ÌâÄ¿ÄÚÈÝ

Çë»Ø´ðÂȼîµÄÈçÏÂÎÊÌ⣺
£¨1£©ÂÈÆø¡¢ÉÕ¼îÊǵç½âʳÑÎˮʱ°´Õչ̶¨µÄ±ÈÂÊk£¨ÖÊÁ¿±È£©Éú³ÉµÄ²úÆ·¡£ÀíÂÛÉÏk£½_______£¨ÒªÇó¼ÆËã±í´ïʽºÍ½á¹û£©¡£
£¨2£©Ô­ÁÏ´ÖÑÎÖг£º¬ÓÐÄàɳºÍCa2£«¡¢Mg2£«¡¢Fe3£«¡¢SO2£­4µÈÔÓÖÊ£¬±ØÐë¾«ÖÆºó²ÅÄܹ©µç½âʹÓ᣾«ÖÆÊ±£¬´ÖÑÎÈÜÓÚË®¹ýÂ˺󣬻¹Òª¼ÓÈëµÄÊÔ¼Á·Ö±ðΪ¢ÙNa2CO3¡¢¢ÚHCl£¨ÑÎËᣩ¢ÛBaCl2£¬Õâ3ÖÖÊÔ¼ÁÌí¼ÓµÄºÏÀí˳ÐòÊÇ             £¨ÌîÐòºÅ£©¡£
£¨3£©ÂȼҵÊǸߺÄÄܲúÒµ£¬Ò»ÖÖ½«µç½â³ØÓëȼÁÏµç³ØÏà×éºÏµÄй¤ÒÕ¿ÉÒÔ½Ú£¨µç£©ÄÜ30£¥ÒÔÉÏ¡£ÔÚÕâÖÖ¹¤ÒÕÉè¼ÆÖУ¬Ïà¹ØÎïÁϵĴ«ÊäÓëת»¯¹ØÏµÈçÏÂͼËùʾ£¬ÆäÖеĵ缫δ±ê³ö£¬ËùÓõÄÀë×ÓÀ°¶¼Ö»ÔÊÐíÑôÀë×Óͨ¹ý¡£

¢ÙͼÖÐX¡¢Y·Ö±ðÊÇ_____¡¢_______£¨Ìѧʽ£©£¬·ÖÎö±È½ÏͼʾÖÐÇâÑõ»¯ÄÆÖÊÁ¿·ÖÊýa£¥Óëb£¥µÄ´óС_________;
¢Ú·Ö±ðд³öȼÁÏµç³ØBÖÐÕý¼«¡¢¸º¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦
Õý¼«£º_____________; ¸º¼«£º_______________;
ÕâÑùÉè¼ÆµÄÖ÷Òª½Ú£¨µç£©ÄÜÖ®´¦ÔÚÓÚ£¨Ð´³ö2´¦£©____________¡¢____________¡£
(1£©k=M(Cl2)/2 M(NaOH)=71/80=1:1.13»ò0.89
£¨2£©¢Û¢Ù¢Ú
£¨3£©¢ÙCl2  H2  a£¥Ð¡ÓÚb£¥ ¢ÚO2+4e-+2H2O£½4OH-    H2£­2e-+2OH-£½2H2O ¢ÛȼÁÏµç³Ø¿ÉÒÔ²¹³äµç½â³ØÏûºÄµÄµçÄÜ£»Ìá¸ß²ú³ö¼îÒºµÄŨ¶È£»½µµÍÄܺ썯äËûºÏÀí´ð°¸Ò²¸ø·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨20·Ö£©¹¤ÒµÉú²ú´¿¼îµÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º

Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©´ÖÑÎË®¼ÓÈë³Áµí¼ÁA¡¢B³ýÔÓÖÊ£¨³Áµí¼ÁAÀ´Ô´ÓÚʯ»ÒÒ¤³§£©£¬Ð´³öA¡¢BµÄ»¯Ñ§Ê½A ____________________  B _______________________
£¨2£©ÊµÑéÊÒÌá´¿´ÖÑεÄʵÑé²Ù×÷ÒÀ´ÎΪ£º
È¡Ñù¡¢_______¡¢³Áµí¡¢_______¡¢_______¡¢ÀäÈ´½á¾§¡¢_______¡¢ºæ¸É
£¨3£©¹¤ÒµÉú²ú´¿¼î¹¤ÒÕÁ÷³ÌÖУ¬Ì¼Ëữʱ²úÉúµÄÏÖÏóÊÇ__________________________
̼ËữʱûÓÐÎö³ö̼ËáÄÆ¾§Ì壬ÆäÔ­ÒòÊÇ____________________________________
£¨4£©Ì¼Ëữºó¹ýÂË£¬ÂËÒºD×îÖ÷ÒªµÄ³É·ÖÊÇ__________________________£¨Ìîд»¯Ñ§Ê½£©£¬¼ìÑéÕâÒ»³É·ÖµÄÒõÀë×ӵľßÌå·½·¨ÊÇ£º___________________________________
£¨5£©°±¼î·¨Á÷³ÌÖа±ÊÇÑ­»·Ê¹Óõģ¬Îª´Ë£¬ÂËÒºD¼ÓÈëʯ»ÒË®²úÉú°±¡£¼Óʯ»ÒË®ºóËù·¢ÉúµÄ·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º__________________________________________ÂËÒºD¼Óʯ»ÒˮǰÏÈÒª¼ÓÈÈ£¬Ô­ÒòÊÇ___________________________________________
£¨6£©²úÆ·´¿¼îÖк¬ÓÐ̼ËáÇâÄÆ¡£Èç¹ûÓüÓÈÈ·Ö½âµÄ·½·¨²â¶¨´¿¼îÖÐ̼ËáÇâÄÆµÄÖÊÁ¿·ÖÊý£¬´¿¼îÖÐ̼ËáÇâÄÆµÄÖÊÁ¿·ÖÊý¿É±íʾΪ£º____________________________
£¨×¢Ã÷ÄãµÄ±í´ïʽÖÐËùÓõÄÓйطûºÅµÄº¬Ò壩
£¨14·Ö£©½üÄêÀ´£¬Ëæ×ÅÎÒ¹ú¾­¼ÃµÄ¿ìËÙ·¢Õ¹£¬¶ÔµçÁ¦µÄÐèÇóÔ½À´Ô½¸ß£¬ÕâÒ²´Ù½øÁËÎÒ¹úµçÁ¦¹¤Òµ¸ßËÙ·¢Õ¹£¬µ«ÎÒ¹úµçÁ¦½á¹¹ÖУ¬»ðµç±ÈÖØ·Ç³£´ó£¬Õ¼·¢µç×°»ú×ÜÈÝÁ¿µÄ75%ÒÔÉÏ£¬ÇÒ»ðµç±ÈÖØ»¹ÔÚÖðÄêÉÏÉý¡£»ðÁ¦·¢µç³§Êͷųö´óÁ¿µÄµªÑõ»¯ÎNOx£©¡¢¶þÑõ»¯ÁòºÍ¶þÑõ»¯Ì¼µÈÆøÌå»áÔì³É»·¾³ÎÛȾ¡£¶Ôȼú·ÏÆø½øÐÐÍÑÏõ¡¢ÍÑÁòºÍÍÑ̼µÈ´¦Àí£¬¿ÉʵÏÖÂÌÉ«»·±£¡¢½ÚÄܼõÅÅ¡¢·ÏÎïÀûÓõÈÄ¿µÄ¡£
£¨1£©ÍÑÏõ¡£ÀûÓü×Íé´ß»¯»¹Ô­NOx£º
CH4(g)£«4NO2(g)===4NO(g)£«CO2(g)£«2H2O(g)£»¦¤H1£½£­574kJ¡¤mol£­1
CH4(g)£«4NO(g)===2N2(g)£«CO2(g)£«2H2O(g)£»¦¤H2£½£­1160kJ¡¤mol£­1
¼×ÍéÖ±½Ó½«NO2»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ£º                            ¡£
£¨2£©ÍÑ̼¡£½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ£º
CO2(g)£«3H2(g)CH3OH(g)£«H2O(g)£»¦¤H3
¢ÙÈ¡Îå·ÝµÈÌå»ýCO2ºÍH2µÄ»ìºÏÆøÌå(ÎïÖʵÄÁ¿Ö®±È¾ùΪ1¡Ã3)£¬·Ö±ð¼ÓÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦Ïàͬʱ¼äºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ(CH3OH)Ó뷴ӦζÈTµÄ¹ØÏµÇúÏßÈçÓÒͼËùʾ£¬ÔòÉÏÊöCO2ת»¯Îª¼×´¼µÄ·´Ó¦µÄ¦¤H3¡¡  0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£
¢ÚÔÚÒ»ºãκãÈÝÃܱÕÈÝÆ÷ÖгäÈë1mol CO2ºÍ3mol H2£¬½øÐÐÉÏÊö·´Ó¦¡£²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæÊ±¼ä±ä»¯ÈçÏÂͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ¡¡  ¡¡£¨Ìî×Öĸ´úºÅ£©¡£
A£®µÚ10minºó£¬Ïò¸ÃÈÝÆ÷ÖÐÔÙ³äÈë1mol CO2ºÍ3mol H2£¬ÔòÔٴδﵽƽºâʱc(CH3OH)£½1.5mol¡¤L£­1
B£®0¡«10minÄÚ£¬ÇâÆøµÄƽ¾ù·´Ó¦ËÙÂÊΪ0.075mol/(L¡¤min)
C£®´ïµ½Æ½ºâʱ£¬ÇâÆøµÄת»¯ÂÊΪ0.75
D£®Éý¸ßζȽ«Ê¹n(CH3OH)/n(CO2)¼õС
¢Û¼×´¼¼îÐÔȼÁÏµç³Ø¹¤×÷ʱ¸º¼«µÄµç¼«·´Ó¦Ê½¿É±íʾΪ¡¡                ¡¡¡£
£¨3£©ÍÑÁò¡£Ä³ÖÖÍÑÁò¹¤ÒÕÖн«·ÏÆø¾­´¦Àíºó£¬ÓëÒ»¶¨Á¿µÄ°±Æø¡¢¿ÕÆø·´Ó¦£¬Éú³ÉÁòËá狀ÍÏõËá淋ĻìºÏÎï×÷Ϊ¸±²úÆ·»¯·Ê¡£ÉèÑÌÆøÖеÄSO2¡¢NO2µÄÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã1£¬Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                                      ¡¡¡£
£¨4£©ÁòËá狀ÍÏõËáï§µÄË®ÈÜÒºµÄpH£¼7£¬ÆäÖÐÔ­Òò¿ÉÓÃÒ»¸öÀë×Ó·½³Ìʽ±íʾΪ£º¡¡                               ¡¡£»ÔÚÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÏõËáï§ÈÜÒºÖеμÓÊÊÁ¿µÄNaOHÈÜÒº£¬Ê¹ÈÜÒºµÄpH£½7£¬ÔòÈÜÒºÖУºc(Na£«)£«c(H£«)¡¡¡¡c(NO)£«c(OH£­)£¨Ìîд¡°£¾¡±¡°£½¡±»ò¡°£¼¡±£©¡£
°±ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬°±µÄºÏ³ÉÓëÓ¦ÓÃÊǵ±½ñÖØÒªÑо¿ÄÚÈÝÖ®Ò»¡£²»Í¬Î¶ȡ¢Ñ¹Ç¿Ï£¬ºÏ³É°±Æ½ºâÌåϵÖÐNH3µÄÎïÖʵÄÁ¿·ÖÊý¼ûÏÂ±í£¨N2ºÍH2ÆðʼÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã3£©¡£
ѹǿ(Mpa)
°±µÄƽ
ºâº¬Á¿(%)
ζÈ(¡æ)
0.1
10
20
30
60
100
200
15.3
81.5
86.4
89.9
95.4
98.8
300
2.2
52.0
64.2
71.0
84.2
92.6
400
0.4
25.1
38.2
47.0
65.2
79.8
500
0.1
10.6
19.1
26.4
42.2
57.5
600
0.05
4.5
9.1
13.8
23.1
31.4
»Ø´ðÏÂÁÐÓйØÎÊÌ⣺
£¨1£©ÀûÓñíÖÐÊý¾ÝÍÆ¶ÏµÃ³öºÏ³É°±µÄ·´Ó¦ÊÇ__________·´Ó¦£¨Ìî¡°·ÅÈÈ¡±¡¢¡°ÎüÈÈ¡±»ò¡°ÎÞÈÈÁ¿±ä»¯¡±£©¡£
£¨2£©¸ù¾Ý±íÖÐÊý¾Ý£¬ÔÚ200¡æºÍ100MPaʱ£¬Æ½ºâÌåϵÖÐNH3µÄÎïÖʵÄÁ¿·ÖÊý×î¸ß£¬¶øÊµ¼Ê¹¤ÒµÉú²ú²»Ñ¡ÓøÃÌõ¼þµÄÖ÷ÒªÔ­ÒòÊÇ___________________________________¡£
£¨3£©Ò»¶¨Ìõ¼þÏ£¬¶ÔÔÚÃܱÕÈÝÆ÷ÖнøÐеĺϳɰ±·´Ó¦´ïƽºâºó£¬ÆäËûÌõ¼þ²»±äʱ£¬ÈôͬʱѹËõÈÝÆ÷µÄÌå»ýºÍÉý¸ßζȴïÐÂÆ½ºâºó£¬ÓëԭƽºâÏà±È£¬Ç뽫ÓйØÎïÀíÁ¿µÄ±ä»¯µÄÇé¿öÌîÈëϱíÖУ¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°¿ÉÄÜÔö´óÒ²¿ÉÄܼõС¡±£©£º
 
·´Ó¦ËÙÂÊ
ƽºâ³£ÊýK
°±µÄÌå»ý·ÖÊý
±ä»¯Çé¿ö
 
 
 
£¨4£©½«1molH2ºÍ1molN2ͨÈëÒ»Ìå»ý²»±äµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Î¶Ⱥʹ߻¯¼Á×÷ÓÃÏ£¬·´Ó¦´ïµ½Æ½ºâ£¬²âµÃNH3µÄÎïÖʵÄÁ¿Îª0.3mol£¬´ËʱÈôÒÆ×ß0.5molH2ºÍ0.5molN2£¬Ôò·´Ó¦´ïµ½ÐÂµÄÆ½ºâʱ£¬NH3µÄÎïÖʵÄÁ¿Îª_____________£¨Ñ¡Ìî´ð°¸±àºÅ£©¡£
A£®0.3molB£®0.15molC£®Ð¡ÓÚ0.15molD£®´óÓÚ0.15mol£¬Ð¡ÓÚ0.3mol

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

¾«Ó¢¼Ò½ÌÍø