ÑÇÂÈËáÄÆ£¨NaClO2£©ÊÇÒ»ÖÖÖØÒªµÄº¬ÂÈÏû¶¾¼Á£¬Ö÷ÒªÓÃÓÚË®µÄÏû¶¾ÒÔ¼°É°ÌÇ¡¢ÓÍÖ¬µÄƯ°×Óëɱ¾ú¡£ÒÔÏÂÊǹýÑõ»¯Çâ·¨Éú²úÑÇÂÈËáÄƵŤÒÕÁ÷³Ìͼ£º

ÒÑÖª£º¢ÙNaClO2µÄÈܽâ¶ÈËæζÈÉý¸ß¶øÔö´ó£¬Êʵ±Ìõ¼þÏ¿ɽᾧÎö³öNaClO2?3H2O¡£
¢Ú´¿ClO2Ò׷ֽⱬը£¬Ò»°ãÓÃÏ¡ÓÐÆøÌå»ò¿ÕÆøÏ¡Ê͵½10£¥ÒÔÏ°²È«¡£
¢Û160 g/L NaOHÈÜÒºÊÇÖ¸160 gNaOH¹ÌÌåÈÜÓÚË®ËùµÃÈÜÒºµÄÌå»ýΪ1L¡£
£¨1£©160 g/L NaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ         ¡£ÈôÒª¼ÆËã¸ÃÈÜÒºµÄÖÊÁ¿·ÖÊý£¬»¹ÐèÒªµÄÒ»¸öÌõ¼þÊÇ            £¨ÓÃÎÄ×Ö˵Ã÷£©¡£
£¨2£©·¢ÉúÆ÷ÖйÄÈë¿ÕÆøµÄ×÷ÓÿÉÄÜÊÇ  £¨Ñ¡ÌîÐòºÅ£©¡£
a£®½«SO2Ñõ»¯³ÉSO3£¬ÔöÇ¿ËáÐÔ£»     b£®Ï¡ÊÍClO2ÒÔ·ÀÖ¹±¬Õ¨£»
c£®½«NaClO3Ñõ»¯³ÉClO2
£¨3£©ÎüÊÕËþÄڵķ´Ó¦µÄÀë×Ó·´Ó¦·½³ÌʽΪ              ¡£
ÎüÊÕËþµÄζȲ»Äܳ¬¹ý20¡æ£¬ÆäÄ¿µÄÊÇ                           ¡£
£¨4£©ÔÚ¼îÐÔÈÜÒºÖÐNaClO2±È½ÏÎȶ¨£¬ËùÒÔÎüÊÕËþÖÐӦά³ÖNaOHÉÔ¹ýÁ¿£¬ÅжÏNaOHÊÇ·ñ¹ýÁ¿µÄ¼òµ¥ÊµÑé·½·¨ÊÇ                                ¡£
£¨5£©ÎüÊÕËþÖÐΪ·ÀÖ¹NaClO2±»»¹Ô­³ÉNaCl£¬ËùÓû¹Ô­¼ÁµÄ»¹Ô­ÐÔÓ¦ÊÊÖС£³ýH2O2Í⣬»¹¿ÉÒÔÑ¡ÔñµÄ»¹Ô­¼ÁÊÇ       £¨Ñ¡ÌîÐòºÅ£©¡£
a£®Na2O2             b£®Na2S              c£®FeCl2
£¨6£©´ÓÂËÒºÖеõ½NaClO2?3H2O´Ö¾§ÌåµÄʵÑé²Ù×÷ÒÀ´ÎÊÇ  £¨Ñ¡ÌîÐòºÅ£©
a£®ÕôÁó      b£®Õô·¢    c£®×ÆÉÕ     d£®¹ýÂË      e£®ÀäÈ´½á¾§
ÒªµÃµ½¸ü´¿µÄNaClO2?3H2O¾§Ìå±ØÐë½øÐеIJÙ×÷ÊÇ      £¨Ìî²Ù×÷Ãû³Æ£©

£¨1£©4 mol/L   ÈÜÒºµÄÃܶȣ¨2£©b
£¨3£©2OH-+2ClO2+H2O2£½2ClO2-+O2¡ü+2H2O ·ÀÖ¹H2O2ÊÜÈȷֽ⣬ÓÐÀûÓÚNaClO2?3H2OµÄÎö³ö
£¨4£©Á¬Ðø²â¶¨ÎüÊÕËþÄÚÈÜÒºµÄpHÖµ £¨5£© a£¨6£©bed       Öؽᾧ

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÓÉÐÅÏ¢¢ÛÖª£º160g/LNaOHÈÜÒº±íʾ1LÇâÑõ»¯ÄÆÈÜÒºº¬ÓÐ160gNaOH£®¸ù¾ÝÎïÖʵÄÁ¿Å¨¶ÈµÄ¶¨Òåʽc=n/v¼ÆËã¿ÉµÃ¸ÃÈÜÒºÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿Å¨¶ÈΪ4mol/L£»¸ù¾ÝÎïÖʵÄÁ¿Å¨¶ÈºÍÖÊÁ¿·ÖÊýµÄ»»Ë㹫ʽ ¿ÉÖª£¬Òª¼ÆËãÖÊÁ¿·ÖÊý»¹ÐèÒªÖªµÀÈÜÒºµÄÃܶȡ£
£¨2£©ÓÉÐÅÏ¢¢Ú¿ÉÖª£¬´¿ClO2Ò׷ֽⱬը£¬Ò»°ãÓÃÏ¡ÓÐÆøÌå»ò¿ÕÆøÏ¡Ê͵½10%ÒÔÏ°²È«¡£·¢ÉúÆ÷ÖйÄÈë¿ÕÆøµÄ×÷ÓÃÓ¦ÊÇÏ¡ÊÍClO2ÒÔ·ÀÖ¹±¬Õ¨¡£
£¨3£©¸ù¾Ý¹¤ÒÕÁ÷³Ìͼ¿ÉÖª£¬ÎüÊÕËþÄڵķ´Ó¦ÎïΪClO2¡¢H2O2ºÍNaOHÈÜÒº£¬Éú³ÉÎïÓÐNaClO2£¬ËùÒÔÒ»¶¨ÓÐClO2ת»¯ÎªNaClO2£¬»¯ºÏ¼Û½µµÍ£¬±»»¹Ô­£»ÔòH2O2±Ø¶¨±»Ñõ»¯£¬ÓÐÑõÆø²úÉú£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ2OH-+2ClO2+H2O2£½2ClO2-+O2¡ü+2H2O£»H2O2²»Îȶ¨£¬Î¶ȹý¸ßÈÝÒ׷ֽ⣬ÎüÊÕËþµÄζȲ»Äܳ¬¹ý20¡æ£¬ÆäÄ¿µÄÊÇ·ÀÖ¹H2O2·Ö½â¡£½áºÏÐÅÏ¢¢ÙNaClO2µÄÈܽâ¶ÈËæζÈÉý¸ß¶øÔö´ó£¬ÎüÊÕËþµÄζȲ»Äܳ¬¹ý20¡æµÄÔ­ÒòÓÐÀûÓÚNaClO2?3H2OµÄÎö³ö¡£
£¨4£©NaOH¹ýÁ¿ÔòÈÜÒº³Ê¼îÐÔ£¬²â¶¨ÈÜÒºËá¼îÐÔ×î¼òµ¥·½·¨ÊÇ£ºÁ¬Ðø²â¶¨ÎüÊÕËþÄÚÈÜÒºµÄpH¡£
£¨5£©¸ù¾ÝÌâÒâClO2ת»¯ÎªNaClO2µÄ»¹Ô­¼ÁΪH2O2£¬Na2O2ÈÜÓÚË®Ï൱ÓÚH2O2£¬¶øNa2S¡¢FeCl2»¹Ô­ÐÔ½ÏÇ¿£¬Ñ¡a¡£
£¨6£©´ÓÈÜÒºÖеõ½º¬½á¾§Ë®µÄ¾§Ìå²ÉÈ¡Õô·¢¡¢Å¨Ëõ¡¢ÀäÈ´½á¾§·½·¨£¬Í¨¹ý¹ýÂ˵õ½´Ö¾§Ì壮ËùÒÔ²Ù×÷˳ÐòΪbed£»µÃµ½µÄ´Ö¾§Ìå¾­¹ýÖؽᾧ¿ÉµÃµ½´¿¶È¸ü¸ßµÄ¾§Ìå¡£
¿¼µã£º±¾ÌâÒÔÑÇÂÈËáÄÆÖƱ¸µÄ»¯Ñ§¹¤ÒÕÁ÷³ÌΪÔØÌ壬¿¼²éѧÉúÔĶÁÌâÄ¿»ñÈ¡ÐÅÏ¢µÄÄÜÁ¦¡¢ÎïÖʵÄÁ¿Å¨¶È¼ÆËã¡¢Ñõ»¯»¹Ô­·´Ó¦Ïà¹Ø֪ʶºÍ»¯Ñ§»ù±¾ÊµÑé²Ù×÷¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉϳ£Óú¬ÉÙÁ¿SiO2¡¢Al2O3µÄ¸õÌú¿ó£¨FeOCr2O3£©Ò±Á¶¸õ£¬¼òÒªÁ÷³ÌÈçÏ£º

£¨1£©Íê³ÉÏÂÁл¯Ñ§·½³Ìʽ£¨ÔÚºáÏßÌîдÎïÖʵĻ¯Ñ§Ê½¼°ÏµÊý£©£º
2FeO¡¤Cr2O3£«4Na2CO3£«7NaNO34Na2CrO4£«Fe2O3£«4CO2£«______________¡£
£¨2£©²Ù×÷¢Ù°üÀ¨¹ýÂËÓëÏ´µÓ£¬ÔÚʵÑéÊÒÖнøÐÐÏ´µÓ³ÁµíµÄ²Ù×÷__________________________________¡£²Ù×÷¢Ú¿ÉÑ¡ÓõÄ×°Ö㨲¿·Ö¼Ð³Ö×°ÖÃÂÔÈ¥£©ÊÇ________£¨ÌîÐòºÅ£©

£¨3£©Ð´³öÄܹ»Íê³É²Ù×÷¢ÛµÄÏà¹Ø·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____________________¡£
£¨4£©»¯Ñ§ÐèÑõÁ¿£¨COD£©¿É¶ÈÁ¿Ë®ÔâÊÜÓлúÎïÎÛȾµÄ³Ì¶È¡£ÔÚÇ¿Ëá²¢¼ÓÈȵÄÌõ¼þÏ£¬ÓÃK2Cr2O7×öÇ¿Ñõ»¯¼Á´¦ÀíË®Ñù£¬²¢²â¶¨ÏûºÄµÄK2Cr2O7µÄÁ¿£¬È»ºó»»Ëã³ÉÏ൱ÓÚO2µÄº¬Á¿³ÆΪ»¯Ñ§ÐèÑõÁ¿£¨ÒÔmg/L¼Æ£©¡£»¯Ñ§ÐËȤС×é²â¶¨Ä³Ë®ÑùµÄ»¯Ñ§ÐèÑõÁ¿£¨COD£©¹ý³ÌÈçÏ£º
I.È¡amLË®ÑùÖÃÓÚ׶ÐÎÆ¿£¬¼ÓÈë10£®00mL 0.2500 mol/LµÄK2Cr2O7ÈÜÒº£»
II£®¡­¡­¡£
III£®¼Óָʾ¼Á£¬ÓÃc molµÄÁòËáÑÇÌúï§[ £¨NH4£©2Fe£¨SO4£©]µÎ¶¨£¬ÖÕµãʱÏûºÄb mL£¨´Ë²½ÖèµÄÄ¿µÄÊÇÓÃFe2£«°Ñ¶àÓàµÄCr2O72£­×ª»¯Îª£©Cr3£«¡£
¢ÙIÖÐÁ¿È¡K2Cr2O7ÈÜÒºµÄÒÇÆ÷ÊÇ_____________£»
¢Ú¼ÆËã¸ÃË®ÑùµÄ»¯Ñ§ÐèÑõÁ¿Ê±ÐèÓõ½ÏÂÁйØϵ£ºÒª³ýÈ¥1molCr2O72£­ ÐèÏûºÄ___molFe2£«£¬1molCr2O72£­Ï൱ÓÚ_______ molO2¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

º£Ë®×ÛºÏÀûÓõŤÒÕÁ÷³ÌͼÈçÏ£º

£¨l£©µç½âNaClÈÜÒº£¬ÔÚµç½â²ÛÖпÉÖ±½ÓµÃµ½µÄ²úÆ·ÓÐH2            ¡¢          »òH2¡¢               ¡£
£¨2£©²½ÖèIÖÐÒÑ»ñµÃBr2£¬²½ÖèIIÖÐÓÖ½«Br2»¹Ô­ÎªBr-£¬ÆäÄ¿µÄÊÇ             ¡£
£¨3£©²½ÖèIIÓÃSO2Ë®ÈÜÒºÎüÊÕBr2£¬ÎüÊÕÂÊ¿É´ï95%£¬Óйط´Ó¦µÄÀë×Ó·½³ÌʽΪ            £¬·´Ó¦¿ÉÖª£¬³ý»·¾³±£»¤Í⣬ÔÚ¹¤ÒµÉú²úÖÐÓ¦½â¾öµÄÖ÷ÒªÎÊÌâÊÇ                         ¡£
£¨4£©·ÖÀë³ö´ÖÑκóµÄº£Ë®ÖÐÔ̺¬×ŷḻµÄþ×ÊÔ´£¬¾­×ª»¯ºó¿É»ñµÃMgCl2´Ö²úÆ·¡£´Óº£Ë®ÖÐÌáȡþµÄ²½ÖèΪ£º
a£®½«º£±ß´óÁ¿´æÔڵı´¿Ç¶ÍÉÕ³Éʯ»Ò£¬²¢½«Ê¯»ÒÖƳÉʯ»ÒÈ飻
b£®½«Ê¯»ÒÈé¼ÓÈ˺£Ë®³Áµí³ØÖо­¹ýÂ˵õ½Mg(OH)2³Áµí£»
c£®ÔÚMg(OH)2³ÁµíÖмÓÈËÑÎËáµÃµ½MgCl2ÈÜÒº£¬ÔÙ¾­Õô·¢½áÆ·µÃµ½MgCl2¡¤ 6H2O;
d£®½«MgCl2¡¤6H2OÔÚÒ»¶¨Ìõ¼þϼÓÈȵõ½ÎÞË®MgCl2£®µç½âÈÛÈÚµÄÂÈ»¯Ã¾¿ÉµÃµ½Mg¡£
¢Ù²½ÖèdÖеġ°Ò»¶¨Ìõ¼þ¡±Ö¸µÄÊÇ                      £¬Ä¿µÄÊÇ               ;
¢ÚÓÐͬѧÈÏΪ£º²½Öèbºó¿É¼ÓÈÈMg(OH)2µÃµ½MgO£¬ÔÙµç½âÈÛÈÚµÄMgOÖƽðÊôþ£¬ÕâÑù¿É¼ò»¯Éú²ú²½Ö裬ÌåÏÖ¼òÔ¼ÐÔÔ­Ôò£ºÄãÊÇ·ñͬÒâ¸ÃͬѧµÄÏë·¨²¢ËµÃ÷ÀíÓÉ£º                    ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂÁи÷×éÎïÖʵķÖÀë»òÌá´¿£¬Ó¦Ñ¡ÓÃÏÂÊö·½·¨µÄÄÄÒ»ÖÖ£¿£¨ÌîÑ¡Ïî×Öĸ£©

A£®·ÖÒº B£®¹ýÂË C£®ÝÍÈ¡ D£®ÕôÁó¡¡¡¡E£®Õô·¢½á¾§¡¡¡¡F£®¸ßηֽâ
¢Ù·ÖÀëCCl4ºÍH2O£º¡¡¡¡   ¡¡    ¡¡¡¡£»
¢Ú³ýÈ¥³ÎÇåʯ»ÒË®ÖÐÐü¸¡µÄCaCO3£º¡¡¡¡   ¡¡    ¡¡¡¡£»
¢Û³ýÈ¥CaO¹ÌÌåÖÐÉÙÁ¿µÄCaCO3¹ÌÌ壺¡¡¡¡   ¡¡    ¡¡¡¡£»
¢Ü´ÓµâË®ÖÐÌáÈ¡µâ£º¡¡¡¡   ¡¡    ¡¡¡¡£» 
¢Ý·ÖÀëCCl4£¨·ÐµãΪ76.75¡æ£©ºÍ¼×±½£¨·ÐµãΪ110.6¡æ£©µÄÒºÌå»ìºÏÎ¡¡¡¡   ¡¡    ¡¡¡¡¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂͼËùʾÊÇ·ÖÀë»ìºÏÎïʱ³£ÓõÄÒÇÆ÷£¬»Ø´ðÏÂÁÐÎÊÌâ:

£¨1£©Ð´³öÒÇÆ÷C¡¢EµÄÃû³Æ                 ¡¢                
£¨2£©·ÖÀëÒÔÏ»ìºÏÎïÓ¦¸ÃÖ÷ҪѡÓÃÉÏÊöʲôÒÇÆ÷£¿£¨Ìî×Öĸ·ûºÅ£©
¢ÙNaCl¹ÌÌåºÍÄàɳ£º                  ¢Ú»¨ÉúÓͺÍË®£º             
£¨3£©ÏÂÁйØÓÚÒÇÆ÷µÄʹÓÃ˵·¨ÕýÈ·µÄÊÇ                              
A£®AÒÇÆ÷¿ÉÒÔÓþƾ«µÆÖ±½Ó¼ÓÈÈ
B£®BÒÇÆ÷¿ÉÒÔÓÃÓÚÏò¾Æ¾«µÆÖÐÌí¼Ó¾Æ¾«ks5u
C£®CÒÇÆ÷ÔڷųöÒºÌåʱӦ´ò¿ªÉϱߵÄÆ¿Èû
D£®ÔÚʵÑéÊÒÓ¦ÓÃDÒÇÆ÷½øÐÐʵÑéʱҪ²»¶ÏÓò£Á§°ô½Á°è
E£®ÖÆÕôÁóˮʱEÒÇÆ÷ÖÐË®µÄÁ÷ÏòÊÇÉϽøϳö
£¨4£©ÈôÏòC×°ÖÃÖмÓÈëµâË®ºÍ×ãÁ¿CCl4£¬³ä·ÖÕñµ´ºó¾²Ö㬻á¹Û²ìµ½Ê²Ã´ÏÖÏó£¿
                                                                               
£¨5£©µâµ¥ÖʺÍäåµ¥ÖÊÓÐÏàÀàËƵÄÐÔÖÊ£¬¶¼¿ÉÒÔÓÃÓлúÈܼÁÝÍÈ¡Ë®ÈÜÒºÖеĵ¥ÖÊ£¬ÈôÀûÓÃCÒÇÆ÷ÌáÈ¡äåË®ÖеÄäåµ¥ÖÊ£¬ÏÂÁÐÓлúÈܼÁÖв»ÄÜÑ¡ÓõÄÊÇ£º           
A£®ÆûÓÍ          B£®CCl4           C£®¾Æ¾«          D£®´×Ëá

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂͼÊÇÖÐѧ»¯Ñ§Öг£ÓÃÓÚ»ìºÏÎïµÄ·ÖÀëºÍÌá´¿µÄ×°Öã¬Çë¸ù¾Ý×°ÖûشðÎÊÌ⣺

£¨1£©´ÓÂÈ»¯¼ØÈÜÒºÖеõ½ÂÈ»¯¼Ø¹ÌÌ壬ѡÔñ×°Öà      (Ìî´ú±í×°ÖÃͼµÄ×Öĸ£¬ÏÂͬ)£»³ýÈ¥×ÔÀ´Ë®ÖеÄCl£­µÈÔÓÖÊ£¬Ñ¡Ôñ×°Öà       ¡£
£¨2£©´ÓµâË®ÖзÖÀë³öI2£¬Ñ¡Ôñ×°Öà       £¬¸Ã·ÖÀë·½·¨µÄÃû³ÆΪ             ¡£
£¨3£©×°ÖÃAÖТٵÄÃû³ÆÊÇ       £¬½øË®µÄ·½ÏòÊÇ´Ó     ¿Ú½øË®¡£×°ÖÃBÔÚ·ÖҺʱΪʹҺÌå˳ÀûÏµΣ¬Ó¦½øÐеľßÌå²Ù×÷ÊÇ                                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÌúËáп£¨ZnFe2O4£©ÊǶԿɼû¹âÃô¸ÐµÄ°ëµ¼Ìå´ß»¯¼Á£¬ÆäʵÑéÊÒÖƱ¸Ô­ÀíΪ£º
Zn2++2Fe2++3C2O42£­+6H2O ZnFe2(C2O4)3¡¤6H2O¡ý¡­¡­¡­¡­¡­(a)
ZnFe2(C2O4)3¡¤6H2O  ZnFe2O4+2CO2¡ü+4CO¡ü+6H2O ¡­¡­¡­¡­¡­(b)
£¨1£©ÉÏÊöÖƱ¸Ô­ÀíÖÐÊôÓÚÑõ»¯»¹Ô­·´Ó¦µÄÊÇ         £¨Ñ¡Ì¡°a¡±»ò¡°b¡±£©¡£
£¨2£©ÖƱ¸ZnFe2(C2O4)3¡¤6H2Oʱ£¬¿ÉÑ¡ÓõÄÒ©Æ·ÓУº
¢ñ¡¢(NH4)2Fe(SO4)2¡¤6H2O£¬¢ò¡¢ZnSO4¡¤7H2O¼°¢ó¡¢(NH4)2C2O4¡¤7H2O¡£
¢Ù³ÆÁ¿Ò©Æ·Ê±£¬±ØÐëÑϸñ¿ØÖÆn(Fe2+)/n(Zn2+)=   ¡£
¢ÚÑ¡ÓõļÓÁÏ·½Ê½ÊÇ           (Ìî×Öĸ)¡£
a£®°´Ò»¶¨¼ÆÁ¿±È£¬¢ñ¡¢¢ò¡¢¢óͬʱ¼ÓÈë·´Ó¦Æ÷²¢¼ÓË®½Á°è£¬È»ºóÉýÎÂÖÁ75¡æ¡£
b£®°´Ò»¶¨¼ÆÁ¿±È£¬¢ñ¡¢¢óͬʱ¼ÓÈë·´Ó¦Æ÷¼ÓË®Åä³ÉÈÜÒº£¬È»ºó¼ÓÈë¢ò£¬È»ºóÉýε½75¡æ¡£
c£®°´Ò»¶¨¼ÆÁ¿±È£¬½«¢ñ¡¢¢ò»ìºÏ²¢Åä³ÉÈÜÒº¼×£¬¢óÁíÅä³ÉÅäÖÆÈÜÒºÒÒ£¬¼×¡¢ÒÒͬʱ¼ÓÈȵ½75¡æ£¬È»ºó½«ÒÒÈÜÒº»ºÂý¼ÓÈë¼×ÈÜÒºÖУ¬²¢³ÖÐø½Á°è¡£
£¨3£©´ÓÈÜÒºÖзÖÀë³öZnFe2(C2O4)3¡¤6H2OÐè¹ýÂË¡¢Ï´µÓ¡£ÒÑÏ´µÓÍêÈ«µÄÒÀ¾ÝÊÇ         ¡£
£¨4£©ZnFe2(C2O4)3¡¤6H2OÈÈ·Ö½âÐèÓþƾ«ÅçµÆ£¬»¹Óõ½µÄ¹èËáÑÎÖÊÒÇÆ÷ÓР         ºÍ        ¡£
£¨5£©Ä³»¯Ñ§¿ÎÍâС×éÄâÓ÷Ͼɸɵç³ØпƤ£¨º¬ÔÓÖÊÌú£©£¬½áºÏÏÂͼÐÅÏ¢ÀûÓÃʵÑé¿ÉÌṩµÄÊÔ¼ÁÖÆÈ¡´¿¾»µÄZnSO4ÈÜÒº¡£

ʵÑéÖпÉÑ¡ÓõÄÊÔ¼ÁÈçÏ£º
£¨a£©30%H2O2£»£¨b£©ÐÂÖÆÂÈË®£»
£¨c£©1.0 mol¡¤L£­1NaOHÈÜÒº£»
£¨d£©3mol¡¤L-1Ï¡ÁòË᣻
£¨e£©´¿ZnO·ÛÄ©£»
£¨f£©´¿Ð¿·Û¡£
ʵÑé²½ÖèÒÀ´ÎΪ£º¢Ù½«Ð¿Æ¬ÍêÈ«ÈÜÓÚÉÔ¹ýÁ¿µÄ3mol¡¤L£­1Ï¡ÁòËᣬ¼ÓÈë     £¨Ñ¡Ìî×Öĸ£¬ÏÂͬ£©£»¢Ú¼ÓÈë    £»¢Û¼ÓÈȵ½60¡æ×óÓÒ²¢²»¶Ï½Á°è£»
¢Ü¹ýÂ˵ÃZnSO4ÈÜÒº¡£ÆäÖв½Öè¢Û¼ÓÈȵÄÖ÷ҪĿµÄÓР      ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

³ýÈ¥ÏÂÁÐÎïÖÊÖÐËùº¬ÉÙÁ¿ÔÓÖÊ£¬ÌîдËùÑ¡ÓõÄÊÔ¼ÁºÍ·ÖÀë·½·¨

 
»ìºÏÎï
(À¨ºÅÄÚΪÉÙÁ¿ÔÓÖÊ)
ÊÔ¼Á
£¨×ãÁ¿£©
·ÖÀë·½·¨
A
±½£¨±½·Ó£©
 
 
B
ÒÒÏ©£¨SO2£©
 
 
C
ÒÒËáÒÒõ¥£¨ÒÒËᣩ
 
 
D
ÒÒ´¼£¨Ë®£©
 
 
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

Îíö²ÑÏÖØÓ°ÏìÈËÃǵÄÉú»îÓ뽡¿µ¡£Ä³µØÇøµÄÎíö²ÖпÉÄܺ¬ÓÐÈçÏ¿ÉÈÜÐÔÎÞ»úÀë×Ó£ºNa+¡¢NH4+¡¢Mg2+¡¢Al3+¡¢SO42£­¡¢NO3£­¡¢Cl£­ ¡£Ä³Í¬Ñ§ÊÕ¼¯Á˸õØÇøµÄÎíö²£¬¾­±ØÒªµÄÔ¤´¦ÀíºóÊÔÑùÈÜÒº£¬Éè¼Æ²¢Íê³ÉÁËÈçϵÄʵÑ飺

ÒÑÖª£º3NO3£­+ 8Al + 5OH£­ + 2H2O3NH3 + 8AlO2£­
¸ù¾ÝÒÔÉϵÄʵÑé²Ù×÷ÓëÏÖÏ󣬸ÃͬѧµÃ³öµÄ½áÂÛ²»ÕýÈ·µÄÊÇ
ÊÔÑùÖп϶¨´æÔÚNH4+¡¢Mg2+¡¢SO42£­ºÍNO3£­
ÊÔÑùÖÐÒ»¶¨²»º¬Al3+
ÊÔÑùÖпÉÄÜ´æÔÚNa+¡¢Cl£­
¸ÃÎíö²ÖпÉÄÜ´æÔÚNaNO3¡¢NH4ClºÍMgSO4

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸