³µÓÃÒÒ´¼ÆûÓÍÊÇÖ¸ÔÚÆûÓÍ×é·ÖÓÍÖа´Ìå»ý»ìºÏ±È¼ÓÈë10%µÄ±äÐÔȼÁÏÒÒ´¼ºó×÷ΪÆûÓͳµÈ¼ÁÏÓõÄÆûÓÍ¡£

£¨1£©4.6 g CH3CH2OHÍêȫȼÉÕÉú³ÉË®ºÍ¶þÑõ»¯Ì¼£¬ÓÐ___________molµç×ÓתÒÆ¡£

£¨2£©ÌþAÊÇÆûÓ͵ÄÖ÷Òª³É·ÖÖ®Ò»¡£È¡11.4 gÌþA£¬ÖÃÓÚÒ»¶¨Ìå»ýµÄÑõÆøÖУ¬µãȼʹ֮³ä·Ö·´Ó¦ºó£¬»Ö¸´µ½±ê×¼×´¿öÆøÌåÌå»ý¼õÉÙx L¡£½«Ê£ÓàÆøÌå¾­¹ý¼îʯ»ÒÎüÊÕ£¬¼îʯ»ÒÖÊÁ¿Ôö¼Óy g¡£Êý¾Ý¼ûÏÂ±í£¨±íÖÐËùÓÐÌå»ý¾ùÔÚ±ê×¼×´¿öϲⶨ£©¡£

 

ÑõÆøÌå»ýΪ20 L

ÑõÆøÌå»ýΪ30 L

ÑõÆøÌå»ýΪ40 L

x

2.22

10.08

10.08

y

4.3

35.2

35.2

11.4 gÌþAÖк¬Ì¼µÄÎïÖʵÄÁ¿ÊÇ_______________mol£¬º¬ÇâµÄÎïÖʵÄÁ¿ÊÇ_____________mol¡£¸ÃÌþµÄ·Ö×ÓʽÊÇ_____________¡£

£¨1£©1.2  (2)0.8  1.8  C8H18

½âÎö£º

(1)C2H6O+3O22CO2+3H2O  תÒƵç×Ó12e-

       46 g                                                     12 mol

      4.6 g                                                     1.2 mol

 (2)±ê×¼×´¿öÏ£¬Ë®ÎªÒºÌ¬£¬¼îʯ»ÒÔö¼ÓÖÊÁ¿¾ÍÊÇCO2µÄÖÊÁ¿£¬30 LO2ÒѹýÁ¿£¬ÌþÄÜÍêȫȼÉÕ£¬Ôòn(C)=n(CO2)==0.8 mol¡£º¬HµÄÖÊÁ¿Îª11.4 g-(0.8¡Á12) g=1.8 g,n(H)==1.8 mol¡£ÓÉÒÔÉϼÆËã¿ÉÖªn(C)¡Ãn(H)=0.8¡Ã1.8,¿ÉÖª¸ÃÌþ·Ö×ÓʽΪC8H18¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§ÓëÉú»î¡¢Éú²úÃܲ»¿É·Ö£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³µÓÃÒÒ´¼ÆûÓÍÊÇÖ¸ÔÚÆûÓÍ×é·ÖÖа´Ìå»ý»ìºÏ±È¼ÓÈë10%µÄ±äÐÔȼÁÏÒÒ´¼ºó×÷ΪÆûÓͳµÈ¼ÁÏÓõÄÆûÓÍ¡£?

(1)4.6 g CH3CH2OHÍêȫȼÉÕÉú³ÉË®ºÍ¶þÑõ»¯Ì¼£¬ÓР     molµç×ÓתÒÆ¡£?

(2)ÌþAÊÇÆûÓ͵ÄÖ÷Òª³É·ÖÖ®Ò»¡£È¡11.4gÌþA£¬ÖÃÓÚÒ»¶¨Ìå»ýµÄÑõÆøÖУ¬µãȼʹ֮³ä·Ö·´Ó¦ºó£¬»Ö¸´µ½±ê×¼×´¿öÆøÌåÌå»ý¼õÉÙx L¡£½«Ê£ÓàÆøÌå¾­¹ý¼îʯ»ÒÎüÊÕ£¬¼îʯ»ÒÖÊÁ¿Ôö¼Óyg¡£Êý¾Ý¼ûϱí(±íÖÐËùÓÐÌå»ý¾ùÔÚ±ê×¼×´¿öϲⶨ)¡£

 

ÑõÆøÌå»ýΪ20 L

ÑõÆøÌå»ýΪ30 L

ÑõÆøÌå»ýΪ40 L

x

2.22

10.08

10.08

y

4.3

35.2

35.2

11.4 gÌþAÖк¬Ì¼µÄÎïÖʵÄÁ¿ÊÇ      mol£¬º¬ÇâµÄÎïÖʵÄÁ¿ÊÇ      mol¡£¸ÃÌþµÄ·Ö×ÓʽÊÇ     ¡£?

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

(¢ñ)Ö¸³öÔÚʹÓÃÏÂÁÐÒÇÆ÷(ÒѾ­Ï´µÓ¸É¾»)»òÓÃƷʱµÄµÚÒ»²½²Ù×÷£º

¢ÙʯÈïÊÔÖ½(¼ìÑéÆøÌå)£º¡£?

¢ÚµÎ¶¨¹Ü£º                ¡£?

¢Û¼¯ÆøÆ¿(ÊÕ¼¯°±Æø)£º                 ¡£?

¢ÜÍÐÅÌÌìƽ                           ¡£?

(¢ò)Ä¿Ç°¹ú¼ÊÓͼ۾Ӹ߲»Ï¡£ÒÒ´¼ÆûÓÍÊÇÖ¸ÔÚÆûÓÍÖмÓÈë10%µÄÒÒ´¼(Ìå»ý±È)¡£ÒÒ´¼ÆûÓ;ßÓÐÐÁÍéÖµ¸ß¡¢¿¹±¬ÐԺõÄÌص㣬ÍƹãʹÓÿɻº½âÄÜÔ´½ôȱ¡¢´Ù½ø¹úÃñ¾­¼Ã·¢Õ¹¡£ÆûÓÍÖÐÌí¼ÓµÄÒÒ´¼²»Äܺ¬Ë®£¬·ñÔò»áÓ°Ïì·¢¶¯»úÕý³£ÔËתºÍʹÓÃÊÙÃü¡£ÖƱ¸ÎÞË®ÒÒ´¼¿É²ÉÈ¡ÈçÏ·½·¨£º??

¢ÙÔÚ250 mLÔ²µ×ÉÕÆ¿ÖмÓÈë95%µÄÒÒ´¼100 mLºÍÐÂÖƵÄÉúʯ»Ò30 g£¬ÔÚˮԡÖмÓÈÈ»ØÁ÷1ÖÁ2Сʱ¡£(Èçͼ1Ëùʾ)?

¢ÚÈ¡ÏÂÀäÄý¹Ü£¬¸Ä³ÉÈçͼ2ËùʾµÄ×°Öã¬ÔÙ½«AÖÃÓÚˮԡÖÐÕôÁó¡£?

¢Û°Ñ×î³õÕô³öµÄ5 mLÁó³öÒºÁíÍâ»ØÊÕ¡£?

¢ÜÓúæ¸ÉµÄÎüÂËÆ¿×÷Ϊ½ÓÊÜÆ÷£¬Æä²à¹Ü½Óһ֧װÓÐCaCl2µÄ¸ÉÔï¹ÜC£¬Ê¹ÆäÓë´óÆøÏàͨ£¬ÕôÖÁÎÞÒºµÎ³öÀ´ÎªÖ¹£¬¼´µÃ99.5%µÄ¾Æ¾«¡£ÊԻشð£º?

(1)ÒÑÖªÔÚ101 kPa¡¢25 ¡æʱ£¬?

C2H5OH(l)+3O2(g)=2CO2(g)+3H2O(l);¦¤H =-1 367 kJ¡¤mol-1??

1 gÒÒ´¼ÍêȫȼÉÕÉú³ÉҺ̬ˮʱ·Å³ö            ÈÈÁ¿¡£?

(2)ͼ2ÖеĸÉÔï¹ÜCµÄ×÷ÓÃÊÇ            ¡£ÒÇÆ÷BµÄÃû³ÆÊÇ            £¬ÀäÄýË®ÊÇ´Ó¿Ú            (Ìî¡°a¡±»ò¡°b¡±)½øÈëÀäÄý¹Ü¡£?

(3)ÎÞË®CaCl2³£ÓÃ×÷ÎüË®¼Á£¬ÔÚÉÕÆ¿AÖÐÄÜ·ñÓÃÎÞË®CaCl2´úÌæÉúʯ»Ò£¿            (Ìîд¡°ÄÜ¡±¡°²»ÄÜ¡±)£¬Ô­ÒòÊÇ                           ¡£?

(4)¼ìÑéËùµÃ²úÆ·ÖÐÊÇ·ñº¬Ë®µÄ²Ù×÷·½·¨ÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

(5)ijͬѧÏëÒª¼ìÑéͼ2×°ÖõÄÆøÃÜÐÔ£¬Ëû¿É´Óͼ3ÖÐÑ¡ÔñÄļ¸ÖÖÒÇÆ÷(ÌîÒÇÆ÷´úºÅ)            ¡£Çë¼òÊö¼ìÑéÆøÃÜÐԵIJÙ×÷¹ý³Ì            ¡£?

(6)д³öÓÉÆÏÌÑÌÇת»¯Îª¾Æ¾«µÄ»¯Ñ§·½³Ìʽ£º                                 ¡£?

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2009Äêºþ±±Ê¡»Æ¸ÔÊлÆÖÝÖÐѧ¸ß¿¼»¯Ñ§¶þÄ£ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

»¯Ñ§ÓëÉú»î¡¢Éú²úÃܲ»¿É·Ö£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨ £©
A£®´¿¾»Ë®¼¸ºõ²»º¬ÈκÎÔÓÖÊ£¬ºÈ´¿¾»Ë®±ÈºÈ¿óȪˮ¶ÔÈËÌå¸üÓªÑø¡¢¸ü½¡¿µ
B£®¡°ÒÒ´¼ÆûÓÍ¡±ÒòÓͼÛÉÏÕǶø±»¹ã·ºÊ¹Óã®ËüÊÇÖ¸ÔÚÆûÓÍÀï¼ÓÈëÊÊÁ¿ÒÒ´¼»ìºÏ¶ø³ÉµÄÒ»ÖÖȼÁÏ£¬ËüÊÇÒ»ÖÖÐÂÐ͵Ļ¯ºÏÎï
C£®Ä³ÓêË®ÑùÆ·²É¼¯ºó·ÅÖÃÒ»¶Îʱ¼ä£¬pHÖµÓÉ4.68±äΪ4.28£¬ÊÇÒòΪˮÖÐÈܽâÁ˽϶àµÄCO2
D£®ÃɹÅÄÁÃñϲ»¶ÓÃÒøÆ÷Ê¢·ÅÏÊÅ£ÄÌÓÐÆä¿ÆѧµÀÀí£ºÓÃÒøÆ÷Ê¢·ÅÏÊÅ£ÄÌ£¬ÈÜÈëµÄ¼«Î¢Á¿µÄÒøÀë×Ó£¬¿ÉɱËÀÅ£ÄÌÖеÄϸ¾ú£¬·ÀֹţÄ̱äÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸