ÒÑÖª»¯ºÏÎïAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª182.5£¬Æä·Ö×Ó×é³É¿ÉÒÔ±íʾΪCxHyOCl£®ÓйØת»¯¹ØϵÈçͼËùʾ£º
CÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢Éú·´Ó¦£¬Éú³ÉÒ»Öָ߷Ö×Ó»¯ºÏÎïE£¬EµÄ½á¹¹¼òʽΪ£¬ÆäÖÐR¡¢R¡äΪÌþ»ù£®Çë»Ø´ðÒÔÏÂÎÊÌ⣺

£¨1£©AÖк¬Óеĺ¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇ
 
£»¢ÛµÄ·´Ó¦ÀàÐÍΪ
 
£»
£¨2£©ÒÑÖªº¬ÓÐÌþ»ùRµÄÓлúÎïR-OHº¬Ñõ50%£¬ÔòAµÄ»¯Ñ§Ê½Îª
 
£»
£¨3£©ÒÑÖªEÖÐR¡äµÄÁ½¸öÈ¡´ú»ù³Ê¶Ô룬ÔòEµÄ½á¹¹¼òʽΪ
 
£»
£¨4£©Ð´³ö·´Ó¦¢ÙµÄ»¯Ñ§·½³Ìʽ
 
£»
£¨5£©DÔÚŨÁòËá¡¢¼ÓÈȵÄÌõ¼þÏ·´Ó¦Éú³ÉF£¨C11H12O2£©£¬FÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔ·¢Éú¼Ó¾Û·´Ó¦£¬Ð´³ö¸Ã¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£»
£¨6£©ÔÚBµÄ¶àÖÖͬ·ÖÒìÌåÖУ¬Ð´³ö·Ö×ӽṹÖк¬ÓР£¬ÇÒÊôÓÚõ¥µÄËùÓÐͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ
 
£®
¿¼µã£ºÓлúÎïµÄÍƶÏ
רÌ⣺ÓлúÎïµÄ»¯Ñ§ÐÔÖʼ°ÍƶÏ
·ÖÎö£º»¯ºÏÎïAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª182.5£¬Æä·Ö×Ó×é³É¿ÉÒÔ±íʾΪCxHyOCl£¬Ôò12x+y=182.5-16-35.5=131£¬CÔ­×Ó×î´óÊýÄ¿=
131
12
=10¡­11£¬¹ÊAµÄ·Ö×ÓʽӦΪC10H11OCl£¬BÄÜ·¢ÉúÒø¾µ·´Ó¦£¬¿ÉÖªBº¬ÓÐ-CHO£¬µÃµ½CÖÐ-COOH£¬¶øCÓëR-OHµÃµ½D£¬º¬ÓÐÌþ»ùRµÄÓлúÎïR-OHº¬Ñõ50%£¬ÔòR-OHµÄÏà¶Ô·Ö×ÓÖÊÁ¿=
16
50%
=32£¬¹ÊRΪ¼×»ù£¬R-OHΪCH3OH£¬¿ÉÍÆÖªC·Ö×ÓʽΪC10H12O3£¬C¿ÉÒԵõ½¸ß·Ö×Ó»¯ºÏÎïE£¬ÓÉEµÄ½á¹¹¼òʽ¿ÉÖªCÖк¬ÓÐ-COOH¡¢-OH£¬¶øAË®½âµÃµ½B£¬×ۺϷÖÎö¿ÉÖªAÖк¬ÓÐ-CHO¡¢-OH£¬AµÄ±¥ºÍ¶È=
2¡Á10+2-11-1
2
=5£¬¿¼ÂÇAÖк¬Óб½»·£¬½áºÏEÖÐR¡äµÄÁ½¸öÈ¡´ú»ù³Ê¶Ô룬½áºÏEµÄ½á¹¹¼òʽ¿ÉÖªCΪ£¬ÔòBΪ£¬AΪ£¬ÔòDΪ£¬½áºÏÓлúÎïµÄ½á¹¹ºÍÐÔÖÊÒÔ¼°Ìâ¸ø·´Ó¦ÐÅÏ¢½â´ð¸ÃÌ⣮
½â´ð£º ½â£º»¯ºÏÎïAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª182.5£¬Æä·Ö×Ó×é³É¿ÉÒÔ±íʾΪCxHyOCl£¬Ôò12x+y=182.5-16-35.5=131£¬CÔ­×Ó×î´óÊýÄ¿=
131
12
=10¡­11£¬¹ÊAµÄ·Ö×ÓʽӦΪC10H11OCl£¬BÄÜ·¢ÉúÒø¾µ·´Ó¦£¬¿ÉÖªBº¬ÓÐ-CHO£¬µÃµ½CÖÐ-COOH£¬¶øCÓëR-OHµÃµ½D£¬º¬ÓÐÌþ»ùRµÄÓлúÎïR-OHº¬Ñõ50%£¬ÔòR-OHµÄÏà¶Ô·Ö×ÓÖÊÁ¿=
16
50%
=32£¬¹ÊRΪ¼×»ù£¬R-OHΪCH3OH£¬¿ÉÍÆÖªC·Ö×ÓʽΪC10H12O3£¬C¿ÉÒԵõ½¸ß·Ö×Ó»¯ºÏÎïE£¬ÓÉEµÄ½á¹¹¼òʽ¿ÉÖªCÖк¬ÓÐ-COOH¡¢-OH£¬¶øAË®½âµÃµ½B£¬×ۺϷÖÎö¿ÉÖªAÖк¬ÓÐ-CHO¡¢-OH£¬AµÄ±¥ºÍ¶È=
2¡Á10+2-11-1
2
=5£¬¿¼ÂÇAÖк¬Óб½»·£¬½áºÏEÖÐR¡äµÄÁ½¸öÈ¡´ú»ù³Ê¶Ô룬½áºÏEµÄ½á¹¹¼òʽ¿ÉÖªCΪ£¬ÔòBΪ£¬AΪ£¬ÔòDΪ£¬
£¨1£©ÓÉÒÔÉÏ·ÖÎö¿ÉÖª£¬Aº¬Óеĺ¬Ñõ¹ÙÄÜÍÅΪȩ»ù£¬¢ÛµÄ·´Ó¦ÀàÐÍΪõ¥»¯·´Ó¦£¬
¹Ê´ð°¸Îª£ºÈ©»ù£»õ¥»¯·´Ó¦£»
£¨2£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬AµÄ»¯Ñ§Ê½Îª£ºC10H11OCl£¬¹Ê´ð°¸Îª£ºC10H11OCl£»
£¨3£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬ÔòEµÄ½á¹¹¼òʽΪ£º£¬
¹Ê´ð°¸Îª£º£»
£¨4£©·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽΪ£º£¬
¹Ê´ð°¸Îª£º£»
£¨5£©DΪ£¬ÔÚŨÁòËá¡¢¼ÓÈȵÄÌõ¼þÏ·´Ó¦ÏûÈ¥·´Ó¦Éú³ÉF£¨C11H12O2£©£¬FΪ£¬F·¢Éú¼Ó¾Û·´Ó¦µÄ·½³ÌʽΪ£º£¬
¹Ê´ð°¸Îª£º£»

£¨6£©BΪ£¬·Ö×ӽṹÖк¬ÓÐÇÒÊôÓÚõ¥µÄͬ·ÖÒì¹¹ÌåÓУº£¬
¹Ê´ð°¸Îª£º£®
µãÆÀ£º±¾Ì⿼²éÓлúÎïµÄÍƶϣ¬¼ÆËãÈ·¶¨AµÄ·Ö×Óʽ£¬½áºÏEµÄ½á¹¹Ìص㼰·´Ó¦Ìõ¼þΪͻÆÆ¿Ú½øÐÐÍƶϣ¬½ÏºÃµÄ¿¼²éѧÉúµÄ·ÖÎöÍÆÀíÄÜÁ¦£¬ÌâÄ¿ÄѶȽϴó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

25¡æʱ£¬ÏÂÁÐÈÜÒºµÄpH»ò΢Á£µÄÎïÖʵÄÁ¿Å¨¶È¹ØϵÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢½«0.2mol/LµÄijһԪËáHAÈÜÒººÍ0.1mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬»ìºÏÒºpH´óÓÚ7£¬Ôò·´Ó¦ºóµÄ»ìºÏÒº£º2c£¨OH-£©+c£¨A-£©=2c£¨H+£©+c£¨HA£©
B¡¢pH¾ùΪ9µÄÈýÖÖÈÜÒº£ºCH3COOH¡¢Na2CO3¡¢NaOH£¬ÆäÎïÖʵÄÁ¿Å¨¶ÈµÄ´óС˳ÐòÊÇNaOH ÈÜÒº£¾CH3COOHÈÜÒº£¾Na2CO3ÈÜÒº
C¡¢pH=3µÄ¶þÔªÈõËáH2RÈÜÒºÓëpH=11µÄNaOHÈÜÒº»ìºÏºó£¬»ìºÏÒºµÄpHµÈÓÚ7£¬Ôò·´Ó¦ºóµÄ»ìºÏÒº£ºc£¨R2-£©+c£¨HR-£©=c£¨Na+£©
D¡¢0.2mol/L NaHCO3ÈÜÒºÓë0.1mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏ£ºc£¨H+£©=c£¨OH-£©+c£¨HCO3-£©+2c£¨H2CO3£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁйý³Ì»òÊÂʵÉæ¼°Ñõ»¯»¹Ô­·´Ó¦µÄÊÇ£¨¡¡¡¡£©
¢ÙÕáÌÇÌ¿»¯ ¢ÚÓÃÇâ·úËáµñ¿Ì²£Á§ ¢ÛÌúÓöÀäµÄŨÁòËá¶Û»¯ ¢Ü×°¼îÒºµÄÊÔ¼ÁÆ¿²»Óò£Á§Èû ¢ÝÀ×Ó귢ׯ¼Ú ¢Þʵ¼ÊʹÓõÄŨÏõËáÏÔ»ÆÉ«£®
A¡¢¢Ù¢Û¢Ý¢ÞB¡¢¢Ú¢Ü
C¡¢¢Ú¢Û¢Ü¢ÞD¡¢¢Ù¢Ú¢Û¢Ü¢Ý¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

2013Äê³õ£¬Îíö²ÌìÆø¶à´ÎËÁÅ°Ìì½ò¡¢±±¾©µÈµØÇø£®ÆäÖУ¬È¼ÃººÍÆû³µÎ²ÆøÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»£®
£¨1£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO£¨g£©+2CO£¨g£©
´ß»¯¼Á
2CO2£¨g£©+N2£¨g£©£®¡÷H£¼0
¢Ù¸Ã·´Ó¦Æ½ºâ³£Êý±í´ïʽ
 
£®
¢ÚÈô¸Ã·´Ó¦ÔÚ¾øÈÈ¡¢ºãÈݵÄÃܱÕÌåϵÖнøÐУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦ÔÚ½øÐе½t1ʱ¿Ì´ïµ½Æ½ºâ״̬µÄÊÇ
 
 £¨Ìî´úºÅ£©£®

£¨2£©Ö±½ÓÅÅ·ÅúȼÉÕ²úÉúµÄÑÌÆø»áÒýÆðÑÏÖصĻ·¾³ÎÊÌ⣮
úȼÉÕ²úÉúµÄÑÌÆøº¬µªµÄÑõ»¯ÎÓÃCH4´ß»¯»¹Ô­NOX¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ£®
ÒÑÖª£ºCH4£¨g£©+2NO2£¨g£©¨TN2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H=-867kJ/mol
2NO2£¨g£©?N2O4£¨g£©¡÷H=-56.9kJ/mol
H2O£¨g£©¨TH2O£¨l£©¡÷H=-44.0kJ/mol
д³öCH4´ß»¯»¹Ô­N2O4£¨g£©Éú³ÉN2ºÍH2O£¨l£©µÄÈÈ»¯Ñ§·½³Ìʽ£º
 
£®
£¨3£©¼×ÍéȼÁϵç³Ø¿ÉÒÔÌáÉýÄÜÁ¿ÀûÓÃÂÊ£®ÈçͼÊÇÀûÓü×ÍéȼÁϵç³Øµç½â100mL 1mol/LʳÑÎË®£¬µç½âÒ»¶Îʱ¼äºó£¬ÊÕ¼¯µ½±ê×¼×´¿öϵÄÇâÆø2.24L£¨Éèµç½âºóÈÜÒºÌå»ý²»±ä£©£®
¢Ù¼×ÍéȼÁϵç³ØµÄ¸º¼«·´Ó¦Ê½£º
 
£®
¢Úµç½âºóÈÜÒºµÄpH=
 
£¨ºöÂÔÂÈÆøÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£©£®
¢ÛÑô¼«²úÉúÆøÌåµÄÌå»ýÔÚ±ê×¼×´¿öÏÂÊÇ
 
 L£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

±ûÏ©¿ÉÓÃÓںϳÉÊÇɱ³ý¸ùÁöÏß³æµÄÅ©Ò©B£¨·Ö×ÓʽΪC3H5Br2Cl£©£¬¸Ã·Ö×ÓÖÐÿ¸ö̼ԭ×ÓÉϾùÁ¬Óбԭ×Ó£¬ÆäºÏ³É·ÏßÈçͼËùʾ£®

£¨1£©±ûÏ©µÄ½á¹¹¼òʽÊÇ
 
£¬Ëüº¬ÓеĹÙÄÜÍÅÃû³ÆÊÇ
 
£®
£¨2£©ÓÉAÉú³ÉBµÄ·´Ó¦ÀàÐÍÊÇ
 
£®
£¨3£©AË®½â¿ÉµÃµ½CH2=CHCH2OH£¬¸ÃË®½â·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓлúÎïA¡¢B¡¢C¡¢D¡¢E¡¢F¡¢GµÄÏ໥¹ØϵÈçͼËùʾ£®

£¨1£©¼ìÑéAÖбԪËصÄʵÑé·½·¨ÊÇ
 
£®
£¨2£©BµÄ½á¹¹¼òʽΪ
 
£»¢ÙµÄ»¯Ñ§·´Ó¦ÀàÐÍÊÇ
 
£®
£¨3£©GÔÚʵÑéÊÒÖпÉͨ¹ýÁ½²½·´Ó¦Ñõ»¯³ÉF£®ÆäÖеÚÒ»²½·´Ó¦µÄÌõ¼þÊÇ
 
£¬·´Ó¦µÃµ½µÄ²úÎï¿ÉÄÜÓУ¨Ìîд½á¹¹¼òʽ£©
 
£®
£¨4£©FÊÇÒ»ÖÖ¶þÔªËᣬËüÔÚÒ»¶¨Ìõ¼þÏ¿ÉÓëG·´Ó¦Éú³É¸ß·Ö×Ó»¯ºÏÎ¸Ã¸ß·Ö×ӵĽṹ¼òʽΪ
 
£¬Ð´³ö·´Ó¦¢ÜµÄ»¯Ñ§·´Ó¦·½³Ìʽ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¼×¡¢ÒÒ¡¢±û¡¢¶¡ËÄÖÖ¶ÌÖÜÆÚÔªËØ¿ÉÒÔ×é³ÉÏÂÁпòͼÖгýBr2ºÍLÒÔÍâµÄÎïÖÊ£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£®¼×ºÍÒÒ¿ÉÐγɳ£¼ûҺ̬»¯ºÏÎïK£¬¹ÌÌåAÖк¬ÓбûÔªËصÄÕýÒ»¼ÛÑôÀë×Ó£¬Æäµç×Ó²ã½á¹¹ÓëÄÊÔ­×ÓÏàͬ£¬¶¡ÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇÆäµç×Ó²ãÊýµÄ2±¶£®ÔÚÒ»¶¨Ìõ¼þÏ£¬ÏÂÁи÷ÎïÖÊ¿É·¢ÉúÈçͼËùʾµÄ±ä»¯£¨·´Ó¦ÖÐÉú³ÉµÄˮûÓÐд³ö£©£º
ÊԻشð£º

£¨1£©AµÄµç×ÓʽΪ
 
£»Ëùº¬»¯Ñ§¼üÀàÐÍΪ
 
¡¢
 
£®
£¨2£©äåÔªËØÔÚÖÜÆÚ±íµÄλÖÃ
 
¶¡ÔªËصÄÒõÀë×ӽṹʾÒâͼΪ
 
£®
£¨3£©·´Ó¦£¨I£©µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨4£©·´Ó¦£¨II£©µÄÀë×Ó·½³ÌʽΪ
 
£®
£¨5£©º¬1mol CµÄÈÜÒºÖлºÂýͨÈë15.68LµÄCO2£¨±ê¿ö£©£¬ËùµÃÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Ö®±ÈΪ
 
£¨Ê½Á¿´óµÄ±ÈʽÁ¿Ð¡µÄ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A¡¢B¡¢C¡¢D¡¢EΪÖÐѧ»¯Ñ§³£¼ûµÄµ¥ÖÊ»ò»¯ºÏÎÏ໥ת»¯¹ØϵÈçͼ¼×Ëùʾ
£¨1£©ÈôAÊÇÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壻C¡¢D¾ùΪ¿ÕÆøµÄÖ÷Òª³É·Ö£»EÊÇÒ»ÖÖÎÞÉ«¡¢ÎÞζµÄÓж¾ÆøÌ壮д³ö·´Ó¦¢òµÄ»¯Ñ§·½³Ìʽ
 

£¨2£©ÈôAÊǵ­»ÆÉ«¹ÌÌ廯ºÏÎ³£ÎÂÏÂDÊÇÎÞÉ«ÆøÌ壻CÖк¬ÓеÄÒõ¡¢ÑôÀë×Ó¾ùΪ10µç×ÓµÄÁ£×Ó£®Ð´³ö·´Ó¦¢óµÄÀë×Ó·½³Ìʽ
 
£®
£¨3£©½«Ò»¶¨Á¿£¨2£©ÖеÄÆøÌåDͨÈë2L CµÄÈÜÒºÖУ¬ÔÚËùµÃÈÜÒºÖÐÖðµÎ¼ÓÈëÏ¡ÑÎËáÖÁ¹ýÁ¿£¬²¢½«ÈÜÒº¼ÓÈÈ£¬²úÉúµÄÆøÌåÓëHClµÄÎïÖʵÄÁ¿µÄ¹ØϵÈçͼÒÒ£¨ºöÂÔÆøÌåµÄÈܽâºÍHClµÄ»Ó·¢£©£®
¢ÙOµãÈÜÒºÖÐËùº¬ÈÜÖʵĻ¯Ñ§Ê½Îª
 
£»³£ÎÂÏÂaµãÈÜÒºµÄpH
 
 7£¨Ñ¡Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
¢Ú±ê¿öÏ£¬Í¨ÈëÆøÌåDµÄÌå»ýΪ
 
L£¬CÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ
 
mol/L£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijѧϰС×éΪÑо¿µç»¯Ñ§Ô­Àí£¬Éè¼ÆÈçͼװÖã®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢KÓëM¡¢N¾ù¶Ï¿ªÊ±£¬¸Ã×°ÖÃÖÐÎÞ·´Ó¦·¢Éú
B¡¢KÓëM¡¢N¾ù¶Ï¿ªÊ±£¬Zn±íÃæÓÐCuÎö³ö
C¡¢KÓëMÏàÁ¬Ê±£¬Cu2+ÍùZn¼«Òƶ¯
D¡¢KÓëNÏàÁ¬Ê±£¬ZnÈܽ⣬CuƬÖÊÁ¿Ôö¼Ó

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸