ÌâÄ¿ÄÚÈÝ

12£®ÒÑÖª³£ÎÂÏ£¬0.1mol•L?1CH3COONH4ÈÜÒº³ÊÖÐÐÔ£®ÏÂÁÐÓйØÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¸ÃÈÜÒº¿ÉÓɵÈÌå»ý¡¢Å¨¶È¾ùΪ0.1mol•L?1µÄ´×ËáºÍ°±Ë®·´Ó¦µÃµ½
B£®CH3COONH4¼ÓÈë´¿Ë®ÖУ¬Ë®µÄµçÀë³Ì¶È²»±ä
C£®³£ÎÂÏ£¬µçÀë³£ÊýK£¨CH3COOH£©=K£¨NH3•H2O£©
D£®¸ÃÈÜÒºÖÐc£¨CH3COO?£©´óÓÚͬŨ¶ÈCH3COONaÈÜÒºÖеÄc£¨CH3COO?£©

·ÖÎö 0.1mol•L-1 CH3COONH4ÈÜÒº³ÊÖÐÐÔ£¬ËµÃ÷´×Ëá¸ùÀë×ÓÓë笠ùÀë×ÓµÄË®½â³Ì¶ÈÏàͬ£¬ÏàͬÎïÖʵÄÁ¿µÄ´×ËáÓëһˮºÏ°±ÖкÍÉú³É´×Ëá泥¬´×Ëá¸ùÀë×ÓÓë笠ùÀë×ÓË®½âÏ໥´Ù½ø£¬Ë®µÄµçÀë³Ì¶ÈÔö´ó£¬K£¨CH3COOH£©=$\frac{´×ËáµçÀëƽºâ³£Êý}{Kw}$£¬K£¨NH3•H2O£©=$\frac{笠ùË®½âƽºâ³£Êý}{Kw}$£¬¾Ý´Ë·ÖÎö£®

½â´ð ½â£ºA¡¢µÈÌå»ý¡¢Å¨¶È¾ùΪ0.1mol•L-1µÄ´×ËáºÍ°±Ë®·´Ó¦µÃµ½0.05mol•L-1 CH3COONH4ÈÜÒº£¬¸ùA´íÎó£»
B¡¢´×Ëá¸ùÀë×ÓÓë笠ùÀë×ÓË®½âÏ໥´Ù½ø£¬Ë®µÄµçÀë³Ì¶ÈÔö´ó£¬¹ÊB´íÎó£»
C¡¢K£¨CH3COOH£©=$\frac{Kw}{´×Ëá¸ùµÄË®½âƽºâ³£Êý}$£¬K£¨NH3•H2O£©=$\frac{Kw}{笠ùµÄË®½âƽºâ³£Êý}$£¬0.1mol•L-1 CH3COONH4ÈÜÒº³ÊÖÐÐÔ£¬ËµÃ÷´×Ëá¸ùÀë×ÓÓë笠ùÀë×ÓµÄË®½â³Ì¶ÈÏàͬ£¬¼´´×Ëá¸ùÀë×ÓµÄË®½âƽºâ³£Êý=笠ùÀë×ÓµÄË®½âƽºâ³£Êý£¬Ë®µÄÀë×Ó»ý³£ÊýÏàͬ£¬ËùÒÔK£¨CH3COOH£©=K£¨NH3•H2O£©£¬¹ÊCÕýÈ·£»
D¡¢CH3COONaÈÜÒºÖеÄc£¨CH3COO-£©×ÔȻˮ½â£¬´×Ëáï§ÈÜÒºÖд×Ëá¸ùÀë×ÓÓë笠ùÀë×ÓË®½âÏ໥´Ù½ø£¬¸ÃÈÜÒºÖÐc£¨CH3COO-£©Ð¡ÓÚͬŨ¶ÈCH3COONaÈÜÒºÖеÄc£¨CH3COO-£©£¬¹ÊD´íÎó£»
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²éÁËÑÎÀàË®½âµÄÓ°ÏìÒòËØ¡¢µçÀëƽºâ³£ÊýÓëË®½âƽºâ³£ÊýÒÔ¼°Ë®µÄÀë×Ó»ýµÄ¹Øϵ£¬×¢ÒâÈõËá¸ùÀë×ÓÓëÈõ¼îÑôÀë×ÓË®½âÏ໥´Ù½ø£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
20£®µª»¯ÂÁ£¨AlN£©ÊÇÒ»ÖÖÐÂÐÍÎÞ»ú·Ç½ðÊô²ÄÁÏ£®Ä³AlN ÑùÆ·½öº¬ÓÐAl2O3 ÔÓÖÊ£¬Îª²â¶¨AlN µÄº¬Á¿£¬Éè¼ÆÈçÏÂÈýÖÖʵÑé·½°¸£®£¨ÒÑÖª£ºAlN+NaOH+H2O=NaAlO2+NH3¡ü£© 
·½°¸l£ºÈ¡Ò»¶¨Á¿µÄÑùÆ·£¬ÓÃÒÔÏÂ×°ÖòⶨÑùÆ·ÖÐAlN µÄ´¿¶È£¨¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©£®
£¨1£©Í¼1 ÖÐÇòÐθÉÔï¹ÜµÄ×÷ÓÃÊÇ·ÀÖ¹µ¹Îü£®
£¨2£©Íê³ÉÒÔÏÂʵÑé²½Ö裺×é×°ºÃʵÑé×°Ö㬼ì²é×°ÖÃÆøÃÜÐÔ£¬ÔÙ¼ÓÈëʵÑéÒ©Æ·£®½ÓÏÂÀ´µÄʵÑé²Ù×÷ÊÇ£º¹Ø±ÕK1£¬´ò¿ªK2£¬´ò¿ª·ÖҺ©¶·»îÈû£¬¼ÓÈëNaOH Å¨ÈÜÒº£¬ÖÁ²»ÔÙ²úÉúÆøÌ壮Ȼºó´ò¿ªK1£¬Í¨È뵪ÆøÒ»¶Îʱ¼ä£¬²â¶¨C ×°Ö÷´Ó¦Ç°ºóµÄÖÊÁ¿±ä»¯£®Í¨È뵪ÆøµÄÄ¿µÄÊÇ°Ñ×°ÖÃÖвÐÁôµÄ°±ÆøÈ«²¿¸ÏÈëC×°Öã®
£¨3£©¼ÙÉèÆäËû²Ù×÷¾ùÕýÈ·£¬°´ÕÕͼ1 ×°Öý«µ¼Ö²ⶨ½á¹ûÆ«¸ß£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£®
·½°¸2£ºÓÃͼ2 ×°Öòⶨmg ÑùÆ·ÖÐAlN µÄ´¿¶È£¨²¿·Ö¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©£®
£¨4£©Îª²â¶¨Éú³ÉÆøÌåµÄÌå»ý£¬Á¿Æø×°ÖÃÖеÄX ÒºÌå×îºÃÑ¡Ôña£®
a£®CCl4 b£®H2O c£®±¥ºÍNH4Cl ÈÜÒºd£®Å¨ÁòËá
£¨5£©Èôm g ÑùÆ·ÍêÈ«·´Ó¦£¬²âµÃÉú³ÉÆøÌåµÄÌå»ýΪVmL£¨ÒÑת»»Îª±ê×¼×´¿ö£©£®ÔòAlNµÄÖÊÁ¿·ÖÊýÊÇ$\frac{41V}{22400m}$¡Á100%£¨Ð軯¼ò£©£®
·½°¸3£º°´Í¼3·¾¶²â¶¨ÑùÆ·ÖÐAlN µÄ´¿¶È£º
£¨6£©Ð´³ö²½Öè¢Ù·´Ó¦µÄÀë×Ó·½³ÌʽAlN+OH-+H2O=AlO2-+NH3¡ü¡¢Al2O3+2OH-=2AlO2-+H2O£®
£¨7£©¸Ã·½°¸ÄÜ·ñ²â³öAlN µÄ´¿¶È£¿£¨ÈôÄÜÇëÓÃm1£¬m2Áгö¼ÆËã´¿¶ÈµÄ±í´ïʽ£©$\frac{4.1{m}_{2}-4.1{m}_{1}}{{m}_{1}}$¡Á100%£®

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

¾«Ó¢¼Ò½ÌÍø