ÌâÄ¿ÄÚÈÝ

£¨15·Ö£©
¢ñ£®£¨6·Ö£©¹âºÏ×÷ÓÃÊÇÓɶþÑõ»¯Ì¼ÓëË®ºÏ³ÉÆÏÌÑÌÇ£¬Í¬Ê±·Å³öÑõÆø¡£ÉèÏ뽫¸Ã·´Ó¦µ¹¹ýÀ´£¬¿ÉÉè¼Æ³ÉÒ»¸öÔ­µç³Ø£¬²úÉúµçÁ÷£¬ÕâÑù¾Í½«Ì«ÑôÄÜת±ä³ÉÁ˵çÄÜ¡£
£¨1£©Ð´³ö¸ÃÔ­µç³ØµÄ×Ü·´Ó¦Ê½£º______________________________________
£¨2£©Ð´³ö¸Ãµç³ØÔÚËáÐÔ½éÖÊÖзŵçµÄµç¼«·´Ó¦Ê½£º
¸º¼«£º_____________________________________________
Õý¼«£º_____________________________________________
¢ò£®£¨6·Ö£©Àà±È˼άÊÇѧϰ»¯Ñ§µÄÖØÒª·½·¨£¬µ«½á¹ûÊÇ·ñÕýÈ·±ØÐë¾­ÊܼìÑé¡£ÔÚ½øÐÐÀà±È˼άµÄʱºò£¬²»ÄÜ»úеÀà±È£¬Ò»¶¨Òª×¢ÒâһЩÎïÖʵÄÌØÊâÐÔ£¬ÒÔ·ÀÖ¹Àà±È³ö´íÎóµÄ½áÂÛ¡£Æ¾ÒÑÓеĻ¯Ñ§ÖªÊ¶£¬ÏÂÁÐÀà±È½á¹ûÕýÈ·µÄÊÇ£º£¨Ìî±êÐòºÅ£©                   ¡£Èô´íÎó£¬ÔÚÆäºóд³öÕýÈ·µÄ¡£
¢ÙÔÚÏàͬÌõ¼þÏ£¬Na2CO3Èܽâ¶È±ÈNaHCO3´ó
Àà±È£ºÔÚÏàͬÌõ¼þÏ£¬CaCO3Èܽâ¶È±ÈCa£¨HCO3£©2´ó
ÕýÈ·µÄÓ¦¸ÃΪ£¨ÈôÀà±ÈÕýÈ·£¬´Ë´¦²»Ð´£¬ÏÂͬ¡££©£º                              ¡£
¢ÚÏò´ÎÂÈËá¸ÆÈÜÒºÖÐͨ¹ýÁ¿CO2£ºCO2 + ClO- + H2O = HCO3- + HClO
Àà±È£ºÏò´ÎÂÈËáÄÆÈÜÒºÖÐͨ¹ýÁ¿SO2£ºSO2 + ClO- +H2O = HSO3- + HClO
ÕýÈ·µÄÓ¦¸ÃΪ£º                                           ¡£
¢Û¸ù¾Ý»¯ºÏ¼ÛFe3O4¿É±íʾΪ£ºFeO¡¤Fe2O3    Àà±È£ºFe3I8Ò²¿É±íʾΪFeI2¡¤2FeI3
ÕýÈ·µÄÓ¦¸ÃΪ£º                                ¡£
¢ÜCaC2ÄÜË®½â£ºCaC2+2H2O =Ca£¨OH£©2 + C2H2¡ü
Àà±È£ºAl4C3Ò²ÄÜË®½â£ºAl4C3 + 12H2O = 4Al£¨OH£©3¡ý+ 3CH4¡ü
ÕýÈ·µÄÓ¦¸ÃΪ£º                                ¡£
¢ó£®£¨£³·Ö£©³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓУºNa+¡¢A-¡¢H+¡¢OH-¡£Èô¸ÃÈÜÒºMÓÉ pH=3µÄHAÈÜÒºmLÓëpH=11µÄNaOHÈÜÒºmL»ìºÏ·´Ó¦¶øµÃ£¬ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ                     £¨Ìî×Öĸ£©¡£

A£®ÈôÈÜÒºM³ÊÖÐÐÔ£¬ÔòÈÜÒºMÖÐC£¨H+£©+C£¨OH-£©=2¡Á10-7mol¡¤L-1
B£®ÈôV1=V2£¬ÔòÈÜÒºMµÄpHÒ»¶¨µÈÓÚ7
C£®ÈôÈÜÒºM³ÊËáÐÔ£¬ÔòV1Ò»¶¨´óÓÚV2
D£®ÈôÈÜÒºM³Ê¼îÐÔ£¬ÔòV1Ò»¶¨Ð¡ÓÚV2

£¨15·Ö£©
¢ñ£®£¨6·Ö£©
£¨1£©C6H12O6£«6O26CO2£«6H2O£¨£²·Ö£©
£¨2£©C6H12O6£«6H2O£­24e£­= 6CO2£«24H£«£¨£²·Ö£© 6O2£«24e£­£«24H£«= 12H2O£¨£²·Ö£©
¢ò£®£¨6·Ö£©¢Ü£¨1·Ö£©
¢ÙÔÚÏàͬÌõ¼þÏ£¬CaCO3Èܽâ¶È±ÈCa£¨HCO3£©2С£¨1·Ö£©
¢Ú ClO-­ + SO2 + H2O ==SO42- + Cl+ 2 H+£¨2·Ö£©¢Û3FeI2¡¤I2£¨2·Ö£©
¢ó£®£¨£³·Ö£©AD

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2012?ÉÇͷһģ£©ÓÉ̼µÄÑõ»¯ÎïÖ±½ÓºÏ³ÉÒÒ´¼È¼ÁÏÒѽøÈë´ó¹æÄ£Éú²ú£®ÈçͼÊÇÓɶþÑõ»¯Ì¼ºÏ³ÉÒÒ´¼µÄ¼¼ÊõÁ÷³Ì£º

ÎüÊÕ³ØÖÐÊ¢Óб¥ºÍ̼Ëá¼ØÈÜÒº£¬°Ñº¬ÓжþÑõ»¯Ì¼µÄ¿ÕÆø´µÈëÎüÊÕ³ØÖУ¬ÎüÊÕ³ØÖз´Ó¦Òº½øÈë·Ö½â³Øºó£¬Ïò·Ö½â³ØÖÐͨÈë¸ßÎÂË®ÕôÆû£¬°Ñ¶þÑõ»¯Ì¼´ÓÈÜÒºÖÐÌáÈ¡³öÀ´£¬ÔںϳÉËþÖкÍÇâÆø¾­»¯Ñ§·´Ó¦Ê¹Ö®±äΪ¿ÉÔÙÉúȼÁÏÒÒ´¼£®   »Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÎüÊÕ³ØÖз´Ó¦µÄÀë×Ó·½³Ìʽ
CO2+CO32-+H2O=2HCO3-
CO2+CO32-+H2O=2HCO3-
£®
£¨2£©´Ó·Ö½â³ØÖÐÑ­»·Ê¹ÓõÄÎïÖÊÊÇ
K2CO3
K2CO3
£®
£¨3£©¹¤ÒµÉÏ»¹²ÉÈ¡ÒÔCOºÍH2ΪԭÁϺϳÉÒÒ´¼£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ£º
2CO£¨g£©+4H2£¨g£©?CH3CH2OH£¨g£©+H2O£¨g£© Çëд³ö¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽK=
c(CH3CH2OH)?c(H2O)
c2(CO)?c4(H2)
c(CH3CH2OH)?c(H2O)
c2(CO)?c4(H2)
£®
£¨4£©ÔÚÏàͬÌõ¼þÏ£¬ÓÉCOÖÆÈ¡CH3CH2OHµÄƽºâ³£ÊýÔ¶Ô¶´óÓÚÓÉCO2ÖÆÈ¡CH3CH2OH µÄƽºâ³£Êý£®ÔòÓÉCOÖÆÈ¡CH3CH2OHµÄÓŵãÊÇ
Ô­ÁÏÓнϴóµÄת»¯ÂÊ
Ô­ÁÏÓнϴóµÄת»¯ÂÊ
£¬
ÓÉCO2ÖÆÈ¡CH3CH2OHµÄÓŵãÊÇ
CO2Ô­ÁÏÒ×µÃ
CO2Ô­ÁÏÒ×µÃ
£®£¨Ð´³öÒ»µã¼´¿É£©
£¨5£©ÔÚÒ»¶¨Ñ¹Ç¿Ï£¬²âµÃÓÉCO2ÖÆÈ¡CH3CH2OHµÄʵÑéÊý¾ÝÈçÏÂ±í£º
ζȣ¨K£©
CO2ת»¯ÂÊ£¨%£©
n£¨H2£©/n£¨CO2£©
500 600 700 800
1.5 45 33 20 12
2 60 43 28 15
3 83 62 37 22
¸ù¾Ý±íÖÐÊý¾Ý·ÖÎö£º
¢ÙζÈÉý¸ß£¬¸Ã·´Ó¦µÄƽºâ³£ÊýKÖµ
¼õС
¼õС
£¨Ñ¡Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£®
¢ÚÌá¸ßÇâ̼±È(
n(H2)
n(CO2)
)
£¬¶ÔÉú³ÉÒÒ´¼
ÓÐÀû
ÓÐÀû
£¨Ñ¡Ìî¡°²»Àû¡±¡¢¡°ÓÐÀû¡±»ò¡°ÎÞÓ°Ï족£©

£¨15·Ö£©

¢ñ£®£¨6·Ö£©¹âºÏ×÷ÓÃÊÇÓɶþÑõ»¯Ì¼ÓëË®ºÏ³ÉÆÏÌÑÌÇ£¬Í¬Ê±·Å³öÑõÆø¡£ÉèÏ뽫¸Ã·´Ó¦µ¹¹ýÀ´£¬¿ÉÉè¼Æ³ÉÒ»¸öÔ­µç³Ø£¬²úÉúµçÁ÷£¬ÕâÑù¾Í½«Ì«ÑôÄÜת±ä³ÉÁ˵çÄÜ¡£

£¨1£©Ð´³ö¸ÃÔ­µç³ØµÄ×Ü·´Ó¦Ê½£º______________________________________

£¨2£©Ð´³ö¸Ãµç³ØÔÚËáÐÔ½éÖÊÖзŵçµÄµç¼«·´Ó¦Ê½£º

¸º¼«£º_____________________________________________

Õý¼«£º_____________________________________________

¢ò£®£¨6·Ö£©Àà±È˼άÊÇѧϰ»¯Ñ§µÄÖØÒª·½·¨£¬µ«½á¹ûÊÇ·ñÕýÈ·±ØÐë¾­ÊܼìÑé¡£ÔÚ½øÐÐÀà±È˼άµÄʱºò£¬²»ÄÜ»úеÀà±È£¬Ò»¶¨Òª×¢ÒâһЩÎïÖʵÄÌØÊâÐÔ£¬ÒÔ·ÀÖ¹Àà±È³ö´íÎóµÄ½áÂÛ¡£Æ¾ÒÑÓеĻ¯Ñ§ÖªÊ¶£¬ÏÂÁÐÀà±È½á¹ûÕýÈ·µÄÊÇ£º£¨Ìî±êÐòºÅ£©                   ¡£Èô´íÎó£¬ÔÚÆäºóд³öÕýÈ·µÄ¡£

            ¢ÙÔÚÏàͬÌõ¼þÏ£¬Na2CO3Èܽâ¶È±ÈNaHCO3´ó

            Àà±È£ºÔÚÏàͬÌõ¼þÏ£¬CaCO3Èܽâ¶È±ÈCa£¨HCO3£©2´ó

            ÕýÈ·µÄÓ¦¸ÃΪ£¨ÈôÀà±ÈÕýÈ·£¬´Ë´¦²»Ð´£¬ÏÂͬ¡££©£º                              ¡£

            ¢ÚÏò´ÎÂÈËá¸ÆÈÜÒºÖÐͨ¹ýÁ¿CO2£ºCO2 +ClO- + H2O = HCO3- + HClO

            Àà±È£ºÏò´ÎÂÈËáÄÆÈÜÒºÖÐͨ¹ýÁ¿SO2£ºSO2 + ClO- + H2O= HSO3- + HClO

            ÕýÈ·µÄÓ¦¸ÃΪ£º                                           ¡£

            ¢Û¸ù¾Ý»¯ºÏ¼ÛFe3O4¿É±íʾΪ£ºFeO¡¤Fe2O3    Àà±È£ºFe3I8Ò²¿É±íʾΪFeI2¡¤2FeI3

            ÕýÈ·µÄÓ¦¸ÃΪ£º                                ¡£

            ¢ÜCaC2ÄÜË®½â£ºCaC2+2H2O =Ca£¨OH£©2 + C2H2¡ü

            Àà±È£ºAl4C3Ò²ÄÜË®½â£ºAl4C3 + 12H2O = 4Al£¨OH£©3¡ý+ 3CH4¡ü

            ÕýÈ·µÄÓ¦¸ÃΪ£º                                ¡£

¢ó£®£¨£³·Ö£©³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓУºNa+¡¢A-¡¢H+¡¢OH-¡£Èô¸ÃÈÜÒºMÓÉ pH=3µÄHAÈÜÒºmLÓëpH=11µÄNaOHÈÜÒºmL»ìºÏ·´Ó¦¶øµÃ£¬ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ                     £¨Ìî×Öĸ£©¡£

            A£®ÈôÈÜÒºM³ÊÖÐÐÔ£¬ÔòÈÜÒºMÖÐC£¨H+£©+C£¨OH-£©=2¡Á10-7mol¡¤L-1

            B£®ÈôV1=V2£¬ÔòÈÜÒºMµÄpHÒ»¶¨µÈÓÚ7

            C£®ÈôÈÜÒºM³ÊËáÐÔ£¬ÔòV1Ò»¶¨´óÓÚV2

            D£®ÈôÈÜÒºM³Ê¼îÐÔ£¬ÔòV1Ò»¶¨Ð¡ÓÚV2

 

£¨15·Ö£©

¢ñ£®£¨6·Ö£©¹âºÏ×÷ÓÃÊÇÓɶþÑõ»¯Ì¼ÓëË®ºÏ³ÉÆÏÌÑÌÇ£¬Í¬Ê±·Å³öÑõÆø¡£ÉèÏ뽫¸Ã·´Ó¦µ¹¹ýÀ´£¬¿ÉÉè¼Æ³ÉÒ»¸öÔ­µç³Ø£¬²úÉúµçÁ÷£¬ÕâÑù¾Í½«Ì«ÑôÄÜת±ä³ÉÁ˵çÄÜ¡£

£¨1£©Ð´³ö¸ÃÔ­µç³ØµÄ×Ü·´Ó¦Ê½£º______________________________________

£¨2£©Ð´³ö¸Ãµç³ØÔÚËáÐÔ½éÖÊÖзŵçµÄµç¼«·´Ó¦Ê½£º

¸º¼«£º_____________________________________________

Õý¼«£º_____________________________________________

¢ò£®£¨6·Ö£©Àà±È˼άÊÇѧϰ»¯Ñ§µÄÖØÒª·½·¨£¬µ«½á¹ûÊÇ·ñÕýÈ·±ØÐë¾­ÊܼìÑé¡£ÔÚ½øÐÐÀà±È˼άµÄʱºò£¬²»ÄÜ»úеÀà±È£¬Ò»¶¨Òª×¢ÒâһЩÎïÖʵÄÌØÊâÐÔ£¬ÒÔ·ÀÖ¹Àà±È³ö´íÎóµÄ½áÂÛ¡£Æ¾ÒÑÓеĻ¯Ñ§ÖªÊ¶£¬ÏÂÁÐÀà±È½á¹ûÕýÈ·µÄÊÇ£º£¨Ìî±êÐòºÅ£©                    ¡£Èô´íÎó£¬ÔÚÆäºóд³öÕýÈ·µÄ¡£

             ¢ÙÔÚÏàͬÌõ¼þÏ£¬Na2CO3Èܽâ¶È±ÈNaHCO3´ó

             Àà±È£ºÔÚÏàͬÌõ¼þÏ£¬CaCO3Èܽâ¶È±ÈCa£¨HCO3£©2´ó

             ÕýÈ·µÄÓ¦¸ÃΪ£¨ÈôÀà±ÈÕýÈ·£¬´Ë´¦²»Ð´£¬ÏÂͬ¡££©£º                               ¡£

             ¢ÚÏò´ÎÂÈËá¸ÆÈÜÒºÖÐͨ¹ýÁ¿CO2£ºCO2 + ClO- + H2O = HCO3- + HClO

             Àà±È£ºÏò´ÎÂÈËáÄÆÈÜÒºÖÐͨ¹ýÁ¿SO2£ºSO2 + ClO- + H2O = HSO3- + HClO

             ÕýÈ·µÄÓ¦¸ÃΪ£º                                            ¡£

             ¢Û¸ù¾Ý»¯ºÏ¼ÛFe3O4¿É±íʾΪ£ºFeO¡¤Fe2O3     Àà±È£ºFe3I8Ò²¿É±íʾΪFeI2¡¤2FeI3

             ÕýÈ·µÄÓ¦¸ÃΪ£º                                 ¡£

             ¢ÜCaC2ÄÜË®½â£ºCaC2+2H2O =Ca£¨OH£©2 + C2H2¡ü

             Àà±È£ºAl4C3Ò²ÄÜË®½â£ºAl4C3 + 12H2O = 4Al£¨OH£©3¡ý+ 3CH4¡ü

             ÕýÈ·µÄÓ¦¸ÃΪ£º                                 ¡£

¢ó£®£¨£³·Ö£©³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓУºNa+¡¢A-¡¢H+¡¢OH-¡£Èô¸ÃÈÜÒºMÓÉ pH=3µÄHAÈÜÒºmLÓëpH=11µÄNaOHÈÜÒºmL»ìºÏ·´Ó¦¶øµÃ£¬ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ                      £¨Ìî×Öĸ£©¡£

             A£®ÈôÈÜÒºM³ÊÖÐÐÔ£¬ÔòÈÜÒºMÖÐC£¨H+£©+C£¨OH-£©=2¡Á10-7mol¡¤L-1

             B£®ÈôV1=V2£¬ÔòÈÜÒºMµÄpHÒ»¶¨µÈÓÚ7

             C£®ÈôÈÜÒºM³ÊËáÐÔ£¬ÔòV1Ò»¶¨´óÓÚV2

             D£®ÈôÈÜÒºM³Ê¼îÐÔ£¬ÔòV1Ò»¶¨Ð¡ÓÚV2

 

Ì«Ñô¹âÈëÉäµØÇòºó,±»´óÆø²ãÓëµØ±íÎüÊÕµÄÄÜÁ¿,ƽ¾ù´óԼΪ235 W/m2(ÈçÏÂͼ).Èô´óÆø²»ÎüÊÕÌ«×èÓëµØ±í·¢³öµÄ·øÉä,ÔòµØ±íζÈÖ»ÄÜά³ÖÔÚ-18¡æ×óÓÒ£»µ«ÒòµØ±í·¢³öµÄ³¤²¨·øÉä,´ó²¿·Ö»á±»´óÆøÖеÄË®Æø¡¢CO2¡¢¼×ÍéÓëÆäËüһЩÆøÌåËùÎüÊÕ,Òò¶ø²úÉúÎÂÊÒЧӦ,ʹµØ±íζȿɱ£³ÖÔÚÊʺÏÉúÎïÉú´æµÄ15¡æ×óÓÒ.´óÆøÖÐÓÉÓÚÈËÀà»î¶¯Ëù´øÀ´µÄCO2Ũ¶ÈÔö¼Ó,»áÔöÇ¿µØÇòµÄÎÂÊÒЧӦ,ÇÒ¿ÉÄÜÖú³¤È«ÇòÓúÇ÷ů»¯µÄÏÖÏó.
¿Æѧ¼Ò×î½ü¿ªÊ¼½øÐÐÒ»ÏîʵÑé,ÊÔ×ÅÔö¼Óº£Ë®ÖеÄÌúÖÊ,¿´¿´ÄÜ·ñ½å´Ë½«´óÆøÖеÄCO2:Ũ¶È¼õÉÙ,ÒÔ¼õ»ºÈ«Çòů»¯µÄËÙÂÊ.ÓÉÓÚÌúÖÊ¿ÉÒÔ°ïÖúÖ²ÎïÐÔ¸¡ÓÎÉúÎïµÄÉú³¤,¶øµØÇòÉÏÓÖÓнüºõÒ»°ëµÄ¹âºÏ×÷ÓÃÊÇÓÉÖ²ÎïÐÔ¸¡ÓÎÉúÎïËù½øÐеÄ,Òò´ËÒÀ¾ÝÀíÂÛÍƲâ,ÈôÔÚº£ÑóÖÐÊ©ÒÔ´óÁ¿Ìú·Ê,ʹֲÎïÐÔ¸¡ÓÎÉúÎï´óÁ¿·±Ö³Óë½øÐйâºÏ×÷ÓÃ,±ã¿É½«´óÆøÖеÄCO2·â´æһЩÔÚº£µ×,×îºó½«ÓÐÖúÓÚ¼õ»ºÈ«Çòů»¯µÄËÙÂÊ.ÖµµÃ×¢ÒâµÄÊÇ, ÉÏÊöÀíÂÛ±ØÐëÔÚÖ²ÎïÐÔ¸¡ÓÎÉúÎïËÀºó±ã³ÁÈ뺣µ×µÄÇ°ÌáϲŻá³ÉÁ¢,ΨÓÐÈç´Ë²ÅÄÜÒ»ÀÍÓÀÒݵĽ«CO2´Ó´óÆøÖÐÒƳý,·ñÔòÒ»µ©ÕâЩֲÎïÐÔ¸¡ÓÎÉúÎï±»¶¯ÎïÐÔ¸¡ÓÎÉúÎïËùʳ,¶ø¶¯ÎïÐÔ¸¡ÓÎÉúÎïÓÖ±»´óÐͺ£Ñó¶¯Îï³Ôµô,ÕâЩ±»ÒƳýµÄCO2:»¹ÊÇ»á½åÓɺôÎü×÷¾É·µ»ØµØÇò´óÆøÖÐ,Èç´ËÒ»À´,º£ÖÐÌúÖʵÄÔö¼Ó²¢²»»á¼õÉÙ´óÆøÖÐCO2µÄŨ¶È¡£
½øÐÐÕâÏîʵÑéµÄ¿Æѧ¼Ò,±ãÊÇÔÚÌúÖÊÔö¼ÓºóµÄº£ÓòÖÐ(½üÄϼ«µÄº£Óò),³¤ÆÚ×·×ÙÖ²ÎïÐÔ¸¡ÓÎÉúÎï,ÒÔ¼°ÆäËüº£ÖÐÉúÎïµÄ·±Ö³Çé¿ö,ÊÔͼÕÒ³ö´ð°¸.È»¶øÒ²ÓÐһЩÉú̬ѧ¼ÒÌá³öÈçϵľ¯¸æ£º¼´Ê¹ÕâÏîʵÑéÖ¤Ã÷ÔÚº£ÖÐÊ©ÒÔÌú·Ê¿ÉÒÔ¼õµÍ´óÆøÖеÄCO2Ũ¶È,µ«¸ÉÈÅ»òÆÆ»µº£ÑóÖеÄʳÎïÁ´,Ò²¿ÉÄܶԺ£ÑóÉú̬Ôì³É¼±åáÇÒ¸º¶øµÄÓ°Ïì.ÉÏÊöÕâЩÔÚ¿ÆѧÉϵÄÕùÂÛ,¶¼ÐèÒª¸ü¶àµÄ¿ÆѧÑо¿,²ÅÄܽøÐнÏÉîÈëµÄ̽ÌÖÓëÁ˽⡣
ÊÔ¸ù¾Ý±¾ÎÄÐðÊö,»Ø´ðÏÂÁÐÎÊÌâ¡£
£¨1£©½üÄêÀ´,´óÆøÖÐCO2µÄŨ¶ÈLÉýÒѳÉΪȫÇòÐÔµÄÎÊÌâ,Òò¶øÐËÆð½ÚÄܼõ̼Ô˶¯,¹úÄڵĻ·±£ÍÅÌåÒ²Ðûµ¼ÖÐÇï½Ú²»¿¾Èâ.¼ÙÈô³¬ÊÐÂôµÄ¿¾ÈâÓÃľ̿,Æ京̼Á¿Îª90%,ÔòÒ»°ü10¹«½ïµÄľ̿ÍêȫȼÉÕºó,»á²úÉú¼¸¹«½ïµÄCO2
[     ]
A £®44
B£®33
C£®22
D£®11
E£®5.5
£¨2£©ÏÂÁÐÈý¸ö·´Ó¦Ê½ÖеÄxÓëY·Ö±ðΪÁ½ÖÖ½ðÊôÔªËصĴúºÅ,µ«ClΪÂȵÄÔªËØ·ûºÅ.
ÒÑÖªÈý¸ö·´Ó¦¾ùÄÜÏòÓÒ½øÐÐ,ÊÔÒÀ¾ÝÒÔÉÏÈý¸ö·´Ó¦Ê½,ÍƲâÏÂÁÐÎïÖÊÖÐÄÄÒ»¸öÊÇ×îÇ¿µÄÑõ»¯¼Á
[     ]
A£®XCl3
B£®XCl2
C£®Cl2
D£®Y
E£®YCl2
£¨3£©ÏÂͼÖмס¢ÒÒ¶þÌõÇúÏßÊÇÒÔ¼îÐÔË®ÈÜÒº,·Ö±ðµÎ¶¨¶þ¸öËáÐÔË®éFÒºµÄpHÖµ±ä»¯Í¼,¶ø±û¡¢¶¡¶þÌõÇúÏßÔòÊÇÒÔËáÐÔË®ÈÜÒº·Ö±ðµÎ¶¨¶þ¸ö¼îÐÔË®ÈÜÒº.ÊÔÎÊÉÏÊöÄÄЩËá¼îµÎ¶¨·´Ó¦,ÊʺÏÒÔ·Ó̪×÷Ϊָʾ¼Á
[     ]
A£®¼×ÒÒ±û¶¡
B£®ÒÒ±û¶¡
C£®¼×ÒÒ¶¡
D£®¼×±û¶¡
E£®¼×ÒÒ±û
£¨4£©²Ð´æÓÚÉ°ÍÁ¡¢×©ÍߵȽ¨²ÄµÄ΢Á¿Ó˵ÈÔªËØ,ÓÚË¥±äºó»áÊͷųöÒ»ÖÖÎÛȾÆøÌå.´ËÆøÌåµÄÁ¿ËäÉõÉÙ,µ«»á³öÏÖÓÚ²»Í¨·çµÄÊÒÄÚ¿ÕÆøÖÐ,ÇÒÆäÎȶ¨Ö®Í¬Î»ËØÒà¾ß·ÅÉäÐÔ. ËüÒ²Êǽö´ÎÓÚÎüÑ̶øµ¼Ö·ΰ©µÄµÚ¶þ´óÔªÐ×.ÊÔÎÊ´ËÎÛȾÆøÌå,ΪÏÂÁÐÄÄÒ»ÖÖ¶èÐÔÆøÌå(¶ÛÆø¡¢Ï¡ÓÐÆøÌå)
[     ]
A£®º¤
B£®ÄÊ
C£®ë²
D£®ë´
E£®ë±
£¨5£©Ï±íËùÁÐΪÎåÖÖ¿ÉÈÜÐÔÑÎÔÚ30¡æµÄÈܽâ¶È(g/100 g H2O):
ÈôÔÚ30¡æµÄ±¥ºÍʳÑÎË®ÖÐͨÈë°±ÆøÖÁ±¥ºÍºó,ÔÙͨÈë¶þÑõ»¯Ì¼¾Í»áÓо§ÌåÎö³ö. ÊԲο¼±íÖеÄÊý¾Ý,ÍƲâÎö³öµÄ¾§ÌåÊÇÏÂÁеÄÄÄÒ»ÖÖ
[     ]
A£®NaCl
B£®NaHCO3
C£®Na2CO3
D£®(NH4)2CO3
E£®NH4Cl

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø