£¨14·Ö£©½«0.8 mol I2(g)ºÍ1.2 mol H2(g)ÖÃÓÚij1LÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Î¶ÈÏ·¢Éú·´Ó¦£ºI2(g)£«H2(g) 2HI(g)²¢´ïµ½Æ½ºâ¡£HIµÄÌå»ý·ÖÊýËæÊ±¼äµÄ±ä»¯Èç±í¸ñËùʾ£º

HIÌå»ý·ÖÊý

1min

2min

3min

4min

5min

6min

7min

Ìõ¼þI

26%

42%

52%

57%

60%

60%

60%

Ìõ¼þII

20%

33%

43%

52%

57%

65%

65%

£¨1£©ÔÚÌõ¼þIµ½´ïƽºâʱ£¬¼ÆËã¸Ã·´Ó¦µÄƽºâ³£ÊýK£¬ÒªÇóÁгö¼ÆËã¹ý³Ì¡£

£¨2£©ÔÚÌõ¼þI´Ó¿ªÊ¼·´Ó¦µ½µ½´ïƽºâʱ£¬H2µÄ·´Ó¦ËÙÂÊΪ____________¡£

£¨3£©Îª´ïµ½Ìõ¼þIIµÄÊý¾Ý£¬¶ÔÓÚ·´Ó¦Ìåϵ¿ÉÄܸıäµÄ²Ù×÷ÊÇ_______________¡£

£¨4£©¸Ã·´Ó¦µÄ¡÷H_____0£¨Ìî">"£¬"<"»ò"="£©

£¨5£©ÔÚÌõ¼þIÏ´ﵽƽºâºó£¬ÔÚ7minʱ½«ÈÝÆ÷Ìå»ýѹËõΪԭÀ´µÄÒ»°ë¡£ÇëÔÚͼÖл­³öc(HI)ËæÊ±¼ä±ä»¯µÄÇúÏß¡£

£¨1£©ÉèI2ÏûºÄŨ¶ÈΪx

I2(g) + H2(g) 2HI(g)

ÆðʼŨ¶È£¨mol/L£©£º 0.8 1.2 0

ת»¯Å¨¶È£¨mol/L£©£º x x 2x

ƽºâŨ¶È£¨mol/L£©£º0.8-x 1.2-x 2x £¨2·Ö£©

(ÈýÐÐÖеÄÊýÖµÎÞŨ¶È»òÁ¿µÄµ¥Î»Ê¾Òâ¿Û1·Ö)

HIµÄÌå»ý·ÖÊýΪ60%£¬Ôò£º2x/2=60%£¬x=0.6 mol/L £¨1·Ö£©

K=c2 (HI) /[c(H2)¡¤c(I2)]=1.22/(0.2¡Á0.6)=12 £¨1·Ö£©

£¨2£©0.12 mol/(L¡¤min) £¨2·Ö£¬µ¥Î»´í¿Û1·Ö£©

£¨3£©½µµÍÎÂ¶È (2·Ö)

£¨4£©< £¨2·Ö£©

£¨5£© £¨4·Ö£¬Á½¶Î¸÷2·Ö£©

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©¸ù¾Ý»¯Ñ§Æ½ºâ¼ÆËãÆ½ºâʱ¸÷ÎïÖʵÄŨ¶È£¬ÉèI2ÏûºÄŨ¶ÈΪx

I2(g) + H2(g) 2HI(g)

ÆðʼŨ¶È£¨mol/L£©£º 0.8 1.2 0

ת»¯Å¨¶È£¨mol/L£©£º x x 2x

ƽºâŨ¶È£¨mol/L£©£º0.8-x 1.2-x 2x

HIµÄÌå»ý·ÖÊýΪ60%£¬Ôò£º2x/2=60%£¬x=0.6 mol/L£¬ÔÙ¸ù¾Ýƽºâ³£ÊýµÄ¼ÆË㹫ʽK=c2 (HI) /[c(H2)¡¤c(I2)]½øÐмÆË㣬K=1.22/(0.2¡Á0.6)=12¡££¨2£©ÇâÆøµÄ·´Ó¦ËÙÂÊ=0.6/5=0.12mol/(L¡¤min)¡££¨3£©¸ù¾Ý±í¸ñ·ÖÎö£¬Ïàͬʱ¼äÄÚ£¬µâ»¯ÇâµÄÌå»ý·ÖÊý±ä»¯Ð¡£¬ËµÃ÷ËÙÂÊÂý£¬Î¶ȵ͡££¨4£©Æ½ºâʱµâ»¯ÇâµÄÌå»ý·ÖÊý´ó£¬ËµÃ÷½µÎÂÆ½ºâÕýÏòÒÆ¶¯£¬¡÷H <0¡££¨5£© Ìå»ýѹËõµ½Ô­À´Ò»°ë£¬Ôòµâ»¯ÇâµÄŨ¶È±ä³ÉÔ­À´2±¶£¬Æ½ºâ²»Òƶ¯£¬ËùÒÔͼÏñΪ£º

¿¼µã£º»¯Ñ§Æ½ºâ״̬µÄ¼ÆËãºÍÅж¨¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014-2015¹ã¶«Ê¡ÕØÇìÊиßÒ»ÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐÎïÖʵĴæÖü·½·¨´íÎóµÄÊÇ

A£®½ðÊôÄÆ±£´æÔÚʯÀ¯ÓÍ»òúÓÍÖÐ

B£®ÓÃÌúÖÆ»òÂÁÖÆÈÝÆ÷ÔËÊäŨÁòËᡢŨÏõËá

C£®FeCl2ÈÜÒºµÄÊÔ¼ÁÆ¿ÖÐÒª·ÅÌú¶¤

D£®ÇâÑõ»¯ÄÆÈÜҺʢװÔÚ²£Á§ÈûµÄÊÔ¼ÁÆ¿ÖÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014-2015¹ã¶«Ê¡¹ãÖÝÊиßÒ»ÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐÓйØÄÆÔªËصȵÄÐðÊöÖУ¬´íÎóµÄÊÇ

A£®µ¥ÖÊÄÆÈÛµãµÍÓÚ100¡æ

B£®ÄÆÔªËØÖ»ÄÜÒÔ»¯ºÏ̬´æÔÚÓÚ×ÔÈ»½ç

C£®NaÓëNa£«¶¼¾ßÓÐÇ¿µÄ»¹Ô­ÐÔ

D£®½ðÊôÄÆ¾ßÓкõĵ¼µçÐÔ£¬µ«ÈÕ³£Öв»ÓÃ×÷µ¼Ïß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014-2015¹ã¶«Ê¡¹ãÖÝÊÐÎåУ¸ß¶þÉÏѧÆÚÆÚÄ©Áª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ijÎÞÉ«ÈÜÒºÖÐÁ£×ÓÄÜ´óÁ¿¹²´æ£¬Í¨ÈëCO2ºóÈÔÄÜ´óÁ¿¹²´æµÄÒ»×éÊÇ

A£®K+¡¢SO42-¡¢Br-¡¢SiO32- B£®H+¡¢Fe2+¡¢Cl- ¡¢NH4+

C£®Na+¡¢Ba2+¡¢NO3-¡¢Cl- D£®Na+¡¢Ag+¡¢NH3H2O¡¢NO3-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014-2015¹ã¶«Ê¡¹ãÖÝÊÐÎåУ¸ß¶þÉÏѧÆÚÆÚÄ©Áª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ

A£®Ç¿µç½âÖÊÈÜÒºµÄµ¼µçÐÔ²»Ò»¶¨±ÈÈõµç½âÖÊÈÜÒºµÄµ¼µçÐÔÇ¿

B£®³£ÎÂÏ£¬½«pH=3µÄ´×ËáÈÜҺϡÊ͵½Ô­Ìå»ýµÄ10±¶ºó£¬ÈÜÒºµÄpH=4

C£®Èô²âµÃÓêË®µÄpHСÓÚ7£¬ÔòϵÄÊÇËáÓê

D£®ÔÚͨ·ç³÷ÖнøÐÐÓж¾ÆøÌåʵÑé·ûºÏ¡°ÂÌÉ«»¯Ñ§¡±Ë¼Ïë

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014-2015¹ã¶«Ê¡¹ãÑŵÈËÄУ¸ß¶þÉÏѧÆÚÆÚÄ©Áª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ

A£®KSp[MgF2]>KSp[Mg(OH)2]£¬²»ÄÜʵÏÖMg(OH)2ת»¯ÎªMgF2¡£

B£®³£ÎÂÏ£¬Í¬Å¨¶ÈµÄNa2SÓëNaHSÈÜÒºÏà±È£¬Na2SÈÜÒºµÄpH´ó

C£®µÈÎïÖʵÄÁ¿Å¨¶ÈµÄNH4ClÈÜÒººÍNH4HSO4ÈÜÒº£¬ºóÕßµÄc(NH4£«)´ó

D£®FeCl3ÓëKSCN·´Ó¦´ïµ½Æ½ºâʱ£¬¼ÓÈëKClÈÜÒº£¬ÔòÈÜÒºÑÕÉ«±äÉî

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014-2015¹ã¶«Ê¡¹ãÑŵÈËÄУ¸ß¶þÉÏѧÆÚÆÚÄ©Áª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÓÃÌúƬÓëÏ¡ÁòËá·´Ó¦ÖÆÈ¡ÇâÆøÊ±£¬ÏÂÁдëÊ©²»ÄÜʹÇâÆøÉú³ÉËÙÂʼӴóµÄÊÇ

A£®¼ÓÈÈ B£®²»ÓÃÏ¡ÁòËᣬ¸ÄÓÃ98%ŨÁòËá

C£®µÎ¼ÓÉÙÁ¿CuSO4ÈÜÒº D£®²»ÓÃÌúƬ£¬¸ÄÓÃÌú·Û

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014-2015ѧÄêÔÆÄÏÊ¡¸ßÈýÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

½öÓÃϱíÌṩµÄ²£Á§ÒÇÆ÷(·Ç²£Á§ÒÇÆ÷ÈÎÑ¡)¾ÍÄÜʵÏÖÏàӦʵÑéÄ¿µÄµÄÊÇ

Ñ¡Ïî

ʵÑéÄ¿µÄ

²£Á§ÒÇÆ÷

A

·ÖÀëÒÒ´¼ºÍÒÒËáÒÒõ¥µÄ»ìºÏÎï

·ÖҺ©¶·¡¢ÉÕ±­

B

ÓÃpH=1µÄÑÎËáÅäÖÆ100 mL,pH=2µÄÑÎËá

100 mLÈÝÁ¿Æ¿¡¢ÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü

C

ÓÃäåË®¡¢µí·ÛKIÈÜÒº±È½ÏBr2ÓëI2µÄÑõ»¯ÐÔÇ¿Èõ

ÊԹܡ¢½ºÍ·µÎ¹Ü

D

ÓÃNH4Cl¹ÌÌåºÍCa(OH)2¹ÌÌåÖÆ±¸²¢ÊÕ¼¯NH3

¾Æ¾«µÆ¡¢ÉÕ±­¡¢µ¼¹Ü¡¢¼¯ÆøÆ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014-2015ѧÄêºþÄÏÊ¡ÒæÑôÊиßÒ»ÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ

A£®1molNaOHµÄÖÊÁ¿ÊÇ40g B£®1molCH4Ìå»ýԼΪ22.4L

C£®CO2µÄĦ¶ûÖÊÁ¿Îª44g D£®1molH2OÖÐÔ¼º¬6.02¡Á1023¸öH

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸