19£®£¨1£©ÒÑÖª³£ÎÂÏÂHCOOHµÄµçÀë³£ÊýΪ${K_a}=2¡Á{10^{-4}}$£¬ÔòHCOO-µÄË®½â·´Ó¦HCOO-+H2O?HCOOH+OH-µÄƽºâ³£ÊýΪKh=5¡Á10-11£®
£¨2£©³£ÎÂÏ£¬½«a mol/LµÄHCOOHÈÜÒºÓë$\frac{a}{2}$mol/LµÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇc£¨HCOO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©£®
£¨3£©³£ÎÂÏ£¬ÔÚa mol/LµÄHCOOHÈÜÒºÖмÓÈëµÈÌå»ýµÄbmol/LµÄNaOHÈÜÒºÖÁÈÜÒº³ÊÖÐÐÔ£¬´ËʱÈÜÒºÖÐHCOOHµÄÎïÖʵÄÁ¿Å¨¶È $\frac{a-b}{2}$£®

·ÖÎö £¨1£©Ð´³ö´×ËáµçÀëƽºâ³£Êý¡¢´×Ëá¸ùÀë×ÓË®½âƽºâ³£Êý£¬Ë®µÄÀë×Ó»ý³£Êý½øÐбȽϣ¬´Ó¶øµÃ³ö¹Øϵʽ£¬½øÐмÆËã¼´¿É£»
£¨2£©½«amol/LµÄHCOOHÈÜÒºÓë$\frac{a}{2}$mol/LµÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬µÃµ½µÄÊǼ×ËáºÍ¼×ËáÄƵĻìºÏÎ¾Ý´Ë»Ø´ðÅжϣ»
£¨3£©¸ù¾ÝÈÜÒºÖеĵçºÉÊغãÒÔ¼°ÎïÁÏÊغã֪ʶÀ´¼ÆË㣮

½â´ð ½â£º£¨1£©Ka=$\frac{c£¨C{H}_{3}CO{O}^{-}£©•c£¨{H}^{+}£©}{c£¨C{H}_{3}COOH£©}$£¬Kh=$\frac{c£¨C{H}_{3}COOH£©•c£¨O{H}^{-}£©}{c£¨C{H}_{3}CO{O}^{-}£©}$£¬Kw=C£¨H+£©•C£¨OH-£©£¬ËùÒÔKa•Kh=Kw£¬Kh=$\frac{1{0}^{-14}}{2¡Á1{0}^{-4}}$=5¡Á10-11£¬
¹Ê´ð°¸Îª£º5¡Á10-11£»
£¨2£©amol/LµÄHCOOHÈÜÒºÓë$\frac{a}{2}$mol/LµÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬µÃµ½µÄÊǵÈŨ¶ÈµÄ¼×ËáÄƺͼ×ËáµÄ»ìºÏÎÈÜÒºÏÔʾËáÐÔ£¬¼×ËáµÄµçÀë³Ì¶È´óÓÚ¼×Ëá¸ùÀë×ÓµÄË®½â³Ì¶È£¬Àë×ÓŨ¶È´óС˳ÐòÊÇ£ºc£¨HCOO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©£»¡¡
¹Ê´ð°¸Îª£ºc£¨HCOO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©£»¡¡
£¨3£©¸ù¾ÝÈÜÒºÏÔʾÖÐÐÔ£¬ËùÒÔc£¨OH-£©=c£¨H+£©£¬¸ù¾ÝµçºÉÊغ㣬µÃµ½c£¨HCOO-£©=c£¨Na+£©=$\frac{b}{2}$mol/L£¬¸ù¾ÝÎïÁÏÊغ㣬ÈÜÒºÖÐc£¨HCOOH£©=$\frac{a}{2}$-c£¨HCOO-£©=$\frac{a-b}{2}$mol/L£¬
¹Ê´ð°¸Îª£º$\frac{a-b}{2}$mol/L£®

µãÆÀ ±¾Ì⿼²éÁ˵çÀëƽºâ³£Êý¸ÅÄî¡¢µçÀëƽºâÓ°ÏìÒòËغͼÆËãµÄÀí½âÓ¦Óá¢ÈÜÒºÖÐÀë×ÓŨ¶È´óС±È½ÏµÈ֪ʶ£¬ÕÆÎÕ»ù´¡ÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

14£®Â±´úÌþÔÚÉú²úÉú»îÖоßÓй㷺µÄÓ¦Ó㬻شðÏÂÁÐÎÊÌ⣺
£¨1£©¶àÂÈ´ú¼×Í鳣ΪÓлúÈܼÁ£¬Æä·Ö×ӽṹΪÕýËÄÃæÌåµÄÃû³ÆΪËÄÂȼ×Í飻̼ԭ×Ó¸öÊý²»´óÓÚ10µÄÍéÌþ·Ö×ÓÖУ¬ÆäÒ»ÂÈ´úÎïÖ»ÓÐÒ»ÖÖµÄÍéÌþµÄ¸öÊýΪ4¸ö£®
£¨2£©Èý·úÂÈäåÒÒÍ飨CF3CHClBr£©ÊÇÒ»ÖÖÂé×í¼Á£¬Ð´³öÆäËùÓÐͬ·ÖÒì¹¹ÌåµÄ½á¹¹Ê½CHFClCF2Br¡¢CHFBrCF2Cl¡¢CFClBrCHF2£¨²»¿¼ÂÇÁ¢ÌåÒì¹¹£©£®
£¨3£©¾ÛÂÈÒÒÏ©ÊÇÉú»îÖг£ÓõÄËÜÁÏ£¬¹¤ÒµÉú²ú¾ÛÂÈÒÒÏ©µÄÒ»ÖÖ¹¤ÒÕ·ÏßÈçÏ£º
ÒÒÏ©$¡ú_{¢Ù}^{Cl_{2}}$1£¬2-¶þÂÈÒÒÍé$¡ú_{¢Ú}^{480-530¡æ}$ÂÈÒÒÏ©$\stackrel{¾ÛºÏ}{¡ú}$¾ÛÂÈÒÒÏ©
·´Ó¦¢ÙµÄ·´Ó¦ÀàÐÍΪ¼Ó³É·´Ó¦£»
·´Ó¦¢ÚµÄ·´Ó¦ÀàÐÍΪÏûÈ¥·ÖÒº£®
£¨4£©ÊµÑéÊÒÒÔÒÒ´¼¡¢Å¨ÁòËáºÍä廯ÄÆΪÊÔ¼Á£¬ÓÃÈçͼµÄ×°ÖÃÖƱ¸äåÒÒÍ飬ͼÖÐÊ¡È¥Á˼ÓÈÈ×°Öã®
ÓйØÊý¾Ý¼û±í£º
  ÒÒ´¼äåÒÒÍé  äå
 ×´Ì¬ ÎÞÉ«ÒºÌå ÎÞÉ«ÒºÌå Éîºì×ØÉ«ÒºÌå
 Ãܶȣ¨g/cm3£© 0.79 1.44 3.1
 ·Ðµã78.5  38.459 
¢ÙÒÇÆ÷BÃû³ÆÊÇÇòÐÎÀäÄý¹Ü£¬ÒÇÆ÷BµÄ×÷ÓÃÊÇÀäÄý»ØÁ÷£®
¢ÚÖƱ¸¹ý³ÌÖУ¬¼ÓÈëµÄŨÁòËá±ØÐë½øÐÐÏ¡ÊÍ£¬ÆäÄ¿µÄÊÇAB£¨ÌîÐòºÅ£©
A£®¼õÉÙ¸±²úÎïÏ©ºÍÃѵÄÉú³É   B£®¼õÉÙBr2µÄÉú³É   C£®Ë®ÊÇ·´Ó¦µÄ´ß»¯¼Á   D£®ÁòËáµÄŨ¶ÈԽϡ·´Ó¦Ô½¿ì
¢Û¼ÓÈÈÓ¦²ÉÈ¡µÄ·½Ê½ÎªË®Ô¡¼ÓÈÈ£»¼ÓÈȵÄÄ¿µÄÊÇÉý¸ßζȣ¬¼Ó¿ì·´Ó¦ËÙÂÊ£¬Í¬Ê±Ê¹Éú³ÉµÄäåÒÒÍé·ÖÀë³öÀ´´Ù½øƽºâÓÒÒÆ£®
¢ÜΪ³ýÈ¥ÊÕ¼¯²úÆ·ÖеÄÖ÷ÒªÔÓÖÊ£¬Ó¦Ñ¡È¡¼ÓÈëµÄºÏÀíÊÔ¼ÁΪNa2SO3ÈÜÒº»ò±¥ºÍNaHSO3ÈÜÒº£»²ÉÓõIJÙ×÷ÊÇ·ÖÒº£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

15£®L-¶à°ÍÊÇÒ»ÖÖÓлúÎËü¿ÉÓÃÓÚÅÁ½ðÉ­×ÛºÏÖ¢µÄÖÎÁÆ£¬Æä½á¹¹¼òʽÈçͼ£ºÕâÖÖÒ©ÎïµÄÑÐÖÆÊÇ»ùÓÚ»ñµÃ2000Äêŵ±´¶ûÉúÀíѧ»òҽѧ½±¡¢2001Äêŵ±´¶û»¯Ñ§½±µÄÑо¿³É¹û£®ÏÂÁйØÓÚL-¶à°ÍµÄÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Ö»ÄÜÓë¼î·´Ó¦£¬²»ÄÜÓëËá·´Ó¦
B£®1mol¸ÃÎïÖÊ×î¶àÓë4molNaOH·´Ó¦
C£®¸ÃÎïÖʲ»ÄÜʹËáÐÔKMnO4ÍÊÉ«
D£®1mol¸ÃÎïÖÊ×î¶à¿ÉÓë1.5molHBr·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

7£®×î¼òʽÏàͬ£¬µ«¼È²»ÊÇͬϵÎïÓÖ²»ÊÇͬ·ÖÒì¹¹ÌåµÄÊÇ£¨¡¡¡¡£©
A£®CH¡ÔCHºÍB£®±ûÏ©ºÍ»·±ûÍéC£®ºÍD£®¼×ÃѺͼ״¼

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

14£®ÒÑ֪ˮÔÚ25¡æºÍ95¡æʱ£¬ÆäµçÀëƽºâÇúÏßÈçͼËùʾ£º
£¨1£©Ôò25¡æʱˮµÄµçÀëƽºâÇúÏßӦΪA£¨Ìî¡°A¡±»ò¡°B¡±£©£¬Çë˵Ã÷ÀíÓÉË®µÄµçÀëÊÇÎüÈȹý³Ì£¬Î¶ȵÍʱ£¬Ë®µÄµçÀë³Ì¶ÈС£¬c£¨H+£©¡¢c£¨OH-£©Ð¡£®
£¨2£©25¡æʱ£¬½«pH=9µÄNaOHÈÜÒºÓëpH=4µÄH2SO4ÈÜÒº»ìºÏ£¬ÈôËùµÃ»ìºÏÈÜÒºµÄp=7£¬ÔòNaOHÈÜÒºÓëH2SO4ÈÜÒºµÄÌå»ý±ÈΪ10£º1£®
£¨3£©95¡æʱ£¬Èô100Ìå»ýpH1=aµÄijǿËáÈÜÒºÓë1Ìå»ýpH2=bµÄijǿ¼îÈÜÒº»ìºÏºóÈÜÒº³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°£¬¸ÃÇ¿ËáµÄpH1ÓëÇ¿¼îµÄpH2Ö®¼äÓ¦Âú×ãµÄ¹ØϵÊÇpH1+pH2=14£®
£¨4£©ÇúÏßB¶ÔӦζÈÏ£¬pH=2µÄijHAÈÜÒººÍpH=10µÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬»ìºÏÈÜÒºµÄpH=5£®Çë·ÖÎöÔ­Òò£ºÇúÏßB¶ÔÓ¦95¡æ£¬´ËʱˮµÄÀë×Ó»ýΪ10-12£¬HAΪÈõËᣬHAÖкÍNaOHºó£¬»ìºÏÈÜÒºÖл¹Ê£Óà½Ï¶àµÄHA·Ö×Ó£¬¿É¼ÌÐøµçÀë³öH+£¬Ê¹ÈÜÒºpH=5£®
£¨5£©ÒÑÖª25¡æʱ£¬0.1L 0.1mol•L-1µÄNaAÈÜÒºµÄpH=10£¬ÔòNaAÈÜÒºÖÐËù´æÔÚµÄÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨Na+£©£¾c£¨A-£©£¾c£¨OH-£©£¾c£¨H+£©£®
£¨6£©ÒÑÖªAnBmµÄÀë×Ó»ýKw=[c£¨Am+£©]n©q[c£¨Bn-£©]m£¬Ê½ÖÐ[c£¨Am+£©]nºÍ[c£¨Am+£©]n±íʾÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈµÄÊýÖµ£®ÔÚijζÈÏ£¬Ca£¨OH£©2µÄÈܽâ¶ÈΪ0.74g£¬Æä±¥ºÍÈÜÒºµÄÃܶÈÉèΪ1g?mL-1£¬ÔòÆäÀë×Ó»ýΪ4¡Á10-3£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

4£®ÒÑÖª1£¬3-¶¡¶þÏ©ºÍÒÒÏ©¿ÉÒÔ·¢Éú·´Ó¦Éú³É»·¼ºÏ©£¬ÇëÉè¼ÆÒ»Ìõ´Ó1£¬3-¶¡¶þÏ©ºÏ³É×èȼ¼Á  µÄºÏ³É·Ïߣ¨ÆäËûÎÞ»úÊÔ¼ÁÈÎÑ¡£©£®ÒÑÖª£®
£¨ºÏ³É·Ïß³£Óõıíʾ·½Ê½Îª£ºA$¡ú_{·´Ó¦Ìõ¼þ}^{·´Ó¦ÊÔ¼Á}$B¡­$¡ú_{·´Ó¦Ìõ¼þ}^{·´Ó¦ÊÔ¼Á}$Ä¿±ê²úÎ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

11£®»¯ºÏFÊÇÒ»ÖÖÖØÒªµÄÓлúºÏ³ÉÖмäÌ壬ËüµÄºÏ³É·ÏßÈçÏ£º

ÇëÈÏÕæ·ÖÎöºÍÍÆÀí»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÉBÉú³ÉCµÄ»¯Ñ§·´Ó¦ÀàÐÍÊǼӳɷ´Ó¦£»»¯ºÏÎïFÖк¬Ñõ¹ÙÄÜÍŵÄÃû³ÆÊÇõ¥»ù¡¢ôÊ»ù£»
£¨2£©Ð´³ö»¯ºÏÎïCÓëÒÒËá·´Ó¦Éú³Éõ¥µÄ»¯Ñ§·½³Ìʽ£º£®
£¨3£©Ð´³ö»¯ºÏÎïBµÄ½á¹¹¼òʽ£º£®Í¬Ê±Âú×ãÏÂÁÐÌõ¼þµÄBµÄͬ·ÖÒì¹¹Ì壨²»°üÀ¨B£©¹²ÓÐ11ÖÖ£º
¢Ù±½»·ÉÏÖ»ÓÐÁ½¸öÈ¡´ú»ù     ¢ÚÄÜÓëFeCl3ÈÜÒºÏÔ×ÏÉ«£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®ÒÑÖª25¡æʱ²¿·ÖÈõµç½âÖʵĵçÀëƽºâ³£ÊýÊý¾ÝÈç±íËùʾ£¬»Ø´ðÏÂÁÐÎÊÌ⣺
»¯Ñ§Ê½CH3COOHH2CO3HClO
µçÀëƽºâ³£ÊýKa=1.8¡Á10-5Ka1=4.3¡Á10-7
Ka2=5.6¡Á10-11
Ka=3.0¡Á10-8
£¨1£©ÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1mol•L-1µÄËÄÖÖÈÜÒº£º
a£®CH3COONa¡¡¡¡¡¡¡¡¡¡¡¡¡¡b£®Na2CO3         c£®NaClO          d£®NaHCO3
pHÓÉСµ½´óÅÅÁеÄ˳ÐòÊÇa£¼d£¼c£¼b£¨ÓñàºÅÌîд£©£®
£¨2£©³£ÎÂÏ£¬0.1mol•L-1 CH3COOHÈÜÒº¼ÓˮϡÊ͹ý³ÌÖУ¬ÏÂÁбí´ïʽµÄÊý¾Ý±ä´óµÄÊÇBD£¨Ìî×Öĸ£©£®
A£®c£¨H+£©   B.$\frac{c£¨{H}^{+}£©}{c£¨C{H}_{3}COOH£©}$   C£®c£¨H+£©•c£¨OH-£©    D.$\frac{c£¨O{H}^{-}£©}{c£¨{H}^{+}£©}$     E.$\frac{c£¨{H}^{+}£©•c£¨C{H}_{3}CO{O}^{-}£©}{c£¨C{H}_{3}COOH£©}$
£¨3£©Ð´³öÏò´ÎÂÈËáÄÆÈÜÒºÖÐͨÈëÉÙÁ¿¶þÑõ»¯Ì¼µÄÀë×Ó·½³Ìʽ£ºClO-+H2O+CO2=HCO3-+HClO£®
£¨4£©25¡æʱ£¬CH3COOHÓëCH3COONaµÄ»ìºÏÈÜÒº£¬Èô²âµÃ»ìºÏÒºpH=6£¬ÔòÈÜÒºÖÐc£¨CH3COO-£©-c£¨Na+£©=9.9¡Á10-7mol•L-1£¨Ìî׼ȷÊýÖµ£©£®
£¨5£©25¡æʱ£¬½«a mol•L-1µÄ´×ËáÓëb mol•L-1ÇâÑõ»¯ÄƵÈÌå»ý»ìºÏ£¬·´Ó¦ºóÈÜҺǡºÃÏÔÖÐÐÔ£¬ÓÃa¡¢b±íʾ´×ËáµÄµçÀëƽºâ³£ÊýΪ$\frac{b¡Á10-7}{a-b}$£®
£¨6£©Ìå»ý¾ùΪ100mL pH=2µÄCH3COOHÓëÒ»ÔªËáHX£¬¼ÓˮϡÊ͹ý³ÌÖÐpHÓëÈÜÒºÌå»ýµÄ¹ØϵÈçͼËùʾ£¬ÔòHXµÄµçÀëƽºâ³£Êý´óÓÚ£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©CH3COOHµÄµçÀëƽºâ³£Êý£®
£¨7£©±ê×¼×´¿öÏ£¬½«1.12L CO2ͨÈë100mL 1mol•L-1µÄNaOHÈÜÒºÖУ¬ÓÃÈÜÒºÖÐ΢Á£µÄŨ¶È·ûºÅÍê³ÉÏÂÁеÈʽ£º
¢Ùc£¨OH-£©=2c£¨H2CO3£©+c£¨HCO3-£©+c£¨H+£©£®
¢Úc£¨H+£©+c£¨Na+£©=2c£¨CO32-£©+c£¨HCO3-£©+c£¨OH-£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

9£®·Ö×ÓʽΪC9H11Cl£¬ÇÒ±½»·ÉÏÓÐÁ½¸öÈ¡´ú»ùµÄ·¼Ïã×廯ºÏÎÆä¿ÉÄܵĽṹÓÐ15£¨²»¿¼ÂÇÁ¢ÌåÒì¹¹£©ÖÖ£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸