12£®ÊÒÎÂÏ£¬½«0.1mol£®L-1ÑÎËáµÎÈë20mL 0.1mol£®L-1°±Ë®ÖУ¬ÈÜÒºpHËæ¼ÓÈëÑÎËáÌå»ýµÄ±ä»¯ÇúÏßÈçͼ1Ëùʾ£®

£¨1£©NH3£®H2OµÄµçÀë·½³ÌʽÊÇNH3•H2O?NH4++OH-£®
£¨2£©bµãËùʾÈÜÒºÖÐc£¨Cl-£©=c£¨NH4+£©£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
£¨3£©cµãËùʾÈÜÒºpH£¼7£¬Ô­ÒòÊÇNH4++H2O?NH3•H2O+H+£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£®
£¨4£©dµãËùʾÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄÅÅÐòÊÇc£¨Cl-£©£¾c£¨H+£©£¾c£¨NH4+£©£¾c£¨OH-£©£®
ËÑË÷Ïà¹Ø×ÊÁÏ 25£®£¨7·Ö£©¶þÔªÈõËáÊÇ·Ö²½µçÀëµÄ£¬25ʱ̼ËáºÍ²ÝËáµÄKaÈç±í£º
H2CO3   Ka1=4.3¡Á10-7         H2C2O4  Ka1=5.6¡Á10-2
Ka2=5.6¡Á10-11                Ka2=5.42¡Á10-5
£¨5£©ÉèÓÐÏÂÁÐËÄÖÖÈÜÒº£º
A£®0.1mol£®L-1µÄNa2C2O4ÈÜÒº          B£®0.1mol£®L-1µÄNaHC2O4ÈÜÒº
C£®0.1mol£®L-1µÄNa2CO3ÈÜÒº           D£®0.1mol£®L-1µÄNaHCO3ÈÜÒº
ÆäÖУ¬c£¨H+£©×î´óµÄÊÇB£¬c£¨OH-£©×î´óµÄÊÇC£®
£¨6£©Ä³»¯Ñ§ÊµÑéÐËȤС×éͬѧÏòÓôóÀíʯºÍÏ¡ÑÎËáÖÆ±¸CO2ºó²ÐÁôÒºÖеμÓ̼ËáÄÆÈÜÒº£¬ÔÚÈÜÒºÖвåÈËpH´«¸ÐÆ÷£¬²âµÃpH±ä»¯ÇúÏßÈçͼ2Ëùʾ£®
¸Õ¿ªÊ¼µÎÈË̼ËáÄÆÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ2H++CO32-=H2O+CO2¡ü£¬BC¶Î·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪCa2++CO32-=CaCO3¡ý£¬Dµãʱ»ìºÏÈÜÒºÖÐÓÉË®µçÀë²úÉúµÄc£¨OH-£©=10-4 mol£®L-1£®

·ÖÎö £¨1£©Ò»Ë®ºÏ°±ÊÇÈõµç½âÖÊ£¬ÔÚË®ÈÜÒºÀï´æÔÚµçÀëÆ½ºâ£»
£¨2£©pH=7ʱ£¬ÈÜÒºÖÐcc£¨H+£©=c£¨OH-£©£¬ÈÜÒºÖдæÔÚµçºÉÊØºã£¬ÔÙ½áºÏµçºÉÊØºãÅжÏc£¨Cl-£©ºÍc£¨NH4+£©µÄÏà¶Ô´óС£»
£¨3£©cµãʱ£¬¶þÕßÇ¡ºÃ·´Ó¦Éú³ÉÂÈ»¯ï§£¬ÂÈ»¯ï§ÊÇÇ¿ËáÈõ¼îÑΣ¬ï§¸ùÀë×ÓË®½â¶øÊ¹ÆäÈÜÒº³ÊËáÐÔ£»
£¨4£©dµãʱ£¬ÈÜÒºÖеÄÈÜÖÊΪµÈÎïÖʵÄÁ¿Å¨¶ÈµÄÂÈ»¯ï§ºÍÑÎËᣬÈÜÒº³ÊËáÐÔ£¬ÂÈ»¯ÇâÍêÈ«µçÀ룬笠ùÀë×ÓÄÜË®½â£¬¾Ý´ËÅжÏÈÜÒºÖÐÀë×ÓŨ¶È´óС£»
£¨5£©ÏàͬÌõ¼þÏ£¬ËáµÄµçÀëÆ½ºâ³£ÊýÔ½´ó£¬ÔòËáµÄËáÐÔԽǿ£¬Ëá¸ùÀë×ÓµÄË®½â³Ì¶ÈԽС£¬ÔòÈÜÒºµÄ¼îÐÔÔ½Èõ£»
£¨6£©ÈÜÒºpH£¼7£¬ËµÃ÷ÈÜÒº³ÊËáÐÔ£¬ËáºÍ̼ËáÄÆÈÜÒº·´Ó¦Éú³ÉÂÈ»¯ÄƺÍË®¡¢¶þÑõ»¯Ì¼£¬µ±pH²»±äʱ£¬Ì¼ËáÄÆºÍÂÈ»¯¸Æ·¢Éú¸´·Ö½â·´Ó¦£¬Ì¼ËáÄÆÎªÇ¿¼îÈõËáÑΣ¬ÆäÈÜÒº³Ê¼îÐÔ£¬¸ù¾ÝË®µÄÀë×Ó»ý³£Êý¼ÆËãÇâÑõ¸ùÀë×ÓŨ¶È£®

½â´ð ½â£º£¨1£©Ò»Ë®ºÏ°±ÊÇÈõµç½âÖÊ£¬ÔÚË®ÈÜÒºÀïÖ»Óв¿·ÖµçÀ룬µçÀë³öÇâÑõ¸ùÀë×ÓºÍ笠ùÀë×Ó£¬Ò»Ë®ºÏ°±µÄµçÀë·½³ÌʽΪ£ºNH3•H2O?NH4++OH-£¬
¹Ê´ð°¸Îª£ºNH3•H2O?NH4++OH-£»       
£¨2£©bµãʱpH=7£¬ÔòÈÜÒºÖÐcc£¨H+£©=c£¨OH-£©£¬ÈÜÒºÖдæÔÚµçºÉÊØºã£¬¸ù¾ÝµçºÉÊØºãµÃc£¨Cl-£©+c£¨OH-£©=c£¨NH4+£©+cc£¨H+£©£¬ËùÒÔµÃc£¨Cl-£©=c£¨NH4+£©£¬¹Ê´ð°¸Îª£º=£»
£¨3£©cµãʱ¶þÕßÇ¡ºÃ·´Ó¦Éú³ÉÂÈ»¯ï§£¬ÂÈ»¯ï§ÊÇÇ¿ËáÈõ¼îÑΣ¬ï§¸ùÀë×ÓË®½â¶øÊ¹ÈÜÒºÖÐÇâÀë×ÓŨ¶È´óÓÚÇâÑõ¸ùÀë×ÓŨ¶È£¬ÔòÈÜÒº³ÊËáÐÔ£¬Ë®½âÀë×Ó·½³ÌʽΪ£ºNH4++H2O?NH3•H2O+H+£¬
¹Ê´ð°¸Îª£ºNH4++H2O?NH3•H2O+H+£»     
£¨4£©dµãʱ£¬ËáµÄÎïÖʵÄÁ¿Êǰ±Ë®µÄ2±¶£¬¶þÕß»ìºÏʱ£¬ÈÜÒºÖеÄÈÜÖÊΪµÈÎïÖʵÄÁ¿Å¨¶ÈµÄÂÈ»¯ï§ºÍÑÎËᣬÈÜÒº³ÊËáÐÔ£¬ÂÈ»¯ÇâÍêÈ«µçÀ룬笠ùÀë×ÓË®½âµ«Ë®½â³Ì¶È½ÏС£¬½áºÏÎïÁÏÊØºãÖª£¬ÈÜÒºÖÐÀë×ÓŨ¶È´óС˳ÐòÊÇ
c £¨Cl-£©£¾c £¨H+£©£¾c £¨NH4+£©£¾c £¨OH-£©£¬
¹Ê´ð°¸Îª£ºc£¨Cl-£©£¾c£¨H+£©£¾c£¨NH4+£©£¾c£¨OH-£©£»
£¨5£©ÏàͬÌõ¼þÏ£¬ËáµÄµçÀëÆ½ºâ³£ÊýÔ½´ó£¬ÔòËáµÄËáÐÔԽǿ£¬Ëá¸ùÀë×ÓµÄË®½â³Ì¶ÈԽС£¬ÔòÈÜÒºÖÐÇâÑõ¸ùÀë×ÓŨ¶ÈԽС£¬ÇâÀë×ÓŨ¶ÈÔ½´ó£¬¸ù¾ÝµçÀëÆ½ºâ³£ÊýÖª£¬ËáÐÔÇ¿Èõ˳ÐòÊÇ£º²ÝË᣾²ÝËáÇâ¸ùÀë×Ó£¾Ì¼Ë᣾̼ËáÇâ¸ùÀë×Ó£¬Àë×ÓË®½âÇ¿Èõ˳ÐòÊÇ£ºÌ¼Ëá¸ùÀë×Ó£¾Ì¼ËáÇâ¸ùÀë×Ó£¾²ÝËá¸ùÀë×Ó£¾´×ËáÇâ¸ùÀë×Ó£¬
¸ù¾ÝÀë×ÓË®½â³Ì¶ÈÖª£¬²ÝËáÇâ¸ùÀë×ÓË®½â³Ì¶È×îС£¬ÔòÆäÈÜÒº¼îÐÔ×îÈõ£¬ÇâÀë×ÓŨ¶È×î´ó£¬ËùÒÔÇâÀë×ÓŨ¶È×î´óµÄÊÇB£¬Ë®½â³Ì¶È×îÇ¿µÄÊÇC£¬ÔòÈÜÒºCÖмîÐÔ×îÇ¿£¬ÇâÑõ¸ùÀë×ÓŨ¶È×î´ó£¬
¹Ê´ð°¸Îª£ºB£»C£»
£¨6£©ÈÜÒºpH£¼7£¬ËµÃ÷ÈÜÒº³ÊËáÐÔ£¬ËáºÍ̼ËáÄÆÈÜÒº·´Ó¦Éú³ÉÂÈ»¯ÄƺÍË®¡¢¶þÑõ»¯Ì¼£¬Àë×Ó·½³ÌʽΪ£º2H++CO32-=H2O+CO2¡ü£¬µ±pH²»±äʱ£¬Ì¼ËáÄÆºÍÂÈ»¯¸Æ·¢Éú¸´·Ö½â·´Ó¦£¬Àë×Ó·½³ÌʽΪCa2++CO32-=CaCO3¡ý£¬Ì¼ËáÄÆÎªÇ¿¼îÈõËáÑΣ¬ÆäÈÜÒº³Ê¼îÐÔ£¬Dµãʱ»ìºÏÈÜÒºÖÐÓÉË®µçÀë²úÉúµÄc£¨OH-£©=$\frac{1{0}^{-14}}{1{0}^{-10}}$mol/L=10-4 mol/L£¬
¹Ê´ð°¸Îª£º2H++CO32-=H2O+CO2¡ü£»Ca2++CO32-=CaCO3¡ý£»10-4£®

µãÆÀ ±¾Ì⿼²éÁËËá¼î»ìºÏÈÜÒº¶¨ÐÔÅжÏÓë¼ÆË㣬ÌâÄ¿ÄѶÈÖеȣ¬Ã÷È·ÈÜÒºÖеÄÈÜÖʼ°ÆäÐÔÖÊÊǽⱾÌâ¹Ø¼ü£¬ÔÙ½áºÏÈÜÒºµÄËá¼îÐÔ¼°ÎïÁÏÊØºãºÍµçºÉÊØºãÀ´·ÖÎö½â´ð£¬ÊÔÌâÅàÑøÁËѧÉúµÄ·ÖÎöÄÜÁ¦¼°»¯Ñ§¼ÆËãÄÜÁ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

2£®ÏÂÁйØÓÚÄÆµÄ»¯ºÏÎïµÄ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®µÈÎïÖʵÄÁ¿NaHCO3¡¢Na2CO3·Ö±ðÓëͬŨ¶ÈÑÎËáÍêÈ«·´Ó¦£¬ÏûºÄµÄÑÎËáÌå»ýNa2CO3ÊÇNaHCO3µÄ¶þ±¶
B£®Na2O2ºÍNa2O¾ù¿ÉÒÔºÍÑÎËá·´Ó¦Éú³ÉÏàÓ¦µÄÑΣ¬¶¼ÊôÓÚ¼îÐÔÑõ»¯Îï
C£®½«³ÎÇåʯ»ÒË®·Ö±ð¼ÓÈëNaHCO3ºÍNa2CO3Á½ÖÖÑÎÈÜÒºÖУ¬Ö»ÓÐNa2CO3ÈÜÒº²úÉú³Áµí
D£®Na2O2ºÍNa2OÖÐNa2O¸üÎȵ±£¬Na2O2ÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔת»¯ÎªNa2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

3£®Ê¹ÓÃʯÓÍÈÈÁѽâµÄ¸±²úÎïCH4À´ÖÆÈ¡COºÍH2£¬ÆäÉú²úÁ÷³ÌÈçͼ1£º

´ËÁ÷³ÌµÄµÚ¢ñ²½·´Ó¦Îª£ºCH4+H2O?CO+3H2£¬Ò»¶¨Ìõ¼þÏÂCH4µÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØÏµÈçͼ2£®ÔòP1£¼P2£®100¡æÊ±£¬½«1mol CH4ºÍ2mol H2OͨÈëÈÝ»ýΪ100LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬´ïµ½Æ½ºâʱCH4µÄת»¯ÂÊΪ0.5£®´Ëʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK=2.25¡Á10-4£¨£®
´ËÁ÷³ÌµÄµÚ¢ò²½·´Ó¦µÄƽºâ³£ÊýËæÎ¶ȵı仯Èç±í£º
ζÈ/¡æ400500830
ƽºâ³£ÊýK1091
´Ó±íÖпÉÒÔÍÆ¶Ï£º¸Ã·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬ÈôÔÚ500¡æÊ±½øÐУ¬ÉèÆðʼʱCOºÍH2OµÄÆðʼŨ¶È¾ùΪ0.020mol/L£¬ÔÚ¸ÃÌõ¼þÏ£¬·´Ó¦´ïµ½Æ½ºâʱ£¬COµÄת»¯ÂÊΪ75%£®Èçͼ3±íʾ¸Ã·´Ó¦ÔÚt1ʱ¿Ì´ïµ½Æ½ºâ¡¢ÔÚt2ʱ¿ÌÒò¸Ä±äij¸öÌõ¼þÒýÆðŨ¶È±ä»¯µÄÇé¿ö£ºÍ¼ÖÐt2ʱ¿Ì·¢Éú¸Ä±äµÄÌõ¼þÊǽµµÍζȣ¬»òÔö¼ÓË®ÕôÆûµÄÁ¿£¬»ò¼õÉÙÇâÆøµÄÁ¿£®
¹¤ÒµÉϳ£ÀûÓ÷´Ó¦¢ñ²úÉúµÄCOºÍH2ºÏ³É¿ÉÔÙÉúÄÜÔ´¼×´¼£®
¢ÙÒÑÖªCO¡¢CH3OHµÄȼÉÕÈÈ·Ö±ðΪ283.0kJ•mol-1ºÍ726.5kJ•mol-1£¬ÔòCH3OH²»ÍêȫȼÉÕÉú³ÉCOºÍH2OµÄÈÈ»¯Ñ§·½³ÌʽΪCH3OH£¨l£©+O2£¨g£©=CO£¨g£©+2 H2O£¨l£©¡÷H=-443.5kJ•mol-1£®
¢ÚºÏ³É¼×´¼µÄ·½³ÌʽΪ£ºCO+2H2?CH3OH¡÷H£¼0£®ÔÚ230¡ãC〜270¡æ×îΪÓÐÀû£®ÎªÑо¿ºÏ³ÉÆø×îºÏÊÊµÄÆðʼ×é³É±Èn£ºn£¬·Ö±ðÔÚ230¡æ¡¢250¡æºÍ270¡æ½øÐÐʵÑ飬½á¹ûÈçͼ4Ëùʾ£®ÆäÖÐ270¡æµÄʵÑé½á¹ûËù¶ÔÓ¦µÄÇúÏßÊÇZ£»µ±ÇúÏßX¡¢Y¡¢Z¶ÔÓ¦µÄͶÁϱȴﵽÏàͬµÄCOƽºâת»¯ÂÊʱ£¬¶ÔÓ¦µÄ·´Ó¦Î¶ÈÓëͶ½Ï±ÈµÄ¹ØÏµÊÇͶÁϱÈÔ½¸ß£¬¶ÔÓ¦µÄ·´Ó¦Î¶ÈÔ½¸ß£®
¢Ûµ±Í¶ÁϱÈΪ1£º1£¬Î¶ÈΪ230¡æ£¬Æ½ºâ»ìºÏÆøÌåÖУ¬CH3OHµÄÎïÖʵÄÁ¿·ÖÊýΪ33.3%£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®±ØÐë¼ÓÈëÑõ»¯¼Á²ÅÄÜʵÏֵı仯ÊÇ£¨¡¡¡¡£©
A£®KClO3¡úO2B£®CO2¡úCOC£®Fe¡úFe3O4D£®CuO¡úCuSO4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

7£®³ãÈȵįÌÅÄÚÓз´Ó¦£ºC£¨s£©+O2£¨g£©¨TCO2£¨g£©¡÷H=-392KJ/mol£¬Íù¯ÌÅÄÚͨÈëË®ÕôÆøÊ±£¬ÓÐÈçÏ·´Ó¦£ºC£¨s£©+H2O£¨g£©¨TH2£¨g£©+CO£¨g£©¡÷H=+131KJ/mol£¬CO£¨g£©+$\frac{1}{2}$O2£¨g£©¨TCO2£¨g£©¡÷H=-282KJ/mol£¬H2£¨g£©+$\frac{1}{2}$O2£¨g£©¨TH2O£¨g£©¡÷H=-241KJ/mol£¬ÓÉÒÔÉÏ·´Ó¦ÍƶÏÍù³ãÈȵįÌÅÄÚͨÈëË®ÕôÆøÊ±£¨¡¡¡¡£©
A£®¼È²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬ÓÖ²»ÄܽÚʡȼÁÏ
B£®Ëä²»ÄÜʹ¯»ð˲¼ä¸üÍú£¬µ«¿ÉÒÔ½ÚʡȼÁÏ
C£®¼ÈÄÜʹ¯»ð˲¼ä¸üÍúÓÖ¿ÉÒÔ½ÚʡȼÁÏ
D£®²»ÄܽÚʡȼÁÏ£¬µ«ÄÜʹ¯»ð˲¼ä¸üÍú

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

17£®ÏÂÁÐʵÑé×°ÖÃͼËùʾµÄʵÑé²Ù×÷£¬ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®     
¸ÉÔïCl2
B£®       
ÅäÖÆ90ml 0.1mol•L-1ÁòËáÈÜÒº
C£®        
·ÖÀë·ÐµãÏà²î½Ï´óµÄ»¥ÈÜÒºÌå»ìºÏÎï
D£®
·ÖÀ뻥²»ÏàÈܵÄÁ½ÖÖÒºÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

4£®ÏÂÁÐÊÇһЩ³£¼ûÎïÖÊÖ®¼äµÄת»¯¹ØÏµ£®AÊÇÒ»ÖÖºìÉ«½ðÊô£¬FΪ²»ÈÜÓÚËáµÄ°×É«³Áµí£¬BºÍCÊÇÁ½ÖÖÇ¿Ëᣮ£¨Í¼Öв¿·Ö²úÎï¼°·´Ó¦Ìõ¼þÂÔÈ¥£©

Ôò£º¢ÙAºÍB·Ö±ðΪ£ºACu BH2SO4£»
¢ÚAÓëB·¢Éú·´Ó¦µÄÌõ¼þÊÇŨÁòËá¡¢¼ÓÈÈ£»
¢Û·´Ó¦¢ÞµÄ»¯Ñ§·½³ÌʽΪ3NO2+H2O=2HNO3+NO£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®NAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®39g±½Öк¬ÓÐC¨TC¼üÊýΪ1.5NA
B£®0.5molFeBr2Óë±ê×¼×´¿öÏÂ33.6LÂÈÆø·´Ó¦Ê±×ªÒƵç×ÓÊýΪ3NA
C£®1L0.5mol•L-1µÄNaClOÈÜÒºÖк¬ÓеÄClO-Àë×ÓÊýΪ0.5NA
D£®³£Î³£Ñ¹Ï£¬14gÓÉC2H4ºÍC3H6×é³ÉµÄ»ìºÏÆøÌåÖк¬ÓеÄÔ­×Ó×ÜÊýΪ3NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

9£®¿ÉÄæ·´Ó¦A£¨g£©+3B£¨g£©?2C£¨g£©¡÷H=-QkJ/mol£¨Q£¾0£©£®Óмס¢ÒÒÁ½¸öÈÝ»ýÏàͬÇÒ²»±äµÄÃܱÕÈÝÆ÷£¬Ïò¼×ÈÝÆ÷ÖмÓÈë1molAºÍ3molB£¬ÔÚÒ»¶¨Ìõ¼þÏ´ﵽƽºâʱ·Å³öÈÈÁ¿ÎªQ1kJ£»ÔÚÏàͬÌõ¼þÏ£¬ÏòÒÒÈÝÆ÷ÖмÓÈë2molC´ïµ½Æ½ºâºóÎüÊÕÈÈÁ¿ÎªQ2kJ£¬ÒÑÖªQ1=3Q2£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¼×Öз´Ó¦´ïµ½Æ½ºâʱ£¬Q1=Q
B£®´ïµ½Æ½ºâºó£¬¼×ÖÐCµÄÌå»ý·ÖÊý±ÈÒÒ´ó
C£®´ïµ½Æ½ºâºó£¬ÔÙÏòÒÒÖмÓÈë0.25molA¡¢0.75molB¡¢1.5molC£¬Æ½ºâÏòÉú³ÉCµÄ·½ÏòÒÆ¶¯
D£®ÒÒÖеÄÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪ2C£¨g£©?A£¨g£©+3B£¨g£©¡÷H=+Q2kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸