1£®Ä³»¯Ñ§Ð¡×éÊÔÀûÓ÷ÏÂÁм£¨º¬ÔÓÖÊÌú£©ÖƱ¸ÁòËáÂÁ¾§Ì壬²¢¶ÔÁòËáÂÁ¾§Ìå½øÐÐÈÈÖØ·ÖÎö£¬ÆäÖ÷ҪʵÑéÁ÷³ÌÈçͼ1£º

£¨1£©ÏòÈÜÒºAÖÐͨÈë¹ýÁ¿µÄCO2£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Al+2NaOH+2H2O¨T2NaAlO2+3H2¡ü£®
£¨2£©²Ù×÷IIËù°üº¬µÄʵÑé²Ù×÷µÄÃû³ÆÒÀ´ÎΪÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ
£¨3£©Èô¿ªÊ¼Ê±³ÆÈ¡µÄ·ÏÂÁмµÄÖÊõ¼Îª5.00g£¬µÃµ½¹ÌÌåAµÄÖÊÁ¿Îª0.95g£¬ÁòËáÂÁ¾§ÌåµÄÖÊÁ¿Îª49.95g£¨¼ÙÉèÿһ²½µÄת»¯ÂʾùΪ100%£©£¬ÔòËùµÃÁòËáÂÁ¾§ÌåµÄ»¯Ñ§Ê½ÎªAl2£¨SO4£©3.18H2O£®
£¨4£©È¡ÒÔÉÏÖƵõÄÁòËáÂÁ¾§Ìå½øÐÐÈÈÖØ·ÖÎö£¬ÆäÈÈ·Ö½âÖ÷Òª·ÖΪÈý¸ö½×¶Î£ºµÚÒ»½×¶ÎʧÖØ40.54%£¬µÚ¶þ½×¶ÎʧÖØ48.65%£¬µÚÈý½×¶ÎʧÖØ84.68%£¬ÒÔºó²»ÔÙʧÖØ£®ÆäÈÈ·Ö½âµÄÇúÏßÈçͼ2Ëùʾ£®
ÒÑÖª£ºÊ§ÖØ%=$\frac{¼ÓÈȼõÉÙµÄÖÊÁ¿}{Ô­¾§ÌåÑùÆ·µÄ×ÜÖÊÁ¿}$¡Á100%
¢ÙʧÖصÚÒ»½×¶Î·´Ó¦µÄ»¯Ñ§·½³ÌʽΪAl2£¨SO4£©3.18H2O$\frac{\underline{\;\;¡÷\;\;}}{\;}$Al2£¨SO4£©3.3H2O+15H2O£®
¢ÚʧÖصÚÈý½×¶Î²ÐÁô¹ÌÌåµÄ»¯Ñ§Ê½ÎªAl2£¨SO4£©3$\frac{\underline{\;\;¡÷\;\;}}{\;}$Al2O3+3SO3¡ü£®

·ÖÎö Ïò·ÏÂÁм£¨º¬ÔÓÖÊÌú£©ÖмÓÈë×ãÁ¿ÇâÑõ»¯ÄÆÈÜÒº£¬·¢Éú·´Ó¦£º2Al+2NaOH+2H2O¨T2NaAlO2+3H2¡ü£¬Ìú²»·´Ó¦£¬¹ÌÌåAΪFe£¬²ÉÓùýÂ˵ķ½·¨½øÐзÖÀ룬ÈÜÒºAΪNaAlO2ÈÜÒº£¬ÏòÆäÖÐͨÈë¶þÑõ»¯Ì¼£¬·¢Éú·´Ó¦£ºNaAlO2+CO2+2H2O¨TAl£¨OH£©3¡ý+NaHCO3£¬ÔÙͨ¹ý¹ýÂ˽øÐзÖÀ룬¹ÌÌåBΪÇâÑõ»¯ÂÁ£¬ÇâÑõ»¯ÂÁÓëÁòËá·´Ó¦µÃµ½ÁòËáÂÁÈÜÒº£¬ÔÙ¾­¹ýÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢Ï´µÓ¡¢¸ÉÔïµÃµ½ÁòËáÂÁ¾§Ì壻
£¨1£©AlºÍÇâÑõ»¯ÄÆÈÜÒºÉú³É¿ÉÈÜÐÔµÄÆ«ÂÁËáÄÆ£»
£¨2£©´ÓÈÜÒºÖлñµÃ¾§Ì壬ÐèÒª¾­¹ýÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïµÈ²Ù×÷£»
£¨3£©AlµÄÖÊÁ¿Îª5g-0.95g=4.05g£¬ÉèÁòËáÂÁ¾§Ì廯ѧʽΪ£ºAl2£¨SO4£©3£®nH2O£¬¸ù¾ÝAlÔªËØÊغã¼ÆËãÁòËáÂÁ¾§ÌåµÄÎïÖʵÄÁ¿£¬ÔÙ¼ÆËãÁòËáÂÁ¾§ÌåµÄÏà¶Ô·Ö×ÓÖÊÁ¿£¬½ø¶ø¼ÆËãnµÄÖµ£¬È·¶¨»¯Ñ§Ê½£»
£¨4£©¸ù¾Ý£¨3£©ÖмÆËã¿ÉÖª£¬¾§ÌåÖнᾧˮµÄÖÊÁ¿·ÖÊý£¬µÍμÓÈÈ£¬Ê×ÏÈʧȥ½á¾§Ë®£¬¸ßÎÂÏ£¬×îÖÕÁòËáÂÁ·Ö½â£¬¸ù¾ÝʧÖØ%¼ÆËãÅжϸ÷½×¶Î·Ö½â²úÎÔÙÊéд»¯Ñ§·½³Ìʽ£®

½â´ð ½â£º£¨1£©AlºÍÇâÑõ»¯ÄÆÈÜÒºÉú³É¿ÉÈÜÐÔµÄÆ«ÂÁËáÄÆÓëÇâÆø£¬Ã¾²»·´Ó¦£¬·´Ó¦·½³ÌʽΪ2Al+2NaOH+2H2O¨T2NaAlO2+3H2¡ü£»
¹Ê´ð°¸Îª£º2Al+2NaOH+2H2O¨T2NaAlO2+3H2¡ü£»
£¨2£©´ÓÈÜÒºÖлñµÃ¾§Ì壬ÐèÒª¾­¹ýÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïµÈ²Ù×÷£»
¹Ê´ð°¸Îª£ºÕô·¢Å¨Ëõ£»¹ýÂË£»
£¨3£©AlµÄÖÊÁ¿Îª5g-0.95g=4.05g£¬ÆäÎïÖʵÄÁ¿Îª4.05g¡Â27g/mol=0.15mol£¬ÉèÁòËáÂÁ¾§Ì廯ѧʽΪ£ºAl2£¨SO4£©3£®nH2O£¬¸ù¾ÝAlÔªËØÊغ㣬ÁòËáÂÁ¾§ÌåµÄÎïÖʵÄÁ¿Îª0.15mol¡Â2=0.075mol£¬¹ÊÁòËáÂÁ¾§ÌåµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª49.95¡Â0.075=666£¬Ôò54+96¡Á3+18n=666£¬½âµÃn=18£¬¹Ê¸ÃÁòËáÂÁ¾§ÌåµÄ»¯Ñ§Ê½Îª£ºAl2£¨SO4£©3.18H2O£»
¹Ê´ð°¸Îª£ºAl2£¨SO4£©3.18H2O£»
£¨4£©¾§ÌåÖнᾧˮµÄº¬Á¿Îª$\frac{18¡Á18}{666}$=48.65%£¬¹ÊµÚ¶þ½×¶ÎÍêȫʧȥ½á¾§Ë®£¬µÃµ½ÎïÖÊΪAl2£¨SO4£©3£¬µÚÒ»½×¶Îʧȥ²¿·Ö½á¾§Ë®£¬Ê§È¥½á¾§Ë®ÊýĿΪ$\frac{666¡Á40.54%}{18}$=15£¬¹ÊµÚÒ»½×¶ÎµÃµ½µÄÎïÖÊΪAl2£¨SO4£©3.3H2O£¬·´Ó¦·½³ÌʽΪ£ºAl2£¨SO4£©3.18H2O$\frac{\underline{\;\;¡÷\;\;}}{\;}$Al2£¨SO4£©3.3H2O+15H2O£»
µÚÈý½×¶ÎÊ£ÓàÎïÖʵÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª666¡Á£¨1-84.68%£©=102£¬Ó¦ÊÇAl2O3£¬¹ÊÁòËáÂÁ·´Ó¦Éú³ÉÑõ»¯ÂÁÓëÈýÑõ»¯Áò£¬·´Ó¦·½³ÌʽΪ£ºAl2£¨SO4£©3$\frac{\underline{\;\;¡÷\;\;}}{\;}$Al2O3+3SO3¡ü£»
¹Ê´ð°¸Îª£ºAl2£¨SO4£©3.18H2O$\frac{\underline{\;\;¡÷\;\;}}{\;}$Al2£¨SO4£©3.3H2O+15H2O£»Al2£¨SO4£©3$\frac{\underline{\;\;¡÷\;\;}}{\;}$Al2O3+3SO3¡ü£®

µãÆÀ ±¾Ì⿼²éÎïÖʵÄÖƱ¸ÊµÑ飬Ϊ¸ßƵ¿¼µã£¬°ÑÎÕÖƱ¸ÊµÑéÁ÷³Ì¼°ÎïÖÊ×é³É¡¢ÐÔÖÊ¡¢·¢ÉúµÄ·´Ó¦Îª½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓëʵÑé¡¢¼ÆËãÄÜÁ¦µÄ¿¼²é£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

11£®µç½âÔ­ÀíÔÚ»¯Ñ§¹¤ÒµÖÐÓй㷺ӦÓã®

£¨1£©µç½âʳÑÎË®ÊÇÂȼҵµÄ»ù´¡£®Ä¿Ç°±È½ÏÏȽøµÄ·½·¨ÊÇÑôÀë×Ó½»»»Ä¤·¨£¬µç½âʾÒâͼÈçͼ1Ëùʾ£¬Í¼ÖеÄÑôÀë×Ó½»»»Ä¤Ö»ÔÊÐíÑôÀë×Óͨ¹ý£¬Çë»Ø´ðÒÔÏÂÎÊÌ⣺
¢ÙͼÖÐA¼«ÒªÁ¬½ÓµçÔ´µÄÕý£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©¼«£®
¢Ú¾«ÖƱ¥ºÍʳÑÎË®´ÓͼÖÐaλÖò¹³ä£¬ÇâÑõ»¯ÄÆÈÜÒº´ÓͼÖÐdλÖÃÁ÷³ö£®£¨Ñ¡Ìî¡°a¡±¡¢¡°b¡±¡¢¡°c¡±¡¢¡°d¡±¡¢¡°e¡±»ò¡°f¡±£©
¢Ûµç½â×Ü·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ2Cl-+2H2O$\frac{\underline{\;ͨµç\;}}{\;}$Cl2¡ü+H2¡ü+2OH-£®
£¨2£©µç½â·¨´¦Àíº¬µªÑõ»¯Îï·ÏÆø£¬¿É»ØÊÕÏõËᣬ¾ßÓнϸߵĻ·¾³Ð§ÒæºÍ¾­¼ÃЧÒ森ʵÑéÊÒÄ£Äâµç½â·¨ÎüÊÕNOxµÄ×°ÖÃÈçͼ2Ëùʾ£¨Í¼Öе缫¾ùΪʯīµç¼«£©£®ÈôÓÃNO2ÆøÌå½øÐÐÄ£Äâµç½â·¨ÎüÊÕʵÑ飮
¢Ùд³öµç½âʱNO2·¢Éú·´Ó¦µÄµç¼«·´Ó¦NO2-e-+H2O=NO3-+2H+£®
¢ÚÈôÓбê×¼×´¿öÏÂ2.24L NO2±»ÎüÊÕ£¬Í¨¹ýÑôÀë×Ó½»»»Ä¤£¨Ö»ÔÊÐíÑôÀë×Óͨ¹ý£©µÄH+Ϊ0.1mol£®
£¨3£©ÎªÁ˼õ»º¸ÖÖÆÆ·µÄ¸¯Ê´£¬¿ÉÒÔÔÚ¸ÖÖÆÆ·µÄ±íÃæ¶ÆÂÁ£®µç½âÒº²ÉÓÃÒ»ÖÖ·ÇË®ÌåϵµÄÊÒÎÂÈÛÈÚÑΣ¬ÓÉÓлúÑôÀë×Ó¡¢A12C17-ºÍAlCl4-×é³É£®
¢Ù¸ÖÖÆÆ·Ó¦½ÓµçÔ´µÄ¸º¼«£®
¢Ú¼ºÖªµç¶Æ¹ý³ÌÖв»²úÉúÆäËûÀë×ÓÇÒÓлúÑôÀë×Ó²»²ÎÓëµç¼«·´Ó¦£¬Òõ¼«µç¼«·´Ó¦Ê½Îª4Al2Cl7-+3e-=Al+7AlCl4-£®
¢ÛÈô¸ÄÓÃAlCl3Ë®ÈÜÒº×÷µç½âÒº£¬ÔòÒõ¼«µç¼«·´Ó¦Ê½Îª2H++2e-=H2¡ü£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®ÏÂÁÐÎïÖʵĵç×ÓʽÊéдÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®´ÎÂÈËáµÄµç×ÓʽB£®·ú»¯ÇâµÄµç×Óʽ£ºC£®OH-D£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

9£®Ä³Î¶ÈÏÂÔڼס¢ÒÒ¡¢±ûÈý¸öºãÈÝÃܱÕÈÝÆ÷ÖУ¬Í¶ÈëH2£¨g£©ºÍCO2£¨g£©£¬·¢ÉúH2£¨g£©+CO2£¨g£©?CO£¨g£©+H2O£¨g£©£¬ÆäÆðʼŨ¶ÈÈç±íËùʾ£®ÒÑÖª£ºÆ½ºâʱ¼×ÖÐCOÆøÌåµÄŨ¶ÈΪ0.006mol/L£®ÏÂÁÐÅжϲ»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
ÆðʼŨ¶È£¨mol/L£©¼×ÒÒ±û
C£¨H2£©0.0100.0200.020
C£¨CO2£©0.0100.0100.020
A£®·´Ó¦¿ªÊ¼Ê±£¬±ûÖеķ´Ó¦ËÙÂÊ×î¿ì£¬¼×Öеķ´Ó¦ËÙÂÊ×îÂý
B£®Æ½ºâʱ£¬¼×¡¢ÒÒ¡¢±ûÖÐCO2µÄת»¯ÂÊÓÐÈçϹØϵ£ºÒÒ£¾¼×=±û=60%
C£®Æ½ºâʱ£¬±ûÖÐc£¨CO2£©ÊǼ×ÖеÄ2±¶£¬ÊÇ0.012mol/L
D£®¸Ä±äÌõ¼þʹÒÒÖÐζȽµµÍ£¬ÐÂƽºâÖÐH2µÄŨ¶ÈÔö´ó£¬ÔòÕý·´Ó¦µÄ¡÷H£¾0

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

16£®Á˽âһЩÓÃÒ©³£Ê¶£¬ÓÐÀûÓÚ×ÔÎÒ±£½¡£®ÏÖÓÐÏÂÁÐÒ©Î
a£®µâ¾Æ    b£®ÇàùËØ    c£®°¢Ë¾Æ¥ÁÖ    d£®ÇâÑõ»¯ÂÁ¸´·½ÖƼÁ
ÇëÑ¡ÔñÕýÈ·´ð°¸£¬½«×Öĸ±àºÅÌîÔÚÏàÓ¦¿Õ¸ñÀ
¢ÙÆäÖÐÊôÓÚ½âÈÈÕòÍ´Ò©µÄÊÇc£»
¢Ú¾ßÓÐÏûÑס¢ÒÖ¾ú×÷ÓõĿ¹ÉúËØÊÇb£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®ÏÂÁÐÊÔ¼ÁÖü´æ·½·¨´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®Å¨ÁòËáÃÜ·â±£´æ
B£®Çâ·úËá±£´æÔÚ²£Á§Æ¿ÖÐ
C£®ÐÂÖÆÂÈË®±£´æÔÚ×ØÉ«ÊÔ¼ÁÆ¿ÖÐ
D£®ÉÕ¼îÈÜÒº¿ÉÓôøÏðƤÈûµÄ²£Á§Æ¿ÔÝʱ±£´æ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

13£®µ¨·¯£¨CuSO4•5H2O£©Óй㷺µÄÓÃ;£®Ä³Ñо¿ÐÔѧϰС×éÀûÓÃij´ÎʵÑéºóµÄÏ¡ÁòËᡢϡÏõËá»ìºÏÒºÖƱ¸µ¨·¯£®ÊµÑéÁ÷³ÌÈçͼ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²Ù×÷XΪÕô·¢Å¨Ëõ¡¢ÀäÈ´½á¾§£®¹ýÂËÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷ÓЩ¶·¡¢ÉÕ±­¡¢²£Á§°ô£®
£¨2£©NOÐèÒª»ØÊÕÀûÓã¬NOÓë¿ÕÆø¡¢H2O·´Ó¦Éú³ÉÏõËáµÄ»¯Ñ§·½³Ìʽ4NO+3O2+2H2O=4HNO3£®
£¨3£©ÏÖÓÐ48gÍ­·Û£¨ÆäÖк¬CuOÖÊÁ¿·ÖÊýΪ20%£©£¬½«Æä¼ÓÈëÒ»¶¨Á¿Ï¡ÁòËᡢϡÏõËá»ìºÏҺǡºÃÍêÈ«·´Ó¦Éú³ÉCuSO4
¢ÙÀíÂÛÉϵõ½µ¨·¯µÄÖÊÁ¿Îª180g£®
¢ÚÔ­»ìºÏÒºÖÐÁòËáºÍÏõËáµÄÎïÖʵÄÁ¿Ö®±ÈΪ9£º5£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

10£®ÓÐЩ¹ÅÎÄ»òÑèÓï°üº¬Á˷ḻµÄ»¯Ñ§ÖªÊ¶£¬´Ó»¯Ñ§½Ç¶È½âÊÍÏÂÁйÅÎÄ»òÑèÓÆäÖв»ÕýÈ·µÄÊǵÄÊÇ£¨¡¡¡¡£©
 Ñ¡Ïî ¹ÅÎÄ»òÑèÓï »¯Ñ§½âÊÍ
 A À¯¾æ³É»ÒÀáʼ¸É¡°Àᡱ¾ÍÊÇÀ¯ÖòȼÉÕ²úÉúµÄË®
 B ÒÔÔøÇàÍ¿Ìú£¬Ìú³àÉ«ÈçÍ­ Öû»·´Ó¦
 C Õæ½ð²»Å»ðÁ¶ ½ðµÄ»î¶¯ÐÔ˳ÐòÅÅ×îºó£¬ÐÔÖÊÎȶ¨
 D Ò°»ðÉÕ²»¾¡£¬´º·ç´µÓÖÉú Éæ¼°Ñõ»¯»¹Ô­·´Ó¦
A£®AB£®BC£®CD£®D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

5£®10gijÆø̬ÍéÌþºÍÆø̬µ¥Ï©Ìþ×é³ÉµÄ»ìºÏÆøÌåÊÇÏàͬ״¿öÏÂH2ÃܶȵÄ12.5±¶£¬Í¨Èë×ãÁ¿äåË®ÖУ¬äåË®ÔöÖØ8.4g£¬´ËÁ½ÖÖÌþÊÇ£¨¡¡¡¡£©
A£®¼×ÍéºÍÒÒÏ©B£®¼×ÍéºÍ¶¡Ï©C£®ÒÒÍéºÍÒÒÏ©D£®ÒÒÍéºÍ¶¡Ï©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸