11£®³£ÎÂÏ£¬Ka £¨HCOOH£©=1.77¡Á10-4£¬Ka £¨CH3COOH£©=1.75¡Á10-5£¬Kb £¨NH3•H2O£©=1.76¡Á10-5£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Å¨¶È¾ùΪ0.1 mol•L-1µÄ HCOONaºÍNH4Cl ÈÜÒºÖÐÑôÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÖ®ºÍ£ºÇ°ÕßСÓÚºóÕß
B£®ÓÃÏàͬŨ¶ÈµÄNaOHÈÜÒº·Ö±ðµÎ¶¨µÈÌå»ýpH¾ùΪ3µÄHCOOHºÍCH3COOHÈÜÒºÖÁÖյ㣬ÏûºÄNaOHÈÜÒºµÄÌå»ýÏàµÈ
C£®0.2 mol•L-1 HCOOH Óë 0.1 mol•L-1 NaOH µÈÌå»ý»ìºÏºóµÄÈÜÒºÖУºc£¨HCOO-£©+c£¨OH-£©=c£¨HCOOH£©+c£¨H+£©
D£®0.10 mol•L-1CH3COONaÈÜÒºÖÐͨÈëHClÖÁÈÜÒº³ÊÖÐÐÔ£ºc£¨Na+£©£¾c£¨Cl-£©=c£¨CH3COOH£©

·ÖÎö A£®µçÀëƽºâ³£ÊýÔ½´ó£¬ÆäÀë×ÓË®½â³Ì¶ÈԽС£¬¸ù¾ÝµçÀëƽºâ³£ÊýÖª£¬ÆäÀë×ÓË®½â³Ì¶È£ºCH3COO-£¾NH4+£¾HCOO-£¬Èκεç½âÖÊÈÜÒºÖж¼´æÔÚµçºÉÊغ㣬ËùÒԵóöc£¨HCOO-£©+c£¨OH-£©=c£¨Na+£©+c£¨H+£©=0.1mol/L+c£¨H+£©¡¢c£¨NH4+£©+c£¨H+£©=c£¨Cl-£©+c£¨OH-£©=0.1mol/L+c£¨OH-£©£¬Ë®½â³Ì¶ÈNH4+£¾HCOO-£¬ËùÒÔÇ°Õßc£¨H+£©´óÓÚºóÕßc£¨OH-£©£»
B£®pHÏàͬµÄHCOOHºÍCH3COOH£¬Å¨¶È£ºc£¨HCOOH£©£¼c£¨CH3COOH£©£¬ÓÃÏàͬŨ¶ÈµÄNaOHÈÜÒº·Ö±ðµÎ¶¨µÈÌå»ýpH¾ùΪ3µÄHCOOHºÍCH3COOHÈÜÒºÖÁÖÕµãʱ£¬ËáµÄŨ¶ÈÔ½´ó£¬ÏûºÄµÄ¼îÌå»ýÔ½´ó£»
C£®Èκεç½âÖÊÈÜÒºÖж¼´æÔÚµçºÉÊغãºÍÎïÁÏÊغ㣬¸ù¾ÝµçºÉÊغãµÃc£¨HCOO-£©+c£¨OH-£©=c£¨Na+£©+c£¨H+£©£¬»ìºÏÈÜÒºÖÐÈÜÖÊΪµÈÎïÖʵÄÁ¿Å¨¶ÈµÄHCOOHºÍHCOONa£»
D£®Ïò0.10mol•L-1CH3COONaÈÜÒºÖÐͨÈëHCl£¬ÖÁÈÜÒºpH=7£¬·´Ó¦ºóÈÜÒºÖÐÈÜÖÊΪNaCl¡¢´×ËáºÍ´×ËáÄÆ£®

½â´ð ½â£ºA£®µçÀëƽºâ³£ÊýÔ½´ó£¬ÆäÀë×ÓË®½â³Ì¶ÈԽС£¬¸ù¾ÝµçÀëƽºâ³£ÊýÖª£¬ÆäÀë×ÓË®½â³Ì¶È£ºCH3COO-£¾NH4+£¾HCOO-£¬Èκεç½âÖÊÈÜÒºÖж¼´æÔÚµçºÉÊغ㣬ËùÒԵóöc£¨HCOO-£©+c£¨OH-£©=c£¨Na+£©+c£¨H+£©=0.1mol/L+c£¨H+£©¡¢c£¨NH4+£©+c£¨H+£©=c£¨Cl-£©+c£¨OH-£©=0.1mol/L+c£¨OH-£©£¬Ë®½â³Ì¶ÈNH4+£¾HCOO-£¬ËùÒÔÇ°Õßc£¨H+£©´óÓÚºóÕßc£¨OH-£©£¬ËùÒÔŨ¶È¾ùΪ0.1 mol•L-1µÄ HCOONaºÍNH4Cl ÈÜÒºÖÐÑôÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÖ®ºÍ£ºÇ°Õß´óÓÚºóÕߣ¬¹ÊAÕýÈ·£»
B£®pHÏàͬµÄHCOOHºÍCH3COOH£¬Å¨¶È£ºc£¨HCOOH£©£¼c£¨CH3COOH£©£¬ÓÃÏàͬŨ¶ÈµÄNaOHÈÜÒº·Ö±ðµÎ¶¨µÈÌå»ýpH¾ùΪ3µÄHCOOHºÍCH3COOHÈÜÒºÖÁÖÕµãʱ£¬ËáµÄŨ¶ÈÔ½´ó£¬ÏûºÄµÄ¼îÌå»ýÔ½´ó£¬pH¡¢Ìå»ýÏàͬµÄHCOOHºÍCH3COOH£¬ÎïÖʵÄÁ¿Ç°ÕßСÓÚºóÕߣ¬ËùÒÔºóÕßÏûºÄµÄNaOHÌå»ý¶à£¬¹ÊB´íÎó£»
C£®Èκεç½âÖÊÈÜÒºÖж¼´æÔÚµçºÉÊغãºÍÎïÁÏÊغ㣬¸ù¾ÝµçºÉÊغãµÃc£¨HCOO-£©+c£¨OH-£©=c£¨Na+£©+c£¨H+£©£¬»ìºÏÈÜÒºÖÐÈÜÖÊΪµÈÎïÖʵÄÁ¿Å¨¶ÈµÄHCOOHºÍHCOONa£¬¼×ËáµçÀë³Ì¶È´óÓÚË®½â³Ì¶È£¬ËùÒÔc£¨HCOOH£©£¼c£¨Na+£©£¬ËùÒÔµÃc£¨HCOO-£©+c£¨OH-£©£¾c£¨HCOOH£©+c£¨H+£©£¬¹ÊC´íÎó£»
D£®Ïò0.10mol•L-1CH3COONaÈÜÒºÖÐͨÈëHCl£¬ÖÁÈÜÒºpH=7£¬c£¨H+£©=c£¨OH-£©£¬HCl²»×㣬·´Ó¦ºóÈÜÒºÖÐÈÜÖÊΪNaCl¡¢´×ËáºÍ´×ËáÄÆ£¬ÓɵçºÉÊغã¿ÉÖª£¬c£¨Na+£©=c£¨CH3COO-£©+c£¨Cl-£©£¬ÓÉÎïÁÏÊغã¿ÉÖª£¬c£¨Na+£©=c£¨CH3COOH£©+c£¨CH3COO-£©£¬Ôòc£¨Na+£©£¾c£¨CH3COOH£©=c£¨Cl-£©£¬¹ÊDÕýÈ·£»
¹ÊÑ¡D£®

µãÆÀ ±¾Ì⿼²éÈõµç½âÖʵĵçÀ룬Ϊ¸ßƵ¿¼µã£¬²àÖØ¿¼²éѧÉú·ÖÎöÅжÏÄÜÁ¦£¬Ã÷È·µçÀëƽºâ³£ÊýÓëË®½â³Ì¶È¹Øϵ¡¢ÈÜÒºÖдæÔÚµÄÊغãÊǽⱾÌâ¹Ø¼ü£¬×¢ÒâCÖеÈÁ¿´ú»»£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÏÂÁл¯Ñ§Ê½±íʾµÄÎïÖÊÒ»¶¨ÊÇ´¿¾»ÎïµÄÊÇ£¨¡¡¡¡£©
A£®C4H10B£®C5H8C£®CH2Br2D£®C2H4O2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

2£®Ä³ÌþXÍêȫȼÉÕʱÉú³ÉµÈÎïÖʵÄÁ¿µÄCO2ÓëH2O£¬1mol XÐèÒªÏûºÄ7.5mol O2£¬XÄÜʹäåµÄCCl4ÈÜÒºÍÊÉ«£¬ÔòX½á¹¹£¨²»¿¼ÂÇÁ¢ÌåÒì¹¹£©×î¶àÓУ¨¡¡¡¡£©
A£®5ÖÖB£®3ÖÖC£®7ÖÖD£®4ÖÖ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÏÂÁÐÎïÖÊÒ»¶¨ÊôÓÚ´¿¾»ÎïµÄÊÇ£¨¡¡¡¡£©
A£®C4H10B£®¾ÛÒÒÏ©C£®ÆûÓÍD£®±½

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

6£®ÒÔµí·ÛºÍÓÍ֬ΪԭÁÏ£¬ÖƱ¸Éú»îÖÐijЩÎïÖÊ£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µí·ÛµÄ·Ö×ÓʽΪ£¨C6H10O5£©n£®
£¨2£©FµÄ»¯Ñ§Ê½ÊÇC17H35COONa£®
£¨3£©E¿ÉÄܵĽṹ¼òʽΪ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

16£®¶ÔÓÚ·´Ó¦A2+3B2¨T2C+2D£¨A¡¢B¡¢C¡¢DȫΪÆøÌ壩À´Ëµ£¬ÒÔÏ»¯Ñ§·´Ó¦ËÙÂʵıíʾÖУ¬·´Ó¦ËÙÂÊ×î¿ìµÄÊÇ£¨¡¡¡¡£©
A£®v£¨B2£©=0.8 mol/£¨L•s£©B£®v£¨A2£©=0.4 mol/£¨L•s£©C£®v£¨C£©=0.6 mol/£¨L•s£©D£®v£¨D£©=1.2 mol/£¨L•s£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

3£®äå±½ÊÇÒ»ÖÖÎÞÉ«ÓÍ×´ÒºÌ壬ʵÑéÊÒÖÆÈ¡äå±½µÄ»¯Ñ§·½³ÌʽΪ£»·ÖÀëäå±½ºÍ±½»ìºÏÎïµÄ·½·¨ÎªÕôÁó£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®»¯Ñ§ÓëÉú²ú¡¢Éú»îÃÜÇÐÏà¹Ø£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÃºµÄ¸ÉÁóºÍʯÓ͵ķÖÁó¾ùÊô»¯Ñ§±ä»¯
B£®ÓýþÅݹý¸ßÃÌËá¼ØÈÜÒºµÄ¹èÔåÍÁÎüÊÕË®¹ûÊͷŵÄÒÒÏ©£¬¿Éµ½´ïË®¹û±£ÏʵÄÄ¿µÄ
C£®ÌìȻҩÎïÎÞÈκζ¾¸±×÷Ó㬿ɳ¤ÆÚ·þÓÃ
D£®¿ÉÒÔÓÃ×ÆÉյķ½·¨¼ø±ðÑòëºÍ²ÏË¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

5£®µç½âÄòËØ[CO£¨NH2£©2]ÊÇÒ»ÖÖÄܺĺܵ͵ÄÖÆÇâ·½·¨£¬Æ乤×÷Ô­ÀíÈçͼËùʾ£¬×Ü·´Ó¦ÎªCO£¨NH2£©2+2OH-$\frac{\underline{\;ͨµç\;}}{\;}$N2¡ü+3H2¡ü+CO32-£®ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®Òõ¼«µÄ²ÄÁÏ¿ÉÑ¡Ôñ¸ÖÌúÖÆÆ·
B£®µç½â³Ø¹¤×÷ʱ£¬Ñô¼«ÇøÓòÈÜÒºµÄpH¼õС
C£®Ñô¼«µÄµç¼«·´Ó¦Ê½ÎªCO£¨NH2£©2+8OH--6e-¨TN2¡ü+CO32-+6H2O
D£®Èôµç·ÖÐͨ¹ý3 mol µç×Ó£¬ÔòÉú³ÉÆøÌåµÄ×ÜÌå»ýΪ33.6L£¨±ê×¼×´¿ö£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸