ÌâÄ¿ÄÚÈÝ

ÀûÓ÷ϾɶÆпÌúƤ¿ÉÖƱ¸´ÅÐÔFe3O4½ºÌåÁ£×Ó¼°¸±²úÎïZnO¡£ÖƱ¸Á÷³ÌÈçÏ£º

ÒÑÖª£ºZn¼°Æ仯ºÏÎïµÄÐÔÖÊÓëAl¼°Æ仯ºÏÎïµÄÐÔÖÊÏàËÆ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃNaOHÈÜÒº´¦Àí·Ï¾É¶ÆпÌúƤ¿ÉÒÔÈ¥³ýÓÍÎÛ£¬»¹¿ÉÒÔ____________________¡£
£¨2£©µ÷½ÚÈÜÒºAµÄpH¿É²úÉúZn(OH)2³Áµí£¬ÎªÖƵÃZnO£¬ºóÐø²Ù×÷²½ÖèÊÇ_____¡úÏ´µÓ¡ú_____¡£
£¨3£©ÓÉÈÜÒºBÖƵÃFe3O4½ºÌåÁ£×ӵĹý³ÌÖУ¬Ðë³ÖÐøͨÈëN2£¬ÆäÔ­ÒòÊÇ________________¡£
£¨4£©ÓÃÖظõËá¼Ø·¨£¨Ò»ÖÖÑõ»¯»¹Ô­µÎ¶¨·¨£©¿É²â¶¨²úÎïFe3O4ÖеĶþ¼ÛÌúº¬Á¿¡£Çëд³öËáÐÔÌõ¼þÏÂK2Cr2O7ÓëFe2+·´Ó¦µÄÀë×Ó·½³Ìʽ              £¨K2Cr2O7±»»¹Ô­ÎªCr3+£©¡£
£¨5£©ÈôÐèÅäÖÆŨ¶ÈΪ0.01000 mol·L-1µÄK2Cr2O7±ê×¼ÈÜÒº250mL£¬Ó¦×¼È·³ÆÈ¡K2Cr2O7   g£¨±£ÁôËÄλÓÐЧÊý×Ö£¬ÒÑÖªM(K2Cr2O7)="294.0" g·mol-1£©¡£ÅäÖƸñê×¼ÈÜҺʱ£¬ÏÂÁÐÒÇÆ÷Ò»¶¨²»ÒªÓõ½µÄÓР           £¨ÓñàºÅ±íʾ£©¡£
¢Ùµç×ÓÌìƽ ¢ÚÉÕ±­ ¢ÛÁ¿Í² ¢Ü²£Á§°ô ¢Ý250 mLÈÝÁ¿Æ¿ ¢Þ½ºÍ·µÎ¹Ü ¢ßÍÐÅÌÌìƽ
£¨6£©ÈôÅäÖÆK2Cr2O7±ê×¼ÈÜҺʱ£¬¸©Êӿ̶ÈÏߣ¬Ôò²â¶¨½á¹û_______£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£¬ÏÂͬ£©£»µÎ¶¨²Ù×÷ÖУ¬ÈôµÎ¶¨Ç°×°ÓÐK2Cr2O7±ê×¼ÈÜÒºµÄµÎ¶¨¹Ü¼â×첿·ÖÓÐÆøÅÝ£¬¶øµÎ¶¨½áÊøºóÆøÅÝÏûʧ£¬ÔòµÎ¶¨½á¹û½«________¡£
£¨1£©Èܽâ¶Æп²ã £¨2£©¹ýÂË¡¢×ÆÉÕ £¨3£©N2Æø·ÕÏ£¬·ÀÖ¹Fe2+±»Ñõ»¯
£¨4£©6Fe2+ + Cr2O72- + 14H+ = 6Fe3+ + 2Cr3+ + 7H2O    £¨5£©0.7350   ¢Û¢ß £¨6£©Æ«Ð¡   Æ«´ó

ÊÔÌâ·ÖÎö£º£¨1£©NaOHÈÜÒºÄÜÓëп·¢Éú·´Ó¦£¬ËùÒÔÓÃNaOHÈÜÒº´¦Àí·Ï¾É¶ÆпÌúƤ¿ÉÒÔÈ¥³ýÓÍÎÛ£¬»¹¿ÉÒÔÈܽâ¶Æп²ã¡££¨2£©·´Ó¦²úÉúZn(OH)2³Áµí£¬¼ÓÈÈ·Ö½â²úÉúÑõ»¯Ð¿ºÍË®¡£ËùÒÔΪÖƵÃZnO£¬ºóÐø²Ù×÷²½ÖèÊǹýÂË¡úÏ´µÓ¡ú×ÆÉÕ¡££¨3£©µªÆø»¯Ñ§ÐÔÖʲ»»îÆã¬ÔÚÕâÖÖ¶èÐÔ»·¾³ÖпɷÀÖ¹Fe2+±»Ñõ»¯¡££¨4£©·´Ó¦µÄÁ½ÖÖ·½³ÌʽÊÇ£º6Fe2+ + Cr2O72- + 14H+ = 6Fe3+ + 2Cr3+ + 7H2O £¨5£©m=n·M="0.01mol/L¡Á0.25L¡Á294.0" g/mol=0.7350g.ÅäÖÆÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒºÓõ½µÄÒÇÆ÷ÓТٵç×ÓÌìƽ ¢ÚÉÕ±­ ¢ÛÁ¿Í² ¢Ü²£Á§°ô ¢Ý250 mLÈÝÁ¿Æ¿ ¢Þ½ºÍ·µÎ¹Ü¡££¨6£©ÈôÅäÖÆK2Cr2O7±ê×¼ÈÜҺʱ£¬¸©Êӿ̶ÈÏߣ¬ÓÉÓÚÈÜÒºµÄÌå»ýƫСËùÒÔK2Cr2O7±ê×¼ÈÜҺŨ¶ÈÆ«´ó£¬ÓÃÕâÖÖ±ê×¼ÈÜÒº²â¶¨£¬ÏûºÄµÄ±ê×¼ÈÜÒº¾ÍƫС½á¹ûƫС£»µÎ¶¨²Ù×÷ÖУ¬ÈôµÎ¶¨Ç°×°ÓÐK2Cr2O7±ê×¼ÈÜÒºµÄµÎ¶¨¹Ü¼â×첿·ÖÓÐÆøÅÝ£¬¶øµÎ¶¨½áÊøºóÆøÅÝÏûʧ£¬ÏûºÄµÄ±ê×¼ÈÜÒºµÄÌå»ý¶ÁÊýÆ«´óÔòµÎ¶¨½á¹û½«Æ«´ó¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ʵÑéÊÒ³£ÀûÓü×È©£¨HCHO£©·¨²â¶¨(NH4)2SO4ÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý£¬Æä·´Ó¦Ô­ÀíΪ£º4NH4£« £«6HCHO =3H£«£«6H2O£«(CH2)6N4H£« £ÛµÎ¶¨Ê±£¬1 mol (CH2)6N4H£«Óë l mol H£«Ï൱£¬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¶¨´ïµ½ÖÕµãʱÏÖÏó£º__________________________________________________¡£
£¨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
 
ÈôNaOH±ê×¼ÈÜÒºµÄŨ¶ÈΪ0£®1010 mol¡¤L£­1£¬Ôò¸ÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊýΪ___________¡£
ʵÑéÊÒÐèÒª0£®2mol/L CuSO4ÈÜÒº250ml£¬ÊµÑéÊÒ¿ÉÌṩÅäÖÆÈÜÒºµÄÊÔ¼ÁÓУº¢ÙÀ¶É«µ¨·¯¾§Ìå(CuSO4¡¤5H2O) ¢Ú4mol/L CuSO4ÈÜÒº
(1)ÎÞÂÛ²ÉÓúÎÖÖÊÔ¼Á½øÐÐÅäÖÆ£¬ÊµÑé±ØÐëÓõ½µÄ²£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜÍ⣬ÖÁÉÙ»¹ÐèÒªµÄÒ»ÖÖÒÇÆ÷ÊÇ________£¬ÔÚʹÓøÃÒÇÆ÷Ç°±ØÐë½øÐеIJÙ×÷ÊÇ              ¡£
(2)ÈôÓõ¨·¯¾§Ìå½øÐÐÅäÖÆ£¬ÐèÓÃÍÐÅÌÌìƽ³ÆÈ¡CuSO4¡¤ 5H2OµÄÖÊÁ¿Îª________g£»Èç¹ûÓÃ4mol/LµÄCuSO4ÈÜҺϡÊÍÅäÖÆ£¬ÐèÓÃÁ¿Í²Á¿È¡___________ml4mol/L CuSO4ÈÜÒº¡£
(3)ʵÑéÊÒÓÃ4mol/LµÄÁòËáÍ­ÈÜҺϡÊÍÅäÖÆÈÜÒºËùÐèµÄʵÑé²½ÖèÓУº
ÆäÖÐÕýÈ·µÄ²Ù×÷˳ÐòΪ                             
¢ÙÍùÉÕ±­ÖмÓÈëÔ¼100mlË®½øÐгõ²½Ï¡ÊÍ£¬ÀäÈ´ÖÁÊÒÎÂ
¢ÚÓÃÁ¿Í²Á¿È¡Ò»¶¨Ìå»ý4mol/L µÄÁòËáÍ­ÈÜÒºÓÚÒ»ÉÕ±­ÖÐ
¢Û¼ÆËãËùÐè4mol/L ÁòËáÍ­ÈÜÒºµÄÌå»ý
¢Ü½«ÈÜÒºµßµ¹Ò¡ÔȺóת´æÓÚÊÔ¼ÁÆ¿
¢Ý¼ÓË®ÖÁÒºÃæÀëÈÝÁ¿Æ¿1-2cm´¦¸ÄÓýºÍ·µÎ¹Ü½øÐж¨ÈÝ
¢ÞÏ´µÓÉÕ±­ºÍ²£Á§°ô2-3´Î²¢½«Ï´µÓҺעÈëÈÝÁ¿Æ¿£¬ÇáÇáÒ¡¶¯ÈÝÁ¿Æ¿£¬Ê¹ÈÜÒº»ìºÏ¾ùÔÈ
¢ß½«ÈÜҺתÒÆÈëÈÝÁ¿Æ¿
(4)ÅäÖÆÈÜÒº¹ý³ÌÖУ¬Èç¹û³öÏÖÒÔÏÂÇé¿ö£¬¶Ô½á¹ûÓкÎÓ°Ï죨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©
£Á£®¶¨ÈݺóÈûÉÏÆ¿Èû·´¸´Ò¡ÔÈ£¬¾²Öúó£¬ÒºÃæ²»µ½¿Ì¶ÈÏߣ¬ÔÙ¼ÓË®ÖÁ¿Ì¶ÈÏß           
B£®¶¨ÈÝʱ¸©Êӿ̶ÈÏß                
C£®ÈÝÁ¿Æ¿Î´¸ÉÔï¼´ÓÃÀ´ÅäÖÆÈÜÒº             
£¨5£©ÊµÑé¹ý³ÌÖÐÓõ½ÕôÁóË®¡£ÈçͼΪʵÑéÊÒÖÆÈ¡ÕôÁóË®µÄ×°ÖÃʾÒâͼ¡£Í¼ÖеÄÁ½´¦Ã÷ÏԵĴíÎóÊÇ ______________________________£»_______________________________¡£ÊµÑéʱAÖгý¼ÓÈëÉÙÁ¿×ÔÀ´Ë®Í⣬»¹Ðè¼ÓÈëÉÙÁ¿___         ____£¬Æä×÷ÓÃÊÇ·ÀÖ¹±©·Ð¡£

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

¾«Ó¢¼Ò½ÌÍø