ÌâÄ¿ÄÚÈÝ

ÏÂÁÐ˵·¨¿ÉÒÔÖ¤Ã÷¿ÉÄæ·´Ó¦N2£«3H22NH3ÒѴﵽƽºâ״̬µÄÊÇ

¢ÙÒ»¸öN¡ÔN¼ü¶ÏÁѵÄͬʱ£¬ÓÐ6¸öN£­H¼ü¶ÏÁÑ

¢Úv(H2)£½0.6 mol¡¤L£­1¡¤min£­1£¬v(NH3)£½0.4 mol¡¤L£­1¡¤min£­1

¢Û±£³ÖÆäËûÌõ¼þ²»±äʱ£¬Ìåϵѹǿ²»Ôٸıä

¢ÜNH3¡¢N2¡¢H2µÄÌå»ý·ÖÊý¶¼²»Ôٸıä

¢ÝºãκãÈÝʱ£¬»ìºÏÆøÌåÖÊÁ¿±£³Ö²»±ä

A. ¢Ú¢Û¢Ü B. ¢Ù¢Û¢Ü C. ¢Ù¢Ú¢Ü D. ¢Û¢Ü¢Ý

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

îÜÖÜÆÚ±íµÚËÄÖÜÆÚµÚ¢ø×åÔªËØ£¬Æ仯ºÏÎïÓÃ;¹ã·º£¬È磺LiCoO2×ö﮵ç³ØµÄÕý¼«²ÄÁÏ£»²ÝËáîÜ¿ÉÓÃÓÚָʾ¼ÁºÍ´ß»¯¼ÁÖƱ¸¡£

¢ñ.£¨1£©LiCoO2ÖÐîÜÔªËصĻ¯ºÏ¼ÛΪ_______

£¨2£©¹¤ÒµÉϽ«·Ï﮵ç³ØµÄÕý¼«²ÄÁÏÓë¹ýÑõ»¯ÇâÈÜÒº£¬Ï¡ÁòËá»ìºÏ¼ÓÈÈ£¬¿ÉµÃµ½CoSO4»ØÊÕ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º______________£»¿ÉÓÃÑÎËá´úÌæH2SO4ºÍH2O2µÄ»ìºÏÒº£¬µ«È±µãÊÇ____________

¢ò.ÀûÓÃÒ»ÖÖº¬îÜ¿óʯ[Ö÷Òª³É·ÖΪCo2O3£¬º¬ÉÙÁ¿Fe2O3¡¢Al2O3¡¢MnO¡¢MgO¡¢CaOµÈ]ÖÆÈ¡CoC2O4¡¤2H2O¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢Ù½þ³öÒºº¬ÓеÄÑôÀë×ÓÖ÷ÒªÓÐH+¡¢Co2+¡¢Fe2+¡¢Mn2+¡¢Ca2+¡¢Mg2+¡¢Al3+µÈ£»

¢Ú²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º

³ÁµíÎï

Fe(OH)3

Fe(OH£©2

Co(OH)2

Al(OH)3

Mn(OH)2

ÍêÈ«³Áµí

3.7

9.6

9.2

5.2

9.8

£¨3£©½þ³öÒºÖмÓNaClO3µÄÄ¿µÄÊÇ£º______________£®

£¨4£©ÇëÓÃƽºâÒƶ¯Ô­Àí˵Ã÷¼ÓNa2CO3µ÷pHÖÁ5.2ËùµÃ³ÁµíµÄÔ­Òò£º____________

£¨5£©ÂËÒº¢òÖмÓÈëÝÍÈ¡¼ÁµÄ×÷ÓÃÊÇ___________

£¨6£©¡°³ý¸Æ¡¢Ã¾¡±Êǽ«ÈÜÒºÖÐCa2+ÓëMg2+ת»¯ÎªMgF2¡¢CaF2³Áµí¡£ÒÑÖªKsp(MgF2)£½7.35¡Á10-11¡¢Ksp(CaF2)£½1.05¡Á10-10¡£µ±¼ÓÈë¹ýÁ¿NaFºó£¬ËùµÃÂËÒºc(Mg2+)/c(Ca2+)£½___________¡£

¼×´¼£¨CH3OH£©ÊÇÖØÒªµÄÈܼÁºÍÌæ´úȼÁÏ£¬¹¤ÒµÉÏÓÃCOºÍH2ÔÚÒ»¶¨Ìõ¼þÏÂÖƱ¸CH3OHµÄ·´Ó¦ÎªCO(g)+2H2(g)CH3OH(g) ¡÷H¡£

£¨1£©ÔÚÌå»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬³äÈë2molCOºÍ4molH2£¬Ò»¶¨Ìõ¼þÏ·¢ÉúÉÏÊö·´Ó¦£¬²âµÃCO(g)ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼äµÄ±ä»¯ÈçÏÂͼ¼×Ëùʾ¡£

¢Ù´Ó·´Ó¦¿ªÊ¼µ½5min£¬ÓÃÇâÆø±íʾµÄƽ¾ù·´Ó¦ËÙÂÊv(H2)=________¡£

¢ÚÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________£¨ÌîÐòºÅ£©¡£

A. ´ïµ½Æ½ºâʱ£¬H2µÄת»¯ÂÊΪ75%

B. 5minºóÈÝÆ÷ÖÐѹǿ²»Ôٸıä

C. ´ïµ½Æ½ºâºó£¬ÔÙ³äÈëë²Æø£¬·´Ó¦ËÙÂÊÔö´ó

D. 2minÇ°v(Õý)£¾v(Äæ)£¬2minºóv(Õý)£¼v(Äæ)

£¨2£©Ä³Î¶ÈÏ£¬ÔÚÒ»ºãѹÈÝÆ÷Öзֱð³äÈë1.2molCOºÍ1molH2£¬´ïµ½Æ½ºâʱÈÝÆ÷Ìå»ýΪ2L£¬ÇÒº¬ÓÐ0.4molCH3OH(g)£¬Ôò¸Ã·´Ó¦Æ½ºâ³£ÊýµÄֵΪ_______£¬´ËʱÏòÈÝÆ÷ÖÐÔÙͨÈë0.35molCOÆøÌ壬Ôò´Ëƽºâ½«___£¨Ìî¡°ÕýÏòÒƶ¯¡±¡°²»Òƶ¯¡±»ò¡°ÄæÏòÒƶ¯¡±£©

£¨3£©Èôѹǿ¡¢Í¶ÁϱÈx [n£¨CO£©/n£¨H2£©]¶Ô·´Ó¦µÄÓ°ÏìÈçͼÒÒËùʾ£¬ÔòͼÖÐÇúÏßËùʾµÄѹǿ¹Øϵ£ºp1___p2£¨Ìî¡°=¡±¡°£¾¡±»ò¡°£¼¡±£©£¬ÆäÅжÏÀíÓÉÊÇ__________¡£

£¨4£©¼×´¼ÊÇÒ»ÖÖÐÂÐ͵ÄÆû³µ¶¯Á¦È¼ÁÏ¡£ÒÑÖªH2(g)¡¢CO(g)¡¢CH3OH(l)µÄȼÉÕÈÈ·Ö±ðΪ285.8kJ/mol¡¢283.0kJ/molºÍ726.5kJ/mol£¬Ôò¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ_____¡£

£¨5£©ÏÖÓÐÈÝ»ý¾ùΪ1LµÄa¡¢b¡¢cÈý¸öÃܱÕÈÝÆ÷£¬ÍùÆäÖзֱð³äÈë1molCOºÍ2molH2µÄ»ìºÏÆøÌ壬¿ØÖÆζȣ¬½øÐз´Ó¦£¬²âµÃÏà¹ØÊý¾ÝµÄ¹ØϵÈçÏÂͼËùʾ¡£bÖм״¼Ìå»ý·ÖÊý´óÓÚaÖеÄÔ­ÒòÊÇ_____¡£´ïµ½Æ½ºâʱ£¬a¡¢b¡¢cÖÐCOµÄת»¯ÂÊ´óС¹ØϵΪ___________¡£

£¨6£©¼×´¼×÷ΪһÖÖȼÁÏ»¹¿ÉÓÃÓÚȼÁϵç³Ø¡£ÔÚζÈΪ650¡æµÄÈÛÈÚÑÎȼÁϵç³ØÖÐÓü״¼¡¢¿ÕÆøÓëCO2µÄ»ìºÏÆøÌå×÷·´Ó¦ÎÄø×÷µç¼«£¬ÓÃLi2CO3ºÍNa2CO3»ìºÏÎï×÷µç½âÖÊ¡£¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½Îª____¡£

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

¾«Ó¢¼Ò½ÌÍø