¡¾ÌâÄ¿¡¿ÊµÑéÊÒÖƱ¸1£¬2¡ª¶þäåÒÒÍ飬¿ÉÓÃ×ãÁ¿µÄÒÒ´¼ÏÈÖƱ¸ÒÒÏ©£¬ÔÙÓÃÒÒÏ©ºÍÉÙÁ¿µÄäåÖƱ¸1£¬2¡ª¶þäåÒÒÍ飬װÖÃÈçͼËùʾ¡£ÓйØÊý¾ÝÁбíÈç±íËùʾ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

ÒÒ´¼

1£¬2¡ª¶þäåÒÒÍé

ÒÒÃÑ

״̬

ÎÞÉ«ÒºÌå

ÎÞÉ«ÒºÌå

ÎÞÉ«ÒºÌå

ÃܶÈ/gcm-3

0.79

2.2

0.71

·Ðµã/¡æ

78.5

132

34.6

ÈÛµã/¡æ

-130

9

-116

£¨1£©ÔÚ´ËÖƱ¸ÊµÑéÖУ¬Òª¾¡¿ÉÄÜѸËٵذѷ´Ó¦Î¶ÈÌá¸ßµ½170¡æ×óÓÒ£¬Æä×îÖ÷ҪĿµÄÊÇ__¡£

a.Òý·¢·´Ó¦ b.¼Ó¿ì·´Ó¦ËÙ¶È

c.·ÀÖ¹ÒÒ´¼»Ó·¢ d.¼õÉÙ¸±²úÎïÒÒÃÑÉú³É

£¨2£©ÔÚ×°ÖÃAÖгýÁËŨÁòËáºÍÒÒ´¼Í⣬»¹Ó¦¼ÓÈë__£¬ÆäÄ¿µÄÊÇ__¡£×°ÖÃAÖÐÉú³É¸±²úÎïÒÒÃѵĻ¯Ñ§·´Ó¦·½³ÌʽΪ__¡£

£¨3£©ÊµÑéÊÒÖÆÈ¡ÒÒÏ©£¬³£Òòζȹý¸ß¶øʹÒÒ´¼ºÍŨH2SO4·´Ó¦Éú³ÉÉÙÁ¿µÄSO2£¬ÎªÁËÑéÖ¤SO2µÄ´æÔÚ²¢³ýÈ¥SO2¶ÔºóÐø·´Ó¦µÄ¸ÉÈÅ£¬Ä³Í¬Ñ§ÔÚAºÍDÖ®¼ä¼ÓÈëÁËB¡¢CÁ½¸ö×°Öã¬ÆäÖÐBºÍCÖпɷֱðÊ¢·Å___¡£

a.ËáÐÔKMnO4ºÍË® b.Æ·ºìºÍNaOHÈÜÒº

c.ËáÐÔKMnO4ºÍNaOHÈÜÒº d.Æ·ºìºÍËáÐÔKMnO4

£¨4£©¼×¡¢ÒÒÁ½×°Öþù¿ÉÓÃ×÷ʵÑéÊÒÓÉÎÞË®ÒÒ´¼ÖÆÈ¡ÒÒÏ©£¬ÒÒͼ²ÉÓøÊÓÍÔ¡¼ÓÈÈ(¸ÊÓͷеã290¡æ£¬ÈÛµã18.17¡æ)£¬µ±¸ÊÓÍζȴﵽ·´Ó¦Î¶Èʱ£¬½«Ê¢ÓÐÎÞË®ÒÒ´¼ºÍŨÁòËá»ìºÏÒºµÄÉÕÆ¿·ÅÈë¸ÊÓÍÖУ¬ºÜ¿ì´ïµ½·´Ó¦Î¶ȡ£¼×¡¢ÒÒÁ½×°ÖÃÏà±È½Ï£¬ÒÒ×°ÖÃÓÐÄÄЩÓŵã__£¬Ð´³ö¸ÃʵÑéÖÐÓÉÎÞË®ÒÒ´¼ÖÆÈ¡ÒÒÏ©µÄ»¯Ñ§·½³Ìʽ___¡£

£¨5£©½«1£¬2¡ª¶þäåÒÒÍé´Ö²úÆ·ÖÃÓÚ·ÖҺ©¶·ÖмÓË®£¬Õñµ´ºó¾²Öú󣬲úÎïÓ¦ÔÚ__²ã£»Èô²úÎïÖÐÓÐÉÙÁ¿¸±²úÎïÒÒÃÑ¡£¿ÉÓÃ__µÄ·½·¨³ýÈ¥¡£

¡¾´ð°¸¡¿d ·Ðʯ»òËé´ÉƬ·ÀÖ¹±¬·Ð ·ÀÖ¹±¬·Ð 2CH3CH2OHC2H5¡ªO¡ªC2H5+H2O b ÓÐÀûÓÚ¿ØÖÆζȣ¬ÊÜÈȾùÔÈ£»¼õÉÙ¸±·´Ó¦µÄ·¢Éú CH3CH2OHCH2=CH2+H2O Ï ÕôÁó

¡¾½âÎö¡¿

ÂÔ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÓÃȼÉÕ·¨²â¶¨Ä³ÖÖ°±»ùËáµÄ·Ö×Ó×é³É¡£È¡Wg¸ÃÖÖ°±»ùËá·ÅÔÚ´¿ÑõÆøÖÐȼÉÕ£¬È¼ÉÕºóÉú³ÉµÄµÄÌå»ýÓÃF×°ÖýøÐвâÁ¿£¬ËùÐè×°ÖÃÈçÏÂͼ¼Ð³ÖÒÇÆ÷¼°²¿·Ö¼ÓÈÈ×°ÖÃÒÑÂÔÈ¥£º

(1)¸ÃʵÑé×°ÖõĺÏÀíÁ¬½Ó˳ÐòΪ£ºA¡¢C¡¢____¡¢F¡£(×°ÖÿÉÒÔÖظ´Ñ¡ÓÃ)

(2)ʵÑ鿪ʼʱ£¬Ê×ÏÈ´ò¿ªÖ¹Ë®¼Ða£¬¹Ø±Õֹˮ¼Ðb£¬Í¨Ò»¶Îʱ¼äµÄ´¿Ñõ£¬ÕâÑù×öµÄÄ¿µÄÊÇ_____¡£

(3)ȼÉÕ¹ÜDÖзÅÈëCuOµÄ×÷ÓÃÊÇ___________________£¬×°ÖÃBµÄ×÷ÓÃÊÇ____________________ ¡£

(4)ΪÁËÈ·¶¨´Ë°±»ùËáµÄ·Ö×Óʽ£¬³ýÁË׼ȷ²âÁ¿N2µÄÌå»ýÍ⣬»¹ÐèµÃµ½µÄÊý¾ÝÓÐ______________(Ìî×Öĸ)¡£

A.¸Ã°±»ùËáµÄĦ¶ûÖÊÁ¿ B.Éú³É¶þÑõ»¯Ì¼ÆøÌåµÄÖÊÁ¿

C.Éú³ÉË®µÄÖÊÁ¿ D.ͨÈëO2µÄÌå»ý

(5)ÔÚ¶ÁÈ¡F×°ÖÃÖÐËùÅÅË®µÄÌå»ýʱ£¬ÒºÃæ×óµÍÓҸߣ¬ÔòËù²âÆøÌåµÄÌå»ý________(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족)¡£

(6)ÒÑÖª·Ö×ÓʽΪµÄÓлúÎïÒ²º¬Óа±»ùËáÖеÄij¸ö¹ÙÄÜÍÅ£¬ÇëÉè¼ÆʵÑéÖ¤Ã÷¸Ã¹ÙÄÜÍÅÊÔ¼ÁÈÎÑ¡_____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ( )

A.̼ËáÇâÄÆÈÜÒºÖмÓÈëÇâÑõ»¯ÄÆÈÜÒº£ºHCO3-+OH-=CO2+H2O

B.Ïò´ÎÂÈËá¼ØÈÜÒºÖеÎÈëÉÙÁ¿FeSO4ÈÜÒº£»2Fe2++ClO-+2H+=Cl-+2Fe3++H2O

C.Na2S2O3ÓëÏ¡H2SO4»ìºÏ£ºS2O32-+2H+=S¡ý+SO2¡ü+H2O

D.ÓÃʯīµç¼«µç½âMgCl2ÈÜÒº£º2Cl-+2H2O2OH-+Cl2¡ü+H2¡ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏõËṤҵβÆøÖеÄNO¡¢NO2½øÈë´óÆøºó£¬»áÐγɹ⻯ѧÑÌÎí¡¢ÆÆ»µ³ôÑõ²ãµÈ¡£¿ÉÓÃÇâÑõ»¯ÄÆÈÜÒº¶Ôº¬µªÑõ»¯ÎïµÄβÆø½øÐд¦Àí£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÈçÏ£ºNO2+NO+2NaOH=2NaNO2+H2O£»2NO2+2NaOH=NaNO2+NaNO3+H2O¡£°±ÆøÒ²¿ÉÒÔÓÃÀ´´¦ÀíµªÑõ»¯Îï¡£ÀýÈ磬°±ÆøÓëÒ»Ñõ»¯µª¿É·¢ÉúÈçÏ·´Ó¦£º4NH3+6NO=5N2+6H2O¡£½«Ò»¶¨Á¿NOºÍNO2µÄ»ìºÏÆøÌåͨÈë300mL5mol/LNaOHÈÜÒºÖУ¬Ç¡ºÃ±»ÍêÈ«ÎüÊÕ¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A.ÔÚÓð±Æø´¦ÀíµªÑõ»¯Îïʱ£¬µªÑõ»¯Îï·¢Éú»¹Ô­·´Ó¦

B.ËùµÃÈÜÒºÖÐNaNO3ºÍNaNO2µÄÎïÖʵÄÁ¿Ö®±È¿ÉÄÜΪ2£º1

C.ÈôÓð±Æø´¦Àí£¬ËùÐè°±ÆøÔÚ±ê×¼×´¿öϵÄÌå»ý¿ÉÄÜΪ39.2L

D.Ô­»ìºÏÆøÌåÖÐNOÔÚ±ê×¼×´¿öϵÄÌå»ý¿ÉÄÜΪ16.8L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ºãÎÂÏ£¬ÈÝ»ý¾ùΪ2LµÄÃܱÕÈÝÆ÷M¡¢NÖУ¬·Ö±ðÓÐÒÔÏÂÁÐÁ½ÖÖÆðʼͶÁϽ¨Á¢µÄ¿ÉÄæ·´Ó¦ µÄ»¯Ñ§Æ½ºâ״̬£¬Ïà¹ØÊý¾ÝÈçÏ£ºM£º£» 2min´ïµ½Æ½ºâ£¬Éú³É£¬²âµÃ´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâCµÄËÙÂÊΪ£»N£º£»´ïµ½Æ½ºâʱÏÂÁÐÍƶϵĽáÂÛÖв»ÕýÈ·µÄÊÇ£¨ £©

A.B.ƽºâʱMÖÐ

C.D.

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿º£ÑóÖ²ÎïÈ纣´ø¡¢º£ÔåÖк¬ÓзḻµÄµâÔªËØ£¬ÇÒËùº¬µâÔªËØÒÔµâÀë×ÓµÄÐÎʽ´æÔÚ¡£ÊµÑéÊÒÀï´Óº£ÔåÖÐÌáÈ¡µâµÄÁ÷³ÌÈçͼËùʾ£º

(1)Ö¸³öÖÆÈ¡µâµÄ¹ý³ÌÖÐÓйØʵÑé²Ù×÷µÄÃû³Æ£º¢Ù__________£¬¢Ú__________¡£

(2)²Ù×÷¢ÚÖпɹ©Ñ¡ÔñµÄÓлúÊÔ¼ÁÊÇ_________(ÌîÐòºÅ)¡£

A.¼×±½¡¢¾Æ¾« B.ËÄÂÈ»¯Ì¼¡¢±½ C.ÆûÓÍ¡¢ÒÒËá D.ÆûÓÍ¡¢¸ÊÓÍ

(3)ÏÖÐè×öʹº£Ôå»ÒÖеĵâÀë×Óת»¯ÎªÓлúÈÜÒºÖеĵⵥÖʵÄʵÑ飬ʵÑéÊÒÀïÓÐÉÕ±­¡¢²£Á§°ô¡¢¼¯ÆøÆ¿¡¢¾Æ¾«µÆ¡¢µ¼¹Ü¡¢Ô²µ×ÉÕÆ¿¡¢Ê¯ÃÞÍøÒÔ¼°±ØÒªµÄ¼Ð³ÖÒÇÆ÷¡¢Ò©Æ·£¬ÉÐȱÉٵIJ£Á§ÒÇÆ÷ÊÇ__________¡¢___________¡£

(4)Òª´ÓµâµÄ±½ÈÜÒºÖÐÌáÈ¡µâºÍ»ØÊÕ±½£¬»¹ÐèÒª¾­¹ýÕôÁó²Ù×÷¡£½øÐÐÕôÁó²Ù×÷ʱ£¬ÐèʹÓÃˮԡ¼ÓÈÈ£¬Ä¿µÄÊÇ__________£¬×îºó¾§Ì¬µâÔÚ________Öоۼ¯¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Æû³µÎ²ÆøµÄÖ÷Òª³É·ÖÓÐCO¡¢SO2¡¢µªÑõ»¯ÎïµÈ£¬¿Æѧ¼ÒÃÇ-Ö±ÖÂÁ¦ÓÚÎÛȾÎïµÄÓÐЧÏû³ý¡£

(1)ÀûÓð±Ë®¿ÉÒÔ½«SO2¡¢µªÑõ»¯ÎïÎüÊÕ£¬Ô­ÀíÈçÏÂͼËùʾ¡£

Çëд³öNO2ºÍNO°´Ìå»ý±È1£º1±»ÎüÊÕʱ·´Ó¦µÄÀë×Ó·½³Ìʽ_________________________¡£

(2)¿ÆÑй¤×÷ÕßÄ¿Ç°ÕýÔÚ³¢ÊÔÒÔ¶þÑõ»¯îÑ(TiO2)´ß»¯·Ö½âÆû³µÎ²ÆøµÄÑо¿¡£

¢Ù¼ºÖª£º2NO(g)+O2(g)=2NO2(g) ¡÷H1=-113.0kJ¡¤ mol-1

2SO2(g)+O2(g)=2SO3(l) ¡÷H2=-288.4kJ¡¤ mol-1

N2(g)+O2(g)2NO(g) ¡÷H3=+180.5kJ¡¤ mol-1

ÇëÅжϷ´Ó¦NO2(g)+SO2(g)=NO(g)+SO3(l) ¡÷H4£¬ÔÚµÍÎÂÏÂÄÜ·ñ×Ô·¢½øÐÐ_______(Ìî¡°ÄÜ¡±»ò¡°·ñ¡±)£¬ÀíÓÉÊÇ__________________________¡£

¢Ú¼ºÖªTiO2´ß»¯Î²Æø½µ½âÔ­ÀíΪ£º

2CO(g)+O2(g)2CO2(g)£»2H2O(g)+4NO(g)+3O2(g)4HNO3(g)

i¡¢ÔÚÒ»¶¨Ìõ¼þÏ£¬Ä£ÄâCO¡¢NOµÄ½µ½â£¬µÃµ½½µ½âÂÊ(¼´×ª»¯ÂÊ)Ëæʱ¼ä±ä»¯ÈçͼËùʾ£¬

·´Ó¦40Ãëºó¼ì²âµ½»ìºÏÆøÌåÖÐN2Ũ¶ÈÉÏÉý£¬HNO3ÆøÌåŨ¶ÈÓÐËù½µµÍ£¬ÇëÓû¯Ñ§·½³Ìʽ²¢½áºÏ»¯Ñ§·´Ó¦Ô­Àí֪ʶ½âÊÍ¿ÉÄܵÄÔ­Òò____________________________________________¡£

ii£¬Á¤Çà»ìÄýÍÁÒ²¿É½µ½âCO¡£ÈçͼΪÔÚ²»Í¬¿ÅÁ£¼ä϶µÄÁ¤Çà»ìÄýÍÁ(¦Á¡¢¦ÂÐÍ)ÔÚ²»Í¬Î¶ÈÏ£¬·´Ó¦Ïàͬʱ¼ä£¬²âµÃCO½µ½âÂʱ仯¡£½áºÏͼ±í»Ø´ðÏÂÁÐÎÊÌ⣺

ÒÑÖªÔÚ50¡æʱÔÚ¦ÁÐÍÁ¤Çà»ìÄýÍÁÈÝÆ÷ÖУ¬Æ½ºâʱO2Ũ¶ÈΪ0.01mol¡¤L-1£¬Çó´ËζÈÏÂCO½µ½â·´Ó¦µÄƽºâ³£Êý____________________(Óú¬xµÄ´úÊýʽ±íʾ)£»ÒÔ¦ÂÐÍÁ¤Çà»ìÄýÍÁ¿ÅÁ£ÎªÔØÌ壬½«TiO2¸ÄΪ´ß»¯Ð§¹û¸üºÃµÄTiO2ÄÉÃ׹ܣ¬ÔÚ10¡«60¡æ·¶Î§ÄÚ½øÐÐʵÑ飬ÇëÔÚͼÖÐÓÃÏ߶ÎÓëÒõÓ°£¬·ÂÕÕ¡°Ê¾Àý¡±Ãè»æ³öCO½µ½âÂÊËæζȱ仯µÄÇúÏß¿ÉÄܳöÏÖµÄ×î´óÇøÓò·¶Î§(ʾÀý£º)_____________________¡£

(3)ÀûÓÃÈçͼËùʾװÖÃ(µç¼«¾ùΪ¶èÐԵ缫)Ò²¿ÉÎüÊÕSO2£¬²¢ÓÃÒõ¼«ÅųöµÄÈÜÒºÎüÊÕNO2£¬b¼«µÄµç¼«·´Ó¦Ê½Îª_______________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¸ßÌúËáÄÆ(Na2FeO4)ÊÇÒ»ÖÖÐÂÐÍ¡¢¸ßЧµÄË®´¦Àí¼Á£¬ÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ4Na2FeO4£«10H2O===4Fe(OH)3¡ý£«3O2¡ü£«8NaOH¡£µç½âÖƱ¸Na2FeO4×°ÖÃʾÒâͼÈçͼËùʾ¡£

(1)aÊǵçÔ´µÄ________(Ìî¡°Õý¡±»ò¡°¸º¡±)¼«¡£µç½âʱ£¬Ê¯Ä«µç¼«¸½½üÈÜÒºµÄ¼îÐÔ________(Ìî¡°ÔöÇ¿¡±¡°¼õÈõ¡±»ò¡°²»±ä¡±)¡£

(2)Ìúµç¼«µÄ·´Ó¦Ê½Îª_________________________________________________¡£

(3)ά³ÖÒ»¶¨µÄµçÁ÷Ç¿¶ÈºÍµç½âζȣ¬NaOHÆðʼŨ¶È¶ÔNa2FeO4Ũ¶ÈÓ°ÏìÈçͼ(µç½âÒºÌå»ýÏàͬµÄÇé¿öϽøÐÐʵÑé)¡£

¢Ùµç½â3.0 hÄÚ£¬ËæNaOHÆðʼŨ¶ÈÔö´ó£¬Na2FeO4Ũ¶È±ä»¯Ç÷ÊÆÊÇ________(Ìî¡°Ôö´ó¡±¡°²»±ä¡±»ò¡°¼õС¡±)¡£

¢Úµ±NaOHÆðʼŨ¶ÈΪ16 mol¡¤L£­1ʱ£¬1.0¡«2.0 hÄÚÉú³ÉNa2FeO4µÄËÙÂÊÊÇ__________mol¡¤L£­1¡¤h£­1¡£

(4)Ìá´¿µç½âËùµÃNa2FeO4£¬²ÉÓÃÖؽᾧ¡¢¹ýÂË¡¢Ï´µÓ¡¢µÍκæ¸ÉµÄ·½·¨£¬ÔòÏ´µÓ¼Á×îºÃÑ¡ÓÃ________(Ìî±êºÅ)ÈÜÒººÍÒì±û´¼¡£

A£®Fe(NO3)3¡¡ ¡¡B£®NH4Cl¡¡ ¡¡C£®CH3COONa

(5)´ÎÂÈËáÄÆÑõ»¯·¨Ò²¿ÉÒÔÖƵÃNa2FeO4¡£

ÒÑÖª£º2H2(g)£«O2(g)===2H2O(l)¡¡¦¤H£½a kJ¡¤mol£­1

NaCl(aq)£«H2O(l)===NaClO(aq)£«H2(g)¡¡¦¤H£½b kJ¡¤mol£­1

4Na2FeO4(aq)£«10H2O(l)===4Fe(OH)3(s)£«3O2(g)£«8NaOH(aq)¡¡¦¤H£½c kJ¡¤mol£­1

·´Ó¦2Fe(OH)3(s)£«3NaClO(aq)£«4NaOH(aq)===2Na2FeO4(aq)£«3NaCl(aq)£«5H2O(l)µÄ¦¤H£½_______kJ¡¤mol£­1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÓйØÎÄÏ×¼ÇÔØÖÐÉæ¼°µÄ»¯Ñ§ÖªÊ¶±íÊö²»ÕýÈ·µÄÊÇ£¨ £©

A.¡°ÒÔÔøÇàÍ¿Ìú£¬Ìú³àÉ«ÈçÍ­¡±ËµÃ÷ÎÒ¹ú¹Å´ú¾ÍÕÆÎÕÁË¡°Êª·¨Ò±½ð¡±¼¼Êõ

B.¡°Ç½ËúѹÌÇ£¬È¥ÍÁ¶øÌǰס±ÖеÄÍÑÉ«¹ý³Ì·¢ÉúÁË»¯Ñ§±ä»¯

C.¡°µ¤É°ÉÕÖ®³ÉË®Òø£¬»ý±äÓÖ»¹³Éµ¤É°¡±½²µÄÊǵ¥ÖÊÓ뻯ºÏÎïÖ®¼äµÄ»¥±ä

D.¡¶±¾²Ý¸ÙÄ¿¡·ÖмÇÔØ£º¡°ÉվƷǹŷ¨Ò²£¬¡­¡­£¬ÓÃŨ¾ÆºÍÔãÈëêµ£¬ÕôÁîÆøÉÏ£¬ÓÃÆ÷³ÐÈ¡µÎ¶¡£¡±Éæ¼°µÄ²Ù×÷·½·¨ÊÇÕôÁó

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸