¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£ºCO2(g)£«H2(g)CO(g)£«H2O(g)£¬Æ仯ѧƽºâ³£ÊýKºÍζÈTµÄ¹ØϵÈçÏÂ±í£º

T/¡æ

700

800

830

1000

1200

K

0.6

0.9

1.0

1.7

2.6

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¸Ã·´Ó¦Îª____________·´Ó¦(Ìî¡°ÎüÈÈ¡±¡¢¡°·ÅÈÈ¡±)¡£

(2)ÄÜʹ¸Ã·´Ó¦µÄ·´Ó¦ËÙÂÊÔö´ó£¬ÇÒƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯µÄÊÇ___________¡£

a.¼°Ê±·ÖÀë³öCOÆøÌå b.Êʵ±Éý¸ßζÈ

c.Ôö´óCO2µÄŨ¶È d.Ñ¡Ôñ¸ßЧ´ß»¯¼Á

(3)ijζȣ¬Æ½ºâŨ¶È·ûºÏÏÂʽ£º c(CO2)¡¤c(H2)£½c(CO)¡¤c(H2O)£¬ÊÔÅжϴËʱµÄζÈΪ______¡æ¡£

(4)ÈôÔÚ(3)Ëù´¦µÄζÈÏ£¬ÔÚ1LµÄÃܱÕÈÝÆ÷ÖУ¬¼ÓÈë2molCO2ºÍ3molH2³ä·Ö·´Ó¦´ïƽºâʱ£¬H2µÄÎïÖʵÄÁ¿Îª__________£¬CO2µÄÎïÖʵÄÁ¿Îª__________¡£

a.µÈÓÚ1.0mol b.´óÓÚ1.0mol c.´óÓÚ0.5mol£¬Ð¡ÓÚ1.0mol d.ÎÞ·¨È·¶¨

(5)¸ù¾ÝÑо¿±¨µÀ£¬ÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬N2ÔÚ´ß»¯¼Á(²ôÓÐÉÙÁ¿Fe2O3µÄTiO2)±íÃæÓëË®·¢ÉúÏÂÁз´Ó¦£º2N2(g)£«6H2O(l)4NH3(g)£«3O2(g)¡¡¦¤H£½akJ¡¤mol-1

½øÒ»²½Ñо¿NH3Éú³ÉÁ¿ÓëζȵĹØϵ£¬³£Ñ¹Ï´ﵽƽºâʱ²âµÃ²¿·ÖʵÑéÊý¾ÝÈç±í£º

T/K

303

313

323

NH3Éú³ÉÁ¿/(10-6 mol)

4.8

5.9

6.0

¢Ù´ËºÏ³É·´Ó¦µÄ________0£¬¦¤S________0£»(Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±)

¢ÚÒÑÖª£ºN2(g)£«3H2(g)=2NH3(g)¡¡¦¤H=-92.4kJ¡¤mol-1

2H2(g)£«O2(g)=2H2O(l)¡¡¦¤H=-571.6kJ¡¤mol-1

Ôò³£ÎÂϵªÆøÓëË®·´Ó¦Éú³É°±ÆøÓëÑõÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ£º_____________________________¡£

¡¾´ð°¸¡¿ÎüÈÈ bc 830 b c £¾ £¾ 2N2(g)£«6H2O(l)===4NH3(g)£«3O2(g) ¦¤H£½£«1530kJ¡¤mol-1

¡¾½âÎö¡¿

(1)ͼ±íÊý¾Ý·ÖÎö£¬Æ½ºâ³£ÊýËæζÈÉý¸ßÔö´ó£¬ËµÃ÷ζÈÉý¸ßƽºâÕýÏò½øÐУ¬Õý·´Ó¦ÊÇÎüÈÈ·´Ó¦£»

(2)¸Ã·´Ó¦CO2(g)£«H2(g)CO(g)£«H2O(g)ÊÇÎüÈÈ·´Ó¦£¬ÇÒ·´Ó¦¹ý³ÌÖÐÆøÌåÌå»ý²»±ä¡£ÎªÁËʹ¸Ã·´Ó¦µÄ·´Ó¦ËÙÂÊÔö´óÇÒƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬¿ÉÒÔÉý¸ßζȻòÔö´ó·´Ó¦ÎïCO2µÄŨ¶È£»

(3)ijζȣ¬Æ½ºâŨ¶È·ûºÏÏÂʽ£º c(CO2)¡¤c(H2)£½c(CO)¡¤c(H2O)£¬ÔòK= =1£»

(4)ÁС°Èý¶Îʽ¡±½â´ð£»

(5) ¢Ù¡÷G=¡÷H-T¡÷S£¬Èç¹û¡÷G<0£¬·´Ó¦¿ÉÒÔ×Ô·¢½øÐУ»

¢Úa N2(g)£«3H2(g)=2NH3(g)¡¡¦¤H=-92.4kJ¡¤mol-1£»b2H2(g)£«O2(g)=2H2O(l) ¦¤H=-571.6kJ¡¤mol-1£»¸ù¾Ý¸Ç˹¶¨ÂÉ2a-3b¿ÉµÃ2N2(g)£«6H2O(l)===4NH3(g)£«3O2(g)¡£

(1)ͼ±íÊý¾Ý·ÖÎö£¬Æ½ºâ³£ÊýËæζÈÉý¸ßÔö´ó£¬ËµÃ÷ζÈÉý¸ßƽºâÕýÏò½øÐУ¬Õý·´Ó¦ÊÇÎüÈÈ·´Ó¦£»

(2)¸Ã·´Ó¦CO2(g)£«H2(g)CO(g)£«H2O(g)ÊÇÎüÈÈ·´Ó¦£¬ÇÒ·´Ó¦¹ý³ÌÖÐÆøÌåÌå»ý²»±ä¡£ÎªÁËʹ¸Ã·´Ó¦µÄ·´Ó¦ËÙÂÊÔö´óÇÒƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬¿ÉÒÔÉý¸ßζȻòÔö´ó·´Ó¦ÎïCO2µÄŨ¶È£»´ß»¯¼Á²»Äܸı䷴Ӧ½ø³Ì£¬¼°Ê±·ÖÀë³öCOÆøÌåËäÈ»ÄÜʹ·´Ó¦ÕýÏòÒƶ¯£¬µ«²»ÄÜʹ¸Ã·´Ó¦µÄ·´Ó¦ËÙÂÊÔö´ó£¬¹ÊÑ¡bc£»

(3)ijζȣ¬Æ½ºâŨ¶È·ûºÏÏÂʽ£º c(CO2)¡¤c(H2)£½c(CO)¡¤c(H2O)£¬ÔòK= =1£¬Óɱí¿ÉÖª£¬Î¶ÈΪ830¡æ£»

(4) CO2(g)£«H2(g)CO(g)£«H2O(g)

³õʼ£¨mol£©2 3 0 0

ת»¯£¨mol£©x x x x

ƽºâ£¨mol£©2-x 3-x x x

K===1£¬½âµÃx=1.2mol£¬ÔòƽºâʱH2µÄÎïÖʵÄÁ¿Îª3mol-1.2mol=1.8mol£¬´óÓÚ1.0mol£¬¹ÊÑ¡b£»CO2µÄÎïÖʵÄÁ¿Îª2mol-1.2mol=0.8mol£¬´óÓÚ0.5mol£¬Ð¡ÓÚ1.0mol£¬¹ÊÑ¡c£»

(5) ¢ÙÓɱí¸ñ¿ÉÖª£¬Éý¸ßζȣ¬NH3Éú³ÉÁ¿Ôö´ó£¬·´Ó¦ÕýÏòÒƶ¯£¬ËµÃ÷¸Ã·´Ó¦ÎüÈÈ£¬¦¤H£½akJ¡¤mol-1£¾0£¬Ôòa£¾0£»¸Ã·´Ó¦ÄÜ×Ô·¢½øÐÐÇÒ¦¤H£¾0£¬¡÷G=¡÷H-T¡÷S£¬Èç¹û¡÷G<0 ·´Ó¦¿ÉÒÔ×Ô·¢½øÐУ¬¹Ê¡÷S£¾0£»

¢Úa N2(g)£«3H2(g)=2NH3(g)¡¡¦¤H=-92.4kJ¡¤mol-1£»b2H2(g)£«O2(g)=2H2O(l) ¦¤H=-571.6kJ¡¤mol-1£»¸ù¾Ý¸Ç˹¶¨ÂÉ2a-3b¿ÉµÃ2N2(g)£«6H2O(l)===4NH3(g)£«3O2(g) ¦¤H= -92.4kJ¡¤mol-1¡Á2-3¡Á£¨-571.6kJ¡¤mol-1£©=£«1530 kJ¡¤mol-1£¬¹Ê´ð°¸Îª£º2N2(g)£«6H2O(l)=4NH3(g)£«3O2(g) ¦¤H£½£«1530kJ¡¤mol-1¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿·´Ó¦C2H6(g)C2H4(g)+H2(g) ¡÷H>0£¬ÔÚÒ»¶¨Ìõ¼þÏÂÓÚÃܱÕÈÝÆ÷Öдﵽƽºâ¡£ÏÂÁи÷Ïî´ëÊ©ÖУ¬²»ÄÜÌá¸ßÒÒÍéƽºâת»¯ÂʵÄÊÇ( )

A. Ôö´óÈÝÆ÷ÈÝ»ýB. Éý¸ß·´Ó¦Î¶È

C. ·ÖÀë³ö²¿·ÖÇâÆøD. µÈÈÝÏÂͨÈë¶èÐÔÆøÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÐèÒªÅäÖÆ0.50 mol/LNaClÈÜÒº480 mL¡£°´ÏÂÁвÙ×÷²½ÖèÌîÉÏÊʵ±µÄÎÄ×Ö£¬ÒÔʹÕû¸ö²Ù×÷ÍêÕû¡£

(1)Ñ¡ÔñÒÇÆ÷¡£Íê³É±¾ÊµÑéËù±ØÐèµÄÒÇÆ÷ÓУºÍÐÅÌÌìƽ(¾«È·µ½0.1 g)¡¢Ò©³×¡¢ÉÕ±­¡¢²£Á§°ô¡¢________¡¢________ÒÔ¼°µÈÖÊÁ¿µÄÁ½Æ¬ÂËÖ½¡£

(2)¼ÆËã³ÆÁ¿¡£ÅäÖƸÃÈÜÒºÐè³ÆÈ¡NaCl¾§Ìå________g¡£

(3)Èܽ⡢ÀäÈ´£¬¸Ã²½ÊµÑéÖÐÐèҪʹÓò£Á§°ô£¬Ä¿µÄÊÇ___________¡£

(4)תÒÆ¡¢Ï´µÓ¡£

(5)ÏòÈÝÁ¿Æ¿ÖмÓÈëÕôÁóË®£¬Ö±µ½ÒºÃæÀë¿Ì¶ÈÏßÔ¼1-2ÀåÃ×ʱ£¬¸ÄÓÃ___________µÎ¼ÓÕôÁóË®ÖÁÒºÃæÓë¿Ì¶ÈÏßÏàÇС£¸ÇºÃÆ¿Èû£¬Ò¡ÔÈ¡£Èç¹û¼ÓˮʱҺÃ泬¹ý¿Ì¶ÈÏߣ¬½«Ê¹ÅäµÃµÄÈÜҺŨ¶È___________£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©¡£

(6)½«ÅäºÃµÄÈÜÒº¾²ÖÃÒ»¶Îʱ¼äºó£¬µ¹ÈëÖ¸¶¨µÄÊÔ¼ÁÆ¿£¬²¢ÌùºÃ±êÇ©£¬×¢Ã÷ÅäÖƵÄʱ¼ä¡¢ÈÜÒºÃû³Æ¼°Å¨¶È¡£

(7)ÔÚÅäÖƹý³ÌÖУ¬Ä³Ñ§Éú¹Û²ì¶¨ÈÝʱҺÃæÇé¿öÈçͼËùʾ£¬ËùÅäÈÜÒºµÄŨ¶È»á________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª·´Ó¦£º2NO2£¨ºì×ØÉ«£©N2O4£¨ÎÞÉ«£© ¡÷H£¼0£¬ÔÚ100¡æʱ£¬½«0.40mol NO2ÆøÌå³äÈë2LÃܱÕÈÝÆ÷ÖУ¬Ã¿¸ôÒ»¶Îʱ¼ä¶Ô¸ÃÈÝÆ÷µÄÎïÖʽøÐвâÁ¿£¬µÃµ½µÄÊý¾ÝÈçÏÂ±í£º

ʱ¼ä/s

n/mol

0

20

40

60

80

100

n(NO2)

0.40

a

0.26

c

d

e

n(N2O4)

0.00

0.05

b

0.08

0.08

0.08

£¨1£©100sºó½µµÍ·´Ó¦»ìºÏÎïµÄζȣ¬»ìºÏÆøÌåµÄÑÕÉ«_________£¨Ìî¡°±ädz¡±¡¢¡°±äÉ»ò¡°²»±ä¡±£©¡£

£¨2£©20ÖÁ40sÄÚ£¬v(NO2)=__________mol/(L¡¤s)£¬100¡æʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK =_____________¡£

£¨3£©½«Ò»¶¨Á¿µÄNO2³äÈëÃܱÕ×¢ÉäÆ÷ÖУ¬Í¼ÊÇÔÚÀ­ÉìºÍѹËõ×¢ÉäÆ÷µÄ¹ý³ÌÖÐÆøÌå͸¹âÂÊËæʱ¼äµÄ±ä»¯£¨ÆøÌåÑÕÉ«Ô½É͸¹âÂÊԽС£©¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ______________

A.bµãµÄ²Ù×÷ÊÇѹËõ×¢ÉäÆ÷

B.cµãÓëaµãÏà±È£¬c(NO2)Ôö´ó£¬c(N2O4)¼õС

C.ÈôÃܱÕ×¢ÉäÆ÷Ϊ¾øÈÈÈÝÆ÷£¬ÔòT(b)£¾T(c)

D.dµãʱv£¨Õý£©£¾v£¨Ä棩

£¨4£©ÄÜ˵Ã÷·´Ó¦2NO2£¨ºì×ØÉ«£©N2O4£¨ÎÞÉ«£©´ïƽºâµÄÊÇ_________

A. ÌåϵµÄÑÕÉ«²»±ä B. ºãÈÝÌõ¼þÏ£¬ÆøÌåµÄÃܶȲ»±ä

C. 2vÕý£¨NO2£©=vÄ棨N2O4£© D. »ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿²»±ä

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿N2O¡¢NOºÍNO2µÈµªÑõ»¯ÎïÊÇ¿ÕÆøÎÛȾÎº¬ÓеªÑõ»¯ÎïµÄβÆøÐè´¦Àíºó²ÅÄÜÅÅ·Å¡£

£¨1£©N2OµÄ´¦Àí¡£N2OÊÇÏõËáÉú²úÖа±´ß»¯Ñõ»¯µÄ¸±²úÎÓÃÌØÖÖ´ß»¯¼ÁÄÜʹN2O·Ö½â¡£NH3ÓëO2ÔÚ¼ÓÈȺʹ߻¯¼Á×÷ÓÃÏÂÉú³ÉN2OµÄ»¯Ñ§·½³ÌʽΪ________¡£

£¨2£©NOºÍNO2µÄ´¦Àí¡£ÒѳýÈ¥N2OµÄÏõËáβÆø¿ÉÓÃNaOHÈÜÒºÎüÊÕ£¬Ö÷Òª·´Ó¦Îª

NO+NO2+2OH2+H2O

2NO2+2OH++H2O

¢ÙÏÂÁдëÊ©ÄÜÌá¸ßβÆøÖÐNOºÍNO2È¥³ýÂʵÄÓÐ________£¨Ìî×Öĸ£©¡£

A£®¼Ó¿ìͨÈëβÆøµÄËÙÂÊ

B£®²ÉÓÃÆø¡¢ÒºÄæÁ÷µÄ·½Ê½ÎüÊÕβÆø

C£®ÎüÊÕβÆø¹ý³ÌÖж¨ÆÚ²¹¼ÓÊÊÁ¿NaOHÈÜÒº

¢ÚÎüÊÕºóµÄÈÜÒº¾­Å¨Ëõ¡¢½á¾§¡¢¹ýÂË£¬µÃµ½NaNO2¾§Ì壬¸Ã¾§ÌåÖеÄÖ÷ÒªÔÓÖÊÊÇ________£¨Ìѧʽ£©£»ÎüÊÕºóÅŷŵÄβÆøÖк¬Á¿½Ï¸ßµÄµªÑõ»¯ÎïÊÇ________£¨Ìѧʽ£©¡£

£¨3£©NOµÄÑõ»¯ÎüÊÕ¡£ÓÃNaClOÈÜÒºÎüÊÕÏõËáβÆø£¬¿ÉÌá¸ßβÆøÖÐNOµÄÈ¥³ýÂÊ¡£ÆäËûÌõ¼þÏàͬ£¬NOת»¯ÎªµÄת»¯ÂÊËæNaClOÈÜÒº³õʼpH£¨ÓÃÏ¡ÑÎËáµ÷½Ú£©µÄ±ä»¯ÈçͼËùʾ¡£

¢ÙÔÚËáÐÔNaClOÈÜÒºÖУ¬HClOÑõ»¯NOÉú³ÉClºÍ£¬ÆäÀë×Ó·½³ÌʽΪ________¡£

¢ÚNaClOÈÜÒºµÄ³õʼpHԽС£¬NOת»¯ÂÊÔ½¸ß¡£ÆäÔ­ÒòÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿2013Äê3Ô£¬»ÆÆÖ½­ÉÏÓÎË®ÖÊÎÛȾ½ÏΪÑÏÖØ£¬Ïà¹Ø×ÔÀ´Ë®³§²ÉÓöàÖÖ·½·¨²¢Óõķ½Ê½½øÐÐË®ÖÊ´¦Àí£¬ÏÂÁÐ˵·¨ÖдíÎóµÄÊÇ£¨ £©

A. ¼Ó»îÐÔÌ¿¿ÉÎü¸½Ë®ÖÐС¿ÅÁ££¬¾»»¯Ë®Öʵķ½·¨ÊôÓÚÎïÀí·½·¨

B. ¼Ó³ôÑõ¶ÔË®½øÐÐÏû¶¾£¬ÀûÓÃÁ˳ôÑõµÄÑõ»¯ÐÔ

C. ÓÃϸ¾úÓëøµÄÀ©³ä·¨È¥³ýË®ÖеݱµªµÄ·½·¨ÊôÓÚÉúÎï·¨

D. ÓþۺÏÁòËáÌú×÷Ϊ¾»Ë®¼Á£¬¸Ã´¦Àí¹ý³ÌÖнö·¢ÉúÁË»¯Ñ§±ä»¯

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿(1)ÒÑÖªÏÂÁÐÊý¾Ý£º

»¯Ñ§¼ü

H-H

N¡ÔN

¼üÄÜ/kJ¡¤mol-1

435

943

ÈçͼÊÇN2(g)ºÍH2(g)·´Ó¦Éú³É1molNH3(g)¹ý³ÌÖÐÄÜÁ¿±ä»¯Ê¾Òâͼ£¬ÊÔ¸ù¾Ý±íÖм°Í¼ÖÐÊý¾Ý¼ÆËãN-HµÄ¼üÄÜ______________¡£

(2)ë¿É×÷Ϊ»ð¼ý·¢¶¯»úµÄȼÁÏ£¬ÓëÑõ»¯¼ÁN2O4·´Ó¦Éú³ÉN2ºÍË®ÕôÆø¡£

ÒÑÖª£º¢ÙN2(g)£«2O2(g)=N2O4(l) ¦¤H1£½-19.5 kJ¡¤mol-1

¢ÚN2H4(l)£«O2(g)=N2(g)£«2H2O(g) ¦¤H2£½-534.2 kJ¡¤mol-1

д³öëºÍN2O4·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ______________________________¡£

(3)ÈôÓñê×¼×´¿öÏÂ4.48LO2Ñõ»¯N2H4ÖÁN2£¬Õû¸ö¹ý³ÌÖÐתÒƵĵç×Ó×ÜÊýΪ___________(°¢·ü¼ÓµÂÂÞ³£ÊýÓÃNA±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÀûÓÃÈçͼËùʾװÖÿÉÒÔ½«ÎÂÊÒÆøÌåCO2ת»¯ÎªÈ¼ÁÏÆøÌåCO.ÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ( )

A.¸Ã×°Öù¤×÷ʱ£¬H£«´Ób¼«ÇøÏòa¼«ÇøÒƶ¯

B.¸Ã×°ÖÃÖÐÿÉú³É1 mol CO£¬Í¬Ê±Éú³É1 mol O2

C.µç¼«a±íÃæ·¢Éú»¹Ô­·´Ó¦

D.¸Ã¹ý³ÌÊǽ«Ì«ÑôÄÜת»¯Îª»¯Ñ§ÄܵĹý³Ì

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓлúÎïXÊǺϳÉÖÎÁÆ°©Ö¢Ò©ÎïµÄÖмäÌ壬ÆäºÏ³É²¿·Ö·¾¶ÈçÏ£º

(1)·´Ó¦¢ÙµÄ·´Ó¦ÎïÒÔ¼°Ìõ¼þΪ________¡£

(2)ÓÉBÖƱ¸XµÄ¹ý³ÌÖУ¬Óи±²úÎïCÉú³É(ÓëX»¥ÎªÍ¬·ÖÒì¹¹Ìå)£¬CµÄ½á¹¹¼òʽΪ_____¡£

(3)ÏÂÁÐÓйØXµÄ˵·¨ÕýÈ·µÄÊÇ________¡£

A£®¸ÃÎïÖÊÊôÓÚ·¼ÏãÌþ B£®X·Ö×ÓÖк¬ÓÐÒ»¸öÊÖÐÔ̼ԭ×Ó

C£®X¿É·¢Éú»¹Ô­·´Ó¦ D£®ÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒº¿É¼ø±ð»¯ºÏÎïXÓëB

(4)д³ö±½¼×ËáµÄÒ»ÖÖº¬Óб½»·µÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ_________¡£

(5)BÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦¿ÉµÃµ½Ò»ÖÖ¸ßÎüË®ÐÔÊ÷Ö¬£¬Æä½á¹¹¼òʽΪ_________¡£

(6)д³öXÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸