ÑÇÁ×ËᣨH3PO3£©ÊǶþÔªËᣬH3PO3ÈÜÒº´æÔÚµçÀëƽºâ£ºH3PO3H+ + H2PO3£­¡£ÑÇÁ×ËáÓë×ãÁ¿NaOHÈÜÒº·´Ó¦£¬ÉúºÍNa2HPO3¡£
£¨1£©¢Ùд³öÑÇÁ×ËáÓëÉÙÁ¿NaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ_____________________________________¡£
¢ÚijζÈÏ£¬0.1000 mol¡¤L£­1µÄH3PO3ÈÜÒºpHµÄ¶ÁÊýΪ1.6£¬¼´´ËʱÈÜÒºÖÐc (H+) = 2.5¡Á10£­2 mol¡¤L£­1£¬³ýOH¡ªÖ®ÍâÆäËûÀë×ÓµÄŨ¶ÈÓÉСµ½´óµÄ˳ÐòÊÇ                      £¬¸ÃζÈÏÂH3PO3µçÀëƽºâµÄƽºâ³£ÊýK=                ¡££¨H3PO3µÚ¶þ²½µçÀëºöÂÔ²»¼Æ£¬½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö£©
¢ÛÏòH3PO3ÈÜÒºÖеμÓNaOHÈÜÒºÖÁÖÐÐÔ£¬ËùµÃÈÜÒºÖÐc£¨Na+£©_______ c£¨H2PO3-£©+ 2c£¨HPO32-£©£¨Ìî¡°>¡±¡¢ ¡°<¡± »ò¡°=¡±£©¡£
£¨2£©ÑÇÁ×Ëá¾ßÓÐÇ¿»¹Ô­ÐÔ£¬¿ÉʹµâË®ÍÊÉ«£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________¡£
£¨3£©µç½âNa2HPO3ÈÜÒºÒ²¿ÉµÃµ½ÑÇÁ×ËᣬװÖÃʾÒâͼÈçÏ£º

˵Ã÷£ºÑôĤֻÔÊÐíÑôÀë×Óͨ¹ý£¬ÒõĤֻÔÊÐíÒõÀë×Óͨ¹ý¡£
¢ÙÒõ¼«µÄµç¼«·´Ó¦Ê½Îª_____________________________¡£
¢Ú²úÆ·ÊÒÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________¡£

£¨14·Ö£©£¨1£©  ¢Ù H3PO3+OH¡ª£½H2PO3¡ª+H2O  £¨2·Ö£©
¢Úc£¨HPO32-£©< c£¨H2PO3-£©< c£¨H+£© £¨2·Ö£©8.3¡Á10£­3mol/L £¨2·Ö£© ¢Û £½£¨2·Ö£©
£¨2£©H3PO3 + I2 +H2O = 2HI+ H3PO£¨2·Ö£© £¨3£©¢Ù 2H+ + 2e£­£½H2¡ü  £¨2·Ö£©  
¢ÚHPO32£­+ 2H+£½H3PO3£¨2·Ö£©»ò£ºHPO32£­+ H+£½H2PO3£­¡¢H2PO3£­+ H+£½H3PO3£¨¸÷1·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¢ÙÑÇÁ×ËáÊǶþÔªËᣬÑÇÁ×ËáºÍÉÙÁ¿ÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉNaH2PO3¡¢H2O£¬ËùÒԸ÷´Ó¦·½³ÌʽΪH3PO3+OH¡ª£½H2PO3¡ª+H2O¡£
¢Ú0.1000mol?L-1µÄH3PO3ÈÜÒºpHµÄ¶ÁÊýΪ1.6£¬ÇâÀë×ÓŨ¶ÈСÓÚÑÇÁ×ËáŨ¶È£¬ËùÒÔÑÇÁ×ËáÊǶþÔªÈõËᣬÔÚË®ÈÜÒºÖзֲ½µçÀ룬ÇÒµÚÒ»²½µçÀë³Ì¶È´óÓÚµÚ¶þ²½£¬Á½²½µçÀëÖж¼ÓÐÇâÀë×ÓÉú³É£¬ËùÒÔÇâÀë×ÓŨ¶È×î´ó£¬Òò´ËÀë×ÓŨ¶È´óС˳ÐòÊÇc£¨HPO32-£©< c£¨H2PO3-£©< c£¨H+£©¡£
H3PO3       H+  +  H2PO3£­
Æðʼʱ¸÷ÎïÖÊŨ¶È£¨mol?L£­1£©     0.10           0       0
·´Ó¦µÄ¸÷ÎïÖʵÄŨ¶È£¨mol?L£­1£©2.5¡Á10£­2           2.5¡Á10£­2   2.5¡Á10£­2
ƽºâʱ¸÷ÎïÖʵÄŨ¶È£¨mol?L£­1£©0.10£­2.5¡Á10£­2   2.5¡Á10£­2     2.5¡Á10£­2  
K£½£½£½8.3¡Á10£­3mol/L
¢ÛÈÜÒº³ÊÖÐÐÔ£¬Ôòc£¨H+£©£½c£¨OH-£©£¬ÓÖÒòΪÈÜÒº³ÊµçÖÐÐÔ£¬Ôòc£¨Na+£©+C£¨H+£©£½C£¨OH-£©+c£¨H2PO3-£©+2c£¨HPO32-£©£¬ÓÉÓÚC£¨H+£©=C£¨OH-£©£¬ËùÒÔc£¨Na+£©£½c£¨H2PO3-£©+2c£¨HPO32-£©¡£
£¨2£©µâ¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÑÇÁ×Ëá¾ßÓÐÇ¿»¹Ô­ÐÔ£¬ËùÒÔÑÇÁ×ËáºÍµâÄÜ·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉÇâµâËáºÍÁ×Ëᣬ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪH3PO3 + I2 +H2O = 2HI+ H3PO4 ¡£
£¨3£©¢Ùµç½â³ØÖÐÒõ¼«µÃµ½µç×Ó£¬·¢Éú»¹Ô­·´Ó¦£¬ËùÒÔÒõ¼«ÉÏÊÇÇâÀë×ӵõç×Ó·¢Éú»¹Ô­·´Ó¦£¬µç¼«·´Ó¦Ê½Îª2H+ + 2e£­£½H2¡ü¡£
¢Ú²úÆ·ÊÒÖÐHPO32-ºÍÇâÀë×Ó½áºÏÉú³ÉÑÇÁ×ËᣬÒò´Ë·´Ó¦Àë×Ó·½³ÌʽΪHPO32£­+ 2H+£½H3PO3¡£
¿¼µã£ºÈõµç½âÖʵĵçÀë¡¢µçÀëƽºâ³£ÊýµÄ¼ÆËã¡¢ÈÜÒºÖÐÖÐÀë×ÓŨ¶È´óС±È½Ï¡¢Ñõ»¯»¹Ô­·´Ó¦·½³ÌʽµÄÊéдÒÔ¼°µç»¯Ñ§Ô­ÀíµÄÓ¦ÓõÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ËáÐÔÌõ¼þÏ£¬ÎýÔÚË®ÈÜÒºÖÐÓÐSn2+¡¢Sn4+Á½ÖÖÖ÷Òª´æÔÚÐÎʽ¡£SnSO4ÊÇÒ»ÖÖÖØÒªµÄÁòËáÑΣ¬¹ã·ºÓ¦ÓÃÓÚ¶ÆÎý¹¤Òµ£¬ÆäÖƱ¸Â·ÏßÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©SnCl2ÓÃÑÎËá¶ø²»ÓÃˮֱ½ÓÈܽâµÄÔ­ÒòÊÇ                                   
¼ÓÈëÎý·ÛµÄ×÷ÓÃÊÇ                                                           
£¨2£©·´Ó¦IÉú³ÉµÄ³ÁµíΪSnO£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                          
£¨3£©¼ìÑé³ÁµíÒѾ­¡°Ï´µÓ¡±¸É¾»µÄ²Ù×÷ÊÇ£º                                       £¨3·Ö£©
£¨4£©·´Ó¦¢òÁòËáµÄ×÷ÓÃÖ®Ò»ÊÇ¿ØÖÆÈÜÒºµÄpH¡£ÈôÈÜÒºÖÐc(Sn2+)=1.0mol¡¤L¡ª1£¬ÔòÊÒÎÂÏÂÓ¦¿ØÖÆÈÜÒºpH              ¡££¨ÒÑÖª£ºKsp[Sn(OH)2]=1.0¡Á10¡ª26£©
£¨5£©ËáÐÔÌõ¼þÏ£¬SnSO4»¹¿ÉÓÃ×÷Ë«ÑõË®µÄÈ¥³ý¼Á£¬ÊÔд³öËù·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º
                                                                             ¡£
£¨6£©³±Êª»·¾³ÖУ¬¶ÆÎýÍ­¼´Ê¹Îý²ãÆÆËðÒ²ÄÜ·ÀÖ¹ÐγÉÍ­ÂÌ£¬Çë½áºÏÓйصÄÔ­Àí½âÊÍÆäÔ­Òò£º
                                                                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(Ò»)£¨1£©¼×ÍéÒ²ÊÇÒ»ÖÖÇå½àȼÁÏ£¬µ«²»ÍêȫȼÉÕʱÈÈЧÂʽµµÍ²¢»á²úÉúÓж¾ÆøÌåÔì³ÉÎÛȾ¡£
ÒÑÖª£º CH4(g) + 2O2(g) £½ CO2(g) + 2H2O(l)   ¦¤H1£½¨D890.3 kJ/mol
2CO (g) + O2(g) £½ 2CO2(g)            ¦¤H2£½¨D566.0 kJ/mol
Ôò¼×Íé²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮʱµÄÈÈЧÂÊÖ»ÊÇÍêȫȼÉÕʱµÄ________±¶£¨¼ÆËã½á¹û±£Áô1λСÊý£©¡£
£¨2£©¼×ÍéȼÁϵç³Ø¿ÉÒÔÌáÉýÄÜÁ¿ÀûÓÃÂÊ¡£ÏÂͼÊÇÀûÓü×ÍéȼÁϵç³Øµç½â50 mL 2 mol/LµÄÂÈ»¯Í­ÈÜÒºµÄ×°ÖÃʾÒâͼ£º

Çë»Ø´ð£º
¢Ù¼×ÍéȼÁϵç³ØµÄ¸º¼«·´Ó¦Ê½ÊÇ________¡£
¢Úµ±Ïß·ÖÐÓÐ0.1 molµç×Óͨ¹ýʱ£¬________£¨Ìî¡°a¡±»ò¡°b¡±£©¼«ÔöÖØ________g¡£
£¨¶þ£©Ï±íÊǼ¸ÖÖÈõµç½âÖʵĵçÀëƽºâ³£Êý¡¢ÄÑÈܵç½âÖʵĠ  
ÈܶȻýKsp (25¡æ)¡£

µç½âÖÊ
ƽºâ·½³Ìʽ
ƽºâ³£ÊýK
Ksp
CH3COOH
CH3COOHCH3COO-£«H+
1.76¡Á10-5
 
H2CO3
H2CO3H+£«HCO3-
HCO3-H+£«CO32-
K1£½4.31¡Á10-7
K2£½5.61¡Á10-11
 
C6H5OH
C6H5OH  C6H5O-£«H+
1.1¡Á10-10
 
H3PO4
H3PO4H+£«H2PO4-
H2PO4-H+£«HPO32-
HPO42-H+£«PO43-
K1£½7.52¡Á10-3
K2£½6.23¡Á10-8
K3£½2.20¡Á10-13
 
NH3¡¤H2O
NH3¡¤H2O NH4+£«OH-
1.76¡Á10-5
 
BaSO4
BaSO4 Ba2+£«SO42-
 
1.07¡Á10-10
BaCO3
BaCO3 Ba2+£«CO32-
 
2.58¡Á10-9
 
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÉÉϱí·ÖÎö£¬Èô¢ÙCH3COOH ¢ÚHCO3-¢ÛC6H5OH ¢ÜH2PO4- ¾ù¿É¿´×÷ËᣬÔòËüÃÇËáÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòΪ__________________________(Ìî±àºÅ)£»
(2)25¡æʱ£¬½«µÈÌå»ýµÈŨ¶ÈµÄ´×ËáºÍ°±Ë®»ìºÏ£¬»ìºÏÒºÖУºc(CH3COO-)______c(NH4+)£»(Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±)
£¨3£©25¡æʱ£¬Ïò10ml 0.01mol/L±½·ÓÈÜÒºÖеμÓVml 0.01mol/L°±Ë®£¬»ìºÏÈÜÒºÖÐÁ£×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ(     )£»
A£®Èô»ìºÏÒºpH£¾7£¬ÔòV¡Ý10
B£®Èô»ìºÏÒºpH£¼7£¬Ôòc((NH4+) £¾c (C6H5O-) £¾c (H+)£¾c (OH£­)
C£®V=10ʱ£¬»ìºÏÒºÖÐË®µÄµçÀë³Ì¶ÈСÓÚ10ml 0.01mol/L±½·ÓÈÜÒºÖÐË®µÄµçÀë³Ì¶È
D£®V=5ʱ£¬2c(NH3¡¤H2O)+ 2 c (NH4+)=" c" (C6H5O-)+ c (C6H5OH)
£¨4£©ÈçÏÂͼËùʾ£¬ÓÐT1¡¢T2Á½ÖÖζÈÏÂÁ½ÌõBaSO4ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏߣ¬»Ø´ðÏÂÁÐÎÊÌ⣺
ÌÖÂÛT1ζÈʱBaSO4µÄ³ÁµíÈܽâƽºâÇúÏߣ¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ(   )

A£®¼ÓÈëNa2SO4¿ÉʹÈÜÒºÓÉaµã±äΪbµã
B£®ÔÚT1ÇúÏßÉÏ·½ÇøÓò(²»º¬ÇúÏß)ÈÎÒâÒ»µãʱ£¬ ¾ùÓÐBaSO4³ÁµíÉú³É
C£®Õô·¢ÈܼÁ¿ÉÄÜʹÈÜÒºÓÉdµã±äΪÇúÏßÉÏa¡¢ bÖ®¼äµÄijһµã(²»º¬a¡¢b)
D£®ÉýοÉʹÈÜÒºÓÉbµã±äΪdµã

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

µ¨·¯ÊÇÒ»ÖÖ³£¼ûµÄ»¯ºÏÎ¹¤ÒµÉÏËüÒ²ÊÇÒ»ÖÖÖÆÈ¡ÆäËüº¬Í­»¯ºÏÎïµÄÔ­ÁÏ£¬ÏÖÓзÏÍ­£¨Ö÷ÒªÔÓÖÊΪFe£©À´ÖƱ¸µ¨·¯¡£ÓÐÈËÉè¼ÆÁËÈçÏÂÁ÷³Ì£º

pHÖµ¿ØÖƿɲο¼ÏÂÁÐÊý¾Ý

ÎïÖÊ
¿ªÊ¼³ÁµíʱµÄpHÖµ
ÍêÈ«³ÁµíʱµÄpHÖµ
ÇâÑõ»¯Ìú
2.7
3.7
ÇâÑõ»¯ÑÇÌú
7.6
9.6
ÇâÑõ»¯Í­
5.2
6.4
 
Çë¸ù¾ÝÉÏÊöÁ÷³Ì»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©A¿ÉÑ¡ÓÃ________£¨Ìî×Öĸ£©
a£®Ï¡H2SO4    b£®Å¨H2SO4¡¢¼ÓÈÈ    c£®Å¨FeCl3ÈÜÒº    d£®Å¨HNO3
£¨2£©¢ñÖмÓH2O2µÄÄ¿µÄ___________________________________________________¡£
£¨3£©¢òÖмÓCu2£¨OH£©2CO3µÄÄ¿µÄÊÇ___________________________________,
ÆäÓŵãÊÇ_____________________________________________________________¡£
£¨4£©¢ó¼ÓÈÈÖó·Ðʱ·¢ÉúµÄ»¯Ñ§·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________________¡£
£¨5£©VÖмÓH2SO4µ÷½ÚpH£½1ÊÇΪÁË_________________________________________£¬
ij¹¤³ÌʦÈÏΪÉÏÊöÁ÷³ÌÖÐËù¼ÓµÄAÎïÖʲ¢²»ÀíÏ룬Ðè×÷¸Ä½ø£¬ÆäÀíÓÉÊÇ______________£¬
ÈôÄãÊǹ¤³Ìʦ£¬½«¶ÔËù¼ÓµÄAÎïÖÊ×÷ºÎ¸Ä½ø?ÇëÌá³ö½¨Òé______________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉÏÒÔ»ÆÍ­¿ó(Ö÷Òª³É·ÖÊÇCuFeS2£¬ÔÓÖʲ»ÈÜÓÚË®ºÍËá)ΪԭÁÏ£¬ÖƱ¸À¶É«¾§ÌåG£¬Æ仯ѧʽΪ[Cu(NH3)4]SO4¡¤H2O£¬Éæ¼°Á÷³ÌÈçÏ£º

ÒÑÖª25¡æʱ£¬¼¸ÖÖ½ðÊôÇâÑõ»¯ÎïµÄÈܶȻý³£ÊýºÍÍêÈ«³ÁµíµÄpH·¶Î§ÈçÏÂ±í£º

 
Fe(OH)2
Cu(OH)2
Fe(OH)3
Ksp
8.0¡Á10-16
2.2¡Á10-22
4.0¡Á10-38
ÍêÈ«³ÁµípH
¡Ý9.6
¡Ý6.4
¡Ý3.2
 
(1)¼Ó¿ì»ÆÍ­¿ó±ºÉÕËÙÂÊ£¬¿É²ÉÓõĴëÊ©ÓР                 (дÁ½ÖÖ)£º
(2)¼ÓÈëË«ÑõË®¿ÉÄÜ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                           £»
ÊÔ¼ÁXµÄ»¯Ñ§Ê½Îª                    ¡£
(3)³£ÎÂÏ£¬0.1 mol£¯LÊÔ¼ÁYµÄpH=11£¬Ôò¸ÃζÈÏ£¬ÊÔ¼ÁYµÄµçÀë³£ÊýΪ         £»
ÓÃpHÊÔÖ½²â¸ÃÈÜÒºpHÖµµÄ·½·¨ÊÇ                             
(4)ÒÑÖªCu(OH)2+4NH3¡¤H2O[Cu(NH3)4]2++2OH-+4H2O£¬Ð´³ö¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽ£º                   ¡£
(5)ÔÚÈÜÒºNÖмÓÈëÒÒ´¼µÄÄ¿µÄÊÇ                                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

£¨17·Ö£©»¯Ñ§·´Ó¦Ô­ÀíÔÚ¿ÆÑкÍÉú²úÖÐÓй㷺ӦÓá£
£¨1£©Ò»¶¨Ìõ¼þÏ£¬Ä£Äâij¿óʯÐγɵķ´Ó¦aW+bQ¡úcN+dP+eRµÃµ½Á½¸öͼÏñ¡£

¢Ù¸Ã·´Ó¦µÄ¡÷H            0£¨Ìî¡°>¡±¡¢¡°=¡±»ò¡°<¡±£©¡£
¢ÚijζÈÏ£¬Æ½ºâ³£Êý±í´ïʽΪK =c2£¨X£©£¬ÔòÓÉͼ£¨2£©Åж¨X´ú±íµÄÎïÖÊΪ____¡£
£¨2£©½«EºÍF¼ÓÈëÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºE£¨g£©+F£¨s£©2G£¨g£©¡£ºöÂÔ
¹ÌÌåÌå»ý£¬Æ½ºâʱGµÄÌå»ý·ÖÊý£¨%£©ËæζȺÍѹǿµÄ±ä»¯ÈçϱíËùʾ£º

ÔòK£¨915¡æ£©ÓëK£¨810¡æ£©µÄ¹ØϵΪK£¨915¡æ£©____K£¨810¡æ£©£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©£¬a¡¢b¡¢fÈýÕߵĴóС¹ØϵΪ       £¬1000¡æ¡¢3£®0 MPaʱEµÄת»¯ÂÊΪ____¡££¨3£©25¡æʱ£¬H2CO3 HCO3£­+H+µÄµçÀë³£ÊýKa=4¡Á10¡ª7 mo1¡¤L£­1£¬Ôò¸ÃζÈÏ£¬NaHCO3µÄË®½â³£ÊýKh=           £¬ÇëÓÃÊʵ±µÄÊÔ¹ÜʵÑéÖ¤Ã÷Na2CO3ÈÜÒºÖдæÔÚCO32£­+H2O  HCO3£­+OH£­µÄÊÂʵ               ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

£¨15·Ö£©¹¤ÒµÉÏÉè¼Æ½«VOSO4ÖеÄK2SO4¡¢SiO2ÔÓÖʳýÈ¥²¢»ØÊյõ½V2O5µÄÁ÷³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²½Öè¢ÙËùµÃ·ÏÔüµÄ³É·ÖÊÇ          £¨Ð´»¯Ñ§Ê½£©£¬²Ù×÷IµÄÃû³Æ        ¡£
£¨2£©²½Öè¢Ú¡¢¢ÛµÄ±ä»¯¹ý³Ì¿É¼ò»¯Îª£¨ÏÂʽR±íʾVO2+£¬HA±íʾÓлúÝÍÈ¡¼Á£©£º
R2(SO4)n (Ë®²ã)+ 2nHA£¨Óлú²ã£©2RAn£¨Óлú²ã£© + nH2SO (Ë®²ã) 
¢ÚÖÐÝÍȡʱ±ØÐë¼ÓÈëÊÊÁ¿¼î£¬ÆäÔ­ÒòÊÇ                                                     ¡£
¢ÛÖÐXÊÔ¼ÁΪ                            ¡£
£¨3£©¢ÜµÄÀë×Ó·½³ÌʽΪ                                      ¡£
£¨4£©25¡æʱ£¬È¡Ñù½øÐÐÊÔÑé·ÖÎö£¬µÃµ½·°³ÁµíÂʺÍÈÜÒºpHÖ®¼ä¹ØϵÈçÏÂ±í£º

pH
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
·°³ÁµíÂÊ%
88.1
94.8
96.5
98.0
98.8
98.8
96.4
93.1
89.3
½áºÏÉÏ±í£¬ÔÚʵ¼ÊÉú²úÖУ¬¢ÝÖмÓÈ백ˮ£¬µ÷½ÚÈÜÒºµÄ×î¼ÑpHΪ             £»Èô·°³ÁµíÂÊΪ93.1%ʱ²»²úÉúFe(OH)3³Áµí£¬ÔòÈÜÒºÖÐc(Fe3+)<           ¡£¡¼ÒÑÖª£º25¡æʱ£¬Ksp[Fe(OH)3]=2.6¡Á10-39¡½
£¨5£©¸Ã¹¤ÒÕÁ÷³ÌÖУ¬¿ÉÒÔÑ­»·ÀûÓõÄÎïÖÊÓР             ºÍ                      ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¼ÆËãÌâ

£¨17·Ö£©Ëæ×ÅÎÒ¹ú¹¤Òµ»¯Ë®Æ½µÄ²»¶Ï·¢Õ¹£¬½â¾öË®¡¢¿ÕÆøÎÛȾÎÊÌâ³ÉΪÖØÒª¿ÎÌâ¡£
£¨1£©Æû³µÎ²ÆøµÄ´óÁ¿ÅÅ·ÅÊÇÔì³É¿ÕÆøÎÛȾµÄÖØÒªÒòËØÖ®Ò»£¬·¢Õ¹È¼Áϵç³ØÆû³µ¿ÉÒÔÓÐЧµØ½â¾öÉÏÊöÎÊÌâ¡£Ö±½Ó¼×´¼È¼Áϵç³Ø(DMFC)²»»á²úÉúÓк¦²úÎÄÜÁ¿×ª»»Ð§ÂʱÈÄÚȼ»úÒª¸ß2¡«3±¶£¬µç³Ø½á¹¹ÈçͼËùʾ£¬c´¦Í¨ÈëµÄÎïÖÊΪΪ______£¬Íâµç·Öеç×Ó´Ó______µ½______(Ìî¡°A¡±»ò¡°B¡±)Òƶ¯£¬Ð´³öµç³Ø¸º¼«µÄµç¼«·´Ó¦·½³Ìʽ              

£¨2£©¹¤Òµ·ÏË®Öг£º¬ÓÐÒ»¶¨Á¿µÄCr2O72£­£¬»á¶ÔÈËÀ༰Éú̬ϵͳ²úÉúºÜ´óË𺦣¬µç½â·¨ÊÇ´¦Àí¸õÎÛȾµÄ³£Ó÷½·¨¡£¸Ã·¨ÓÃFe×öµç¼«µç½âº¬Cr2O72£­µÄËáÐÔ·ÏË®£¬µç½âʱ£¬ÔÚÒõ¼«ÉÏÓдóÁ¿ÆøÅÝÉú³É£¬²¢²úÉúCr(OH)3¡¢Fe(0H)3³Áµí¡£
¢Ù·´Ó¦ÖУ¬1molCr2O72£­ÍêÈ«Éú³ÉCr(OH)3³Áµí£¬Íâµç·ͨ¹ýµç×ÓµÄÎïÖʵÄÁ¿Îª_________ mol¡£
¢Ú³£ÎÂÏ£¬Cr(OH)3µÄÈܶȻý£¬µ±Cr3£«Å¨¶ÈСÓÚ10molʱ¿ÉÈÏΪÍêÈ«³Áµí£¬µç½âÍêÈ«ºó£¬²âµÃÈÜÒºµÄpH=6£¬Ôò¸ÃÈÜÒº¹ýÂ˺óΪ___________£¨Ìî¡°ÄÜ¡±»ò¡°·ñ¡±£©Ö±½ÓÅÅ·Å¡£
£¨3£©º¬°±·ÏË®Ò×Òý·¢Ë®Ì帻ӪÑø»¯¡£ÏòNH4ClÈÜÒºÖмÓÈëÉÙÁ¿NaOH¹ÌÌ壬ÈÜÒºÖÐ________£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©£»25ʱ£¬NH3?H2OµÄµçÀëƽºâ³£Êý£¬¸ÃζÈÏÂ,1molµÄNH4ClÈÜÒºÖÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ijѧÉúÓûÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÑÎËáÀ´²â¶¨Î´ÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÇâÑõ»¯ÄÆÈÜҺʱ£¬Ñ¡Ôñ¼×»ù³È×÷ָʾ¼Á¡£ÇëÌîдÏÂÁпհףº
(1)Óñê×¼ÑÎËáµÎ¶¨´ý²âµÄÇâÑõ»¯ÄÆÈÜҺʱ£¬×óÊÖÎÕËáʽµÎ¶¨¹ÜµÄ»îÈû£¬ÓÒÊÖÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦×¢ÊÓ         ¡£Ö±µ½¼ÓÈëÒ»µÎÑÎËáºó£¬ÈÜÒºÑÕÉ«ÓÉ»ÆÉ«±äΪ³ÈÉ«£¬²¢   ÎªÖ¹¡£
(2)ÏÂÁвÙ×÷ÖпÉÄÜʹËù²âÇâÑõ»¯ÄÆÈÜÒºµÄŨ¶ÈֵƫµÍµÄÊÇ        ¡£

A£®ËáʽµÎ¶¨¹ÜδÓñê×¼ÑÎËáÈóÏ´¾ÍÖ±½Ó×¢Èë±ê×¼ÑÎËá
B£®µÎ¶¨Ç°Ê¢·ÅÇâÑõ»¯ÄÆÈÜÒºµÄ׶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºóδ¸ÉÔï
C£®ËáʽµÎ¶¨¹ÜÔڵζ¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ
D£®¶ÁÈ¡ÑÎËáÌå»ýʱ£¬¿ªÊ¼ÑöÊÓ¶ÁÊý£¬µÎ¶¨½áÊøºó¸©ÊÓ¶ÁÊý
(3)ijѧÉú¸ù¾Ý3´ÎʵÑé·Ö±ð¼Ç¼ÓйØÊý¾ÝÈç±í£º

ÒÀ¾ÝÉϱíÊý¾ÝÁÐʽ¼ÆËã¸ÃÇâÑõ»¯ÄÆÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸