ÒÑÖªZn(OH)2ÄÜÈÜÓÚNaOHÈÜҺת±ä³ÉNa2[Zn(OH)4]¡£ÎªÁ˲ⶨij°×ÌúƤÖÐпµÄÖÊÁ¿·ÖÊý£¬½øÐÐÈçÏÂʵÑ飺ȡa g°×ÌúƤÑùÆ·ÓÚÉÕ±­ÖУ¬¼ÓÈë¹ýÁ¿Ï¡ÑÎËᣬÓñíÃæÃó¸ÇºÃ£¬¿ªÊ¼Ê±²úÉúÆøÅݵÄËٶȺܿ죬ÒÔºóÖð½¥±äÂý¡£´ýÊÔÑùÈ«²¿Èܽâºó£¬ÏòÉÕ±­ÖмÓÈë¹ýÁ¿NaOHÈÜÒº£¬³ä·Ö½Á°èºó¹ýÂË£¬½«ËùµÃ³ÁµíÔÚ¿ÕÆøÖмÓÇ¿ÈÈÖÁÖÊÁ¿²»±ä£¬³ÆµÃ²ÐÁô¹ÌÌåÖÊÁ¿Îªb g¡£

(1)°×ÌúƤ¶ÆпµÄÄ¿µÄÊÇ

________________________________________________________________________¡£

(2)ÆøÅݲúÉúËÙ¶ÈÏÈ¿ìºóÂýµÄÔ­ÒòÊÇ

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________¡£

(3)ÈÜÒºÖмÓÈë¹ýÁ¿NaOHÈÜÒº£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ(²»ÊÇÀë×Ó·´Ó¦µÄд»¯Ñ§·½³Ìʽ)£º

________________________________________________________________________£»

________________________________________________________________________£»

________________________________________________________________________£»

________________________________________________________________________¡£

(4)×îºó²ÐÁôµÄ¹ÌÌåÊÇ__________£¬µÃµ½¸Ã¹ÌÌåµÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ

________________________________________________________________________¡£

(5)°×ÌúƤº¬Ð¿µÄÖÊÁ¿·ÖÊýÊÇ______________(Óú¬a¡¢bµÄʽ×Ó±íʾ)¡£

 

¡¾´ð°¸¡¿

(1)±£»¤ÄÚ²¿µÄÌú£¬Ôö¼ÓÌúµÄ¿¹¸¯Ê´ÄÜÁ¦

(2)¿ªÊ¼Ê±£¬ÇâÀë×ÓŨ¶È´ó£¬¼ÓÉÏпÓëÌú¹¹³ÉÎÞÊý΢СµÄÔ­µç³Ø£¬·´Ó¦Ëٶȿ졣Ëæ×Å·´Ó¦µÄ½øÐУ¬ÈÜÒºÖÐÇâÀë×ÓŨ¶È¼õС£¬±íÃæпÈܽâºó£¬Ô­µç³Ø×÷ÓÃÖð½¥¼õСÒÔÖÁ²»´æÔÚ£¬·´Ó¦ËÙÂÊÃ÷ÏÔ¼õÂý

(3)Zn2£«£«2OH£­=Zn(OH)2¡ý

Zn(OH)2£«2OH£­=[Zn(OH)4]2£­

Fe2£«£«2OH£­=Fe(OH)2¡ý

4Fe(OH)2£«O2£«2H2O=4Fe(OH)3

(4)Fe2O3¡¡2Fe(OH)3Fe2O3£«3H2O

(5) %

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º(1)°×ÌúƤ¶ÆпµÄÄ¿µÄÊÇΪÁ˱£»¤ÄÚ²¿µÄFe£¬Ôö¼ÓFeµÄ¿¹¸¯Ê´ÄÜÁ¦¡£

(2)Zn¡¢FeÔÚÏ¡ÑÎËáÖÐÐγÉÔ­µç³Ø£¬·´Ó¦ËÙÂʿ죬Ëæ×ÅH£«Å¨¶ÈµÄ²»¶Ï¼õС£¬±íÃæZnÈܽ⣬ԭµç³Ø×÷ÓÃÖð½¥¼õÈõ£¬·´Ó¦ËÙÂÊÖð½¥¼õÂý¡£

(3)½áºÏÌâ¸øÐÅÏ¢ºÍËùѧ»¯Ñ§ÖªÊ¶£¬¿ÉÒÔд³ö(3)ÖеÄÀë×Ó·½³Ìʽ¡£

(4)Fe(OH)3²»Îȶ¨£¬ÊÜÈÈʱÒ×·¢Éú·Ö½â£¬ËùÒԵõ½µÄ¹ÌÌåÊÇFe2O3¡£

(5)¸ù¾Ý²ÐÁô¹ÌÌåÑõ»¯ÌúµÄÖÊÁ¿b g¿ÉµÃÌúÔªËصÄÖÊÁ¿Îª£ºb g£¬ËùÒÔa g°×ÌúƤÖÐпµÄÖÊÁ¿Îª(a£­b) g£¬´Ó¶ø¿ÉÒÔ¼ÆËã³öпµÄÖÊÁ¿·ÖÊýÊÇ%¡£

¿¼µã£º¿¼²é°×ÌúƤÖÐпµÄÖÊÁ¿·ÖÊý²â¶¨µÄʵÑéÉè¼Æ¡¢²Ù×÷¡¢¼ÆËãÒÔ¼°³£¼û»¯Ñ§ÓÃÓïµÄÊéд

µãÆÀ£º¸ÃÌâÒԲⶨij°×ÌúƤÖÐпµÄÖÊÁ¿·ÖÊýΪÔØÌ壬Öص㿼²éÁËѧÉú¹æ·¶ÑϽ÷µÄʵÑéÉè¼ÆºÍ¶¯ÊÖ²Ù×÷ÄÜÁ¦£¬ÒÔ¼°¹æ·¶µÄ´ðÌâÄÜÁ¦£¬ÓÐÀûÓÚÌáÉýѧÉúµÄѧ¿ÆËØÑø£¬¼¤·¢Ñ§ÉúµÄѧϰÐËȤ£¬Ìá¸ßѧÉúµÄÓ¦ÊÔÄÜÁ¦ºÍѧϰЧÂÊ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º¹ã¶«Ê¡¹ãÖÝÊз¬Ø®Çø2010½ì¸ßÈýµÚËÄ´Îͳ²âÀí×Û»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

п±µ°×ÊÇÒ»ÖÖ°×É«ÑÕÁÏ¡£¹¤ÒµÉÏÊÇÓÉZnSO4ÓëBaSÈÜÒº»ìºÏ¶ø³É£ºBaS+ZnSO4  =  ZnS¡ý+BaSO4¡ý¡£ÒÔÏÂÊǹ¤ÒµÉú²úÁ÷³Ì¡£Çë»Ø´ðÓйØÎÊÌ⣺
¢ñ.ZnSO4ÈÜÒºµÄÖƱ¸ÓëÌá´¿
ÓйØ×ÊÁÏ£ºÒÑÖªZn(OH)2ÓëAl(OH)3ÏàËÆ£¬ÄÜÈÜÓÚ¹ýÁ¿µÄNaOHÈÜÒºÉú³ÉNa2ZnO2£»
Áâп¿óµÄÖ÷Òª³É·ÖÊÇZnCO3£¬º¬ÉÙÁ¿SiO2¡¢FeCO3¡¢Cu2(OH)2CO3µÈ¡£
£¨1£©¢ÚÖÐʹÓõÄÑõ»¯¼Á×îºÃÊÇÏÂÁеĠ     £¨ÌîÐòºÅ£©£¬ÀíÓÉÊÇ                       ¡£

A£®Cl2B£®H2O2C£®KMnO4D£®Å¨HNO3
£¨2£©Ð´³ö·´Ó¦¢ÜµÄÀë×Ó·½³Ìʽ£º                                               ¡£
£¨3£©ÎªÁË´ïµ½×ÛºÏÀûÓᢽÚÄܼõÅŵÄÄ¿µÄ£¬ÉÏÊöÁ÷³ÌÖв½Öè    ²úÉúµÄ    ¿ÉÒÔÓÃÓÚ²½
Öè      £¨ÆäÖв½ÖèÑ¡Ìî¢Ù¡¢¢Ú¡¢¢Û¡¢¢Ü¡¢¢Ý£©¡£
¢ò.BaSÈÜÒºµÄÖƱ¸
ÓйØÊý¾Ý£º  Ba£¨s£©£«S£¨s£©£«2O2£¨g£©£½BaSO4£¨s£©£»¡÷H = ¡ª1473.2 kJ?mol-1
C£¨s£©£« ¡ªO2£¨g£©£½CO£¨g£©£»         ¡÷H = ¡ª110.5 kJ?mol-1
Ba£¨s£©£« S£¨s£©£½BaS£¨s£©£»          ¡÷H = ¡ª460 kJ?mol-1

£¨4£©ìÑÉÕ»¹Ô­µÄÈÈ»¯Ñ§·½³ÌʽΪ:                                                ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêÕã½­Ê¡ÎÂÖÝÊиßÈýµÚÒ»´ÎÊÊÓ¦ÐÔ²âÊÔÀí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

·ÏÆúÎïµÄ×ÛºÏÀûÓüÈÓÐÀûÓÚ½ÚÔ¼×ÊÔ´£¬ÓÖÓÐÀûÓÚ±£»¤»·¾³¡£ÊµÑéÊÒÀûÓ÷ϾɻÆÍ­(Cu¡¢ZnºÏ½ð£¬º¬ÉÙÁ¿ÔÓÖÊFe)ÖƱ¸µ¨·¯¾§Ìå(CuSO4¡¤5H2O)¼°¸±²úÎïZnO¡£ÖƱ¸Á÷³ÌͼÈçÏ£º

ÒÑÖª£ºZn¼°»¯ºÏÎïµÄÐÔÖÊÓëAl¼°»¯ºÏÎïµÄÐÔÖÊÏàËÆ£¬pH£¾11ʱZn(OH)2ÄÜÈÜÓÚNaOHÈÜÒºÉú³É[Zn(OH)4]2£­¡£Ï±íÁгöÁ˼¸ÖÖÀë×ÓÉú³ÉÇâÑõ»¯Îï³ÁµíµÄpH(¿ªÊ¼³ÁµíµÄpH°´½ðÊôÀë×ÓŨ¶ÈΪ1.0mol¡¤L£­1¼ÆËã)¡£

 

Fe3£«

Fe2£«

Zn2£«

¿ªÊ¼³ÁµíµÄpH

1.1

5.8

5.9

³ÁµíÍêÈ«µÄpH

3.0

8.8

8.9

 

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÊÔ¼ÁX¿ÉÄÜÊÇ__________£¬Æä×÷ÓÃÊÇ____________________¡£

£¨2£©¼ÓÈëZnOµ÷½ÚpH=3¡«4µÄÄ¿µÄÊÇ____________________¡£

£¨3£©Óɲ»ÈÜÎïÉú³ÉÈÜÒºDµÄ»¯Ñ§·½³ÌʽΪ______________________________¡£

£¨4£©ÓÉÈÜÒºDÖƵ¨·¯¾§Ìå°üº¬µÄÖ÷Òª²Ù×÷²½ÖèÊÇ______________________________¡£

£¨5£©ÏÂÁÐÊÔ¼Á¿É×÷ΪYÊÔ¼ÁµÄÊÇ______¡£

A£®ZnOB£®NaOHC£®Na2CO3D£®ZnSO4

ÈôÔÚÂËÒºCÖÐÖðµÎ¼ÓÈëÑÎËáÖ±µ½¹ýÁ¿£¬Ôò²úÉúµÄÏÖÏóÊÇ______________________________¡£

£¨6£©²â¶¨µ¨·¯¾§ÌåµÄ´¿¶È(²»º¬ÄÜÓëI£­·¢Éú·´Ó¦µÄÑõ»¯ÐÔÔÓÖÊ)£º×¼È·³ÆÈ¡0.5000gµ¨·¯¾§ÌåÖÃÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÊÊÁ¿Ë®Èܽ⣬ÔÙ¼ÓÈë¹ýÁ¿KI£¬ÓÃ0.1000mol¡¤L£­1Na2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNa2S2O3±ê×¼ÈÜÒº19.40mL¡£ÒÑÖª£ºÉÏÊöµÎ¶¨¹ý³ÌÖеÄÀë×Ó·½³ÌʽÈçÏ£º

2Cu2£«£«4I£­2CuI(°×É«)¡ý£«I2£¬I2£«2S2O32£­2I£­£«S4O62£­

¢Ùµ¨·¯¾§ÌåµÄ´¿¶ÈΪ_______________¡£

¢ÚÔڵζ¨¹ý³ÌÖоçÁÒÒ¡¶¯(ÈÜÒº²»Í⽦)׶ÐÎÆ¿£¬ÔòËù²âµÃµÄ´¿¶È½«»á__________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°²»±ä¡±)¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

п±µ°×ÊÇÒ»ÖÖ°×É«ÑÕÁÏ¡£¹¤ÒµÉÏÊÇÓÉZnSO4ÓëBaSÈÜÒº»ìºÏ¶ø³É£ºBaS+ZnSO4   =  ZnS¡ý+BaSO4¡ý¡£ÒÔÏÂÊǹ¤ÒµÉú²úÁ÷³Ì¡£Çë»Ø´ðÓйØÎÊÌ⣺

¢ñ.ZnSO4ÈÜÒºµÄÖƱ¸ÓëÌá´¿

ÓйØ×ÊÁÏ£ºÒÑÖªZn(OH)2ÓëAl(OH)3ÏàËÆ£¬ÄÜÈÜÓÚ¹ýÁ¿µÄNaOHÈÜÒºÉú³ÉNa2ZnO2£»

Áâп¿óµÄÖ÷Òª³É·ÖÊÇZnCO3£¬º¬ÉÙÁ¿SiO2¡¢FeCO3¡¢Cu2(OH)2CO3µÈ¡£

£¨1£©¢ÚÖÐʹÓõÄÑõ»¯¼Á×îºÃÊÇÏÂÁеĠ      £¨ÌîÐòºÅ£©£¬ÀíÓÉÊÇ                        ¡£

     A.Cl2        B.H2O2          C.KMnO4         D.ŨHNO3

£¨2£©Ð´³ö·´Ó¦¢ÜµÄÀë×Ó·½³Ìʽ£º                                                ¡£

£¨3£©ÎªÁË´ïµ½×ÛºÏÀûÓᢽÚÄܼõÅŵÄÄ¿µÄ£¬ÉÏÊöÁ÷³ÌÖв½Öè     ²úÉúµÄ     ¿ÉÒÔÓÃÓÚ²½

Öè       £¨ÆäÖв½ÖèÑ¡Ìî¢Ù¡¢¢Ú¡¢¢Û¡¢¢Ü¡¢¢Ý£©¡£

¢ò.BaSÈÜÒºµÄÖƱ¸

ÓйØÊý¾Ý£º  Ba£¨s£©£«S£¨s£©£«2O2£¨g£©£½BaSO4£¨s£©£»¡÷H = ¡ª1473.2 kJ??mol-1

C£¨s£©£« ¡ªO2£¨g£©£½CO£¨g£©£»         ¡÷H = ¡ª110.5 kJ??mol-1

Ba£¨s£©£« S£¨s£©£½BaS£¨s£©£»           ¡÷H = ¡ª460 kJ??mol-1

£¨4£©ìÑÉÕ»¹Ô­µÄÈÈ»¯Ñ§·½³ÌʽΪ:                                                ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÐ½®×ÔÖÎÇøÔ¿¼Ìâ ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªZn(OH)2ÓëAl(OH)3Ò»Ñù£¬ÊÇÒ»ÖÖÁ½ÐÔÇâÑõ»¯ÎËüÓëÇ¿Ëᡢǿ¼î·´Ó¦µÄÀë×Ó·½³Ìʽ·Ö±ðΪZn(OH)2£«2H£«===Zn2£«£«2H2O£¬Zn(OH)2£«2OH£­===ZnO22£­£«2H2O£»µ«Zn(OH)2ÄÜÈÜÓÚ¹ýÁ¿µÄ°±Ë®ÖжøAl(OH)3È´²»ÄÜ£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽΪZn(OH)2£«4NH3¡¤H2O===Zn(NH3)42£«£«2OH£­£«4H2O¡£ÔòÏÂÃæËÄ×éÎïÖʵÄÈÜÒº£¬²»ÄÜÔËÓõμÓ˳Ðò½øÐмø±ðµÄÊÇ
[     ]
A£®AlCl3¡¢NH3¡¤H2O    
B£®ZnCl2¡¢NH3¡¤H2O
C£®AlCl3¡¢NaOH   
D£®ZnCl2¡¢NaOH

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸