ÌâÄ¿ÄÚÈÝ

ËÜ»¯¼ÁDBP£¨ÁÚ±½¶þ¼×Ëá¶þ¶¡õ¥£©Ö÷ÒªÓ¦ÓÃÓÚPVCµÈºÏ³É²ÄÁÏÖÐ×÷Èí»¯¼Á¡£·´Ó¦Ô­ÀíΪ£º

ʵÑé²½ÖèÈçÏ£º
²½Öè1£ºÔÚÈý¾±Æ¿ÖзÅÈë14.8gÁÚ±½¶þ¼×Ëáôû¡¢25mLÕý¶¡´¼¡¢4µÎŨÁòËᣬ¿ª¶¯½Á°èÆ÷b£¨·´Ó¦×°ÖÃÈçͼ£©¡£
²½Öè2£º»º»º¼ÓÈÈÖÁÁÚ±½¶þ¼×Ëáôû¹ÌÌåÏûʧ£¬ÉýÎÂÖÁ·ÐÌÚ¡£
²½Öè3£ºµÈõ¥»¯µ½Ò»¶¨³Ì¶Èʱ£¬ÉýÎÂÖÁ150¡æ
²½Öè4£ºÀäÈ´£¬½«Èý¾±Æ¿ÖеÄÒºÌåµ¹Èë·Ö©¶·ÖУ¬Óñ¥ºÍʳÑÎË®ºÍ5%̼ËáÄÆÏ´µÓ¡£
²½Öè5£º¼õѹÕôÁó£¬ÊÕ¼¯200~210¡æ2666PaÁó·Ö£¬¼´µÃDBP²úÆ·
£¨1£©Å¨ÁòËáµÄ×÷Óà                     £¬½Á°èÆ÷µÄ×÷Óà                     ¡£
£¨2£©·´Ó¦¹ý³ÌÖÐÕý¶¡´¼¹ýÁ¿µÄÄ¿µÄÊÇ                                           ¡£

£¨3£©Í¼ÖÐÒÇÆ÷aµÄÃû³ÆÊÇ·ÖË®Æ÷£¬ÊÔ·ÖÎöËüµÄ×÷ÓÃÊÇ                             ¡£
²½Öè3ÖÐÈ·¶¨ÓдóÁ¿õ¥Éú³ÉµÄÒÀ¾ÝÊÇ                                       ¡£
£¨4£©Ì¼ËáÄÆÈÜҺϴµÓµÄÄ¿µÄÊÇ                                                 ¡£
ÓüõѹÕôÁóµÄÄ¿µÄÊÇ                                                     ¡£
£¨5£©Ð´³öÕý¶¡´¼ÔÚ135¡æÉú³ÉÃѵķ´Ó¦·½³Ìʽ                                   ¡£
д³öDBPÔÚÇâÑõ»¯ÄÆÈÜÒºÖÐË®½âµÄ·½³Ìʽ                               ¡£
£¨¹²12·Ö£©
£¨1£©´ß»¯¼Á¡¢ÍÑË®¼Á£¬2·Ö£¬¸÷1·Ö       Ê¹·´Ó¦Îï³ä·Ö»ìºÏ1·Ö
£¨2£©Ôö´óÕý¶¡´¼µÄº¬Á¿£¬¿É´Ùʹ·´Ó¦ÕýÏòÒƶ¯£¬Ôö´óÁÚ±½¶þ¼×ËáôûµÄת»¯ÂÊ1·Ö
£¨3£©¼°Ê±·ÖÀë³öõ¥»¯·´Ó¦Éú³ÉµÄË®£¬´Ùʹ·´Ó¦ÕýÏòÒƶ¯£»1·Ö
·ÖË®ÆäÖÐÓдóÁ¿µÄË®Éú³É1·Ö
£¨4£©ÓÃ̼ËáÄƳýÈ¥õ¥ÖеĴ¼ºÍË᣻1·Ö
¼õѹÕôÁó¿É½µµÍÓлúÎïµÄ·Ðµã£¬¿ÉÒÔ·ÀÖ¹ÓлúÎïÍÑˮ̼»¯£¬Ìá¸ß²úÎïµÄ´¿¶È¡£1·Ö
£¨5£©2CH3(CH2)2CH2OH   = CH3(CH2)3O(CH2)3CH3+H2O
+2NaOH 2CH3(CH2)2CH2OH+2H2O+
·½³Ìʽ¸÷2·Ö£¬Ìõ¼þ²»Ð´»ò´íÎó¿Û1·Ö

ÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝÌâÖÐÐÅÏ¢µÃ£¬Å¨ÁòËáµÄ×÷ÓÃΪ´ß»¯¼Á¡¢ÍÑË®¼Á¡£½Á°èÆ÷µÄ×÷ÓÃΪʹ·´Ó¦Îï³ä·Ö»ìºÏ¡£
£¨2£©·´Ó¦¹ý³ÌÖÐÕý¶¡´¼¹ýÁ¿µÄÄ¿µÄÊÇÔö´óÕý¶¡´¼µÄº¬Á¿£¬¿É´Ùʹ·´Ó¦ÕýÏòÒƶ¯£¬Ôö´óÁÚ±½¶þ¼×ËáôûµÄת»¯ÂÊ¡£
£¨3£©·ÖË®Æ÷µÄ×÷ÓÃÊÇʱ·ÖÀë³öõ¥»¯·´Ó¦Éú³ÉµÄË®£¬´Ùʹ·´Ó¦ÕýÏòÒƶ¯¡£È·¶¨ÓдóÁ¿õ¥Éú³ÉµÄÒÀ¾ÝÊÇ·ÖË®ÆäÖÐÓдóÁ¿µÄË®Éú³É¡£
£¨4£©Ì¼ËáÄÆÈÜҺϴµÓµÄÄ¿µÄÊÇÓÃ̼ËáÄƳýÈ¥õ¥ÖеĴ¼ºÍËá¡£ÓüõѹÕôÁóµÄÄ¿µÄÊǼõѹÕôÁó¿É½µµÍÓлúÎïµÄ·Ðµã£¬¿ÉÒÔ·ÀÖ¹ÓлúÎïÍÑˮ̼»¯£¬Ìá¸ß²úÎïµÄ´¿¶È¡£
£¨5£©¸ù¾ÝÌâÖÐÐÅÏ¢µÃ£¬Õý¶¡´¼ÔÚ135¡æÉú³ÉÃѵķ´Ó¦·½³ÌʽΪ2CH3(CH2)2CH2OH   = CH3(CH2)3O(CH2)3CH3+H2O¡£DBPÔÚÇâÑõ»¯ÄÆÈÜÒºÖÐË®½âµÄ·½³ÌʽΪ+2NaOH 2CH3(CH2)2CH2OH+2H2O+¡£
µãÆÀ£º±¾Ì⿼²éµÄÊÇËÜ»¯¼ÁµÄÓ¦ÓúÍÓлú·´Ó¦µÄÏà¹Ø֪ʶ£¬ÌâÄ¿ÄѶȴó£¬ÀûÓúÃÌâÖеÄÐÅÏ¢ÊǽâÌâµÄ¹Ø¼ü¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ﯲúÒµÊǼ«Óз¢Õ¹Ç±Á¦¼°Ç°¾°µÄÐÂÐ˲úÒµ£¬ï¯£¨Zr£©ÔªËØÊǺ˷´Ó¦¶ÑȼÁÏ°ôµÄ°ü¹ü²ÄÁÏ£¬¶þÑõ»¯ï¯£¨ZrO2£©¿ÉÒÔÖÆÔìÄ͸ßÎÂÄÉÃ×ÌÕ´É¡£ÎÒ¹úÓзḻµÄï¯Ó¢Ê¯£¨ZrSiO4£©£¬º¬Al2O3¡¢SiO2¡¢Fe2O3µÈÔÓÖÊ£¬Éú²úï¯Á÷³ÌÖ®Ò»ÈçÏ£º   

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÉÏÊöÁ÷³ÌÖиßÎÂÆø»¯µÄ·´Ó¦·½³Ìʽ£¨Ì¼×ª»¯³ÉCO£©£º                    £»
£¨2£©Ð´³öZrOCl2?8H2OÔÚ900¡æÉú³ÉZrO2µÄ·´Ó¦·½³Ìʽ                       £»
£¨3£©¹ØÓÚ¶þÑõ»¯ï¯ÄÉÃ×ÌմɺÍﯺϽðµÄ˵·¨²»ÕýÈ·µÄÊÇ         £¨µ¥Ñ¡£©£»
A£®¶þÑõ»¯ï¯ÄÉÃ×ÌÕ´ÉÊÇÐÂÐÍÎÞ»ú·Ç½ðÊô²ÄÁÏ
B£®1ÄÉÃ×=10£­10Ã×
C£®ï¯ºÏ½ðµÄÓ²¶È±È´¿ï¯Òª¸ß
D£®ÈÕ±¾¸£µººËµçÕ¾µÄ±¬Õ¨¿ÉÄÜÊÇÓÉﯺϽðÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦²úÉúµÄÇâÆø±¬Õ¨ÒýÆð
£¨4£©Ò»ÖÖÐÂÐÍȼÁϵç³Ø£¬Ò»¼«Í¨Èë¿ÕÆø£¬ÁíÒ»¼«Í¨È붡Í飻µç½âÖÊÊDzôÔÓÑõ»¯îÆ£¨Y2O3£©µÄÑõ»¯ï¯£¨ZrO2£©¾§Ì壬ÔÚÈÛÈÚ״̬ÏÂÄÜ´«µ¼O2£­¡£ÔÚÈÛÈÚµç½âÖÊÖУ¬O2£­Ïò     (ÌîÕý¸º)¼«Òƶ¯¡£µç³ØÕý¼«µç¼«·´Ó¦Îª£º                  £¬¸º¼«µç¼«·´Ó¦Îª£º                    ¡£
ij»¯Ñ§Ð¡×éÒÔ±½¼×ËáΪԭÁÏ£¬ÖÆÈ¡±½¼×Ëá¼×õ¥¡£ÒÑÖªÓйØÎïÖʵķеãÈçÏÂ±í£º
ÎïÖÊ
¼×´¼
±½¼×Ëá
±½¼×Ëá¼×õ¥
·Ðµã£¯¡æ
64.7
249
199.6
I. ºÏ³É±½¼×Ëá¼×õ¥´Ö²úÆ·
ÔÚÔ²µ×ÉÕÆ¿ÖмÓÈë12.2g ±½¼×ËáºÍ20 mL ¼×´¼£¨ÃܶÈÔ¼0.79g ? mL£­1) £¬ÔÙСÐļÓÈë3 mL ŨÁòËᣬ»ìÔȺó£¬Í¶È뼸Á£Ëé´ÉƬ£¬Ð¡ÐļÓÈÈʹ·´Ó¦ÍêÈ«£¬µÃ±½¼×Ëá¼×õ¥´Ö²úÆ·¡£
£¨1£©Å¨ÁòËáµÄ×÷ÓÃÊÇ_________________________________£»Èô·´Ó¦²úÎïË®·Ö×ÓÖÐÓÐͬλËØ18O£¬Ð´³öÄܱíʾ·´Ó¦Ç°ºó18OλÖõĻ¯Ñ§·½³Ìʽ£º_________________________________________________¡£
£¨2£©¼×¡¢ÒÒ¡¢±ûÈýλͬѧ·Ö±ðÉè¼ÆÁËÈçͼÈýÌ×ʵÑéÊÒÖÆÈ¡±½¼×Ëá¼×õ¥µÄ×°Ö㨼гÖÒÇÆ÷ºÍ¼ÓÈÈÒÇÆ÷¾ùÒÑÂÔÈ¥£©¡£¸ù¾ÝÓлúÎïµÄ·Ðµã£¬×îºÃ²ÉÓà     ×°Öã¨Ì¼×¡¢ÒÒ¡¢±û£©¡£

II. ´Ö²úÆ·µÄ¾«ÖÆ
£¨3£©±½¼×Ëá¼×õ¥´Ö²úÆ·ÖÐÍùÍùº¬ÓÐÉÙÁ¿¼×´¼¡¢ÁòËá¡¢±½¼×ËáºÍË®µÈ£¬ÏÖÄâÓÃÏÂÁÐÁ÷³Ìͼ½øÐо«ÖÆ£¬Çë¸ù¾ÝÁ÷³ÌͼÌîÈëÇ¡µ±²Ù×÷·½·¨µÄÃû³Æ£º²Ù×÷IΪ     £¬²Ù×÷IIΪ        ¡£
£¨4£©ÒÔÉÏÁ÷³ÌͼÖмÓÈëNa2CO3ÈÜÒººó£¬·ÅÈë·ÖҺ©¶·ÖÐÕñµ´¡¢¾²Öã¬ÒªµÃµ½Óлú²ã£¬Æä¾ßÌå²Ù×÷ÊÇ
_____________________________________________________¡£
£¨5£©Í¨¹ý¼ÆË㣬±½¼×Ëá¼×õ¥µÄ²úÂÊΪ_________________________¡£
¿×ȸʯµÄÖ÷Òª³É·ÖΪCu2(OH)2CO3£¬»¹º¬ÉÙÁ¿Fe¡¢SiµÄ»¯ºÏÎʵÑéÊÒÒÔ¿×ȸʯΪԭÁÏÖƱ¸CuSO4¡¤5H2OµÄ²½ÖèÈçÏ£º

Ï¡ÁòËá

 

Ϊ½â¾öÓйØÎÊÌ⣬ÐËȤС×éͬѧ²éµÃÓйØÎïÖʳÁµíµÄpHÊý¾ÝÈçÏ£º
ÎïÖÊ
pH (¿ªÊ¼³Áµí)
pH(ÍêÈ«³Áµí)
Fe(OH)3
1.9
3.2
Fe(OH)2
7.0
9.0
Cu(OH)2
4.7
6.7
(1)¡°³ýÔÓ¡±Ê±ÏÈͨÈë×ãÁ¿H2O2½«Fe2£«Ñõ»¯³ÉFe3£«£¬ÔÙ¼ÓÈëCuO¹ÌÌå¡£ÆäÖмÓÈëCuO×÷ÓÃÊÇ
                                                              £¬Ðèµ÷½ÚÈÜÒºpHµÄ·¶Î§Îª                   ¡£
(2)²Ù×÷X°üÀ¨ Õô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂ˺ÍÏ´µÓµÈ¡£ÔÚ½øÐиòÙ×÷ʱ£¬ÐèÒª½«ÈÜÒºBÔÙ
Êʵ±ËữĿµÄ                                                                ¡£
(3)ÈôÏòÈÜÒºBÖÐÖðµÎ¼ÓÈëNaOHÈÜÒº£¬¸ÕºÃ³öÏÖ³Áµíʱ£¬Çëд³ö´æÔÚµÄÄÑÈÜÎïÖʵijÁµíÈܽâƽºâ·½³ÌʽΪ                                                                   
(4) Ϊ׼ȷ²â¶¨ÈÜÒºAÖк¬ÓÐFe2£«µÄÎïÖʵÄÁ¿Å¨¶È£¬ÊµÑéÈçÏ£º
¢ÙÈ¡³ö25.00mLÈÜÒºA£¬ÅäÖƳÉ250 mL ÈÜÒº¡£
¢ÚµÎ¶¨£º×¼È·Á¿È¡25.00mlËùÅäÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬½«0.20mol/LKMnO4ÈÜÒº
×°Èë              £¬¼Ç¼Êý¾Ý¡£Öظ´µÎ¶¨2´Î¡£Æ½¾ùÏûºÄKMnO4ÈÜÒºV mL¡£
(·´Ó¦Ê½£º 5Fe2+ + MnO4¡ª +10 H+ = 5Fe3+ + Mn2+ + 5H2O)
¢Û ¼ÆËãÈÜÒºAÖÐFe2+µÄÎïÖʵÄÁ¿Å¨¶È=             mol/L £¨Ö»ÁгöËãʽ£¬²»×öÔËË㣩¡£

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

¾«Ó¢¼Ò½ÌÍø