ÌâÄ¿ÄÚÈÝ

19£®ÏÂÁÐÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®NaÓëË®·´Ó¦£ºNa+2H2O¨TNa++2OH-+H2¡ü
B£®0.01mol/L NH4Al£¨SO4£©2ÈÜÒºÓë0.02mol/L Ba£¨OH£©2ÈÜÒºµÈÌå»ý»ìºÏ NH4++Al3++2SO42-+2Ba2++4 OH-¨T2 Ba SO4¡ý+Al£¨OH£©3¡ý+NH3•H2O
C£®¹èËáÄÆÈÜÒºÓë´×ËáÈÜÒº»ìºÏ£ºSiO32-+2H+¨TH2SiO3¡ý
D£®Å¨ÏõËáÖмÓÈë¹ýÁ¿Ìú·Û²¢¼ÓÈÈ£ºFe+3NO3-+6H+ $\frac{\underline{\;\;¡÷\;\;}}{\;}$Fe3++3NO2¡ü+3H2O

·ÖÎö A£®µçºÉ²»Êغ㣻
B£®ÎïÖʵÄÁ¿±ÈΪ1£º2£¬·´Ó¦Éú³ÉÁòËá±µ¡¢ÇâÑõ»¯ÂÁºÍһˮºÏ°±£»
C£®·´Ó¦Éú³É¹èËáºÍ´×ËáÄÆ£¬´×ËáÔÚÀë×Ó·´Ó¦Öб£Áô»¯Ñ§Ê½£»
D£®Fe¹ýÁ¿£¬Éú³ÉÏõËáÑÇÌú¡¢¶þÑõ»¯µªºÍË®£®

½â´ð ½â£ºA£®NaÓëË®·´Ó¦µÄÀë×Ó·´Ó¦Îª2Na+2H2O¨T2Na++2OH-+H2¡ü£¬¹ÊA´íÎó£»
B.0.01mol/L NH4Al£¨SO4£©2ÈÜÒºÓë0.02mol/L Ba£¨OH£©2ÈÜÒºµÈÌå»ý»ìºÏµÄÀë×Ó·´Ó¦ÎªNH4++Al3++2SO42-+2Ba2++4OH-¨T2BaSO4¡ý+Al£¨OH£©3¡ý+NH3•H2O£¬¹ÊBÕýÈ·£»
C£®¹èËáÄÆÈÜÒºÓë´×ËáÈÜÒº»ìºÏµÄÀë×Ó·´Ó¦ÎªSiO32-+2CH3COOH¨TH2SiO3¡ý+22CH3COO-£¬¹ÊC´íÎó£»
D£®Å¨ÏõËáÖмÓÈë¹ýÁ¿Ìú·Û²¢¼ÓÈȵÄÀë×Ó·´Ó¦ÎªFe+2NO3-+4H+ $\frac{\underline{\;\;¡÷\;\;}}{\;}$Fe2++2NO2¡ü+2H2O£¬¹ÊD´íÎó£»
¹ÊÑ¡B£®

µãÆÀ ±¾Ì⿼²éÀë×Ó·´Ó¦·½³ÌʽÊéдµÄÕýÎóÅжϣ¬Îª¸ßƵ¿¼µã£¬°ÑÎÕϰ·¢ÉúµÄ·´Ó¦¼°Àë×Ó·´Ó¦µÄÊéд·½·¨Îª½â´ðµÄ¹Ø¼ü£¬²àÖØÑõ»¯»¹Ô­·´Ó¦¡¢¸´·Ö½â·´Ó¦µÄÀë×Ó·´Ó¦¿¼²é£¬×¢ÒâÀë×Ó·´Ó¦Öб£Áô»¯Ñ§Ê½µÄÎïÖʼ°µç×Ó¡¢µçºÉÊØºã£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
9£®SO2µÄº¬Á¿ÊÇ¿ÕÆøÖÊÁ¿ÈÕ±¨ÖÐÒ»ÏîÖØÒª¼ì²âÖ¸±ê£¬Çë½áºÏËùѧ֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©¹¤ÒµÖÆÁòËá¹ý³ÌÖУ¬SO2´ß»¯Ñõ»¯µÄÔ­ÀíΪ£º
2SO2£¨g£©+O2£¨g£©$?_{¡÷}^{´ß»¯¼Á}$2SO3£¨g£©
T¡æÊ±£¬ÏòijÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨SO2£¨g£©ºÍO2£¨g£©£¬·¢ÉúÉÏÊö·´Ó¦£¬²âµÃSO2£¨g£©µÄƽºâת»¯ÂÊ£¨a£©ÓëÌåϵ×Üѹǿ£¨p£©µÄ¹ØÏµÈçͼ1Ëùʾ£®

¢Ùa¡¢bÁ½µã¶ÔÓ¦µÄƽºâ³£ÊýK£¨a£©=K£¨b£©£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£¬ÏÂͬ£©£¬SO2Ũ¶Èc£¨a£©£¾c£¨b£©£®
¢Úcµãʱ£¬·´Ó¦ËÙÂÊv£¨Õý£©£¼v£¨Ä棩£®
£¨2£©µç»¯Ñ§·¨´¦ÀíSO2£®
ÁòËá¹¤ÒµÎ²ÆøÖеÄSO2¾­·ÖÀëºó£¬¿ÉÓÃÓÚÖÆ±¸ÁòËᣬͬʱ»ñµÃµçÄÜ£¬×°ÖÃÈçͼ2Ëùʾ£¨µç¼«¾ùΪ¶èÐÔ²ÄÁÏ£©£º
¢ÙM¼«·¢ÉúµÄµç¼«·´Ó¦Ê½ÎªSO2-2e-+2H2O=4H++SO42-£®
¢ÚÈôʹ¸Ã×°ÖõĵçÁ÷Ç¿¶È´ïµ½2.0A£¬ÀíÂÛÉÏÿ·ÖÖÓÓ¦Ïò¸º¼«Í¨Èë±ê×¼×´¿öÏÂÆøÌåµÄÌå»ýΪ0.014L£¨ÒÑÖª£º1¸öe-Ëù´øµçÁ¿Îª1.6¡Á10-19C£©£®
£¨3£©ÈÜÒº·¨´¦ÀíSO2£®
ÒÑÖª³£ÎÂÏÂH2SO3ºÍH2CO3µÄµçÀë³£ÊýÈç±íËùʾ£º
µçÀë³£Êý
Ëá
K1K2
H2SO31.3¡Á10-26.3¡Á10-8
H2CO34.2¡Á10-75.6¡Á10-11
³£ÎÂÏ£¬½«SO2»ºÂýͨÈë100mL 0.2mol•L-1µÄNa2CO3ÈÜÒºÖУ¬µ±Í¨Èë448mLSO2ʱ£¨ÒÑÕÛËãΪ±ê×¼×´¿öϵÄÌå»ý£¬ÏÂͬ£©£¬·¢ÉúµÄÀë×Ó·½³ÌʽΪSO2+H2O+CO32-=HCO3-+HSO3-£»µ±Í¨Èë896mLSO2ʱ£¬ËùµÃÈÜÒº³ÊÈõËáÐÔ£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨Na+£©£¾c£¨HSO3-£©£¾c£¨H+£©£¾c£¨SO32-£©£¾c£¨OH-£©£®
8£®ÏÖÓк¬ÓÐÉÙÁ¿NaCl¡¢Na2SO4¡¢Na2CO3µÈÔÓÖʵÄNaNO3ÈÜÒº£¬Ñ¡ÔñÊʵ±µÄÊÔ¼Á³ýÈ¥ÔÓÖÊ£¬µÃµ½´¿¾»µÄNaNO3¹ÌÌ壬ʵÑéÁ÷³ÌÈçͼËùʾ£®

£¨1£©³ÁµíAµÄÖ÷Òª³É·ÖÊÇBaSO4¡¢BaCO3£¨Ìѧʽ£©£®
£¨2£©¢ÚÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇAg++Cl-¨TAgCl¡ý£®
£¨3£©¢Ù¢Ú¢ÛÖоù½øÐеķÖÀë²Ù×÷ÊǹýÂË£®
£¨4£©ÈÜÒº3¾­¹ý´¦Àí¿ÉÒԵõ½NaNO3¹ÌÌ壬ÈÜÒº3Öп϶¨º¬ÓеÄÔÓÖÊÊÇNa2CO3£¬ÎªÁ˳ýÈ¥ÔÓÖÊ£¬¿ÉÏòÈÜÒº3ÖмÓÈëÊÊÁ¿µÄHNO3£®
£¨5£©ÊµÑéÊÒÓÃÉÏÊöʵÑé»ñµÃµÄNaNO3¹ÌÌåÅäÖÆ500mL 0.40mol/L NaNO3ÈÜÒº£®
¢ÙÅäÖÆÈÜҺʱ£¬½øÐÐÈçϲÙ×÷£ºa£®¶¨ÈÝ£»b£®¼ÆË㣻c£®Èܽ⣻d£®Ò¡ÔÈ£»e£®×ªÒÆ£»f£®Ï´µÓ£»j£®³ÆÁ¿£®³ÆÈ¡NaNO3¹ÌÌåµÄÖÊÁ¿ÊÇ17.0 g£®°´ÕÕ²Ù×÷˳Ðò£¬µÚ4²½ÊÇe£¨ÌîÐòºÅ£©£®
¢ÚÄ³Í¬Ñ§×ªÒÆÈÜÒºµÄ²Ù×÷ÈçͼËùʾ£¬¸Ãͬѧ²Ù×÷ÖеĴíÎóÊÇδÓò£Á§°ôÒýÁ÷£®
¢ÛÈôÓýºÍ·µÎ¹Ü¶¨ÈÝʱ£¬²»Ð¡ÐĵÎË®µÎ¹ýÁ˿̶ÈÏߣ¬ÄãÈÏΪӦ¸Ã²ÉÈ¡µÄ´¦Àí·½·¨ÊÇ£ºÖØÐÂÅäÖÆ£®
¢ÜÏÂÁвÙ×÷ÖУ¬¿ÉÄÜÔì³ÉËùÅäÖÆÈÜÒºµÄŨ¶ÈÆ«µÍµÄÊÇa£¨ÌîÑ¡Ï£®
a£®Ã»ÓÐÏ´µÓÉÕ±­ºÍ²£Á§°ô
b£®¶¨ÈÝʱ£¬¸©Êӿ̶ÈÏß
c£®Ï´µÓºóµÄÈÝÁ¿Æ¿ÖвÐÁôÉÙÁ¿ÕôÁóË®£®

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

¾«Ó¢¼Ò½ÌÍø