë×Ó·´Ó¦·½³Ìʽ£º
                                                                        ¡£
£¨2£©ÔÚÂËÒºAÖÐÖ±½Ó¼ÓÈëÁòËáËù»ñµÃµÄÁòËáÂÁÈÜÒºÖлẬÓÐÔÓÖÊ£¨K2SO4£©£¬ÇëÉè¼ÆÒ»¸ö¸üºÏÀíµÄʵÑé·½°¸ÓÉÂËÒºAÖƱ¸´¿¾»µÄÁòËáÂÁÈÜÒº¡£·ÂÕÕÉÏͼÐÎʽ»­³öÖƱ¸Á÷³Ìͼ£º£¨Ìáʾ£ºÔÚ¼ýÍ·ÉÏÏ·½±ê³öËùÓÃÊÔ¼Á»òʵÑé²Ù×÷£©

£¨3£©ÒÑÖªFe(OH)3³ÁµíµÄpHÊÇ2~3.2¡£ÈÜÒºCͨ¹ýµ÷½ÚpH¿ÉÒÔʹFe3+³ÁµíÍêÈ«¡£ÏÂÁÐÎï
ÖÊÖУ¬¿ÉÓÃ×÷µ÷ÕûÈÜÒºCµÄpHµÄÊÔ¼ÁÊÇ               £¨ÌîÐòºÅ£©
A£®Í­·ÛB£®°±Ë®C£®Ñõ»¯Í­D£®ÇâÑõ»¯Í­
£¨4£©ÀûÓÃÂËÒºDÖÆÈ¡ÏõËáÍ­¾§Ì壬±ØÐë½øÐеÄʵÑé²Ù×÷²½Ö裺¼ÓÈÈÕô·¢¡¢ÀäÈ´½á¾§¡¢
                £¨Ìî²Ù×÷Ãû³Æ£©¡¢×ÔÈ»¸ÉÔï¡£
£¨5£©ÔÚ0.1LµÄ»ìºÏËáÈÜÒºÖУ¬c(HNO3)=2mol¡¤L-1£¬c(H2SO4)=3mol¡¤L-1¡£½«0.3molµÄÍ­·ÅÈë²¢³ä·Ö·´Ó¦ºó£¬²úÉúµÄÍ­ÑεĻ¯Ñ§Ê½ÊÇ            £¬±»»¹Ô­µÄn(HNO3£©= ¡£            
£¨15·Ö£©ÎªÑо¿ÌúÖʲÄÁÏÓëÈÈŨÁòËáµÄ·´Ó¦£¬Ä³Ð¡×é½øÐÐÁËÒÔÏÂ̽¾¿»î¶¯£º
[̽¾¿Ò»]
£¨1£©³ÆÈ¡Ìú¶¤£¨Ì¼Ëظ֣©12£®0g·ÅÈë30£®0mLŨÁòËáÖУ¬¼ÓÈÈ£¬³ä·ÖÓ¦ºóµÃµ½ÈÜÒºX²¢ÊÕ¼¯µ½ÆøÌåY¡£
¢Ù¼×ͬѧÈÏΪXÖгýFe3+Í⻹¿ÉÄܺ¬ÓÐFe2+¡£ÈôҪȷÈÏÆäÖÐÊÇ·ñº¬ÓÐFe2+£¬Ó¦Ñ¡Ôñ¼ÓÈëµÄÊÔ¼ÁΪ____   £¨Ñ¡ÌîÐòºÅ£©¡£ 
a£®KSCNÈÜÒººÍÂÈË®              b£®Ìú·ÛºÍKSCNÈÜÒº
c£®Å¨°±Ë®    d£®ËáÐÔKMnO4ÈÜÒº
¢ÚÒÒͬѧȡ672 mL£¨±ê×¼×´¿ö£©ÆøÌåYͨÈë×ãÁ¿äåË®ÖУ¬·¢Éú·´Ó¦£º
SO2+Br2+2H2O=2HBr+H2SO4
È»ºó¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬¾­Êʵ±²Ù×÷ºóµÃµ½¸ÉÔï¹ÌÌå4£®66g¡£¾Ý´ËÍÆÖªÆøÌåY   ÖÐSO2µÄÌå»ý·ÖÊýΪ____    ¡££¨Ïà¶ÔÔ­×ÓÖÊÁ¿£ºO¡ª16  S¡ª32  Ba¡ª137£©
[̽¾¿¶þ]
·ÖÎöÉÏÊöʵÑéÖÐSO2Ìå»ý·ÖÊýµÄ½á¹û£¬±ûͬѧÈÏΪÆøÌåYÖл¹¿ÉÄܺ¬Á¿ÓÐH2ºÍCO2ÆøÌ塣Ϊ´ËÉè¼ÆÁËÏÂÁÐ̽¾¿ÊµÑé×°Öã¨Í¼ÖмгÖÒÇÆ÷Ê¡ÂÔ£©¡£

£¨2£©Ð´³ö²úÉúCO2µÄ»¯Ñ§·½³Ìʽ___                   _¡£
£¨3£©×°ÖÃAÖÐÊÔ¼ÁµÄ×÷ÓÃÊÇ____             ¡£
£¨4£©¼òÊöÈ·ÈÏÆøÌåYÖк¬ÓÐCO2µÄʵÑéÏÖÏó                  ¡£
£¨5£©Èç¹ûÆøÌåYÖк¬ÓÐH2£¬Ô¤¼ÆʵÑéÏÖÏóÓ¦ÊÇ               ¡£
£¨14·Ö£©Ä³ÐËȤС×éµÄѧÉú¸ù¾ÝMgÓëCO2·´Ó¦Ô­Àí£¬ÍƲâÄÆÒ²Ó¦ÄÜÔÚCO2ÖÐȼÉÕ£¬ÎªÁËÈ·¶¨ÆäÉú³É²úÎï²¢½øÐÐʵÑéÂÛÖ¤¡£Ä³Í¬Ñ§Éè¼ÆÁËÏÂͼËùʾװÖýøÐÐʵÑ飨ÒÑÖªPdCl2Äܱ»CO»¹Ô­µÃµ½ºÚÉ«µÄPd£©£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÎªÊ¹·´Ó¦Ë濪ËæÓã¬Ëæ¹ØËæÍ££¬·½¿òÄÚӦѡÓÃÏÂͼËùʾ         ×°Öã¨Ìî×Öĸ´úºÅ£©¡£

£¨2£©ÊµÑéÊÒÖÆÈ¡¶þÑõ»¯Ì¼ÆøÌåÒËÓõÄÒ©Æ·ÊÇ               £¨Ìî±êºÅ£©¡£
¢Ùʯ»Òʯ£¬¢Ú´¿¼î£¬¢ÛСËÕ´ò£¬¢Ü18.4 mol¡¤L¡ª1ÁòËᣬ¢Ý11.2mol¡¤L¡ª1ÑÎËᣬ¢ÞÕôÁóË®£¬¢ßľ̿·Û¡£
£¨3£©ÔÚ2×°ÖÃÄÚ½øÐз´Ó¦µÄÀë×Ó·½³ÌʽΪ                                         ¡£¼ì²é×°ÖÃÆøÃÜÐÔ²¢×°ºÃÒ©Æ·ºó£¬µãȼ¾Æ¾«µÆ֮ǰ´ý×°Öà             £¨ÌîÊý×Ö±àºÅ£©ÖгöÏÖ        ÏÖÏóʱ£¬ÔÙµãȼ¾Æ¾«µÆ¡£
£¨4£©¢ÙÈô×°ÖÃ6ÖÐÓкÚÉ«³Áµí£¬×°ÖÃ4ÖвÐÁô¹ÌÌ壨ֻÓÐÒ»ÖÖÎïÖÊ£©¼ÓÑÎËáºóÓÐÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÆøÌå·Å³ö£¬ÔòÄÆÓë¶þÑõ»¯Ì¼·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                         £»
¢ÚÈô×°ÖÃ6ÖÐÈÜÒºÎÞÃ÷ÏÔÏÖÏó£¬×°ÖÃ4ÖвÐÁô¹ÌÌ壨ÓÐÁ½ÖÖÎïÖÊ£©¼ÓÑÎËáºóÓÐÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÆøÌå·Å³ö£¬ÔòÄÆÓë¶þÑõ»¯Ì¼·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                          ¡£

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

¾«Ó¢¼Ò½ÌÍø