4£®µª¡¢Áס¢Éé¼°Æä»¯ºÏÎïÔÚÉú²ú¡¢Éú»îÖÐÓÐÖØÒªµÄÓÃ;£®
£¨1£©´Æ»Æ£¨As2S3£©ºÍÐۻƣ¨As4S4£©¶¼ÊÇ×ÔÈ»½çÖг£¼ûµÄÉ黯Îһ¶¨Ìõ¼þÏ£¬´Æ»ÆºÍÐۻƵÄת»¯¹ØÏµÈçͼlËùʾ£®
¢Ù·´Ó¦IµÄÀë×Ó·½³ÌʽΪ2As2S3+4H++2Sn2+=As4S4+2H2S+2Sn4+£®
¢Ú·´Ó¦¢òÖУ¬Èô1mol As4S4²Î¼Ó·´Ó¦Ê±£¬×ªÒÆ12mol e-£¬ÔòÎïÖÊaΪS£¨Ìѧʽ£©£®

£¨2£©³£ÎÂÏ£¬º¬Á×΢Á£ÔÚË®ÈÜÒºÖеķֲ¼·ÖÊý£¨Æ½ºâʱij΢Á£µÄŨ¶ÈÕ¼¸÷΢Á£Å¨¶ÈÖ®ºÍµÄ·ÖÊý£©ÓëpHµÄ¹ØÏµÈçͼ2Ëùʾ£®
¢ÙÏòNa2HPO4ÈÜÒºÖмÓÈë×ãÁ¿CaCl2ÈÜÒº£¬ËùµÃÈÜÒºÏÔËáÐÔ£¬Ô­ÒòÊÇ3Ca2++2HPO42-=Ca3£¨PO4£©2+2H+£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£®
¢ÚpH=7µÄÁ×ËáÑÎÈÜÒº¿É×÷Ϊ»º³åÈÜÒº£®¸ÃÈÜÒºÖÐÁ×ÔªËØµÄ´æÔÚÐÎ̬Ö÷ҪΪH2PO4-¡¢HPO42-£®ÅäÖÆ´ËÈÜҺʱ£¬ÐèÒªH3PO4ºÍNaOHµÄÎïÖʵÄÁ¿Ö®±ÈΪ2£º3£®
¢ÛÒÑÖªÇâÑõ»¯ÂÁ½ºÁ£´øÕýµçºÉ£¬¿ÉÎü¸½ÒõÀë×Ó£®pHÔÚ2¡«4Ö®¼ä£¬ÓÃÂÁÑδ¦Àíº¬Á×·Ïˮʱ£¬ËæÈÜÒºpHÔö´ó£¬Îü¸½Ð§¹ûÔöÇ¿£¬½áºÏͼ2½âÊÍ¿ÉÄܵÄÔ­ÒòËæPHÔö´ó£¬H2PO4-º¬Á¿±ä´ó£¬Í¬Ê±Éú³É¸ü¶àÇâÑõ»¯ÂÁ½ºÁ££®
£¨3£©°±Ë®¿ÉÒÔÓëºÜ¶à½ðÊôÀë×ÓÐγÉÅäºÏÎÒÑ֪ijζÈÏ£º
Cu+£¨aq£©+2NH3•H2O£¨aq£©?Cu£¨NH3£©2+£¨aq£©+2H2O£¨1£©K=8.0¡Á1010
¢ÙCuI£¨s£©+2NH3•H2O£¨aq£©?Cu£¨NH3£©2+£¨aq£©+I-£¨aq£©+2H2O£¨1£©K=0.16£®[ÒÑÖªKsp£¨CuI£©=2.0¡Á10-12]
¢ÚÈܽâ19.1g CuI£¨s£©£¬ÖÁÉÙÐèÒª5mol•L-1°±Ë®µÄÌå»ýԼΪ40mL£®

·ÖÎö £¨1£©¢ÙÒÀ¾Ýת»¯¹ØÏµ·ÖÎö¿ÉÖª£¬´Æ»Æ£¨As2S3£©ËáÐÔÈÜÒºÖкÍSn2+Àë×Ó·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉAs4S4¡¢H2S¡¢Sn4+£¬½áºÏµçºÉÊØºã¡¢Ô­×ÓÊØºãÅ䯽ÊéдÀë×Ó·½³Ìʽ£»
¢ÚÈô1molAs4S4·´Ó¦×ªÒÆ28mole-£¬Ôò·´Ó¦ÏûºÄ7molÑõÆø£¬¸ù¾ÝÔ­×ÓÊØºãÊéдÀë×Ó·½³Ìʽ£¬½áºÏµç×ÓÊØºã¼ÆË㣻
£¨2£©¢ÙÏòNa2HPO4ÈÜÒºÖмÓÈë×ãÁ¿CaCl2ÈÜÒº£¬ËùµÃÈÜÒºÏÔËáÐÔ£¬ËµÃ÷·´Ó¦Éú³ÉÁ×Ëá¸Æ³Áµí¡¢ÂÈ»¯ÄƺÍÂÈ»¯Ç⣻
¢ÚpH=7µÄÁ×ËáÑÎÈÜÒº¿É×÷Ϊ»º³åÈÜÒº£®¸ÃÈÜÒºÖÐÁ×ÔªËØµÄ´æÔÚÐÎ̬ÒÀ¾ÝͼÏó±ä»¯·ÖÎöµÃµ½£»Í¼ÏóÖпÉÖªH2PO4-ºÍHPO42-ÎïÖʵÄÁ¿Ïàͬ£¬½áºÏÔªËØÊØºã¼ÆËãµÃµ½Á×ËáºÍÇâÑõ»¯ÄÆÎïÖʵÄÁ¿Ö®±È£»
¢ÛpHÔÚ2¡«4Ö®¼ä£¬ÓÃÂÁÑδ¦Àíº¬Á×·Ïˮʱ£¬ËæÈÜÒºpHÔö´ó£¬H2PO4-º¬Á¿±ä´óÎü¸½Ð§¹ûÔöÇ¿£»
£¨3£©¢ÙCuI£¨s£©+2NH3•H2O£¨aq£©?Cu£¨NH3£©2+£¨aq£©+I-£¨aq£©+2H2O£¨1£©£¬Æ½ºâ³£ÊýK=$\frac{c£¨{I}^{-}£©c£¨Cu£¨N{H}_{3}{£©_{2}}^{+}£©}{{c}^{2}£¨N{H}_{3}•{H}_{2}O£©}$=$\frac{c£¨{I}^{-}£©c£¨Cu£¨N{H}_{3}{£©_{2}}^{+}£©}{{c}^{2}£¨N{H}_{3}•{H}_{2}O£©}$¡Á$\frac{c£¨C{u}^{+}£©}{c£¨C{u}^{+}£©}$=Ksp¡ÁK£»
¢ÚÒÀ¾ÝCuI£¨s£©+2NH3•H2O£¨aq£©?Cu£¨NH3£©2+£¨aq£©+I-£¨aq£©+2H2O£¨1£©¶¨Á¿¹ØÏµ¼ÆË㰱ˮÎïÖʵÄÁ¿£¬c=$\frac{n}{V}$£®

½â´ð ½â£º£¨1£©¢ÙÒÀ¾Ýת»¯¹ØÏµ·ÖÎö¿ÉÖª£¬´Æ»Æ£¨As2S3£©ËáÐÔÈÜÒºÖкÍSn2+Àë×Ó·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉAs4S4¡¢H2S¡¢Sn4+£¬½áºÏµçºÉÊØºã¡¢Ô­×ÓÊØºãÅ䯽ÊéдÀë×Ó·½³ÌʽΪ£º2As2S3+4H++2Sn2+=As4S4+2H2S+2Sn4+£¬
¹Ê´ð°¸Îª£º2As2S3+4H++2Sn2+=As4S4+2H2S+2Sn4+£»
¢ÚÈô1molAs4S4·´Ó¦×ªÒÆ12mole-£¬Ôò·´Ó¦ÏûºÄ3molÑõÆø£¬Ôò·´Ó¦µÄ·½³ÌʽΪ£ºAs4S4+3O2$\frac{\underline{\;Ò»¶¨Ìõ¼þ\;}}{\;}$2As2O3+4S£¬ÔòaΪS£¬
¹Ê´ð°¸Îª£ºS£»
£¨2£©¢ÙÏòNa2HPO4ÈÜÒºÖмÓÈë×ãÁ¿CaCl2ÈÜÒº£¬ËùµÃÈÜÒºÏÔËáÐÔ£¬ËµÃ÷·´Ó¦Éú³ÉÁ×Ëá¸Æ³Áµí¡¢ÂÈ»¯ÄƺÍÂÈ»¯Ç⣬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º3Ca2++2HPO42-=Ca3£¨PO4£©2+2H+£¬
¹Ê´ð°¸Îª£º3Ca2++2HPO42-=Ca3£¨PO4£©2+2H+£»
¢ÚͼÏóÖпÉÖªH2PO4-ºÍHPO42-ÎïÖʵÄÁ¿Ïàͬ£¬ÎïÖÊΪNa2HPO4£¬NaH2PO4 £¬½áºÏÔªËØÊØºã¼ÆËãµÃµ½Á×ËáºÍÇâÑõ»¯ÄÆÎïÖʵÄÁ¿Ö®±È2£º3£¬
¹Ê´ð°¸Îª£ºH2PO4-ºÍHPO42-£»2£º3£»
¢ÛÒÑÖªÇâÑõ»¯ÂÁ½ºÁ£´øÕýµçºÉ£¬¿ÉÎü¸½ÒõÀë×Ó£®pHÔÚ2¡«4Ö®¼ä£¬ÓÃÂÁÑδ¦Àíº¬Á×·Ïˮʱ£¬ËæÈÜÒºpHÔö´ó£¬Îü¸½Ð§¹ûÔöÇ¿£¬Í¼2·ÖÎö¿ÉÖªËæPHÔö´ó£¬H2PO4-º¬Á¿±ä´ó£¬Í¬Ê±Éú³É¸ü¶àÇâÑõ»¯ÂÁ½ºÁ££¬
¹Ê´ð°¸Îª£ºËæPHÔö´ó£¬H2PO4-º¬Á¿±ä´ó£¬Í¬Ê±Éú³É¸ü¶àÇâÑõ»¯ÂÁ½ºÁ££»
£¨3£©¢ÙCu+£¨aq£©+2NH3•H2O£¨aq£©?Cu£¨NH3£©2+£¨aq£©+2H2O£¨1£©K=8.0¡Á1010
CuI£¨s£©+2NH3•H2O£¨aq£©?Cu£¨NH3£©2+£¨aq£©+I-£¨aq£©+2H2O£¨1£©£¬
ƽºâ³£ÊýK=$\frac{c£¨{I}^{-}£©c£¨Cu£¨N{H}_{3}{£©_{2}}^{+}£©}{{c}^{2}£¨N{H}_{3}•{H}_{2}O£©}$=$\frac{c£¨{I}^{-}£©c£¨Cu£¨N{H}_{3}{£©_{2}}^{+}£©}{{c}^{2}£¨N{H}_{3}•{H}_{2}O£©}$¡Á$\frac{c£¨C{u}^{+}£©}{c£¨C{u}^{+}£©}$=Ksp¡ÁK=2.0¡Á10-12¡Á8.0¡Á1010=0.16£¬
¹Ê´ð°¸Îª£º0.16£»
¢ÚÈܽâ19.1g CuI£¨s£©ÎïÖʵÄÁ¿=$\frac{19.1g}{191g/mol}$=0.1mol£¬CuI£¨s£©+2NH3•H2O£¨aq£©?Cu£¨NH3£©2+£¨aq£©+I-£¨aq£©+2H2O£¨1£©£¬ÐèÒªÈÜÖÊÎïÖʵÄÁ¿Îª0.2mol£¬ÖÁÉÙÐèÒª5mol•L-1°±Ë®µÄÌå»ý=$\frac{0.2mol}{5mol/L}$=0.04L=40ml£¬
¹Ê´ð°¸Îª£º40£®

µãÆÀ ±¾Ì⿼²éÁËÑõ»¯»¹Ô­·´Ó¦¡¢Àë×Ó·½³ÌʽÊéд¡¢Í¼Ïó·ÖÎöÅжϡ¢Æ½ºâ³£Êý¼ÆËãµÈ֪ʶµã£¬ÕÆÎÕ»ù´¡×¢Òâ·ÖÎöÌâ¸ÉÐÅÏ¢ÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶȽϴó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

11£®³£ÎÂÏ£¬ÏÂÁÐÈÜÒºÖеÄ΢Á£Å¨¶È¹ØÏµÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®pH=5µÄCH3COOHÈÜÒººÍpH=5µÄNH4ClÈÜÒºÖУ¬c£¨H+£©²»ÏàµÈ
B£®pH=8.3µÄNaHCO3ÈÜÒºÖУºc£¨Na+£©£¾c£¨HCO3-£©£¾c£¨CO32-£©£¾c£¨H2CO3£©
C£®0.1 mol AgClºÍ0.1 mol AgI»ìºÏºó¼ÓÈë1 LË®ÖУ¬ËùµÃÈÜÒºÖУºc£¨Cl-£©=c£¨I-£©
D£®0.2 mol/L CH3COOHÈÜÒºÓë0.1 mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏºóËùµÃÈÜÒºÖУº2c£¨H+£©-2c£¨OH-£©=c£¨CH3COO-£©-c£¨CH3COOH£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®ÏÂÁÐÎïÖÊÖк¬Ì¼ÖÊÁ¿·ÖÊý×î¸ßµÄÊÇ£¨¡¡¡¡£©
A£®±½B£®ÐÁÏ©C£®ÐÁÍéD£®Ê®ÁùÍé

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®Ä³¾ùÔÈ»ìºÏµÄ°×É«·ÛÄ©¿ÉÄܺ¬ÓÐÏÂÁÐ7ÖÖ³£¼ûÎïÖÊÖеÄij4ÖÖ£ºNaCl¡¢BaCl2¡¢Mg£¨NO3£©2¡¢K2CO3¡¢CuSO4¡¢NaHCO3¡¢X£¬»ìºÏÎïÖи÷³É·ÖµÄÎïÖʵÄÁ¿ÏàµÈ£¬ÇÒËùº¬Òõ¡¢ÑôÀë×ÓÊýĿ֮±ÈΪ3£º2£¬Ä³Í¬Ñ§×öÁËÒÔÏÂʵÑ飬²¿·ÖÏÖÏóûÓÐÃèÊö£®¸ù¾ÝËùÌṩµÄʵÑéÏÖÏó£¬ÏÂÁжԸð×É«·ÛÄ©µÄÅжÏÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Ò»¶¨º¬ÓÐBaCl2¡¢Mg£¨NO3£©2¡¢NaHCO3¡¢X£¬ÇÒXÊÇNH4HSO4
B£®Ò»¶¨²»º¬K2CO3¡¢CuSO4£¬XÊÇNH4Al£¨SO4£©2
C£®Ò»¶¨º¬ÓÐBaCl2¡¢NaHCO3¡¢XÇÒXÊÇNH4AlO2£¬ÁíÒ»ÖÖΪNaClºÍMg£¨NO3£©2ÖеÄÈÎÒ»ÖÖ
D£®Ò»¶¨º¬ÓÐNaCl¡¢BaCl2¡¢NaHCO3ºÍX£¬µ±X²»ÄÜÈ·¶¨ÊÇʲôÎïÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

19£®ÓÉºÏ³ÉÆøÖƱ¸ÒÒ´¼Ò»°ãÓÐÁ½ÖÖ·½·¨£º
£¨1£©Ö±½ÓºÏ³É·¨
¢Ù2CO2£¨g£©+6H2£¨g£©?CH3CH2OH£¨g£©+3H2O£¨g£©¡÷H1
¢Ú2CO£¨g£©+4H2£¨g£©?CH3CH2OH£¨g£©+H2O£¨g£©¡÷H2=-253.6kJ•mol-1
£¨2£©¼ä½ÓºÏ³É·¨
ºÏ³ÉÆø·´Ó¦Éú³É¶þ¼×ÃÑ£¨CH3OCH3£©£¬¶þ¼×ÃÑôÊ»ù»¯ºÏÉú³ÉÒÒËá¼×õ¥£¬ÒÒËá¼×õ¥¼ÓÇâµÃµ½ÒÒ´¼£®ÆäÉú²úÁ÷³ÌÈçÏÂͼËùʾ£º

¢Û3CO£¨g£©+3H2£¨g£©?CH3OCH3£¨g£©+CO2¡÷H3=-260.2kJ•mol-1
¢ÜCH3OCH3£¨g£©+CO£¨g£©?CH3COOCH3£¨g£©
¢ÝCH3COOCH3£¨g£©+2H2£¨g£©?CH3OH£¨g£©+CH3CH2OH£¨g£©
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÔÚ5MPa¡¢ÆðʼͶÁÏÁ¿n£¨H2£©£ºn£¨CO2£©=3£º1ʱ£¬²âµÃ²»Í¬Î¶Èʱ·´Ó¦¢ÙÖи÷ÎïÖÊµÄÆ½ºâ×é³ÉÈçͼ1Ëùʾ£º

¢Ù·´Ó¦¢ÙµÄ¡÷H1£¼0£¬¡÷S£¼0£®£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©
¢ÚÒÑÖªCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©¡÷H=-41kJ•mol-1£¬Ôò¡÷H1=-171.6kJ•mol-1£®
¢Û500Kʱ£¬·´Ó¦´ïµ½Æ½ºâºó£¬ÔÚt1minʱÉýε½600K£¬·´Ó¦ÔÚt2minÖØÐ´ﵽƽºâ£¬ÇëÔÚͼ2Öл­³öÌåϵÖÐË®µÄÌå»ý·ÖÊýÔÚt1minºóµÄ±ä»¯ÇúÏߣ®
£¨2£©ÔÚ2Mpa¡¢T¡æÊ±1LµÄºãÈÝÈÝÆ÷ÖУ¬³äÈë1.0molÒÒËá¼×õ¥ºÍ2.0molÇâÆø£¬·¢Éú·´Ó¦¢Ý£¬¾­t minºó·´Ó¦´ïµ½Æ½ºâ£¬²âµÃÒÒËá¼×õ¥µÄת»¯ÂÊΪ75%£®ÔòÇâÆøµÄ·´Ó¦ËÙÂÊv£¨H2£©=1.5/tmol•L-1•min-1£¬Æ½ºâ³£ÊýK=9£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

9£®°±µªÊÇÔì³ÉË®Ìå¸»ÓªÑø»¯µÄÖØÒªÒòËØÖ®Ò»£¬ÓôÎÂÈËáÄÆË®½âÉú³ÉµÄ´ÎÂÈËὫˮÖеݱµª£¨ÓÃNH3±íʾ£©×ª»¯ÎªµªÆø³ýÈ¥£¬ÆäÏà¹Ø·´Ó¦µÄÖ÷ÒªÈÈ»¯Ñ§·½³ÌʽÈçÏ£º
·´Ó¦¢Ù£ºNH3£¨aq£©+HClO£¨aq£©¨TNH2Cl£¨aq£©+H2O£¨I£©¡÷H1=akJ•mol-1
·´Ó¦¢Ú£ºNH2Cl£¨aq£©+HClO£¨aq£©¨TNHCl2£¨aq£©+H2O£¨I£©¡÷H2=bkJ•mol-1
·´Ó¦¢Û£º2NHCl2£¨aq£©+H2O£¨I£©¨TN2£¨g£©+HClO£¨aq£©+3HCl£¨aq£©¡÷H3=ckJ•mol-1
£¨1£©2NH3£¨aq£©+HClO£¨aq£©¨TN2£¨g£©+3H2O£¨I£©+3HCl£¨aq£©µÄ¡÷H=£¨2a+2b+c£©kJ/mol£®
£¨2£©ÒÑÖªÔÚË®ÈÜÒºÖÐNH2Cl½ÏÎȶ¨£¬NHCl2²»Îȶ¨Ò×ת»¯ÎªµªÆø£®ÔÚÆäËûÌõ¼þÒ»°ãµÄÇé¿öÏ£¬¸Ä±ä$\frac{n£¨NaClO£©}{n£¨N{H}_{3}£©}$£¨¼´NaClOÈÜÒºµÄͶÈëÒº£©£¬ÈÜÒºÖдÎÂÈËáÄÆÈ¥³ý°±µªÐ§¹ûÓëÓàÂÈ£¨ÈÜÒºÖÐ+1¼ÛÂÈÔªËØµÄº¬Á¿£©Ó°ÏìÈçͼ1Ëùʾ£®aµã֮ǰÈÜÒºÖз¢ÉúµÄÖ÷Òª·´Ó¦ÎªA£®
A£®·´Ó¦¢ÙB£®·´Ó¦¢Ù¢ÚC£®·´Ó¦¢Ù¢Ú¢Û
·´Ó¦Öа±µªÈ¥³ýЧ¹û×î¼ÑµÄ$\frac{n£¨NaClO£©}{n£¨N{H}_{3}£©}$ֵԼΪ1.5£®

£¨3£©ÈÜÒºpH¶Ô´ÎÂÈËáÄÆÈ¥³ý°±µªÓнϴóµÄÓ°Ï죨Èçͼ2Ëùʾ£©£®ÔÚpH½ÏµÍʱÈÜÒºÖÐÓÐÎÞÉ«ÎÞζµÄÆøÌåÉú³É£¬°±µªÈ¥³ýЧÂʽϵͣ¬ÆäÔ­ÒòÊÇÑÎËáÓë´ÎÂÈËáÄÆ·´Ó¦Éú³É´ÎÂÈËᣬ´ÎÂÈËá·Ö½âÉú³ÉÑõÆø£¬Óë°±µª·´Ó¦µÄ´ÎÂÈËá¼õС£®

£¨4£©Óõ绯ѧ·¨Ò²¿ÉÒÔÈ¥³ý·ÏË®Öа±µª£®ÔÚÕôÁóË®ÖмÓÈëÁòËáï§ÓöèÐԵ缫ֱ½Óµç½â·¢ÏÖ°±µªÈ¥³ýЧÂʼ«µÍ£¬µ«ÔÚÈÜÒºÖÐÔÙ¼ÓÈëÒ»¶¨Á¿µÄÂÈ»¯Äƺó£¬È¥³ýЧÂÊ¿ÉÒÔ´ó´óÌá¸ß£®·´Ó¦×°ÖÃÈçͼ3Ëùʾ£¬bΪµç¼«Õý¼«£¬µç½âʱÒõ¼«µÄµç¼«·´Ó¦Ê½Îª2H2O+2NH4++2e-=2NH3•H2O+H2¡ü»ò2H++2e-=H2¡ü£®
£¨5£©ÂÈ»¯Á×ËáÈýÄÆ£¨Na2PO4•0.25NaClO•12H2O£©¿ÉÓÃÓÚ¼õСˮµÄÓ²¶È£¬Ïà¹ØÔ­Àí¿ÉÓÃÏÂÁÐÀë×Ó·½³Ìʽ±íʾ£º
3CaSO4£¨s£©+2PO43-£¨aq£©?Ca3£¨PO4£©2£¨s£©+3SO42-£¨aq£©£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK=3.77¡Á1013£®
[ÒÑÖªKsp[Ca3£¨PO4£©2]=2.0¡Á10-29£¬Ksp£¨CaSO4£©=9.1¡Á10-6]£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

16£®XÊÇÒ»ÖÖÐÂÐÍÎÞ»ú·Ç½ðÊô²ÄÁÏ£¬¾ßÓÐÄÍÄ¥¡¢Ä͸¯Ê´¡¢¿¹ÀäÈȳå»÷ÐÔ£®ÓйØÉú²ú¹ý³ÌÈçÏ£º

ΪÁËÈ·¶¨CµÄ×é³É£¬Ä³Í¬Ñ§½øÐÐÁËÒÔϵÄ̽¾¿¹ý³Ì£®ÒÑÖªF¡¢G¶¼ÊÇÄÑÈÜÓÚË®ºÍÏ¡ÏõËáµÄ°×É«³Áµí£¬I¿É×ö¹âµ¼ÏËά£®

°´ÒªÇ󻨴ðÏÂÁÐÎÊÌ⣺
£¨1£©CµÄ¿Õ¼ä¹¹ÐÍΪÕýËÄÃæÌ壻XµÄ»¯Ñ§Ê½ÎªSi3N4£®
£¨2£©·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽΪSiCl4+8NH3¨TSi£¨NH2£©4+4NH4Cl£®
£¨3£©·´Ó¦¢ßµÄÀë×Ó·½³ÌʽSiO32-+2H2O+2CO2¨TH2SiO3¡ý+2HCO3-£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ

13£®Ä³ÈÜÒºÖпÉÄܺ¬ÓÐÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄÏÂÁÐÀë×ÓÖеļ¸ÖÖ£ºFe3+¡¢Cu2+¡¢Na+¡¢SO42-¡¢CO32-¡¢NO3-¡¢Cl-¡¢I-£¬È¡ÉÏÊöÈÜÒºÉÙÁ¿·Ö±ð½øÐÐÈçÏÂʵÑ飺
¢ÙÒ»·ÝÖмÓÈëÉÙÁ¿Ð¿·Ûºó£¬¹ÌÌåÈܽ⡢ÈÜÒºÑÕÉ«Óб仯£¬µ«ÎÞÆäËüÏÖÏó²úÉú
¢ÚÒ»·Ý¼ÓÈëBaCl2ÈÜÒººóµÃµ½°×É«³Áµí£¬ÏÂÁÐÓйØÅжÏÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÈÜÒºÖÐÖ»º¬Fe3+¡¢SO42-¡¢Cl-
B£®ÈÜÒºÖÐÖ»º¬Cu2+¡¢SO42-
C£®ÐèÒªÀûÓÃÑæÉ«·´Ó¦²ÅÄÜÈ·¶¨ÓÐÎÞNa+
D£®ÈÜÒºÖп϶¨Ã»ÓÐI-£¬µ«ÎÞ·¨È·¶¨ÓÐÎÞCl-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

14£®¹ý¶ÉÔªËØTi¡¢Mn¡¢Fe¡¢CuµÈ¿ÉÓëC¡¢H¡¢OÐγɶàÖÖ»¯ºÏÎÇë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸ù¾ÝÔªËØÔ­×ÓµÄÍâΧµç×ÓÅŲ¼µÄÌØÕ÷£¬¿É½«ÔªËØÖÜÆÚ±í·Ö³ÉÎå¸öÇøÓò£¬ÆäÖÐMnÊôÓÚdÇø£®
£¨2£©Ti£¨BH4£©2ÊÇÒ»ÖÖ¹ý¶ÉÔªËØÅðÇ⻯Îï´¢Çâ²ÄÁÏ£®»ù̬Ti2+Öеç×ÓÕ¼¾ÝµÄ×î¸ßÄܲã·ûºÅΪM£¬¸ÃÄܲã¾ßÓеÄÔ­×Ó¹ìµÀÊýΪ9£®BH4-µÄ¿Õ¼ä¹¹ÐÍÊÇÕýËÄÃæÌåÐÍ£®
£¨3£©ÔÚCuµÄ´ß»¯×÷ÓÃÏ£¬ÒÒ´¼¿É±»¿ÕÆøÖÐÑõÆøÑõ»¯ÎªÒÒÈ©£¬ÒÒÈ©·Ö×ÓÖÐ̼ԭ×ÓµÄÔÓ»¯·½Ê½ÊÇsp2¡¢sp3£¬ÒÒÈ©·Ö×ÓÖÐHCOµÄ¼ü½Ç´óÓÚÒÒ´¼·Ö×ÓÖÐH-C-OµÄ¼ü½Ç£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©£®
£¨4£©µç¶Æ³§ÅŷŵķÏË®Öг£º¬Óо綾µÄCN-£¬¿ÉÔÚTiO2µÄ´ß»¯Ï£¬ÏÈÓÃNaClO½«CN-Ñõ»¯³ÉCNO-£¬ÔÙÔÚËáÐÔÌõ¼þÏÂCNO-¼ÌÐø±»NaClOÑõ»¯³ÉN2ºÍCO2£®
¢ÙH¡¢C¡¢N¡¢OËÄÖÖÔªËØµÄµç¸ºÐÔÓÉСµ½´óµÄ˳ÐòΪH£¼C£¼N£¼O£®
¢ÚÓëCN-»¥ÎªµÈµç×ÓÌå΢Á£µÄ»¯Ñ§Ê½ÎªCO£¨»òN2µÈ£©£¨Ð´³öÒ»ÖÖ¼´¿É£©
£¨5£©µ¥ÖÊÌúÓЦġ¢¦Ã¡¢¦ÁÈýÖÖÍ¬ËØÒìÐÎÌ壬ÈýÖÖ¾§°ûÖÐFeÔ­×ÓµÄÅäλÊýÖ®±ÈΪ4£º6£º3£¬ÈýÖÖ¾§°ûÖÐÀⳤ֮±ÈΪ$\frac{1}{\sqrt{3}}$£º$\sqrt{2}$£º1£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸