11£®£¨1£©Ï¡ÊÍ0.1mol•L-1°±Ë®Ê±£¬Ëæ×ÅË®Á¿µÄÔö¼Ó¶ø¼õСµÄÊÇ¢Ù¢Ú£¨ÌîдÐòºÅ£©£®
¢Ù$\frac{c£¨N{H}_{3}•{H}_{2}O£©}{c£¨O{H}^{-}£©}$    ¢Ú$\frac{c£¨O{H}^{-}£©}{c£¨{H}^{+}£©}$      ¢Ûc£¨H+£©ºÍc£¨OH-£©µÄ³Ë»ý     ¢ÜOH-µÄÎïÖʵÄÁ¿
£¨2£©pHÏàͬµÄµÈÌå»ýµÄA¡¢BÁ½·ÝÈÜÒº£¨AΪÑÎËᣬBΪ´×Ëᣩ·Ö±ðÓëп·Û·´Ó¦£¬Èô×îºó½öÓÐÒ»·ÝÈÜÒºÖдæÔÚп·Û£¬ÇҷųöÇâÆøµÄÖÊÁ¿Ïàͬ£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊǢۢܢݣ¨ÌîдÐòºÅ£©£®
¢Ù·´Ó¦ËùÐèµÄʱ¼äB£¾A
¢Ú¿ªÊ¼·´Ó¦Ê±µÄËÙÂÊA£¾B
¢Û²Î¼Ó·´Ó¦µÄп·ÛÎïÖʵÄÁ¿A=B
¢Ü·´Ó¦¹ý³ÌÖеÄƽ¾ùËÙÂÊB£¾A
¢ÝAÈÜÒºÀïÓÐп·ÛÊ£Óà
¢ÞBÈÜÒºÀïÓÐп·ÛÊ£Óà
£¨3£©½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ°±Ë®ºÍÑÎËá»ìºÏºó£¬Éý¸ßζȣ¨ÈÜÖʲ»»á·Ö½â£©ÈÜÒºµÄpHËæζȱ仯ÊÇͼÖеĢÜÇúÏߣ¨ÌîдÐòºÅ£©£®
£¨4£©ÊÒÎÂÏ£¬Ïò0.1mol•L-1µÄÑÎËáÓë0.2mol•L-1µÄ°±Ë®µÈÌå»ý»ìºÏ£¬ÔòËùµÃÈÜÒºÖÐËùÓÐÀë×ÓÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨NH4+£©£¾c£¨Cl-£©£¾c£¨OH-£©£¾c£¨H+£©£®
£¨5£©ÓɵçÀë³£Êý¿ÉÖªËáÐÔÇ¿Èõ£ºCH3COOH£¾H2CO3£¾HCO3-£¬ÔòŨ¶ÈÏàͬµÄÏÂÁÐÈÜÒº¼îÐÔÓÉÇ¿µ½Èõ˳ÐòΪ£º¢Û¢Ú¢Ù£¨ÌîÐòºÅ£©
¢ÙCH3COONa   ¢ÚNaHCO3   ¢ÛNa2CO3
£¨6£©ÂÈ»¯ÌúÈÜÒº³ÊËáÐÔµÄÔ­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£ºFe3++3H2O?Fe£¨OH£©3+3H+£®
ij¶þÔªËᣨ»¯Ñ§Ê½ÓÃH2B±íʾ£©ÔÚË®ÖеĵçÀë·½³ÌʽÊÇ£ºH2B=H++HB-£»HB-?H++B2-
Íê³ÉÏÂÁÐÎÊÌ⣺ÔÚ0.1mol/L µÄNa2BÈÜÒºÖУ¬ÏÂÁÐÁ£×ÓŨ¶È¹ØϵʽÕýÈ·µÄÊÇC£®
A£®c£¨B2-£©+c£¨HB-£©+c£¨H2B£©=0.1mol/L
B£®c£¨Na+£©+c£¨OH-£©=c£¨H+£©+c£¨HB-£©
C£®c£¨Na+£©+c£¨H+£©=c£¨OH-£©+c£¨HB-£©+2c£¨B2-£©
D£®2c£¨Na+£©=c£¨B2-£©+c£¨HB-£©

·ÖÎö £¨1£©¼ÓˮϡÊÍ´Ù½øһˮºÏ°±µçÀ룬ÈÜÒºÖÐc£¨OH-£©¡¢c£¨NH3£®H2O£©¡¢c£¨NH4+£©¶¼¼õС£¬Î¶Ȳ»±ä£¬Ë®µÄÀë×Ó»ý³£Êý²»±ä£¬Ôòc£¨H+£©Ôö´ó£»
£¨2£©pHÏàͬµÄµÈÌå»ýµÄA¡¢BÁ½·ÝÈÜÒº£¨AΪÑÎËᣬBΪ´×Ëᣩ·Ö±ðÓëп·Û·´Ó¦£¬Èô×îºó½öÓÐÒ»·ÝÈÜÒºÖдæÔÚп·Û£¬ÇҷųöÇâÆøµÄÖÊÁ¿Ïàͬ£¬ÓÉÓÚ´×ËᲿ·ÖµçÀ룬ÆäËáµÄŨ¶È´óÓÚÇâÀë×ÓŨ¶È£¬ÑÎËáÖÐHClµÄŨ¶ÈµÈÓÚÇâÀë×ÓµÄŨ¶È£¬ËùÒÔ´×ËáµÄŨ¶È´óÓÚHClµÄŨ¶È£¬ÔòÑÎËáÖÐп·ÛÊ£Óࣻ
£¨3£©µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ°±Ë®ºÍÑÎËá»ìºÏºóÇ¡ºÃÉú³ÉÂÈ»¯ï§ÈÜÒº£¬ï§¸ùÀë×ÓË®½âÏÔËáÐÔ£¬¼ÓÈÈ´Ù½øË®½â£»
£¨4£©ÊÒÎÂÏ£¬Ïò0.1mol•L-1µÄÑÎËáÓë0.2mol•L-1µÄ°±Ë®µÈÌå»ý»ìºÏµÃµ½ÈÜҺΪµÈŨ¶ÈµÄÂÈ»¯ï§ºÍһˮºÏ°±µÄ»ìºÏÈÜÒº£¬Ò»Ë®ºÏ°±µçÀë´óÓÚ笠ùÀë×ÓË®½â£¬ÈÜÒºÏÔ¼îÐÔ£»
£¨5£©¸ù¾ÝÑεÄË®½â¹æÂÉ£ºÔ½ÈõԽˮ½â½áºÏÎïÖÊÖÐË®½âµÄÔ­ÀíÀ´»Ø´ð£»
£¨6£©ÂÈ»¯ÌúΪǿËáÈõ¼îÑΣ¬ÌúÀë×ÓË®½âµ¼ÖÂÈÜÒº³ÊËáÐÔ£»Èκεç½âÖÊÈÜÒºÖж¼´æÔÚµçºÉÊغãºÍÎïÁÏÊغ㣬¾Ý´Ë·ÖÎö½â´ð£®

½â´ð ½â£º£¨1£©¼ÓˮϡÊÍ´Ù½øһˮºÏ°±µçÀ룬ÈÜÒºÖÐÇâÑõ¸ùÀë×ÓŨ¶È¡¢Ò»Ë®ºÏ°±Å¨¶È¡¢ï§¸ùÀë×ÓŨ¶È¶¼¼õС£¬µ«c£¨H+£©Ôö´ó£¬
¢Ù£®¼ÓˮϡÊÍ´Ù½øһˮºÏ°±µçÀ룬ÇâÑõ¸ùÀë×ÓÊýÄ¿Ôö´ó£¬Ò»Ë®ºÏ°±·Ö×ÓÊý¼õС£¬Í¬ÈÜÒºÖÐÌå»ý²»±ä£¬Ôò$\frac{c£¨N{H}_{3}•{H}_{2}O£©}{c£¨O{H}^{-}£©}$¼õС£¬¹Ê¢ÙÕýÈ·£»
¢Ú£®¼ÓˮϡÊÍ´Ù½øһˮºÏ°±µçÀ룬ÇâÑõ¸ùÀë×ÓŨ¶È¼õС£¬Î¶Ȳ»±ä£¬Ë®µÄÀë×Ó»ý³£Êý²»±ä£¬ÔòÇâÀë×ÓŨ¶ÈÔö´ó£¬ËùÒÔ$\frac{c£¨O{H}^{-}£©}{c£¨{H}^{+}£©}$ ¼õС£¬¹Ê¢ÚÕýÈ·£»
¢Û£®Î¶Ȳ»±ä£¬Ë®µÄÀë×Ó»ý³£Êý²»±ä£¬ËùÒÔc£¨H+£©ºÍc£¨OH-£©µÄ³Ë»ý²»±ä£¬¹Ê¢Û´íÎó£»
¢Ü£®¼ÓˮϡÊÍ´Ù½øһˮºÏ°±µçÀ룬ÔòOH-µÄÎïÖʵÄÁ¿Ôö´ó£¬¹Ê¢Ü´íÎó£»
¹Ê´ð°¸Îª£º¢Ù¢Ú£»
£¨2£©pHÏàͬµÄµÈÌå»ýµÄA¡¢BÁ½·ÝÈÜÒº£¨AΪÑÎËᣬBΪ´×Ëᣩ·Ö±ðÓëп·Û·´Ó¦£¬Èô×îºó½öÓÐÒ»·ÝÈÜÒºÖдæÔÚп·Û£¬ÇҷųöÇâÆøµÄÖÊÁ¿Ïàͬ£¬ÓÉÓÚ´×ËᲿ·ÖµçÀ룬ÆäËáµÄŨ¶È´óÓÚÇâÀë×ÓŨ¶È£¬ÑÎËáÖÐHClµÄŨ¶ÈµÈÓÚÇâÀë×ÓµÄŨ¶È£¬ËùÒÔ´×ËáµÄŨ¶È´óÓÚHClµÄŨ¶È£¬ÔòÑÎËáÖÐп·ÛÊ£Óࣻ
¢ÙÓÉÓÚ´×ËáÖÐËá¹ýÁ¿£¬Ôò·´Ó¦½Ï¿ì£¬ËùÒÔ·´Ó¦ËùÐèµÄʱ¼äA£¾B£»¹Ê´íÎó£»
¢Ú¿ªÊ¼pHÏàͬ£¬ÔòÇâÀë×ÓŨ¶ÈÏàͬ£¬ËùÒÔ¿ªÊ¼Ê±·´Ó¦ËÙÂÊA=B£¬¹Ê´íÎó£»
¢ÛÓÉÓÚÉú³ÉµÄÇâÆøÌå»ýÏàͬ£¬ËùÒԲμӷ´Ó¦µÄп·ÛÎïÖʵÄÁ¿A=B£¬¹ÊÕýÈ·£»
¢Ü´×ËáµÄŨ¶È´óÓÚÑÎËáµÄŨ¶È£¬Ôò´×ËáÖз´Ó¦ËÙÂÊ´ó£¬ËùÒÔ·´Ó¦¹ý³ÌÖеÄƽ¾ùËÙÂÊ B£¾A£¬¹ÊÕýÈ·£»
¢Ý´×ËáµÄŨ¶È´óÓÚÑÎËáµÄŨ¶È£¬´×ËáÓÐÊ£Ó࣬ÔòÑÎËáÖÐÓÐп·ÛÊ£Ó࣬¹ÊÕýÈ·£»
¢Þ´×ËáµÄŨ¶È´óÓÚÑÎËáµÄŨ¶È£¬´×ËáÓÐÊ£Ó࣬ÔòÑÎËáÖÐÓÐп·ÛÊ£Ó࣬¹Ê´íÎó£»
¹Ê´ð°¸Îª£º¢Û¢Ü¢Ý£»
£¨3£©µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ°±Ë®ºÍÑÎËá»ìºÏºóÇ¡ºÃÉú³ÉÂÈ»¯ï§ÈÜÒº£¬NH4+Ë®½âÈÜÒºÏÔËáÐÔ£¬PH£¼7£¬¢Ù¢ÚpH´óÓÚ7£¬¹Ê¢Ù¢Ú´íÎó£»
¼ÓÈÈ´Ù½øË®½â£¬¼ÓÈÈË®½âƽºâÏòÓÒÒƶ¯£¬c£¨H+£©Ôö´ó£¬ËáÐÔÔöÇ¿£¬PH¼õС£¬¢Û´íÎó¡¢¢ÜÕýÈ·£¬
¹Ê´ð°¸Îª£º¢Ü£»
£¨4£©ÊÒÎÂÏ£¬Ïò0.1mol•L-1µÄÑÎËáÓë0.2mol•L-1µÄ°±Ë®µÈÌå»ý»ìºÏµÃµ½ÈÜҺΪµÈŨ¶ÈµÄÂÈ»¯ï§ºÍһˮºÏ°±µÄ»ìºÏÈÜÒº£¬Ò»Ë®ºÏ°±µçÀë´óÓÚ笠ùÀë×ÓË®½â£¬ÈÜÒºÏÔ¼îÐÔ£¬ÈÜÒºÖÐÀë×ÓŨ¶È´óСΪ£ºc£¨NH4+£©£¾c£¨Cl-£©£¾c£¨OH-£©£¾c£¨H+£©£¬
¹Ê´ð°¸Îª£ºc£¨NH4+£©£¾c£¨Cl-£©£¾c£¨OH-£©£¾c£¨H+£©£»
£¨5£©ÓɵçÀë³£Êý¿ÉÖªËáÐÔÇ¿Èõ£ºCH3COOH£¾H2CO3£¾HCO3-£¬ËùÒÔË®½â³Ì¶È£ºÌ¼Ëá¸ùÀë×Ó£¾Ì¼ËáÇâ¸ù£¾´×Ëá¸ùÀë×Ó£¬ËùÒÔ¢ÙCH3COONa ¢ÚNaHCO3¡¡¢ÛNa2CO3µÄ¼îÐÔ˳ÐòÊÇ£º
¢Û£¾¢Ú£¾¢Ù£¬¹Ê´ð°¸Îª£º¢Û£¾¢Ú£¾¢Ù£»
£¨6£©ÂÈ»¯ÌúΪǿËáÈõ¼îÑΣ¬ÌúÀë×ÓË®½âµ¼ÖÂÈÜÒº³ÊËáÐÔ£¬Ë®½â·½³ÌʽΪFe3++3H2O?Fe£¨OH£©3+3H+£»
A£®H2BµÚÒ»²½ÍêÈ«µçÀ룬ËùÒÔÈÜÒºÖв»´æÔÚH2B£¬Ó¦¸ÃΪc£¨B2-£©+c£¨HB- £©=0.1mol/L£¬¹ÊA´íÎó£»
B£®H2BµÚÒ»²½ÍêÈ«µçÀ룬ËùÒÔÈÜÒºÖв»´æÔÚH2B£¬ÈÜÒºÖдæÔÚÖÊ×ÓÊغ㣬¸ù¾ÝÖÊ×ÓÊغãµÃc£¨OH-£©=c£¨H+£©+c£¨HB-£©£¬¹ÊB´íÎó£»
C£®¸ù¾ÝµçºÉÊغãµÃc£¨Na+£©+c£¨H+ £©=c£¨OH- £©+c£¨HB-£©+2c£¨B2- £©£¬¹ÊCÕýÈ·£»
D£®¸ù¾ÝÎïÁÏÊغãµÃc£¨Na+£©=2c£¨B2- £©+2c£¨HB- £©£¬¹ÊD´íÎó£»
¹Ê´ð°¸Îª£ºFe3++3H2O?Fe£¨OH£©3+3H+£¬C£®

µãÆÀ ±¾Ì⿼²éµç½âÖÊÈÜÒºÖ®¼äµÄ·´Ó¦£¬Éæ¼°µ½Ç¿µç½âÖÊ¡¢Èõµç½âÖʵĵçÀë¡¢ÑÎÀàµÄË®½â¼°ÈÜÒºµÄpHÖµ¡¢Àë×ÓŨ¶È´óСµÄ±È½Ï¡¢µçÀëƽºâ³£ÊýµÄ¼ÆËãµÈ֪ʶ£¬ÌâÄ¿Éæ¼°µÄ֪ʶµã½Ï¶à£¬×ÛºÏÐÔ½ÏÇ¿£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®ÏÂÁÐÁ£×ÓÔÚ»¯Ñ§·´Ó¦ÖмÈÄÜÏÔʾÑõ»¯ÐÔÓÖÄÜÏÔʾ»¹Ô­ÐÔµÄÊÇ£¨¡¡¡¡£©
A£®Al3+B£®Fe2+C£®AlD£®Br-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

2£®Ä³Ç¿ËáÐÔÈÜÒºX£¬¿ÉÄܺ¬ÓÐBa2+¡¢Al3+¡¢NH4+¡¢Fe2+¡¢Fe3+¡¢CO32-¡¢SO42-¡¢Cl-¡¢NO3-ÖеÄÒ»ÖÖ»ò¼¸ÖÖÀë×Ó£¬È¡¸ÃÈÜÒº½øÐÐÁ¬ÐøʵÑ飬ʵÑé¹ý³ÌÈçÏ£º

¸ù¾ÝÒÔÉÏÐÅÏ¢£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÏÊöÀë×ÓÖУ¬ÈÜÒºXÖгýH+Íâ¿Ï¶¨º¬ÓеÄÀë×ÓÊÇAl3+¡¢NH4+¡¢Fe2+¡¢SO42-£¬²»ÄÜÈ·¶¨ÊÇ·ñº¬ÓеÄÀë×ÓÊÇFe3+¡¢Cl-£®
£¨2£©Ð´³ö·´Ó¦¢ÚµÄÀë×Ó·½³Ìʽ£ºAlO2-+CO2+2H2O=Al£¨OH£©3¡ý+HCO3-£®
£¨3£©Í¨³£¿ÉÒÔÀûÓÃKClOÔÚÒ»¶¨Ìõ¼þÏÂÑõ»¯GÀ´ÖƱ¸Ò»ÖÖÐÂÐÍ¡¢¸ßЧ¡¢¶à¹¦ÄÜË®´¦Àí¼ÁK2FeO4£®Çëд³öÖƱ¸¹ý³ÌÖеÄÀë×Ó·½³Ìʽ3ClO-+2Fe£¨OH£©3+4OH-¨T3Cl-+2FeO42-+5H2O£®
£¨4£©¼ÙÉè²â¶¨A¡¢F¡¢I¾ùΪ0.1mol£¬10mLXÈÜÒºÖÐn£¨H+£©=0.4mol£¬µ±³ÁµíCÎïÖʵÄÁ¿µÈÓÚ0.8molʱ£¬ÈÜÒºXÖл¹Ò»¶¨º¬ÓÐFe3+£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

19£®¸÷ÖÖ±ØÐèÔªËØÔÚÈËÌåÄڵĺ¬Á¿¶¼ÓÐÒ»¸ö×î¼Ñ·¶Î§£¬¹ý¸ß»ò¹ýµÍ¶¼ÓпÉÄÜÓ°ÏìÈËÕý³£µÄÉúÀí»úÄÜ£®ÔÚÈÕ³£Éú»îÖÐӦעÒâºÏÀíµÄÉãÈ¡ÈËÌå±ØÐèµÄÔªËØ£®
£¨1£©µ°°×ÖÊÊǹ¹³ÉÉúÃüµÄ»ù´¡ÎïÖÊ£¬Ê¹ÈÕ³£ÉÅʳµÄÖØÒª×é³É²¿·Ö£®ÏÂÁÐʳÎïÖи»º¬µ°°×ÖʵÄÊÇ¢Û£®
¢ÙÆ»¹û          ¢ÚÆÏÌѸɠ          ¢ÛÅ£ÄÌ              ¢Ü°×²Ë
£¨2£©µ±ÈËÌåȱÉÙµâÔªËØ»òµâÔªËغ¬Á¿¹ý¸ßʱ£¬¶¼»áÒýÆð¼××´ÏÙ¼²²¡£®ÓÉÓÚÎÒ¹ú¾ÓÃñÿÌìÉãÈëµÄµâÔªËز»×㣬Òò´ËÎÒ¹úÕþ¸®´Ó1994ÄêÆðÍƳöÈ«ÃñʳÓüӵâÑι¤³Ì£¬Ä¿Ç°¡°¼ÓµâÑΡ±Ö÷ÒªÊÇÔÚʵÑéÖмÓÈëµâËá¼Ø£¬µâËá¼ØµÄ»¯Ñ§Ê½ÎªKIO3£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

6£®ÔËÓÃËùѧ֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣮
 £¨1£©0.5mol CH4Öк¬ÓÐ0.5NA¸ö¼×Íé·Ö×Ó£¨°¢·ü¼ÓµÂÂÞ³£ÊýÓÃNA±íʾ£©£¬ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ11.2L£®0.1molH2SO4Öк¬0.2molHÔ­×Ó£¬ÈÜÓÚË®Åä³É100mLË®ÈÜÒººóºóC£¨H+£©=2mol/L£®
 £¨2£©ÔÚ·´Ó¦3Zn+2Fe3+¨T3Zn2++2FeÖУ¬Fe3+ÊÇÑõ»¯£¨Ìî¡°Ñõ»¯¡±»ò¡°»¹Ô­¡±£©¼Á£¬1molZnÍêÈ«·´Ó¦×ªÒƵĵç×ÓÊýΪ2mol£®
£¨3£©Ñ¡ÔñÏÂÁÐʵÑé·½·¨·ÖÀëÎïÖÊ£¬½«·ÖÀë·½·¨µÄÐòºÅÌîÔÚºáÏßÉÏ£¨µ¥Ñ¡£©£®
A£®ÝÍÈ¡·ÖÒº    B£®Éý»ª    C£®½á¾§    D£®·ÖÒº   E£®ÕôÁó    F£®¹ýÂË
 ¢Ù·ÖÀë´ÖÑÎÖлìÓеÄÄàɳF£®¢Ú·ÖÀëµâºÍË®µÄ»ìºÏÎïA£®
 ¢Û·ÖÀëË®ºÍÆûÓ͵ĻìºÏÎïD£®¢Ü·ÖÀë¾Æ¾«ºÍË®µÄ»ìºÏÎïE£®
 £¨4£©ÓÐÏÂÁÐÈý×éÎïÖÊ¢ÙCaO¡¢SO3¡¢SO2¡¢P2O5 ¢ÚHNO3¡¢H2SO4¡¢HCl¡¢NaCl¢ÛKClO3¡¢KCl¡¢KMnO4¡¢HgO¸÷×éÖоùÓÐÒ»ÖÖÎïÖÊËùÊôÀà±ðÓëÆäËûÎïÖʲ»Í¬£¬ÕâÈýÖÖÎïÖÊ·Ö±ðÊÇ£º¢ÙCaO£»¢ÚNaCl£»¢ÛHgO£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

16£®ÏÂÁÐÈÜÒºÂÈÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÓë50mL 1mol•L-1AlCl3ÈÜÒºÖÐÂÈÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄÊÇ£¨¡¡¡¡£©
A£®150 mL 3 mol•L-1ÂÈËá¼ØÈÜÒºB£®75 mL 3 mol•L-1ÂÈ»¯¸ÆÈÜÒº
C£®150 mL 3 mol•L-1ÂÈ»¯¼ØÈÜÒºD£®50 mL 3 mol•L-1ÂÈ»¯Ã¾ÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

3£®ÒÑÖªA¡¢B¡¢C¡¢DÊÇÖÐѧ»¯Ñ§³£¼ûÎïÖÊ£¬ËüÃÇÔÚÒ»¶¨Ìõ¼þÏÂÓУºA+B¡úC+DµÄת»¯¹Øϵ£®
£¨1£©ÈôAΪ½ðÊôÂÁ£¬BΪÑõ»¯Ìú£¬¸Ã·´Ó¦µÄÒ»ÖÖÓÃ;ÊǺ¸½Ó¸Ö¹ì£®
£¨2£©ÈôAÊÇÒ»ÖÖÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬ÇҸ÷´Ó¦Êǹ¤ÒµÉÏÖÆÈ¡ÏõËáµÄÖØÒª·´Ó¦Ö®Ò»£¬¸Ã·´Ó¦µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ4NH3+5O2$\frac{\underline{´ß»¯¼Á}}{¡÷}$4NO+6H2O£®
£¨3£©ÈôAÊǵ­»ÆÉ«·ÛÄ©£¬³£ÓÃ×÷¹©Ñõ¼Á£¬CΪǿ¼î£¬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ2Na2O2+2H2O=4NaOH+O2¡ü£®
£¨4£©ÈôA¡¢B¡¢D¶¼ÊÇÓлú»¯ºÏÎÆäÖÐA¡¢BÊǼÒÍ¥³ø·¿Öг£¼ûµ÷ζƷµÄÖ÷Òª³É·Ö£¬ÇÒAµÄÏà¶Ô·Ö×ÓÖÊÁ¿±ÈB´ó14£®¸Ã·´Ó¦µÄ»¯Ñ§·´Ó¦·½³ÌʽΪCH3COOH+C2H5OH$¡ú_{¡÷}^{ŨH_{2}SO_{4}}$CH3COOC2H5+H2O£¬Æä·´Ó¦ÀàÐÍΪȡ´ú£¨õ¥»¯£©·´Ó¦£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®ÊµÏÖÏÂÁб仯ʱ£¬Ðè¿Ë·þÏàͬÀàÐÍ×÷ÓÃÁ¦µÄÊÇ£¨¡¡¡¡£©
A£®¶þÑõ»¯¹èºÍ¸É±ùµÄÈÛ»¯B£®ÒºäåºÍÒº¹¯µÄÆø»¯
C£®Ê³ÑκͱùµÄÈÛ»¯D£®´¿¼îºÍÉÕ¼îµÄÈÛ»¯

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®ºãÎÂÏ£¬ÔÚÒ»ÃܱÕÈÝÆ÷ÖУ¬N2£¨g£©+3H2£¨g£©?2NH3£¨g£©´ïµ½Æ½ºâºó£¬²âµÃc£¨H2£©=2mol•L-1£¬N2µÄÌå»ý·ÖÊýΪ20%£¬ÏÂÁÐÓйØ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®½«ÈÝÆ÷ÈÝ»ýѹËõÒ»°ë£¬Æ½ºâÏòÓÒÒƶ¯£¬´ïµ½ÐÂƽºâʱc£¨H2£©£¼2mol•L-1
B£®ÏòÈÝÆ÷ÖмÓÈë0.1molN2£¬Æ½ºâÏòÓÒÒƶ¯£¬´ïµ½ÐÂƽºâʱN2µÄÌå»ý·ÖÊýСÓÚ20%
C£®Èô´ïµ½ÐÂƽºâʱc£¨H2£©=2.5mol•L-1£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊǽµÎ»ò¼Óѹ
D£®Èô¸Ä±äÌõ¼þºó£¬H2µÄ°Ù·Öº¬Á¿Ôö´ó£¬ÔòƽºâÒ»¶¨ÏòÄæ·´Ó¦·½ÏòÒƶ¯

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸