(11·Ö)¸ù¾Ý·ÏË®ÖÐËùº¬Óк¦ÎïÖʵIJ»Í¬£¬¹¤ÒµÉÏÓжàÖÖ·ÏË®µÄ´¦Àí·½·¨¡£

£¨1£©¢Ù·ÏË®IÈô²ÉÓÃCO2´¦Àí£¬Àë×Ó·½³ÌʽÊÇ                              ¡£

¢Ú·ÏË®¢ò³£ÓÃÃ÷·¯´¦Àí¡£Êµ¼ùÖз¢ÏÖ·ÏË®ÖеÄc£¨HCO-3£©Ô½´ó£¬¾»Ë®Ð§¹ûÔ½ºÃ£¬ÕâÊÇÒòΪ                                                          ¡£

¢Û·ÏË®IIIÖеĹ¯ÔªËØ´æÔÚÈçÏÂת»¯£º

Hg2++CH4 =CH3Hg++H+£¬ÎÒ¹ú¹æ¶¨£¬Hg2+µÄÅŷűê×¼²»Äܳ¬¹ý0.05 mg£¯L¡£Èôij¹¤³§ÅŷŵķÏË®1 LÖк¬Hg2+ 3¡Á10-7mo1£¬ÊÇ·ñ´ïµ½ÁËÅŷűê×¼      £¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©¡£

¢Ü·ÏË®¢ô³£ÓÃC12Ñõ»¯CN-³ÉCO2ºÍN2£¬Èô²Î¼Ó·´Ó¦µÄC12ÓëCN-µÄÎïÖʵÄÁ¿Ö®±ÈΪ5£º2£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                                            ¡£

£¨2£©»¯Ñ§ÐèÑõÁ¿£¨COD£©¿ÉÁ¿¶ÈË®ÌåÊÜÓлúÎïÎÛȾµÄ³Ì¶È£¬ËüÊÇÖ¸ÔÚÒ»¶¨Ìõ¼þÏ£¬ÓÃÇ¿Ñõ»¯¼Á´¦ÀíË®ÑùʱËùÏûºÄµÄÑõ»¯¼ÁµÄÁ¿£¬»»Ëã³ÉÑõµÄº¬Á¿£¨ÒÔmg£¯L¼Æ£©¡£Ä³Ñо¿ÐÔѧϰС×é²â¶¨Ä³Ë®ÑùµÄ»¯Ñ§ÐèÑõÁ¿£¨COD£©£¬¹ý³ÌÈçÏ£º

I£®È¡V1mLË®ÑùÓÚ׶ÐÎÆ¿£¬¼ÓÈë10.00 mL 0.2500 mol£¯L K2Cr2O7ÈÜÒº¡£

II£®¼ÓËé´ÉƬÉÙÐí£¬È»ºóÂýÂý¼ÓÈëÁòËáËữ£¬»ìºÏ¾ùÔÈ£¬¼ÓÈÈ¡£

III£®·´Ó¦Íê±Ïºó£¬ÀäÈ´£¬¼Óָʾ¼Á£¬ÓÃc mol£¯LÁòËáÑÇÌúï§[(NH4)2Fe(SO4)2]ÈÜÒºµÎ¶¨¡£ÖÕµãʱÏûºÄÁòËáÑÇÌúï§ÈÜÒºV2 mL¡£

¢ÙIÖУ¬Á¿È¡K2Cr207ÈÜÒºµÄÒÇÆ÷ÊÇ            ¡££¨ËáʽµÎ¶¨¹Ü»ò¼îʽµÎ¶¨¹Ü£©

¢Ú¢òÖУ¬Ëé´ÉƬµÄ×÷ÓÃÊÇ                     ¡£

¢ÛIIIÖУ¬·¢ÉúµÄ·´Ó¦Îª£ºCr2O2-7+6Fe2++14H+=2Cr3++6Fe3++7H2O

ÓÉ´Ë¿ÉÖª£¬¸ÃË®ÑùµÄ»¯Ñ§ÐèÑõÁ¿COD=         £¨Óú¬c¡¢V1¡¢V2µÄ±í´ïʽ±íʾ£©¡£

 

£¨1£©  ¢ÙOH-+CO2    HCO3-£¨2·Ö£©

¢ÚHCO3-»á´Ù½øAl3+µÄË®½â£¬Éú³É¸ü¶àµÄAl(OH)3£¬¾»Ë®Ð§¹ûÔöÇ¿£¨2·Ö£©

¢Û·ñ£¨1·Ö£©

¢Ü5Cl2+2CN-+4H2O     10Cl-+2CO2+N2+8H+£¨2·Ö£©

   £¨2£©¢ÙËáʽµÎ¶¨¹Ü£¨1·Ö£©

¢Ú·ÀÖ¹±©·Ð£¨1·Ö£©      ¢Û8000£¨15-cV2£©/V1£¨2·Ö£©

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?Õò½­Ò»Ä££©ÎÛË®´¦Àí·ÖÎöʱ£¬³£ÓÃË«Áòë꣨H2Dz£¬¶þÔªÈõËᣩʹ½ðÊôÀë×ÓÐγɵçÖÐÐÔµÄÂçºÏÎÔÙÓÃCCl4ÝÍÈ¡¸ÃÂçºÏÎ´Ó¶ø°Ñ½ðÊôÀë×Ó´ÓË®ÈÜÒºÖÐÍêÈ«·ÖÀë³öÀ´£®ÈçÓÃË«Áòë꣨H2Dz£©¡«CCl4·ÖÀëÎÛË®ÖеÄCu2+ʱ£¬ÏÈ·¢ÉúCu2++2H2Dz?Cu£¨HDz£©2+2H+·´Ó¦ÔÙ¼ÓÈëCCl4£¬Cu£¨HDz£©2¾ÍºÜÈÝÒ×±»ÝÍÈ¡µ½CCl4ÖУ®
£¨1£©Ð´³öË«ÁòëêºÍFe3+ÂçºÏµÄÀë×Ó·½³Ìʽ£º
Fe3++3H2Dz?Fe£¨HDz£©3+3H+
Fe3++3H2Dz?Fe£¨HDz£©3+3H+
£®ÏÂͼÊÇÓÃË«Áòë꣨H2Dz£©¡«CCl4ÂçºÏÝÍȡijЩ½ðÊôÀë×ÓµÄËá¶ÈÇúÏߣ®Ëü·´Ó³ÁËÝÍȡijЩ½ðÊôÀë×ÓʱÊÊÒ˵ÄpH·¶Î§£®E%±íʾijÖÖ½ðÊôÀë×ÓÒÔÂçºÏÎïÐÎʽ±»ÝÍÈ¡·ÖÀëµÄ°Ù·ÖÂÊ£®

ij¹¤Òµ·ÏË®Öк¬ÓÐHg2+¡¢Bi3+¡¢Zn2+£¬ÓÃË«Áòë꣨H2Dz£©¡«CCl4ÂçºÏÝÍÈ¡·¨´¦Àí·ÏË®£®
Çë¸ù¾ÝÉÏͼ»Ø´ðÎÊÌ⣺
£¨2£©ÓûÍêÈ«½«·ÏË®ÖеÄHg2+·ÖÀë³öÀ´£¬Ðë¿ØÖÆÈÜÒºpH=
1
1

£¨3£©µ±µ÷½ÚpH=2ʱ£¬î飨Bi£©µÄ´æÔÚÐÎʽÓУº
Bi3+¡¢Bi£¨HDz£©3
Bi3+¡¢Bi£¨HDz£©3
£¬ÆäÎïÖʵÄÁ¿Ö®±ÈΪ
3£º2
3£º2
£®
£¨4£©ÝÍÈ¡µ½CCl4ÖеÄZn£¨HDz£©2·ÖÒººó£¬¼ÓÈë×ãÁ¿µÄNaOHÈÜÒº£¬³ä·ÖÕñµ´ºó£¬Ð¿ÓÖתµ½Ë®ÈÜÒºÖУ®Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ£º
Zn£¨HDz£©2+6OH-=Zn£¨OH£©42-+2Dz2-+2H2O
Zn£¨HDz£©2+6OH-=Zn£¨OH£©42-+2Dz2-+2H2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ìÕã½­Ê¡½ð»ªÒ»ÖиßÈý¸ß¿¼Ä£Ä⿼ÊÔ£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

(13·Ö)¢ñ.Ò»Ñõ»¯Ì¼ÊÇÒ»ÖÖÓÃ;Ï൱¹ã·ºµÄ»¯¹¤»ù´¡Ô­ÁÏ¡£
(1)ÀûÓÃÏÂÁз´Ó¦¿ÉÒÔ½«´ÖÄøת»¯Îª´¿¶È´ï99.9%µÄ¸ß´¿Äø¡£
Ni(s)£«4CO(g)Ni(CO)4(g) ¸Ã·´Ó¦µÄ¦¤H  ¡ø 0 (Ñ¡Ìî¡°£¾¡±»ò¡°£½¡±»ò¡°£¼¡±)¡£
(2)ÔÚ¸ßÎÂÏÂÒ»Ñõ»¯Ì¼¿É½«¶þÑõ»¯Áò»¹Ô­Îªµ¥ÖÊÁò¡£ÒÑÖª£º
C(s)£«O2(g)£½CO2(g)          ¦¤H1£½£­393.5 kJ¡¤mol£­1
CO2(g)£«C(s)£½2CO(g)       ¦¤H2£½+ 172.5 kJ¡¤mol£­1
S(s)£«O2(g)£½SO2(g)         ¦¤H3£½£­296.0 kJ¡¤mol£­1
Çëд³öCO³ýSO2µÄÈÈ»¯Ñ§·½³Ìʽ              ¡ø                   ¡£
(3)ÏÂͼÖÐ×óͼÊÇһ̼ËáÑÎȼÁϵç³Ø£¬ËüÒÔCOΪȼÁÏ£¬Ò»¶¨±ÈÀýLi2CO3ºÍNa2CO3µÍÈÛ»ìºÏÎïΪµç½âÖÊ£¬ÓÒͼÊÇ´ÖÍ­¾«Á¶µÄ×°ÖÃͼ£¬ÏÖÓÃȼÁϵç³ØΪµçÔ´½øÐдÖÍ­µÄ¾«Á¶ÊµÑé¡£»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ùд³öA¼«·¢ÉúµÄµç¼«·´Ó¦Ê½              ¡ø                ¡£
¢ÚÒªÓÃȼÁϵç³ØΪµçÔ´½øÐдÖÍ­µÄ¾«Á¶ÊµÑ飬ÔòB¼«Ó¦¸ÃÓë¡ø ¼«(Ì¡°C¡±»ò¡°D¡±)ÏàÁ¬¡£
¢Ûµ±ÏûºÄ2.24 L(±ê¿öÏÂ)COʱ£¬´ÖÍ­µç¼«ÀíÂÛÉϼõÉÙÍ­µÄÖÊÁ¿               ¡ø    (Ì¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡± »ò¡°Ð¡ÓÚ¡±)6.4¿Ë¡£

¢ò.(1)ÒÑÖªNa2CrO4ÈÜÒºËữʱ·¢ÉúµÄ·´Ó¦Îª£º2CrO42£­£«2H£«Cr2O72£­£«H2O£¬Èô1LËữºóËùµÃÈÜÒºÖиõÔªËصÄ×ÜÎïÖʵÄÁ¿Îª0.55 mol£¬CrO42£­ÓÐ10/11ת»¯ÎªCr2O72£­¡£ÓÖÖª£º³£ÎÂʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK£½1014¡£ÉÏÊöËữºóËùµÃÈÜÒºµÄpH£½ ¡ø   ¡£
(2)¸ù¾ÝÓйعú¼Ò±ê×¼£¬º¬CrO42£­µÄ·ÏˮҪ¾­»¯Ñ§´¦Àí£¬Ê¹ÆäŨ¶È½µÖÁ5.0¡Á10£­7 mol¡¤L£­1ÒÔϲÅÄÜÅÅ·Å¡£º¬CrO42£­µÄ·ÏË®´¦Àíͨ³£ÓÐÒÔÏÂÁ½ÖÖ·½·¨¡£
¢Ù³Áµí·¨£º¼ÓÈë¿ÉÈÜÐÔ±µÑÎÉú³ÉBaCrO4³Áµí£ÛKsp(BaCrO4)£½1.2¡Á10£­10£Ý£¬ÔÙ¼ÓÈë¿ÉÈÜÐÔÁòËáÑδ¦Àí¶àÓàµÄBa2£«¡£¼ÓÈë¿ÉÈÜÐÔ±µÑκóµÄ·ÏË®ÖÐBa2£«µÄŨ¶ÈÓ¦²»Ð¡ÓÚ       ¡ø   mol¡¤L£­1£¬È»ºóÔÙ½øÐкóÐø´¦Àí·½ÄÜ´ïµ½¹ú¼ÒÅŷűê×¼¡£
¢Ú»¹Ô­·¨£ºCrO42£­Cr3£«Cr(OH)3¡£Óø÷½·¨´¦Àí10 m3 CrO42£­µÄÎïÖʵÄÁ¿Å¨¶ÈΪ1.0¡Á10¡ª3 mol¡¤L£­1µÄ·ÏË®£¬ÖÁÉÙÐèÒªÂÌ·¯(FeSO4¡¤7H2O£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª278) ¡ø  Kg(±£ÁôÁ½Î»Ð¡Êý)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011ѧÄêÕã½­Ê¡¸ßÈý¸ß¿¼Ä£Ä⿼ÊÔ£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

(13·Ö)¢ñ.Ò»Ñõ»¯Ì¼ÊÇÒ»ÖÖÓÃ;Ï൱¹ã·ºµÄ»¯¹¤»ù´¡Ô­ÁÏ¡£

(1)ÀûÓÃÏÂÁз´Ó¦¿ÉÒÔ½«´ÖÄøת»¯Îª´¿¶È´ï99.9%µÄ¸ß´¿Äø¡£

Ni(s)£«4CO(g)Ni(CO)4(g) ¸Ã·´Ó¦µÄ¦¤H  ¡ø  0 (Ñ¡Ìî¡°£¾¡±»ò¡°£½¡±»ò¡°£¼¡±)¡£

(2)ÔÚ¸ßÎÂÏÂÒ»Ñõ»¯Ì¼¿É½«¶þÑõ»¯Áò»¹Ô­Îªµ¥ÖÊÁò¡£ÒÑÖª£º

C(s)£«O2(g)£½CO2(g)          ¦¤H1£½£­393.5 kJ¡¤mol£­1

CO2(g)£«C(s)£½2CO(g)        ¦¤H2£½+ 172.5 kJ¡¤mol£­1

S(s)£«O2(g)£½SO2(g)          ¦¤H3£½£­296.0 kJ¡¤mol£­1

Çëд³öCO³ýSO2µÄÈÈ»¯Ñ§·½³Ìʽ               ¡ø                    ¡£

(3)ÏÂͼÖÐ×óͼÊÇһ̼ËáÑÎȼÁϵç³Ø£¬ËüÒÔCOΪȼÁÏ£¬Ò»¶¨±ÈÀýLi2CO3ºÍNa2CO3µÍÈÛ»ìºÏÎïΪµç½âÖÊ£¬ÓÒͼÊÇ´ÖÍ­¾«Á¶µÄ×°ÖÃͼ£¬ÏÖÓÃȼÁϵç³ØΪµçÔ´½øÐдÖÍ­µÄ¾«Á¶ÊµÑé¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢Ùд³öA¼«·¢ÉúµÄµç¼«·´Ó¦Ê½                ¡ø                  ¡£

¢ÚÒªÓÃȼÁϵç³ØΪµçÔ´½øÐдÖÍ­µÄ¾«Á¶ÊµÑ飬ÔòB¼«Ó¦¸ÃÓë ¡ø  ¼« (Ì¡°C¡±»ò¡°D¡±)ÏàÁ¬¡£

¢Ûµ±ÏûºÄ2.24 L(±ê¿öÏÂ)COʱ£¬´ÖÍ­µç¼«ÀíÂÛÉϼõÉÙÍ­µÄÖÊÁ¿                ¡ø      (Ì¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡± »ò¡°Ð¡ÓÚ¡±)6.4¿Ë¡£

¢ò.(1)ÒÑÖªNa2CrO4ÈÜÒºËữʱ·¢ÉúµÄ·´Ó¦Îª£º2CrO42£­£«2H£«Cr2O72£­£«H2O£¬Èô1LËữºóËùµÃÈÜÒºÖиõÔªËصÄ×ÜÎïÖʵÄÁ¿Îª0.55 mol£¬CrO42£­ÓÐ10/11ת»¯ÎªCr2O72£­¡£ÓÖÖª£º³£ÎÂʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK£½1014¡£ÉÏÊöËữºóËùµÃÈÜÒºµÄpH£½  ¡ø    ¡£

(2)¸ù¾ÝÓйعú¼Ò±ê×¼£¬º¬CrO42£­µÄ·ÏˮҪ¾­»¯Ñ§´¦Àí£¬Ê¹ÆäŨ¶È½µÖÁ5.0¡Á10£­7 mol¡¤L£­1ÒÔϲÅÄÜÅÅ·Å¡£º¬CrO42£­µÄ·ÏË®´¦Àíͨ³£ÓÐÒÔÏÂÁ½ÖÖ·½·¨¡£

¢Ù³Áµí·¨£º¼ÓÈë¿ÉÈÜÐÔ±µÑÎÉú³ÉBaCrO4³Áµí£ÛKsp(BaCrO4)£½1.2¡Á10£­10£Ý£¬ÔÙ¼ÓÈë¿ÉÈÜÐÔÁòËáÑδ¦Àí¶àÓàµÄBa2£«¡£¼ÓÈë¿ÉÈÜÐÔ±µÑκóµÄ·ÏË®ÖÐBa2£«µÄŨ¶ÈÓ¦²»Ð¡ÓÚ         ¡ø    mol¡¤L£­1£¬È»ºóÔÙ½øÐкóÐø´¦Àí·½ÄÜ´ïµ½¹ú¼ÒÅŷűê×¼¡£

¢Ú»¹Ô­·¨£ºCrO42£­Cr3£«Cr(OH)3¡£Óø÷½·¨´¦Àí10 m3 CrO42£­µÄÎïÖʵÄÁ¿Å¨¶ÈΪ1.0¡Á10¡ª3 mol¡¤L£­1µÄ·ÏË®£¬ÖÁÉÙÐèÒªÂÌ·¯(FeSO4¡¤7H2O£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª278)  ¡ø   Kg(±£ÁôÁ½Î»Ð¡Êý)¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì¸£½¨Ê¡ËĵØÁùУ¸ß¶þÏÂѧÆÚµÚÒ»´ÎÁª¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

(11·Ö)¸ù¾Ý·ÏË®ÖÐËùº¬Óк¦ÎïÖʵIJ»Í¬£¬¹¤ÒµÉÏÓжàÖÖ·ÏË®µÄ´¦Àí·½·¨¡£

£¨1£©¢Ù·ÏË®IÈô²ÉÓÃCO2´¦Àí£¬Àë×Ó·½³ÌʽÊÇ                               ¡£

¢Ú·ÏË®¢ò³£ÓÃÃ÷·¯´¦Àí¡£Êµ¼ùÖз¢ÏÖ·ÏË®ÖеÄc£¨HCO-3£©Ô½´ó£¬¾»Ë®Ð§¹ûÔ½ºÃ£¬ÕâÊÇÒòΪ                                                           ¡£

¢Û·ÏË®IIIÖеĹ¯ÔªËØ´æÔÚÈçÏÂת»¯£º

Hg2++ CH4 =CH3Hg++H+£¬ÎÒ¹ú¹æ¶¨£¬Hg2+µÄÅŷűê×¼²»Äܳ¬¹ý0.05 mg£¯L¡£Èôij¹¤³§ÅŷŵķÏË®1 LÖк¬Hg2+ 3¡Á10-7mo1£¬ÊÇ·ñ´ïµ½ÁËÅŷűê×¼       £¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©¡£

¢Ü·ÏË®¢ô³£ÓÃC12Ñõ»¯CN-³ÉCO2ºÍN2£¬Èô²Î¼Ó·´Ó¦µÄC12ÓëCN-µÄÎïÖʵÄÁ¿Ö®±ÈΪ5£º2£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                                            ¡£

£¨2£©»¯Ñ§ÐèÑõÁ¿£¨COD£©¿ÉÁ¿¶ÈË®ÌåÊÜÓлúÎïÎÛȾµÄ³Ì¶È£¬ËüÊÇÖ¸ÔÚÒ»¶¨Ìõ¼þÏ£¬ÓÃÇ¿Ñõ»¯¼Á´¦ÀíË®ÑùʱËùÏûºÄµÄÑõ»¯¼ÁµÄÁ¿£¬»»Ëã³ÉÑõµÄº¬Á¿£¨ÒÔmg£¯L¼Æ£©¡£Ä³Ñо¿ÐÔѧϰС×é²â¶¨Ä³Ë®ÑùµÄ»¯Ñ§ÐèÑõÁ¿£¨COD£©£¬¹ý³ÌÈçÏ£º

I£®È¡V1mLË®ÑùÓÚ׶ÐÎÆ¿£¬¼ÓÈë10.00 mL 0.2500 mol£¯L K2Cr2O7ÈÜÒº¡£

II£®¼ÓËé´ÉƬÉÙÐí£¬È»ºóÂýÂý¼ÓÈëÁòËáËữ£¬»ìºÏ¾ùÔÈ£¬¼ÓÈÈ¡£

III£®·´Ó¦Íê±Ïºó£¬ÀäÈ´£¬¼Óָʾ¼Á£¬ÓÃc mol£¯LÁòËáÑÇÌúï§[(NH4)2Fe(SO4)2]ÈÜÒºµÎ¶¨¡£ÖÕµãʱÏûºÄÁòËáÑÇÌúï§ÈÜÒºV2 mL¡£

¢ÙIÖУ¬Á¿È¡K2Cr207ÈÜÒºµÄÒÇÆ÷ÊÇ            ¡££¨ËáʽµÎ¶¨¹Ü»ò¼îʽµÎ¶¨¹Ü£©

¢Ú¢òÖУ¬Ëé´ÉƬµÄ×÷ÓÃÊÇ                      ¡£

¢ÛIIIÖУ¬·¢ÉúµÄ·´Ó¦Îª£ºCr2O2-7+6Fe2++14 H+=2Cr3++6Fe3++7H2O

ÓÉ´Ë¿ÉÖª£¬¸ÃË®ÑùµÄ»¯Ñ§ÐèÑõÁ¿COD=         £¨Óú¬c¡¢V1¡¢V2µÄ±í´ïʽ±íʾ£©¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸