ÌâÄ¿ÄÚÈÝ

¹¤ÒµÉϳ£²úÉú´óÁ¿µÄ·ÏÆø¡¢·ÏË®¡¢·ÏÔü¡¢·ÏÈÈ£¬Èç¹û´¦Àí²»ºÃ£¬ËæÒâÅÅ·Å£¬»áÔì³ÉÎÛȾ£¬¶øÈç¹û¿Æѧ»ØÊÕ£¬¿É±ä·ÏΪ±¦¡£
£¨1£©¹¤ÒµÖÆÁòËáµÄβÆøÖк¬ÓеÄÉÙÁ¿SO2£¬¿ÉÏÈÓð±Ë®ÎüÊÕ£¬ÔÙÓÃÏ¡ÁòËá´¦Àí¡£
¢Ù¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________£»
¢ÚÕâÑù´¦ÀíµÄÓŵãÊÇ____________________¡£
£¨2£©ÁòË᳧²úÉúµÄ´óÁ¿ÉÕÔü£¨ÁòÌú¿óìÑÉÕºóµÄ»ÒÔü£©µÄÓÃ;ÊÇ________£¨Ð´Ò»ÖÖ¼´¿É£©£»Ð´³öµç½â¾«Á¶Í­Ñô¼«ÄàµÄÒ»ÖÖÓÃ;________¡£
£¨3£©Ò»×ù´óÐÍÁòË᳧ͬʱÓÖÊÇÒ»×ùÄÜÔ´¹¤³§£¬ÁòË᳧Éú²ú¹ý³ÌÖеÄÓàÈÈÈôÄܳä·Ö»ØÊÕÀûÓ㬲»½ö²»ÐèÒªÍâ½ç¹©Ó¦ÄÜÔ´£¬¶øÇÒ»¹¿ÉÒÔÏòÍâ½çÊä³ö´óÁ¿µÄÈÈÄÜ¡£ÁòË᳧²úÉúÓàÈȵÄÖ÷ÒªÉ豸Ãû³ÆÊÇ________¡£
£¨4£©¸ÉϨ½¹¼¼ÊõÊǽ«Á¶½¹Â¯ÍƳöµÄÔ¼1 000¡æµÄ³àÈȽ¹Ì¿£¬ÔÚϨ½¹ÊÒÖб»ÆäÄæÁ÷µÄÀä¶èÐÔÆøÌ壨Ö÷Òª³É·ÖÊǵªÆø£¬Î¶ÈÔÚ170¡«190¡æ£©Ï¨Ã𣬱»¼ÓÈȵ½700¡«800¡æµÄ¶èÐÔÆøÌå¾­³ý³¾ºó½øÈëÓàÈȹø¯£¬²úÉúµÄ¹ýÈÈÕôÆøËÍÍùÆûÂÖ·¢µç»ú·¢µç¡£¸ÉϨ1 t½¹Ì¿¿É²úÉú500 kg¹ýÈÈÕôÆø£¬¿ÉÕۺϳÉ46 kg±ê׼ú¡£¾Ù³öÁ½Àý²ÉÓÃÄæÁ÷Ô­ÀíµÄ»¯¹¤É豸»ò»¯Ñ§ÒÇÆ÷________¡¢________¡£
£¨1£©¢ÙSO2£«NH3¡¤H2O=NH4HSO3¡¢¡¡2NH4HSO3£«H2SO4=£¨NH4£©2SO4£«2H2O£«2SO2¡ü[»òSO2£«2NH3¡¤H2O=£¨NH4£©2SO3£«H2O£¬£¨NH4£©2SO3£«H2SO4=£¨NH4£©2SO4£«H2O£«SO2¡ü]
¢ÚÉú³ÉµÄSO2¿ÉÓÃ×÷ÖÆÁòËáµÄÔ­ÁÏ£¬ÁòËá刺É×÷»¯·Ê
£¨2£©×÷¸ß¯Á¶ÌúµÄÔ­ÁÏ£¨»ò»ØÊÕÓÐÉ«½ðÊô¡¢ÌáÈ¡½ðÒø¡¢ÖÆשµÈºÏÀí´ð°¸¶¼¿É£©¡¡»ØÊÕ¹ó½ðÊô½ð¡¢ÒøµÈ
£¨3£©·ÐÌÚ¯ºÍ½Ó´¥ÊÒ£¨Ö»»Ø´ð·ÐÌÚ¯Ҳ¿ÉÒÔ£©
£¨4£©ÈȽ»»»Æ÷¡¡ÀäÄý¹Ü£¨»ò¹¤ÒµÖÆÁòËáÖеÄÎüÊÕËþ£©
£¨2£©ÁòË᳧²úÉúµÄ´óÁ¿ÉÕÔüÖк¬ÓзḻµÄÌúÔªËØ£¨ÒÔFe2O3ÐÎʽ´æÔÚ£©£¬¿ÉÓÃÓÚÁ¶Ìú¡££¨3£©Å̵㹤ҵ½Ó´¥·¨ÖÆÁòËáµÄ»¯Ñ§·´Ó¦ÖУ¬¿óʯµÄȼÉպͶþÑõ»¯ÁòµÄ½Ó´¥Ñõ»¯ÊôÓÚ·ÅÈÈ·´Ó¦£¬ÕâÁ½¸ö·´Ó¦·Ö±ðÔÚ·ÐÌÚ¯ºÍ½Ó´¥ÊÒÖнøÐС£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ΪÑо¿Í­ÓëŨÁòËáµÄ·´Ó¦£¬Ä³»¯Ñ§ÐËȤС×é½øÐÐÈçÏÂʵÑé¡£
ʵÑé¢ñ¡¡·´Ó¦²úÎïµÄ¶¨ÐÔ̽¾¿
ʵÑé×°ÖÃÈçͼËùʾ¡££¨¹Ì¶¨×°ÖÃÒÑÂÔÈ¥£©

£¨1£©AÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                            ¡£
£¨2£©FÉÕ±­ÖеÄÈÜҺͨ³£ÊÇ               ¡£
£¨3£©ÊµÑé¹ý³ÌÖУ¬ÄÜÖ¤Ã÷ŨÁòËáÖÐÁòÔªËصÄÑõ»¯ÐÔÇ¿ÓÚÇâÔªËصÄÏÖÏóÊÇ                                                                                               ¡£
£¨4£©ÊµÑé½áÊøºó£¬Ö¤Ã÷A×°ÖÃÊÔ¹ÜÖз´Ó¦ËùµÃ²úÎïÊÇ·ñº¬ÓÐÍ­Àë×ӵIJÙ×÷·½·¨ÊÇ                                                                                                                                                                                               ¡£
£¨5£©ÎªËµÃ÷ŨÁòËáÖеÄË®ÊÇ·ñÓ°ÏìB×°ÖÃÏÖÏóµÄÅжϣ¬»¹Ðë½øÐÐÒ»´ÎʵÑ顣ʵÑé·½°¸Îª                                                                                                                                                                                               ¡£
ʵÑé¢ò¡¡·´Ó¦²úÎïµÄ¶¨Á¿Ì½¾¿
£¨6£©ÔÚÍ­ÓëŨÁòËá·´Ó¦µÄ¹ý³ÌÖУ¬·¢ÏÖÓкÚÉ«ÎïÖʳöÏÖ£¬¾­²éÔÄÎÄÏ×»ñµÃÏÂÁÐ×ÊÁÏ¡£
×ÊÁÏ1£º
ÁòËá/mol¡¤L£­1
ºÚÉ«ÎïÖʳöÏÖµÄζÈ/¡æ
ºÚÉ«ÎïÖÊÏûʧµÄζÈ/¡æ
15
Ô¼150
Ô¼236
16
Ô¼140
Ô¼250
18
Ô¼120
²»Ïûʧ
 
×ÊÁÏ2£ºXÉäÏß¾§Ìå·ÖÎö±íÃ÷£¬Í­ÓëŨÁòËá·´Ó¦Éú³ÉµÄºÚÉ«ÎïÖÊΪCu2S¡¢CuS¡¢Cu7S4ÖеÄÒ»ÖÖ»ò¼¸ÖÖ¡£½öÓÉÉÏÊö×ÊÁϿɵóöµÄÕýÈ·½áÂÛÊÇ       __¡£
a£®Í­ÓëŨÁòËᷴӦʱËùÉæ¼°µÄ·´Ó¦¿ÉÄܲ»Ö¹Ò»¸ö
b£®ÁòËáŨ¶ÈÑ¡ÔñÊʵ±£¬¿É±ÜÃâ×îºó²úÎïÖгöÏÖºÚÉ«ÎïÖÊ
c£®¸Ã·´Ó¦·¢ÉúµÄÌõ¼þÖ®Ò»ÊÇÁòËáŨ¶È¡Ý15  mol¡¤L£­1
d£®ÁòËáŨ¶ÈÔ½´ó£¬ºÚÉ«ÎïÖÊÔ½¿ì³öÏÖ¡¢Ô½ÄÑÏûʧ
£¨7£©Îª²â³öÁòËáÍ­µÄ²úÂÊ£¬½«¸Ã·´Ó¦ËùµÃÈÜÒºÖкͺóÅäÖƳÉ250.00 mLÈÜÒº£¬È¡¸ÃÈÜÒº25.00 mL¼ÓÈë×ãÁ¿KIÈÜÒºÕñµ´£¬ÒÔµí·ÛÈÜҺΪָʾ¼Á£¬ÓÃb  mol¡¤L£­1Na2S2O3ÈÜÒºµÎ¶¨Éú³ÉµÄI2£¬3´ÎʵÑéƽ¾ùÏûºÄ¸ÃNa2S2O3ÈÜÒºV mL¡£Èô·´Ó¦ÏûºÄÍ­µÄÖÊÁ¿Îªa g£¬ÔòÁòËáÍ­µÄ²úÂÊΪ    _¡££¨ÒÑÖª£º2Cu2£«£«4I£­=2CuI£«I2£¬2S2O32-£«I2=S4O62-£«2I£­£©
°²»ÕÊ¡´Ó2013Äê12ÔÂ1ÈÕÁãʱÆ𣬳µÓÃÆûÓÍÉý¼¶Îª¡°¹ú¢ô¡±±ê×¼£¬ÆûÓÍÖеÄÁòº¬Á¿Ï½µÈý·ÖÖ®¶þ£¬¶Ô¶þÑõ»¯ÁòµÄÅÅ·ÅÓÐÁË´ó´óµÄ¸ÄÉÆ£¬SO2¿ÉÒÔÓÃFe( NO3)3ÈÜÒºÎüÊÕ£¬Ä³»¯Ñ§ÐËȤС×é¶ÔSO2ºÍ Fe( NO3)3ÈÜÒºµÄ·´Ó¦[0.1mol/LµÄFe(NO3)3ÈÜÒºµÄ pH£½2]×öÁËÏàӦ̽¾¿¡£
̽¾¿I
£¨1£©Ä³Í¬Ñ§½øÐÐÁËÏÂÁÐʵÑé:È¡12.8gͭƬºÍ20 mL 18 mol?L-1µÄŨÁòËá·ÅÔÚÈý¾±Æ¿Öй²ÈÈ,Ö±ÖÁ·´Ó¦Íê±Ï,×îºó·¢ÏÖÉÕÆ¿Öл¹ÓÐͭƬʣÓ࣬ͬʱ¸ù¾ÝËùѧµÄ֪ʶͬѧÃÇÈÏΪ»¹Óн϶àµÄÁòËáÊ£Óà¡£

¢ÙÅäÖÆ1mol/L µÄFe(NO3)3ÈÜÒº£¬Ðè׼ȷ³ÆÁ¿Ò»¶¨ÖÊÁ¿µÄFe(NO3)3¹ÌÌ壬ÔÚÉÕ±­ÖÐÈܽâ»Ö¸´µ½ÊÒκ󣬽«ÈÜҺתÒƵ½ÈÝÁ¿Æ¿ÖУ¬ÔÙ¾­¹ýÏ´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ¿ÉÅäµÃÈÜÒº£¬Çë»Ø´ð½«ÈÜҺתÒÆÖÁÈÝÁ¿Æ¿ÖеIJÙ×÷·½·¨________________________________¡£
¢Ú×°ÖÃAÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_________________________________________¡£
¢Û¸ÃͬѧÉè¼ÆÇóÓàËáµÄÎïÖʵÄÁ¿ÊµÑé·½°¸ÊDzⶨ²úÉúÆøÌåµÄÁ¿¡£Æä·½·¨ÓжàÖÖ,ÇëÎÊÏÂÁз½°¸Öв»¿ÉÐеÄÊÇ______ (Ìî×Öĸ£©¡£
A£®½«²úÉúµÄÆøÌ建»ºÍ¨¹ýÔ¤ÏȳÆÁ¿µÄÊ¢Óмîʯ»ÒµÄ¸ÉÔï¹Ü,½áÊø·´Ó¦ºóÔٴγÆÖØ
B£®½«²úÉúµÄÆøÌ建»ºÍ¨Èë×ãÁ¿Na2SÈÜÒººó£¬²âÁ¿ËùµÃ³ÁµíµÄÖÊÁ¿
C£®ÓÃÅű¥ºÍNaHSO3ÈÜÒºµÄ·½·¨²â¶¨Æä²úÉúÆøÌåµÄÌå»ý(ÕÛËã³É±ê×¼×´¿ö£©
̽¾¿II
£¨2£©ÎªÅųý¿ÕÆø¶ÔʵÑéµÄ¸ÉÈÅ£¬µÎ¼ÓŨÁòËá֮ǰӦ½øÐеIJÙ×÷ÊÇ____________________¡£
£¨3£©×°ÖÃBÖвúÉúÁË°×É«³Áµí£¬·ÖÎöBÖвúÉú°×É«³ÁµíµÄÔ­Òò£¬Ìá³öÏÂÁÐÈýÖÖ²ÂÏ룺
²ÂÏë1: SO2ÓëFe3+·´Ó¦£»
²ÂÏë2£ºÔÚËáÐÔÌõ¼þÏÂSO2ÓëNO3-·´Ó¦£»
²ÂÏë3£º____________________________________£»
¢Ù°´²ÂÏë1£¬×°ÖÃBÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________________________£¬Ö¤Ã÷¸Ã²ÂÏë1ÖÐÉú³ÉµÄ»¹Ô­²úÎijͬѧȡÉÙÁ¿ÈÜÒºµÎ¼Ó¼¸µÎËáÐÔ¸ßÃÌËá¼ØÈÜÒº£¬×ϺìÉ«ÍÊÈ¥£¬Çë·ÖÎö¸Ãͬѧ×ö·¨ÊÇ·ñÕýÈ·_________£¨Ìî¡°ÕýÈ·¡±»ò¡°²»ÕýÈ·¡±£©£¬ÀíÓÉÊÇ__________________________________¡£
¢Ú°´²ÂÏë2,Ö»Ð轫װÖÃBÖеÄFe(NO3)3ÈÜÒºÌ滻ΪµÈÌå»ýµÄÏÂÁÐijÖÖÈÜÒº£¬ÔÚÏàͬÌõ¼þϽøÐÐʵÑ顣ӦѡÔñµÄÌæ»»ÈÜÒºÊÇ______ (ÌîÐòºÅ£©¡£
A£®1 mol/L Ï¡ÏõËá
B£®pH£½1µÄFeCl3ÈÜÒº 
C£®6.0mol/L NaNO3ºÍ0.2 mol/LÑÎËáµÈÌå»ý»ìºÏµÄÈÜÒº
ÏÖÓÐijÌú̼ºÏ½ð£¨ÌúºÍ̼Á½ÖÖµ¥ÖʵĻìºÏÎ£¬Ä³»¯Ñ§ÐËȤС×éΪÁ˲ⶨÌú̼ºÏ½ðÖÐÌúµÄÖÊÁ¿·ÖÊý£¬²¢Ì½¾¿Å¨ÁòËáµÄijЩÐÔÖÊ£¬Éè¼ÆÁËÏÂͼËùʾµÄʵÑé×°Ö㨼гÖÒÇÆ÷ÒÑÊ¡ÂÔ£©ºÍʵÑé·½°¸½øÐÐʵÑé̽¾¿¡£

I.²â¶¨ÌúµÄÖÊÁ¿·ÖÊý£º
£¨1£©¼ì²éÉÏÊö×°ÖÃÆøÃÜÐÔµÄÒ»ÖÖ·½·¨ÊÇ£º¹Ø±Õ·ÖҺ©¶·µÄ»îÈû£¬ÔÚE×°ÖúóÃæÁ¬ÉÏÒ»¸ù
µ¼¹Ü£¬È»ºó________________________________________£¬ÔòÖ¤Ã÷×°ÖõÄÆøÃÜÐÔÁ¼ºÃ¡£
£¨2£©³ÆÁ¿EµÄÖÊÁ¿£¬²¢½«a gÌú̼ºÏ½ðÑùÆ··ÅÈë×°ÖÃAÖУ¬ÔÙ¼ÓÈë×ãÁ¿µÄŨÁòËᣬ´ýAÖв»ÔÙÒݳöÆøÌåʱ£¬Í£Ö¹¼ÓÈÈ£¬²ðÏÂE²¢³ÆÖØ£¬EÔöÖØbg¡£Ìú̼ºÏ½ðÖÐÌúµÄÖÊÁ¿·ÖÊýΪ_____________________________________________£¨Ð´±í´ïʽ£©¡£
£¨3£©×°ÖÃCÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________________________¡£
£¨4£©¼×ͬѧÈÏΪ£¬ÒÀ¾Ý´ËʵÑé²âµÃµÄÊý¾Ý£¬¼ÆËãºÏ½ðÖÐÌúµÄÖÊÁ¿·ÖÊý¿ÉÄÜ»áÆ«µÍ£¬Ô­ÒòÊÇ¿ÕÆøÖÐCO2¡¢H2O½øÈëE¹ÜʹbÔö´ó¡£ÄãÈÏΪ¸Ä½øµÄ·½·¨ÊÇ____________________________________.
£¨5£©ÒÒͬѧÈÏΪ£¬¼´Ê¹¼×ͬѧÈÏΪµÄÆ«²îµÃµ½¸Ä½ø£¬ÒÀ¾Ý´ËʵÑé²âµÃºÏ½ðÖÐÌúµÄÖÊÁ¿·Ö
ÊýÒ²¿ÉÄÜ»áÆ«Àë¡£ÄãÈÏΪÆäÖеÄÔ­ÒòÊÇ_________________________________________¡£
¢ò.̽¾¿Å¨ÁòËáµÄijЩÐÔÖÊ£º
£¨6£©ÍùAÖеμÓ×ãÁ¿µÄ¹öÁòËᣬδµãȼ¾Æ¾«µÆÇ°£¬A¡¢B¾ùÎÞÃ÷ÏÔÏÖÏó£¬ÆäÔ­ÒòÊÇ£º
_____________________________________________________________________¡£
£¨7£©½«AÖйÌÌ廻Ϊ½ðÊôÍ­ÔòÓëŨÁòËá·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________________
_______________________£¬·´Ó¦ÖÐŨÁòËá±íÏÖ³öµÄÐÔÖÊÊÇ______________________¡£

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

¾«Ó¢¼Ò½ÌÍø