¾üÓùâÃô²ÄÁÏKxFe£¨C2O4£©y¡¤3H2O£¨FeΪ£«3¼Û£©µÄʵÑéÊÒÖƱ¸ºÍ²â¶¨Æä×é³ÉµÄ·½·¨ÈçÏÂËùʾ£º
¢ñ.ÖƱ¸£º

£¨1£©ÓÃK2C2O4ºÍFeCl3ÖƱ¸¹âÃô²ÄÁϵķ´Ó¦ÊôÓÚ________£¨ÌîÐòºÅ£©¡£
¢ÙÀë×Ó·´Ó¦¡¡¢Ú·ÇÑõ»¯»¹Ô­·´Ó¦¡¡¢ÛÑõ»¯»¹Ô­·´Ó¦¡¡¢Ü»¯ºÏ·´Ó¦
£¨2£©½á¾§Ê±Ó¦½«±¥ºÍÈÜÒº·ÅÔÚºÚ°µ´¦µÈ´ý¾§ÌåµÄÎö³ö£¬ÕâÑù²Ù×÷µÄÔ­ÒòÊÇ_________¡£
£¨3£©²Ù×÷4µÄʵÑé²Ù×÷ÓÐ____________¡£
¢ò.×é³É²â¶¨£º
³ÆÈ¡Ò»¶¨ÖÊÁ¿ÊµÑéËùµÃµÄ¾§ÌåÖÃÓÚ׶ÐÎÆ¿ÖУ¬¼Ó×ãÁ¿ÕôÁóË®ºÍÏ¡H2SO4£¬½«C2O42-Íêȫת»¯ÎªH2C2O4£¬ÓÃ0.10 mol¡¤L£­1 KMnO4ÈÜÒº½øÐе樣¬ÏûºÄKMnO4ÈÜÒº24.00 mLʱǡºÃÍêÈ«·´Ó¦£¨ËáÐÔÌõ¼þÏÂMnO4-µÄ»¹Ô­²úÎïÊÇMn2£«£©£»ÔÙ¼ÓÈëÊÊÁ¿µÄ»¹Ô­¼Á£¬½«Fe3£«Íêȫת»¯ÎªFe2£«£¬ÓÃKMnO4ÈÜÒº¼ÌÐøµÎ¶¨£¬µ±Fe2£«ÍêÈ«Ñõ»¯Ê±£¬ÓÃÈ¥KMnO4ÈÜÒº4.00 mL¡£
£¨4£©¸ßÃÌËá¼ØÑõ»¯H2C2O4ºÍFe2£«µÄÀë×Ó·½³Ìʽ·Ö±ðÊÇ___________£» ________¡£
£¨5£©ÅäÖÆ100 mL 0.10 mol¡¤L£­1 KMnO4ÈÜÒº¼°ÔÚÉÏÊöµÎ¶¨ÊµÑéÖÐËùÐèµÄ²£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü¡¢Á¿Í²¡¢×¶ÐÎÆ¿Í⣬»¹ÓÐ________ºÍ________£¨Ð´Ãû³Æ£©¡£
£¨6£©Í¨¹ý¼ÆË㣬¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½ÊÇ____________¡£

¢ñ.£¨1£©¢Ù¢Ú
£¨2£©·ÀÖ¹¾§Ìå¼û¹â·Ö½â
£¨3£©¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï
¢ò.£¨4£©2MnO42-£«5H2C2O4£«6H£«=2Mn2£«£«10CO2¡ü£«8H2O¡¡MnO4-£«5Fe2£«£«8H£«=Mn2£«£«5Fe3£«£«4H2O
£¨5£©100 mLÈÝÁ¿Æ¿¡¡ËáʽµÎ¶¨¹Ü
£¨6£©K3Fe£¨C2O4£©3¡¤3H2O

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓÐÒ»ÎÞɫ͸Ã÷ÈÜÒº£¬ÓûÈ·¶¨ÊÇ·ñº¬ÓÐÏÂÁÐÀë×Ó£º Fe2£«¡¢Mg2£«¡¢Al3£«¡¢Ba2£«¡¢¡¢¡¢Cl£­¡¢I£­¡¢£¬È¡¸ÃÈÜÒº½øÐÐʵÑ飺

ʵÑé²½Öè
ʵÑéÏÖÏó
(1)È¡ÉÙÁ¿¸ÃÈÜÒº£¬¼Ó¼¸µÎ×ÏɫʯÈïÊÔÒº
ÈÜÒº±äºì
(2)È¡ÉÙÁ¿¸ÃÈÜÒº¼ÓÈÈ£¬¼ÓCuƬºÍŨH2SO4£¬¼ÓÈÈ
ÓÐÎÞÉ«ÆøÌå²úÉú£¬ÆøÌåÓö¿ÕÆø±ä³Éºì×ØÉ«
(3)È¡ÉÙÁ¿¸ÃÈÜÒº£¬¼ÓBaCl2ÈÜÒº
Óа×É«³Áµí
(4)È¡(3)ÖÐÉϲãÇåÒº£¬¼ÓAgNO3ÈÜÒº
Óа×É«³Áµí£¬ÇÒ²»ÈÜÓÚÏ¡HNO3
(5)È¡ÉÙÁ¿¸ÃÈÜÒº£¬¼ÓNaOHÈÜÒº
Óа×É«³Áµí£¬NaOH¹ýÁ¿Ê±³Áµí²¿·ÖÈܽâ
 
ÓÉ´ËÅжϣº
(1)ÈÜÒºÖп϶¨²»´æÔÚµÄÀë×ÓÓР      £¬ÈÜÒºÖп϶¨´æÔÚµÄÀë×ÓÊÇ   ¡£
(2)ÇëÉè¼ÆʵÑéÑéÖ¤ÆäÖпÉÄÜ´æÔÚµÄÒõÀë×ӵķ½·¨(дÃ÷²Ù×÷¡¢ÏÖÏó¡¢½áÂÛ)    ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓÃÁâÃÌ¿ó£¨MnCO3£©³£º¬ÓÐFe2O3¡¢FeO¡¢HgCO3¡¤2HgOµÈÔÓÖÊ£¬¹¤Òµ³£ÓÃÁâÃÌ¿óÖÆÈ¡ÃÌ£¬¹¤ÒÕÁ÷³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ïò´ÖÒº1ÖмÓÈëµÄË®×îºóÐèÒª          ·½·¨²ÅÄÜ´ïµ½¼¼ÊõÒªÇó¡£
£¨2£©Á÷³ÌÖÐÓõĿÕÆøÊÇÓÃĤ·ÖÀë·¨ÖƱ¸µÄ¸»Ñõ¿ÕÆø£¬¸Ã·½·¨µÄÔ­ÀíÊÇ           ¡£
£¨3£©¾»»¯¼ÁÖ÷Òª³É·ÖΪ(NH4)2S£¬´ÖÒº2Öз¢ÉúÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ        ¡£
£¨4£©Ð´³öÑô¼«µÄµç¼«·´Ó¦Ê½          ¡£ËµÃ÷µç½âҺѭ»·µÄÔ­Òò              ¡£
£¨5£©Ð´³öÂÁÈÈ·¨Á¶Ã̵Ļ¯Ñ§·½³Ìʽ                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ijÖÖ´ÖÑÎÖк¬ÓÐÄàɳ¡¢Ca2£«¡¢Mg2£«¡¢Fe3£«¡¢SOµÈÔÓÖÊ¡£Ä³Í¬Ñ§ÔÚʵÑéÊÒÖÐÉè¼ÆÁËÓÃÕâÖÖ´ÖÑÎÖƱ¸¾«Ñεķ½°¸ÈçÏÂ(ÓÃÓÚ³ÁµíµÄÊÔ¼ÁÉÔ¹ýÁ¿)£º

Çë»Ø´ðÒÔÏÂÎÊÌ⣺
(1)Ϊ²Ù×÷¢ôÑ¡ÔñËùÐèÒÇÆ÷(ÓñêºÅ×ÖĸÌîд)£º________¡£
A£®ÉÕ±­¡¡B£®ÊԹܡ¡C£®²£Á§°ô¡¡D£®·ÖҺ©¶·¡¡E£®Â©¶·¡¡F£®¾Æ¾«µÆ¡¡
G£®Õô·¢Ãó
(2)²Ù×÷¢óÖг£ÓÃNa2CO3ÈÜÒº¡¢NaOHÈÜÒº¡¢BaCl2ÈÜÒº×÷Ϊ³ýÔÓÊÔ¼Á£¬Ôò¼ÓÈë³ýÔÓÊÔ¼ÁµÄ˳ÐòΪ£ºNaOHÈÜÒº¡ú________¡ú________¡£
(3)²Ù×÷¢óÖУ¬ÅжϼÓÈëBaCl2ÒѹýÁ¿µÄ·½·¨ÊÇ___________________________________
(4)²Ù×÷¢õӦѡÔñµÄËáÊÇ________£¬Èô½«²Ù×÷¢õÓë²Ù×÷¢ôµÄÏȺó˳Ðò¶Ôµ÷£¬½«»á¶ÔʵÑé½á¹û²úÉúµÄÓ°ÏìÊÇ___________________________________
(5)²Ù×÷¢öÊÇ________(Ñ¡ÔñºÏÀí²Ù×÷µÄÃû³Æ£¬ÓñêºÅ×Öĸ°´²Ù×÷ÏȺó˳ÐòÌîд)¡£
a£®¹ýÂË¡¢Ï´µÓ  B£®Õô·¢¡¢Å¨Ëõ      c£®ÝÍÈ¡¡¢·ÖÒº     D£®ÀäÈ´¡¢½á¾§

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨15·Ö£©Ð¿Ã̸ɵç³ØËùº¬µÄ¹¯¡¢Ëá»ò¼îµÈÔÚ·ÏÆúºó½øÈë»·¾³Öн«Ôì³ÉÑÏÖØΣº¦¡£¶Ô·Ï¾Éµç³Ø½øÐÐ×ÊÔ´»¯´¦ÀíÏԵ÷dz£ÖØÒª¡£Ä³»¯Ñ§ÐËȤС×éÄâ²ÉÓÃÈçÏ´¦Àí·½·¨»ØÊշϵç³ØÖеĸ÷ÖÖ×ÊÔ´¡£

£¨1£©¼îÐÔпÃ̸ɵç³ØµÄµç½âÖÊΪKOH£¬×Ü·´Ó¦ÎªZn+2MnO2+2H2O=2MnOOH+Zn(OH)2£¬Æ为¼«µÄµç¼«·´Ó¦Ê½Îª_________________________________________________________¡£
£¨2£©Ìî³äÎïÓÃ60¡æÎÂË®Èܽ⣬ĿµÄÊǼӿìÈܽâËÙÂÊ£¬µ«±ØÐë¿ØÖÆζȲ»ÄÜÌ«¸ß£¬ÆäÔ­ÒòÊÇ___________¡£
£¨3£©²Ù×÷AµÄÃû³ÆΪ_____________ ¡£
£¨4£©ÂËÔüµÄÖ÷Òª³É·ÖΪº¬ÃÌ»ìºÏÎÏòº¬ÃÌ»ìºÏÎïÖмÓÈëÒ»¶¨Á¿µÄÏ¡ÁòËᡢϡ²ÝËᣬ²¢²»¶Ï½Á°èÖÁÎÞÆøÅÝΪֹ¡£ÆäÖ÷Òª·´Ó¦Îª2MnO(OH)+MnO2+2H2C2O4+3H2SO4=2MnSO4+4CO2¡ü+6H2O¡£
¢Ùµ±1 mol MnO2²Î¼Ó·´Ó¦Ê±£¬¹²ÓÐ___________molµç×Ó·¢ÉúתÒÆ¡£
¢ÚMnO(OH)ÓëŨÑÎËáÔÚ¼ÓÈÈÌõ¼þÏÂÒ²¿É·¢Éú·´Ó¦£¬ÊÔд³öÆä·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________¡£
£¨5£©Í­Ã±Èܽâʱ¼ÓÈëH2O2µÄÄ¿µÄÊÇ_______________________________£¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£Í­Ã±ÈܽâÍêÈ«ºó£¬¿É²ÉÓÃ_____________·½·¨³ýÈ¥ÈÜÒºÖйýÁ¿µÄH2O2¡£
£¨6£©Ð¿Ã̸ɵç³ØËùº¬µÄ¹¯¿ÉÓÃKMnO4ÈÜÒºÎüÊÕ¡£ÔÚ²»Í¬pHÏ£¬KMnO4ÈÜÒº¶ÔHgµÄÎüÊÕÂʼ°Ö÷Òª²úÎïÈçÏÂͼËùʾ£º

¸ù¾ÝÉÏͼ¿ÉÖª£º
¢ÙpH¶ÔHgÎüÊÕÂʵÄÓ°Ïì¹æÂÉÊÇ__________________________________________________.
¢ÚÔÚÇ¿ËáÐÔ»·¾³ÏÂHgµÄÎüÊÕÂʸߵÄÔ­Òò¿ÉÄÜÊÇ_____________________________________.

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ëæ×Ų»¶ÏÏò»¯¹¤¡¢Ê¯ÓÍ¡¢µçÁ¦¡¢º£Ë®µ­»¯¡¢½¨Öþ¡¢ÈÕ³£Éú»îÓÃÆ·µÈÐÐÒµÍƹ㣬îѽðÊôÈÕÒæ±»ÈËÃÇÖØÊÓ£¬±»ÓþΪ¡°ÏÖ´ú½ðÊô¡±ºÍ¡°Õ½ÂÔ½ðÊô¡±£¬ÊÇÌá¸ß¹ú·À×°±¸Ë®Æ½²»¿É»òȱµÄÖØÒªÕ½ÂÔÎï×Ê¡£¹¤ÒµÖ÷ÒªÒÔ¶þÑõ»¯îÑΪԭÁÏÒ±Á¶½ðÊôîÑ¡£
¢ñ.¶þÑõ»¯îÑ¿ÉÓÉÒÔÏÂÁ½ÖÖ·½·¨ÖƱ¸£º
·½·¨1£º¿ÉÓú¬ÓÐFe2O3µÄîÑÌú¿ó£¨Ö÷Òª³É·ÖΪFeTiO3£¬ÆäÖÐTiÔªËØ»¯ºÏ¼ÛΪ+4¼Û£©ÖÆÈ¡£¬ÆäÖ÷ÒªÁ÷³ÌÈçÏ£º

£¨1£©ÓÉÂËÒº»ñµÃÂÌ·¯¾§ÌåµÄ²Ù×÷¹ý³ÌÊÇ                 ¡£
£¨2£©¼×ÈÜÒºÖгýº¬TiO2+Ö®Í⻹º¬ÓеĽðÊôÑôÀë×ÓÓР                ¡£
£¨3£©ÒÑÖª10kg¸ÃîÑÌú¿óÖÐÌúÔªËصÄÖÊÁ¿·ÖÊýΪ33.6%£¬Äܹ»µÃµ½ÂÌ·¯¾§Ìå22.24kg£¬ÊÔ¼ÆËã×îÉÙ¼ÓÈëÌú·ÛµÄÖÊÁ¿¡£
·½·¨2£ºTiCl4Ë®½âÉú³ÉTiO2¡¤XH2O£¬¹ýÂË¡¢Ë®Ï´³ýÈ¥ÆäÖеÄCl-£¬ÔÙºæ¸É¡¢±ºÉÕ³ýȥˮ·ÖµÃµ½·ÛÌåTiO2£¬´Ë·½·¨ÖƱ¸µÃµ½µÄÊÇÄÉÃ׶þÑõ»¯îÑ¡£
£¨4£©¢ÙTiCl4Ë®½âÉú³ÉTiO2¡¤XH2OµÄ»¯Ñ§·½³ÌʽΪ                 ¡£
¢Ú¼ìÑéTiO2¡¤XH2OÖÐCl-ÊÇ·ñ±»³ý¾»µÄ·½·¨ÊÇ                 ¡£
¢ò.¶þÑõ»¯îÑ¿ÉÓÃÓÚÖÆÈ¡îѵ¥ÖÊ
£¨5£©TiO2ÖÆÈ¡µ¥ÖÊTi£¬Éæ¼°µ½µÄ²½ÖèÈçÏ£º
 
·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ                 £¬¸Ã·´Ó¦³É¹¦ÐèÒªµÄÆäËûÌõ¼þ¼°Ô­ÒòÊÇ                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Á×ËáÌúï®(LiFePO4)±»ÈÏΪÊÇ×îÓÐǰ;µÄï®Àë×Óµç³ØÕý¼«²ÄÁÏ¡£Ä³ÆóÒµÀûÓø»Ìú½þ³öÒºÉú³ÉÁ×ËáÌúï®,¿ª±ÙÁË´¦ÀíÁòËáÑÇÌú·ÏÒºÒ»ÌõÐÂ;¾¶¡£ÆäÖ÷ÒªÁ÷³ÌÈçÏÂ:

ÒÑÖª:H2TiO3ÊÇÖÖÄÑÈÜÓÚË®µÄÎïÖÊ¡£
(1)îÑÌú¿óÓÃŨÁòËá´¦Àí֮ǰ,ÐèÒª·ÛËé,ÆäÄ¿µÄÊÇ                             ¡¡¡£
(2)TiO2+Ë®½âÉú³ÉH2TiO3µÄÀë×Ó·½³ÌʽΪ                                  ¡¡¡£
(3)¼ÓÈëNaClO·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                     ¡¡¡£
(4)ÔÚʵÑéÖÐ,´ÓÈÜÒºÖйýÂ˳öH2TiO3ºó,ËùµÃÂËÒº»ë×Ç,Ó¦ÈçºÎ²Ù×÷¡¡            ¡£
(5)Ϊ²â¶¨îÑÌú¿óÖÐÌúµÄº¬Á¿,ijͬѧȡ¾­Å¨ÁòËáµÈ´¦ÀíµÄÈÜÒº(´ËʱîÑÌú¿óÖеÄÌúÒÑÈ«²¿×ª»¯Îª¶þ¼ÛÌúÀë×Ó),²ÉÈ¡KMnO4±ê×¼ÒºµÎ¶¨Fe2+µÄ·½·¨:(²»¿¼ÂÇKMnO4ÓëÆäËûÎïÖÊ·´Ó¦)Ôڵζ¨¹ý³ÌÖÐ,ÈôδÓñê×¼ÒºÈóÏ´µÎ¶¨¹Ü,Ôòʹ²â¶¨½á¹û¡¡¡¡¡¡¡¡(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족),µÎ¶¨ÖÕµãµÄÏÖÏóÊÇ¡¡                                 ¡£µÎ¶¨·ÖÎöʱ,³ÆÈ¡a gîÑÌú¿ó,´¦Àíºó,ÓÃc mol/L KMnO4±ê×¼ÒºµÎ¶¨,ÏûºÄV mL,ÔòÌúÔªËصÄÖÊÁ¿·ÖÊýµÄ±í´ïʽΪ¡¡¡¡¡¡¡¡¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹ýÁòËá¼Ø£¨£©¾ßÓÐÇ¿Ñõ»¯ÐÔ£¨³£±»»¹Ô­ÎªÁòËá¼Ø£©£¬80 ¡æÒÔÉÏÒ×·¢Éú·Ö½â¡£ÊµÑéÊÒÄ£Ä⹤ҵºÏ³É¹ýÁòËá¼ØµÄÁ÷³ÌÈçÏ£º

£¨1£©ÁòËá狀ÍÁòËáÅäÖƳɵç½âÒº£¬ÒÔ²¬×÷µç¼«½øÐеç½â£¬Éú³É¹ýÁòËáï§ÈÜÒº¡£Ð´³öµç½âʱ·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ_____________________________________
___________________________________¡£
£¨2£©ÒÑÖªÏà¹ØÎïÖʵÄÈܽâ¶ÈÇúÏßÈçÓÒͼËùʾ¡£ÔÚʵÑéÊÒÖÐÌá´¿¹ýÁòËá¼Ø´Ö²úÆ·µÄʵÑé¾ßÌå²Ù×÷ÒÀ´ÎΪ£º½«¹ýÁòËá¼Ø´Ö²úÆ·ÈÜÓÚÊÊÁ¿Ë®ÖУ¬________________£¬¸ÉÔï¡£

£¨3£©ÑùÆ·ÖйýÁòËá¼ØµÄº¬Á¿¿ÉÓõâÁ¿·¨½øÐвⶨ¡£ÊµÑé²½ÖèÈçÏ£º
²½Öè1£º³ÆÈ¡¹ýÁòËá¼ØÑùÆ·0.300 0 gÓÚµâÁ¿Æ¿ÖУ¬¼ÓÈë30 mLË®Èܽ⡣
²½Öè2£ºÏòÈÜÒºÖмÓÈë4.00 0 g KI¹ÌÌ壨ÂÔ¹ýÁ¿£©£¬Ò¡ÔÈ£¬ÔÚ°µ´¦·ÅÖÃ30 min¡£
²½Öè3£ºÔÚµâÁ¿Æ¿ÖмÓÈëÊÊÁ¿´×ËáÈÜÒºËữ£¬ÒÔµí·ÛÈÜÒº×÷ָʾ¼Á£¬ÓÃ0.100 0 mol¡¤L£­1Na2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬¹²ÏûºÄNa2S2O3±ê×¼ÈÜÒº21.00 mL¡£
£¨ÒÑÖª·´Ó¦£ºI2£«2S2O32-=2I£­£«S4O62-£©
¢ÙÈô²½Öè2ÖÐ佫µâÁ¿Æ¿¡°ÔÚ°µ´¦·ÅÖÃ30 min¡±£¬Á¢¼´½øÐв½Öè3£¬Ôò²â¶¨µÄ½á¹û¿ÉÄÜ________£¨Ñ¡Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°ÎÞÓ°Ï족£©£»ÉÏÊö²½Öè3Öеζ¨ÖÕµãµÄÏÖÏóÊÇ____________________________________________¡£
¢Ú¸ù¾ÝÉÏÊö²½Öè¿É¼ÆËã³ö¸ÃÑùÆ·ÖйýÁòËá¼ØµÄÖÊÁ¿·ÖÊýΪ_______________¡£
¢ÛΪȷ±£ÊµÑé½á¹ûµÄ׼ȷÐÔ£¬ÄãÈÏΪ»¹ÐèÒª____________________________¡£
£¨4£©½«0.40 mol¹ýÁòËá¼ØÓë0.20 molÁòËáÅäÖƳÉ1 LÈÜÒº£¬ÔÚ80 ¡æÌõ¼þϼÓÈȲ¢ÔÚtʱ¿ÌÏòÈÜÒºÖеμÓÈëÉÙÁ¿FeCl3ÈÜÒº£¬²â¶¨ÈÜÒºÖи÷³É·ÖµÄŨ¶ÈÈçͼËùʾ£¨H£«Å¨¶Èδ»­³ö£©¡£Í¼ÖÐÎïÖÊXµÄ»¯Ñ§Ê½Îª________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓÃϼʯÑÒ£¨»¯Ñ§Ê½Îª KNa3[AlSiO4]4£¬Ö÷Òª³É·ÝNa2O¡¢K2O¡¢Al2O3¡¢SiO2£©ÖÆ̼ËáÄÆ¡¢Ì¼Ëá¼ØºÍÑõ»¯ÂÁµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£ºNaHCO3ÈÜÒºµÄpHԼΪ8¡«9£¬Na2CO3ÈÜÒºµÄpHԼΪ11¡«12¡£Èܽâ¹ýÂ˹¤Ðò²úÉúµÄÂËÒºÖк¬ÄÆ¡¢¼ØºÍÂÁµÄ¿ÉÈÜÐÔÑÎÀ࣬¸ÆºÍ¹èµÈÆäËûÔÓÖÊÔÚÂËÔüϼʯÄàÖС£²¿·ÖÎïÖʵÄÈܽâ¶È¼ûÓÒͼ¡£

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©×ÆÉյõ½¹ÌÌåMµÄ»¯Ñ§·½³ÌʽÊÇ________________________________¡£
£¨2£©XÎïÖÊÊÇ___________£¬ÂËÒºWÖÐÖ÷Òªº¬ÓеÄÀë×ÓÓÐ____________¡££¨Ð´ÈýÖÖ£©
£¨3£©²Ù×÷¢ñµÃµ½Ì¼ËáÄƾ§ÌåµÄ²Ù×÷Ϊ¡¡¡¡ ¡¢    ¡¢    ¡¢Ï´µÓ¡¢¸ÉÔï¡£
£¨4£©Ì¼Ëữ¢ñÖз¢ÉúÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ__________________________¡£
£¨5£©Ì¼Ëữ¢òµ÷ÕûpH£½8µÄÄ¿µÄÊÇ_______________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸