3£®50mL 0.55mol/LÑÎËáÓë50mL 0.50mol/LNaOHÈÜÒºÔÚÈçͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦£¬Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÕ±­¼äÌîÂúËéÅÝÄ­ËÜÁϵÄ×÷ÓÃÊÇ»·Ðβ£Á§½Á°è°ô£®
£¨2£©´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýֵƫС£¨Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±¡°ÎÞÓ°Ï족£©£®
£¨3£©ÊµÑéÖиÄÓÃ50gŨÁòËá´úÌæÑÎËá½øÐз´Ó¦£¬ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿²»ÏàµÈ£»²âµÃÖкÍÈȲ»ÏàµÈ£¨Ìî¡°ÏàµÈ¡±¡¢¡°²»ÏàµÈ¡±£©£®

·ÖÎö £¨1£©¸ù¾ÝÁ¿ÈȼƵĹ¹ÔìÀ´ÅжϸÃ×°ÖõÄȱÉÙÒÇÆ÷£»
£¨2£©²»¸ÇÓ²Ö½°å£¬»áÓÐÒ»²¿·ÖÈÈÁ¿É¢Ê§£»
£¨3£©Å¨ÁòËáÏ¡ÊÍʱ·Å³öÈÈÁ¿£®

½â´ð ½â£º£¨1£©ÓÉÁ¿ÈȼƵĹ¹Ôì¿ÉÖª¸Ã×°ÖõÄȱÉÙÒÇÆ÷ÊÇ»·Ðβ£Á§½Á°è°ô£»
¹Ê´ð°¸Îª£º»·Ðβ£Á§½Á°è°ô£»
£¨2£©´óÉÕ±­ÉÏÈç²»¸ÇÓ²Ö½°å£¬»áʹһ²¿·ÖÈÈÁ¿É¢Ê§£¬ÇóµÃµÄÖкÍÈÈÊýÖµ½«»á¼õС£¬
¹Ê´ð°¸Îª£ºÆ«Ð¡£»
£¨3£©Å¨ÁòËáÏ¡ÊÍʱ·Å³öÈÈÁ¿£¬¸ÄÓÃ50gŨÁòËá´úÌæÑÎËá½øÐз´Ó¦£¬ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿Æ«´ó£¬ÎÂ¶È²îÆ«´ó£¬ÖкÍÈÈÊýÖµ²»ÏàµÈ£¬
¹Ê´ð°¸Îª£º²»ÏàµÈ£®

µãÆÀ ±¾Ì⿼²éѧÉúÓйØÖкÍÈȵIJⶨ֪ʶ£¬¿ÉÒÔ¸ù¾ÝËùѧ֪ʶ½øÐлشð£¬×¢Òâ¶ÔÖкÍÈȸÅÄîµÄÀí½â£¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

11£®Ñо¿ÈËÔ±·¢ÏÖÁËÒ»ÖÖ¡°Ë®¡¯¡¯µç³Ø£¬Æä×Ü·´Ó¦Îª£º5Mn02+2Ag+2NaCl=Na2Mn5O10+2AgCl£®ÈçͼÓá°Ë®¡±µç³ØÎªµçÔ´µç½âNaClÈÜÒºµÄʵÑéÖУ¬Xµç¼«ÉÏÓÐÎÞÉ«ÆøÌåÒݳö£®ÏÂÁÐÓйطÖÎöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®IΪ¸º¼«£¬Æäµç¼«·´Ó¦Ê½ÎªAg+Cl--e-=AgCl
B£®¡°Ë®¡±µç³ØÄÚNa+²»¶ÏÏò¸º¼«×÷¶¨ÏòÒÆ¶¯
C£®Ã¿×ªÒÆ1mole-£¬UÐ͹ÜÖÐÏûºÄ0.5mol H2O
D£®¿ªÊ¼Ê±UÐ͹ÜÖÐY¼«¸½½üpHÖð½¥Ôö´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

12£®ÒÑÖª£ºN2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H=-92.4kJ/mol25¡æÊ±£¬ÓÐÌå»ý¾ùΪ2LµÄ¼×¡¢ÒÒÁ½¸öÏàͬµÄÃܱÕÈÝÆ÷£¬Ïò¼×ÈÝÆ÷ÖмÓÈë1mol N2ºÍ3mol H2£¬¾­¹ý3minÔÚ25¡æÏ´ﵽƽºâʱ·Å³öQ1kJµÄÈÈÁ¿£»ÏòÒÒÈÝÆ÷ÖмÓÈë2molNH3£¬Ò»¶Îʱ¼äºóÔÚ25¡æÏ´ﵽƽºâʱÎüÊÕQ2kJµÄÈÈÁ¿£¬ÇÒQ1=3Q2£®
£¨1£©ÏÂÁÐÇé¿ö±íÃ÷·´Ó¦ÒѴﵽƽºâ״̬µÄÓÐBC£®
A£®ÆøÌåµÄÃܶȲ»Ôٱ仯
B£®¶ÏÁÑ1molN¡ÔN¼üµÄͬʱ¶ÏÁÑ6molN-H
C£®ÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»Ôٱ仯
D£®µªÆøµÄÉú³ÉËÙÂÊÓë°±ÆøµÄÉú³ÉËÙÂÊÏàµÈ
£¨2£©Q2=23.1kJ
£¨3£©¼×ÈÝÆ÷Öдﵽƽºâʱ£¬H2µÄƽ¾ù·´Ó¦ËÙÂÊΪ0.0375mol/£¨L•min£©£»´ïµ½Æ½ºâºó£¬ÈôÔÙÏò¼×ÈÝÆ÷ÖмÓÈë0.25mol N2¡¢0.75mol H2¡¢1.5mol NH3£¬¾­¹ýÒ»¶Îʱ¼äÔÚ25¡æÏÂÖØÐ´ﵽƽºâ£¬Ôòƽºâ³£ÊýK=Õý·´Ó¦·½Ïò£®£¨±£Áô1λСÊý£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

9£®»¯ºÏÎïHÊǺϳÉÖ²ÎïÉú³¤µ÷½Ú¼Á³àùËáµÄÖØÒªÖмäÌ壬ÆäºÏ³É·ÏßÈçÏ£º

£¨1£©»¯ºÏÎïHµÄº¬Ñõ¹ÙÄÜÍÅΪôÊ»ùºÍÃѼü£¨Ìî¹ÙÄÜÍŵÄÃû³Æ£©£®
£¨2£©»¯ºÏÎïBºÏ³ÉCʱ»¹¿ÉÄÜÉú³ÉÒ»ÖÖ¸±²úÎ·Ö×ÓʽΪC20H24O2£©£¬¸Ã¸±²úÎïµÄ½á¹¹¼òʽΪ£»ÓÉC¡úDµÄ·´Ó¦ÀàÐÍÊÇÑõ»¯·´Ó¦£®
£¨3£©Ð´³öÒ»ÖÖÂú×ãÏÂÁÐÌõ¼þµÄDµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ »ò£®
¢ñ£®·Ö×Óº¬ÓÐ1¸ö±½»·£»
¢ò£®·Ö×ÓÓÐ4ÖÖ²»Í¬»¯Ñ§»·¾³µÄÇ⣻
¢ó£®ÄÜ·¢ÉúÒø¾µ·´Ó¦£®
£¨4£©¸ù¾ÝÒÑÓÐ֪ʶ²¢½áºÏÏà¹ØÐÅÏ¢£¬Ð´³öÒÔΪÓлúÔ­ÁÏÖÆ±¸µÄºÏ³É·ÏßÁ÷³Ìͼ£¨ÎÞ»úÊÔ¼ÁÈÎÑ¡£¬¿ÉÑ¡ÔñÊʵ±ÓлúÈܼÁ£©£®ºÏ³É·ÏßÁ÷³ÌͼʾÀýÈçÏ£º
CH3CHO$¡ú_{´ß»¯¼Á£¬¡÷}^{O_{2}}$CH3COOH$¡ú_{ŨÁòËᣬ¡÷}^{CH_{3}CH_{2}OH}$CH3COOCH2CH3£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

16£®ÔÚ1LÈÝÆ÷ÖмÓÈë1mol AºÍ1.2mol B£¬·¢ÉúÈçÏ·´Ó¦£ºA£¨g£©+2B£¨g£©?c£¨g£©£®·´Ó¦¾­1minºó´ïµ½Æ½ºâ£¬´Ëʱ²âµÃAµÄŨ¶ÈΪ0.6mol/L£®Çó
£¨1£©1minÄ©B¡¢CµÄŨ¶È£»
£¨2£©ÒÔA±íʾ¸Ã·´Ó¦µÄÔÚ1minÄڵķ´Ó¦ËÙÂÊ£»
£¨3£©BµÄת»¯ÂÊ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®FeSO4•7H2OË׳ơ°ÂÌ·¯¡±£¬ÊÇÒ»ÖÖÀ¶ÂÌÉ«¾§Ì壮ij¿ÎÍâ»î¶¯Ð¡×éͬѧÔÚʵÑéÊÒ·¢ÏÖһƿÂÌ·¯ÒѲ¿·Ö±äÖÊ£¬ÄâÓøÃÊÔ¼ÁΪԭÁÏ£¬ÖØÐÂÖÆÈ¡FeSO4•7H2O¾§Ì壬²¢²â¶¨Æä´¿¶È£®
[²éÔÄ×ÊÁÏ]
¢ÙÂÌ·¯»¯Ñ§ÐÔÖʳ£²»Îȶ¨£¬ÔÚ³±Êª¿ÕÆøÖÐFÖð½¥Ñõ»¯±äÖÊ£¬¾ÃÖõÄÂÌ·¯ÈÜÒºÖð½¥±äΪ»Æ
É«£¬²¢³öÏÖר»ÆÉ«»ë×Ç£»
¢ÚÁòËáÑÇÌúµÄÈܽâ¶ÈËæÎ¶ÈÉý¸ß¶øÔö´ó£®
[ÖÆ±¸¾§Ìå]
¢Ùȡһ¶¨Á¿ÒѱäÖʵġ°ÂÌ·¯¡±ÓÚÉÕ±­ÖУ¬¼ÓÊÊÁ¿Ë®£¬½Á°è£¬µÃµ½×Ø»ÆÉ«Ðü×ÇÒº£®
¢Ú¼ÓÈëÒ»¶¨Á¿Ï¡H2SO4ºÍ¹ýÁ¿Ìúм£¬ÔÚ60¡æ×óÓÒˮԡÖмÓÈÈ£¬³ä·Ö·´Ó¦£¬´ýÈÜÒºÍêÈ«
±äΪdzÂÌɫΪֹ£®
¢Û³ÃÈȹýÂË£¬²¢ÓÃÉÙÁ¿ÈÈˮϴµÓ£¬½«ÂËҺתÈëµ½ÃܱÕÈÝÆ÷ÖУ¬¾²Öá¢ÀäÈ´½á¾§£®
¢Ü´ý½á¾§Íê±Ïºó£¬¹ýÂ˳ö¾§Ì壬ÓÃÉÙÁ¿±ùˮϴµÓ£¬ÔÙÓÃÂËÖ½Îü¸ÉË®·Ö£¬×îºó·ÅÈ˹ã¿ÚÆ¿
ÖÐÃܱձ£´æ£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Íê³É¾ÃÖÃÂÌ·¯ÈÜÒºÔÚ¿ÕÆøÖбäÖÊ·´Ó¦µÄÀë×Ó·½³Ìʽ£º
¡õFe2++¡õO2+¡õ6H2O¨T¡õFe3++¡õ4Fe£¨OH£©3¡ý
£¨2£©ÊµÑé²½Öè¢ÚÖмÓÈëÏ¡ÁòËáµÄÄ¿µÄÊÇÈܽâFe£¨OH£©3³Áµí£¬·ÀÖ¹FeSO4Ë®½â
£¨3£©ÊµÑé²½Öè¢ÛÖгÃÈȹýÂËÄ¿µÄÊǼõÉÙ¹ýÂËʱFeSO4Ëðʧ£®
£¨4£©ÊµÑéÖÐÁ½´ÎÓõ½¹ýÂ˲Ù×÷£¬Íê³É¸Ã²Ù×÷µÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢Â©¶·£®
[²â¶¨´¿¶È]
ÓÐͬѧÈÏΪ£¬ÖƱ¸¹ý³ÌÖпÉÄÜÓÐÉÙÁ¿Fe2+±»Ñõ»¯ÎªFe3+£¬µ¼Ö²úÆ·²»´¿£®ËûÃÇÉè¼ÆÁËÈçÏ·½°¸²â¶¨²úÆ·µÄ´¿¶È£®

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
ÏÞÑ¡ÊÔ¼Á£ºNaOHÈÜÒº¡¢KSCNÈÜÒº¡¢H2O2ÈÜÒº¡¢±½·ÓÏ¡ÈÜÒº¡¢K3[Fe£¨CN£©6]ÈÜÒº¡¢BaCl2ÈÜÒº
£¨5£©¼ìÑé²úÆ·ÖÐÊÇ·ñº¬ÓÐFe3+µÄÎÞ»úÊÔ¼Á×îºÃÊÇ£ºKSCNÈÜÒº£®
£¨6£©ÊÔ¼ÁXÊÇ£ºK3[Fe£¨CN£©6]ÈÜÒº£®
£¨7£©²úÆ·ÖÐFeSO4•7H2O£¨Ä¦¶ûÖÊÁ¿Îª278g/mol£©µÄº¬Á¿Îª£º$\frac{417{m}_{2}}{296{m}_{1}}$¡Á100%£¨ÓÃÖÊÁ¿·ÖÊý±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

15£®Ä³³ÇÊÐÃºÆøÖпÉÈ¼ÆøÌå×é³ÉΪH2 48%¡¢CO 15%¡¢CH4 13%£¨¾ùΪÌå»ý°Ù·Ö±È£¬ÆäÓàΪ²»È¼ÐÔÆøÌ壩£®ÒÑÖª£º1mol H2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³öÈÈÁ¿285.8kJ£¬1mol COÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³öÈÈÁ¿282.8kJ£¬1mol CH4ÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³öÈÈÁ¿890.4kJ£®
£¨1£©Ð´³öH2ºÍCH4ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H=-571.6 kJ/molCH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-890.4 kJ/mol
£¨2£©±ê×¼×´¿öϵÄ1ÉýÕâÖÖÃºÆøÍêȫȼÉÕÄܷųö¶àÉÙÈÈÁ¿£¿
£¨3£©±ê×¼×´¿öϵÄ1ÉýÕâÖÖÃºÆøÍêȫȼÉÕÐèÒªÏûºÄ¶àÉÙÉý±ê×¼×´¿öÏÂµÄ¿ÕÆø£¿£¨¼Ù¶¨¿ÕÆøÖÐO2µÄÌå»ý°Ù·Öº¬Á¿Îª21%£©
£¨4£©ÒÑÖª1¿ËˮζÈÿÉÏÉý1¡æÐèÒªÈÈÁ¿4.2J£®Óû½«1kg 20¡æµÄË®¼ÓÈÈÖÁ·ÐÌÚ£¬¼Ù¶¨¼ÓÈȹý³ÌÖеÄÈÈÁ¿ÀûÓÃÂÊΪ60%£¬ÔòÐèÏûºÄ±ê×¼×´¿öϵÄÕâÖÖÃºÆø¶àÉÙÉý£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

12£®ÀûÓÃÏÂͼװÖòⶨÖкÍÈȵÄʵÑé²½ÖèÈçÏ£º
¢ÙÓÃÁ¿Í²Á¿È¡50mL 0.25mol/LÁòËáµ¹ÈëСÉÕ±­ÖУ¬²â³öÁòËáζȣ»
¢ÚÓÃÁíÒ»Á¿Í²Á¿È¡50mL 0.55mol/L NaOHÈÜÒº£¬²¢ÓÃÁíһζȼƲâ³öÆäζȣ»
¢Û½«NaOHÈÜÒºµ¹ÈëСÉÕ±­ÖУ¬É跨ʹ֮»ìºÏ¾ùÔÈ£¬²â³ö»ìºÏÒº×î¸ßζȣ®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÏ¡ÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ£¨ÖкÍÈÈÊýֵΪ57.3kJ/mol£©£º$\frac{1}{2}$H2SO4£¨aq£©+NaOH£¨aq£©¨T$\frac{1}{2}$Na2SO4£¨aq£©+H2O£¨l£©¡÷H=-57.3kJ/mol£®
£¨2£©µ¹ÈëNaOHÈÜÒºµÄÕýÈ·²Ù×÷ÊÇC£¨´ÓÏÂÁÐÑ¡³ö£©£®
A£®Ñز£Á§°ô»ºÂýµ¹Èë
B£®·ÖÈý´ÎÉÙÁ¿µ¹Èë
C£®Ò»´ÎѸËÙµ¹Èë
£¨3£©Ê¹ÁòËáÓëNaOHÈÜÒº»ìºÏ¾ùÔȵÄÕýÈ·²Ù×÷ÊÇD£¨´ÓÏÂÁÐÑ¡³ö£©£®
A£®ÓÃζȼÆÐ¡ÐĽÁ°è
B£®½Ò¿ªÓ²Ö½Æ¬Óò£Á§°ô½Á°è
C£®ÇáÇáµØÕñµ´ÉÕ±­
D£®ÓÃÌ×ÔÚζȼÆÉϵĻ·Ðβ£Á§°ôÇáÇáµØ½Á¶¯
£¨4£©ÊµÑéÊý¾ÝÈçÏÂ±í£º
¢ÙÇëÌîдϱíÖеĿհףº
ζÈ
ʵÑé´ÎÊý
ÆðʼζÈt1¡æÖÕֹζÈt2/¡æÎÂ¶È²îÆ½¾ùÖµ£¨t2-t1£©/¡æ
H2SO4NaOHƽ¾ùÖµ
126.226.026.129.5 
3.4¡æ 
227.027.427.232.3
325.925.925.929.2
426.426.226.329.8
¢Ú½üËÆÈÏΪ0.55mol/L NaOHÈÜÒººÍ0.25mol/LÁòËáÈÜÒºµÄÃܶȶ¼ÊÇ1g/cm3£¬ÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc=4.18J/£¨g•¡æ£©£®ÔòÖкÍÈÈ¡÷H=-56.8kJ/mol£¨È¡Ð¡Êýµãºóһ룩£®
¢ÛÉÏÊöʵÑéÊýÖµ½á¹ûÓë57.3kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔ­Òò¿ÉÄÜÊÇ£¨Ìî×Öĸ£©acd£®
a£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î
b£®Á¿È¡NaOHÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý
c£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±­ÖÐ
d£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζȣ®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

13£®ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙH2£¨g£©+$\frac{1}{2}$O2£¨g£©¨TH2O£¨l£©¡÷H=-285.8kJ•mol-1
¢ÚH2£¨g£©+$\frac{1}{2}$O2£¨g£©¨TH2O£¨g£©¡÷H=-241.8kJ•mol-1
¢ÛC£¨s£©+$\frac{1}{2}$O2£¨g£©¨TCO£¨g£©¡÷H=-110.5kJ•mol-1
¢ÜC£¨s£©+O2£¨g£©¨TCO2£¨g£©¡÷H=-393.5kJ•mol-1
»Ø´ðÏÂÁи÷ÎÊÌ⣺
£¨1£©ÉÏÊö·´Ó¦ÖÐÊôÓÚ·ÅÈÈ·´Ó¦µÄÊǢ٢ڢۢܣ®
£¨2£©È¼ÉÕ10g H2Éú³ÉҺ̬ˮ£¬·Å³öµÄÈÈÁ¿Îª1 429.0 kJ£®
£¨3£©COµÄȼÉÕÈÈΪ283.0 kJ•mol-1£»ÆäÈÈ»¯Ñ§·½³ÌʽΪCO£¨g£©+$\frac{1}{2}$O2£¨g£©¨TCO2£¨g£©¡÷H=-283.0 kJ•mol-1£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸