º£Ë®×ÊÔ´·á¸»£¬º£Ë®ÖÐÖ÷Òªº¬ÓеÈÀë×Ó¡£ºÏÀíÀûÓú£Ë®×ÊÔ´ºÍ±£»¤»·¾³ÊÇÎÒ¹ú¿É³ÖÐø·¢Õ¹µÄÖØÒª±£Ö¤¡£

   ¢ñ£®»ðÁ¦·¢µçȼúÅŷŵĻáÔì³ÉһϵÁл·¾³ºÍÉú̬ÎÊÌâ¡£ÀûÓú£Ë®ÍÑÁòÊÇÒ»ÖÖÓÐЧµÄ·½·¨£¬Æ乤ÒÕÁ÷³ÌÈçÏÂͼËùʾ£º

   (1)ÌìÈ»º£Ë®µÄpH¡Ö8£¬³ÊÈõ¼îÐÔ¡£ÓÃÀë×Ó·½³Ìʽ½âÊÍÔ­Òò                   ¡£

   (2)ÌìÈ»º£Ë®ÎüÊÕÁ˺¬ÁòÑÌÆøºó£¬ÒªÓýøÐÐÑõ»¯´¦Àí£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ       

                  £»Ñõ»¯ºóµÄ¡°º£Ë®¡±ÐèÒªÓôóÁ¿µÄÌìÈ»º£Ë®ÓëÖ®»ìºÏºó²ÅÄÜÅÅ·Å£¬¸Ã²Ù×÷µÄÖ÷ҪĿµÄÊÇ                           ¡£

¢ò£®ÖؽðÊôÀë×Ó¶ÔºÓÁ÷¼°º£ÑóÔì³ÉÑÏÖØÎÛȾ¡£Ä³»¯¹¤³§·ÏË®(pH=2.O£¬p¡Ö1g¡¤mL-1)Öк¬ÓеÈÖؽðÊôÀë×Ó£¬ÆäŨ¶È¸÷ԼΪ0.01mol¡¤L-1¡£ÅÅ·ÅÇ°ÄâÓóÁµí·¨³ýÈ¥ÕâÁ½ÖÖÀë×Ó£¬²éÕÒÓйØÊý¾ÝÈçÏ£º

ÄÑÈܵç½âÖÊ

8.3¡Á10-17

5.6¡Á10-8

6.3¡Á10-50

7.1¡Á10-9

1.2¡Á10-15

3.4¡Á10-28

(3)ÄãÈÏΪÍù·ÏË®ÖÐͶÈë         (Ìî×ÖĸÐòºÅ)£¬³ÁµíЧ¹û×îºÃ¡£

   A£®           B£®           C£®              D£®

   (4)Èç¹ûÓÃÉúʯ»Ò´¦ÀíÉÏÊö·ÏË®£¬Ê¹ÈÜÒºµÄpH=8.0£¬´¦ÀíºóµÄ·ÏË®ÖÐ=         ¡£

   (5)Èç¹ûÓÃʳÑδ¦ÀíÆäÖ»º¬µÄ·ÏË®£¬²âµÃ´¦ÀíºóµÄ·ÏË®ÖеÄÖÊÁ¿·ÖÊýΪ0.117©‡¡£Èô»·¾³ÒªÇóÅŷűê׼ΪµÍÓÚ1.O¡ÁlO-8mol¡¤L-1£¬Îʸù¤³§´¦ÀíºóµÄ·ÏË®ÖÐ=   £¬ÊÇ·ñ·ûºÏÅŷűê×¼      (Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)¡£ÒÑÖª=1.8¡Ál-10mol2¡¤L2¡£

(1)¡¡ ¡¡ »ò¡¡¡¡ (1·Ö)

(2) (1·Ö)

¡¡ Öк͡¢Ï¡Ê;­ÑõÆøÑõ»¯ºóº£Ë®ÖÐÉú³ÉµÄËá(1·Ö)

£¨3)B (1·Ö)

£¨4) (1·Ö)

(5) (2·Ö)¡¡ ÊÇ(1·Ö)

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

2010ÄêÑÇÔË»áÔÚÎÒ¹ú¹ãÖݳɹ¦¾Ù°ì£¬Õû¸öÑÇÔË»áÌåÏÖÁË»·±£ÀíÄ
£¨1£©¹ãÖÝÑÇÔË»á»ð¾æ¡°³±Á÷¡±²ÉÓÃÁ˱ûÍ飨C3H8£©×÷ΪȼÁÏ£¬È¼ÉÕºóµÄ²úÎïΪˮºÍ¶þÑõ»¯Ì¼£®ÔÚ298Kʱ£¬1mol±ûÍéÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ·Å³ö2221.5kJµÄÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
C3H8£¨g£©+5O2£¨g£©=3CO2£¨g£©+4H2O£¨l£©£¬¡÷H=-2221.5kJ?mol-1
C3H8£¨g£©+5O2£¨g£©=3CO2£¨g£©+4H2O£¨l£©£¬¡÷H=-2221.5kJ?mol-1
£®
ÓÖÖª£ºÔÚ298KʱC3H8£¨g£©=C3H6£¨g£©+H2£¨g£©¡÷H=124.2kJ?mol-1£¬H2£¨g£©+
1
2
O2£¨g£©=H2O£¨I£©¡÷H=-285.8kJ?mol-1£¬Ôò1 molC3H6ÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ
2059.9
2059.9
kJ£®
£¨2£©¹ãÖÝÊÇÒ»×ùÃÀÀöµÄº£±õ³ÇÊУ¬º£Ë®×ÊÔ´·Ç³£·á¸»£®
¢Ùº£Ñóµç³ØÊÇÒÔÂÁΪ¸º¼«£¬²¬ÍøΪÕý¼«£¬º£Ë®Îªµç½âÖÊÈÜÒº£¬¿ÕÆøÖÐÑõÆøÓëÂÁ·´Ó¦²úÉúµçÁ÷£¬µç³Ø×Ü·´Ó¦Îª4Al+3O2+6H2O=4Al£¨OH£©3£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
bc
bc
£¨ÌîдÐòºÅ×Öĸ£©£»
a£®µç³Ø¹¤×÷ʱ£¬µçÁ÷ÓÉÂÁµç¼«Ñص¼ÏßÁ÷Ïò²¬µç¼«
b£®²¬µç¼«²ÉÓÃÍø×´±È¿é×´¸üÀûÓÚO2·Åµç
c£®º£Ë®ÖеÄOH-ÏòÂÁµç¼«·½ÏòÒƶ¯
¢ÚÓöèÐԵ缫µç½â200mL l.5mol/LʳÑÎË®£»µç½â2minʱ£¬Á½¼«¹²ÊÕ¼¯µ½448mLÆøÌ壨±ê×¼×´¿öÏ£©£®Ð´³ö¸Ãµç½â·´Ó¦µÄÀë×Ó·½³Ìʽ
2Cl-+2H2O
 Í¨µç 
.
 
2OH-+H2¡ü+Cl2¡ü
2Cl-+2H2O
 Í¨µç 
.
 
2OH-+H2¡ü+Cl2¡ü
£®¼ÙÉèµç½âÇ°ºóÈÜÒºµÄÌå»ý²»±ä£¬Ôòµç½âºó¸ÃÈÜÒºµÄpHΪ
13
13
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2009?´óÁ¬Ä£Ä⣩ÎÒ¹úÓнϳ¤µÄº£°¶Ïߣ¬ºÆ嫵ĺ£ÑóÊÇÒ»¸ö¾Þ´óµÄÎïÖÊ×ÊÔ´ºÍÄÜÁ¿±¦¿â£®Ä¿Ç°£¬ÊÀ½ç¸÷¹ú¶¼ÔÚÑо¿ÈçºÎ³ä·ÖÀûÓú£Ñó×ÊÔ´£®
£¨1£©Èçͼ1£¬Ôڵ糡ÖÐÀûÓÃĤ¼¼Êõ£¨ÑôÀë×Ó½»»»Ä¤Ö»ÔÊÐíÑôÀë×Óͨ¹ý£¬ÒõÀë×Ó½»»»Ä¤Ö»ÔÊÐíÒõÀë×Óͨ¹ý£©µ­»¯º£Ë®£¬¸Ã·½·¨³ÆΪµçÉøÎö·¨£®
¢Ùͼ1ÖÐĤaӦѡÔñ
ÒõÀë×Ó½»»»Ä¤
ÒõÀë×Ó½»»»Ä¤
Ĥ£»
¢ÚµçÉøÎö·¨»¹¿ÉÒÔÓÃÀ´´¦Àíµç¶Æ·ÏÒº£¬Ð´³öÓø÷½·¨´¦Àíº¬ÁòËáÍ­·ÏҺʱ£¨Ê¹ÓöèÐԵ缫£©Ëù·¢ÉúµÄµç¼«·´Ó¦£ºÒõ¼«
2Cu2++4e-=2Cu
2Cu2++4e-=2Cu

Ñô¼«
4OH--4e-=O2¡ü+2H2O
4OH--4e-=O2¡ü+2H2O

£¨2£©º£Ë®ÖеÄäåµÄ´¢Á¿·á¸»£¬Ô¼Õ¼µØÇòäå×Ü´¢Á¿µÄ99%£¬¹ÊäåÓС°º£ÑóÔªËØ¡±Ö®³Æ£¬º£Ë®ÖÐä庬Á¿Îª65mg?L-1£®Æ乤ҵÌáÈ¡·¨ÓУº
£¨I£©¿ÕÆø´µ³ö´¿¼îÎüÈë·¨£®¸Ã·¨Êǽ«ÂÈÆøͨÈëµ½º¬äåÀë×ӵĺ£Ë®ÖУ¬Ê¹äåÖû»³öÀ´£¬ÔÙÓÿÕÆø½«äå´µ³ö£¬Óô¿¼îÈÜÒºÎüÊÕ£¬×îºóÓÃÁòËáËữ£¬¼´¿ÉµÃµ½µ¥ÖÊä壮¸Ã·½·¨Éæ¼°µÄ·´Ó¦ÓУº
¢Ù
Cl2+2Br-=Br2+2Cl-
Cl2+2Br-=Br2+2Cl-
£¨Ð´³ö»¯Ñ§·´Ó¦µÄÀë×Ó·½³Ìʽ£©
¢Ú3Br2+3CO32-=BrO3-+5Br-+3CO2¡ü
¢ÛBrO3-+5Br-+6H+=3Br2+3H2O
ÆäÖз´Ó¦ÖТ۵ÄÑõ»¯¼ÁÊÇ
BrO3-
BrO3-
£¬»¹Ô­¼ÁÊÇ
Br-
Br-
£®
£¨II£©ÈܼÁÝÍÈ¡·¨£®¸Ã·¨ÊÇÀûÓõ¥ÖÊäåÔÚË®ÖкÍÝÍÈ¡¼ÁÖÐÈܽâ¶È²»Í¬µÄÔ­ÀíÀ´½øÐеģ®ÊµÑéÊÒÖÐÝÍÈ¡Óõ½µÄÖ÷Òª²£Á§ÒÇÆ÷Ãû³ÆÊÇ
·ÖҺ©¶·£¬ÉÕ±­
·ÖҺ©¶·£¬ÉÕ±­
£®ÏÂÁпÉÒÔÓÃÓÚ´Óº£Ë®ÖÐÝÍÈ¡äåµÄÊÔ¼ÁÊÇ
¢Ú
¢Ú
£¨Ìî±àºÅ£©£®¢ÙÒÒ´¼ ¢ÚËÄÂÈ»¯Ì¼ ¢ÛÏõËᣮ
£¨3£©º£Ë®ÖеÄ뮣¨º¬HDO 0.03¡ë£©·¢Éú¾Û±äµÄÄÜÁ¿£¬×ãÒÔ±£Ö¤ÈËÀàÉÏÒÚÄêµÄÄÜÔ´Ïû·Ñ£¬¹¤ÒµÉϿɲÉÓá°Áò»¯Çâ-ˮ˫ν»»»·¨¡±¸»¼¯HDO£®ÆäÔ­ÀíÊÇÀûÓÃH2S¡¢HDS¡¢H2OºÍHDOËÄÖÖÎïÖÊ£¬ÔÚ25¡æºÍ100¡æÁ½ÖÖ²»Í¬Î¶ÈÏ·¢ÉúµÄÁ½¸ö²»Í¬·´Ó¦µÃµ½½Ï¸ßŨ¶ÈµÄHDO£® Èçͼ2Ϊ¡°Áò»¯Çâ-ˮ˫ν»»»·¨¡±Ëù·¢ÉúµÄÁ½¸ö·´Ó¦ÖÐÉæ¼°µÄËÄÖÖÎïÖÊÔÚ·´Ó¦ÌåϵÖеÄÎïÖʵÄÁ¿Ëæζȵı仯ÇúÏߣ®Ð´³ö100¡æʱËù·¢ÉúµÄ·´Ó¦µÄ»¯Ñ§·½³Ìʽ
H2S+HDO=HDS+H2O
H2S+HDO=HDS+H2O
£»¹¤ÒµÉϸ»¼¯HDOµÄÉú²ú¹ý³ÌÖУ¬¿ÉÒÔÑ­»·ÀûÓõÄÒ»ÖÖÎïÖÊÊÇ
H2S
H2S
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìÁÉÄþÊ¡ÉòÑôÊÐÌú·ʵÑéÖÐѧ¸ßÈýÄ£Ä⿼ÊÔÀí¿Æ×ۺϻ¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(15·Ö)»¯Ñ§ÊÇÈËÀà½ø²½µÄ¹Ø¼ü£¬»¯Ñ§ÎªÈËÀàµÄÉú²ú¡¢Éú»îÌṩÁËÎïÖʱ£Ö¤¡£
(1)µç¶Æʱ£¬ÓöƲã½ðÊô×÷Ñô¼«µÄ×÷ÓÃÊÇ                             £®ÎªÁËʹ¶Æ²ãºñ¶È¾ùÔÈ¡¢¹â»¬ÖÂÃÜ¡¢Óë¶Æ¼þµÄ¸½×ÅÁ¦Ç¿£¬³ý¿ØÖÆÈÜÒºÖÐÀë×ÓŨ¶ÈÍ⣬ͨ³£»¹¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÓР                                              
(2)±ˮÖÐÔ̺¬×ŷḻµÄþ×ÊÔ´£¬¾­×ª»¯ºó¿É»ñµÃMgCl2´Ö²úÆ·¡£´Ó±ˮÖÐÌáȡþµÄ²½ÖèΪ£º
a£®½«º£±ß´óÁ¿´æÔڵı´¿ÇìÑÉÕ³Éʯ»Ò£¬²¢½«Ê¯»ÒÖƳÉʯ»ÒÈ飻
b£®½«Ê¯»ÒÈé¼ÓÈëµ½º£Ë®³Áµí³ØÖо­¹ýÂ˵õ½Mg(OH)2³Áµí£»
c.ÔÚMg(OH)2³ÁµíÖмÓÈëÑÎËáµÃµ½MgCl2ÈÜÒº£¬ÔÙ¾­Õô·¢½á¾§µÃµ½MgCl2¡¤6H2O£»
d£®½«MgCl2¡¤6H2OÔÚÒ»¶¨Ìõ¼þϼÓÈȵõ½ÎÞË®MgCl2£»
e£®µç½âÈÛÈÚµÄÂÈ»¯Ã¾¿ÉµÃµ½Mg¡£
¢Ù²½ÖèdÖеġ°Ò»¶¨Ìõ¼þ¡±Ö¸µÄÊÇ                       £¬Ä¿µÄÊÇ         ¡£
¢ÚÉÏÊöÌáȡþµÄÁ÷³ÌÖÐ,ΪÁ˽µµÍ³É±¾£®¼õÉÙÎÛȾ£¬¿ÉÒÔ²ÉÈ¡ºÜ¶à´ëÊ©£¬Çëд³öÆäÖÐÒ»µã
                                                                
¢ÛÓÐͬѧÈÏΪ£º²½Öèbºó¿É¼ÓÈÈMg(0H)2µÃµ½Mg0£¬ÔÙµç½âÈÛÈÚµÄMgOÖƽðÊôþ£¬ÕâÑù¿É¼ò»¯ÊµÑé²½Ö裬ÄãͬÒâ¸ÃͬѧµÄÏë·¨Âð?Ϊʲô?
                                                                        ¡£
(3)ÓËÊǺ˷´Ó¦×îÖØÒªµÄȼÁÏ£¬ÒѾ­ÑÐÖƳɹ¦Ò»ÖÖó§ºÏÐÍÀë×Ó½»»»Ê÷Ö¬£¬ËüרÃÅÎü¸½º£Ë®ÖеÄU4+£¬¶ø²»Îü¸½ÆäËûÔªËØ¡£Æä·´Ó¦Ô­ÀíΪ                          (Ê÷Ö¬ÓÃHR´úÌæ)£¬·¢ÉúÀë×Ó½»»»ºóµÄÀë×Ó½»»»Ä¤ÓÃËá´¦Àí»¹¿ÉÔÙÉú²¢µÃµ½º¬Ó˵ÄÈÜÒº£¬Æä·´Ó¦Ô­Àí Îª                                                           ¡£
(4)°¢Ë¾Æ¥ÁÖ()ÔÚ³±Êª¿ÕÆøÖпɷֽâ³ÉË®ÑîËáºÍ´×Ëá¶øÂÔ´øËá³ô棬¹ÊÃÜ·â±£´æ£¬Óû¯Ñ§·½³Ìʽ±íʾ°¢Ë¾Æ¥ÁÖ±ØÐëÖü²ØÓÚÃܱա¢¸ÉÔï´¦µÄÔ­Òò£º                       £¬´Ë·´Ó¦µÄÀàÐÍÊôÓÚ                                                                ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÁÉÄþÊ¡ÉòÑôÊиßÈýÄ£Ä⿼ÊÔÀí¿Æ×ۺϻ¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(15·Ö)»¯Ñ§ÊÇÈËÀà½ø²½µÄ¹Ø¼ü£¬»¯Ñ§ÎªÈËÀàµÄÉú²ú¡¢Éú»îÌṩÁËÎïÖʱ£Ö¤¡£

(1)µç¶Æʱ£¬ÓöƲã½ðÊô×÷Ñô¼«µÄ×÷ÓÃÊÇ                              £®ÎªÁËʹ¶Æ²ãºñ¶È¾ùÔÈ¡¢¹â»¬ÖÂÃÜ¡¢Óë¶Æ¼þµÄ¸½×ÅÁ¦Ç¿£¬³ý¿ØÖÆÈÜÒºÖÐÀë×ÓŨ¶ÈÍ⣬ͨ³£»¹¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÓР                                              

(2)±ˮÖÐÔ̺¬×ŷḻµÄþ×ÊÔ´£¬¾­×ª»¯ºó¿É»ñµÃMgCl2´Ö²úÆ·¡£´Ó±ˮÖÐÌáȡþµÄ²½ÖèΪ£º

a£®½«º£±ß´óÁ¿´æÔڵı´¿ÇìÑÉÕ³Éʯ»Ò£¬²¢½«Ê¯»ÒÖƳÉʯ»ÒÈ飻

b£®½«Ê¯»ÒÈé¼ÓÈëµ½º£Ë®³Áµí³ØÖо­¹ýÂ˵õ½Mg(OH)2³Áµí£»

c.ÔÚMg(OH)2³ÁµíÖмÓÈëÑÎËáµÃµ½MgCl2ÈÜÒº£¬ÔÙ¾­Õô·¢½á¾§µÃµ½MgCl2¡¤6H2O£»

d£®½«MgCl2¡¤6H2OÔÚÒ»¶¨Ìõ¼þϼÓÈȵõ½ÎÞË®MgCl2£»

e£®µç½âÈÛÈÚµÄÂÈ»¯Ã¾¿ÉµÃµ½Mg¡£

¢Ù²½ÖèdÖеġ°Ò»¶¨Ìõ¼þ¡±Ö¸µÄÊÇ                        £¬Ä¿µÄÊÇ          ¡£

¢ÚÉÏÊöÌáȡþµÄÁ÷³ÌÖÐ,ΪÁ˽µµÍ³É±¾£®¼õÉÙÎÛȾ£¬¿ÉÒÔ²ÉÈ¡ºÜ¶à´ëÊ©£¬Çëд³öÆäÖÐÒ»µã

                                                                 

¢ÛÓÐͬѧÈÏΪ£º²½Öèbºó¿É¼ÓÈÈMg(0H)2µÃµ½Mg0£¬ÔÙµç½âÈÛÈÚµÄMgOÖƽðÊôþ£¬ÕâÑù¿É¼ò»¯ÊµÑé²½Ö裬ÄãͬÒâ¸ÃͬѧµÄÏë·¨Âð?Ϊʲô?

                                                                         ¡£

(3)ÓËÊǺ˷´Ó¦×îÖØÒªµÄȼÁÏ£¬ÒѾ­ÑÐÖƳɹ¦Ò»ÖÖó§ºÏÐÍÀë×Ó½»»»Ê÷Ö¬£¬ËüרÃÅÎü¸½º£Ë®ÖеÄU4+£¬¶ø²»Îü¸½ÆäËûÔªËØ¡£Æä·´Ó¦Ô­ÀíΪ                           (Ê÷Ö¬ÓÃHR´úÌæ)£¬·¢ÉúÀë×Ó½»»»ºóµÄÀë×Ó½»»»Ä¤ÓÃËá´¦Àí»¹¿ÉÔÙÉú²¢µÃµ½º¬Ó˵ÄÈÜÒº£¬Æä·´Ó¦Ô­Àí  Îª                                                            ¡£

(4)°¢Ë¾Æ¥ÁÖ()ÔÚ³±Êª¿ÕÆøÖпɷֽâ³ÉË®ÑîËáºÍ´×Ëá¶øÂÔ´øËá³ô棬¹ÊÃÜ·â±£´æ£¬Óû¯Ñ§·½³Ìʽ±íʾ°¢Ë¾Æ¥ÁÖ±ØÐëÖü²ØÓÚÃܱա¢¸ÉÔï´¦µÄÔ­Òò£º                        £¬´Ë·´Ó¦µÄÀàÐÍÊôÓÚ                                                                 ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º¹óÖÝÊ¡Ä£ÄâÌâ ÌâÐÍ£ºÌî¿ÕÌâ

2010ÄêÑÇÔË»áÔÚÎÒ¹ú¹ãÖݳɹ¦¾Ù°ì£¬Õû¸öÑÇÔË»áÌåÏÖÁË»·±£ÀíÄî
(1)¹ãÖÝÑÇÔË»á»ð¾æ¡°³±Á÷¡±²ÉÓñûÍé(C3H8)×÷ȼÁÏ£¬È¼ÉÕºóµÄ²úÎïΪˮºÍ¶þÑõ»¯Ì¼¡£ÔÚ298Kʱ¡£1mol±ûÍéÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ·Å³ö2221. 5 kJµÄÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ__________________£»
ÓÖÖª£ºÔÚ298Kʱ£¬
C3H8(g)=C3H6(g) +H2(g)£»¡÷H= +124.2 kJ/mol£¬
H2(g)+O2(g)=H2O(l); ¡÷H= -285.8 kJ/mol£¬
Ôòl mol C3H8ÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ___________kJ¡£
(2)¹ãÖÝÊÇÒ»×ùÃÀÀöµÄº£±õ³ÇÊУ¬º£Ë®×ÊÔ´·Ç³£·á¸»¡£
¢Ùº£Ñóµç³ØÊÇÒÔÂÁΪ¸º¼«£¬²¬ÍøΪÕý¼«£¬º£Ë®Îªµç½âÖÊÈÜÒº£¬¿ÕÆøÖеÄÑõÆøÓëÂÁ·´Ó¦²úÉúµçÁ÷£¬µç³Ø×Ü·´Ó¦Îª£º4Al+3O2+6H2O=4Al(OH)3£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_________£¨ÌîдÐòºÅ×Öĸ£©£»
a£®µç³Ø¹¤×÷ʱ£¬µçÁ÷ÓÉÂÁµç¼«Ñص¼ÏßÁ÷Ïò²¬µç¼«
b£®²¬µç¼«²ÉÓÃÍø×´±È¿é×´¸üÀûÓÚO2·Åµç
c£®º£Ë®ÖеÄOH-ÏòÂÁµç¼«·½ÏòÒƶ¯
¢ÚÊÒÎÂÏ£¬ÓöèÐԵ缫µç½â200mL l.5 mol/LʳÑÎË®£¬ µç½â2minʱ£¬Á½¼«¹²ÊÕ¼¯µ½448 mLÆøÌ壨±ê×¼×´¿öÏ£©£¬Ð´³ö¸Ãµç½â·´Ó¦µÄÀë×Ó·½³Ìʽ_________________£¬¼ÙÉèµç½âÇ°ºóÈÜÒºµÄÌå»ý²»±ä£¬Ôòµç½âºó¸ÃÈÜÒºµÄpHΪ_____¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸