ΪÑé֤±Ëص¥ÖÊÑõ»¯ÐÔµÄÏà¶ÔÇ¿Èõ£¬Ä³Ð¡×éÓÃÏÂͼËùʾװÖýøÐÐʵÑé(¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£¬ÆøÃÜÐÔÒѼìÑé)¡£

ʵÑé¹ý³Ì£º

¢ñ.´ò¿ªµ¯»É¼Ð£¬´ò¿ª»îÈûa£¬µÎ¼ÓŨÑÎËá¡£

¢ò.µ±BºÍCÖеÄÈÜÒº¶¼±äΪ»Æɫʱ£¬¼Ð½ôµ¯»É¼Ð¡£

¢ó.µ±BÖÐÈÜÒºÓÉ»ÆÉ«±äΪºì×Øɫʱ£¬¹Ø±Õ»îÈûa¡£

¢ô.¡­¡­

(1)AÖвúÉú»ÆÂÌÉ«ÆøÌ壬Æäµç×ÓʽÊÇ______________________________________¡£

(2)ÑéÖ¤ÂÈÆøµÄÑõ»¯ÐÔÇ¿ÓÚµâµÄʵÑéÏÖÏóÊÇ__________________________________¡£

(3)BÖÐÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________________________________¡£

(4)ΪÑéÖ¤äåµÄÑõ»¯ÐÔÇ¿Óڵ⣬¹ý³Ì¢ôµÄ²Ù×÷ºÍÏÖÏóÊÇ_________________________

________________________________________________________________________¡£

(5)¹ý³Ì¢óʵÑéµÄÄ¿µÄÊÇ__________________________________________________¡£

(6)ÂÈ¡¢äå¡¢µâµ¥ÖʵÄÑõ»¯ÐÔÖð½¥¼õÈõµÄÔ­Òò£ºÍ¬Ö÷×åÔªËØ´ÓÉϵ½ÏÂ____________________£¬µÃµç×ÓÄÜÁ¦Öð½¥¼õÈõ¡£


´ð°¸¡¡(1)

(2)ʪÈóµÄµí·Û­KIÊÔÖ½±äÀ¶

(3)Cl2£«2Br£­===Br2£«2Cl£­

(4)´ò¿ª»îÈûb£¬½«ÉÙÁ¿CÖÐÈÜÒºµÎÈëDÖУ¬¹Ø±Õ»îÈûb£¬È¡ÏÂD£¬Õñµ´£¬¾²ÖúóCCl4²ãÈÜÒº±äΪ×ϺìÉ«

(5)È·ÈÏCµÄ»ÆÉ«ÈÜÒºÖÐÎÞCl2£¬ÅųýCl2¶ÔäåÖû»µâʵÑéµÄ¸ÉÈÅ

(6)Ô­×Ӱ뾶Öð½¥Ôö´ó

½âÎö¡¡AÖеμÓŨÑÎËáºó£¬·¢Éú·´Ó¦£º2KMnO4£«16HCl(Ũ)===2KCl£«2MnCl2£«5Cl2¡ü£«8H2O£¬Éú³É»ÆÂÌÉ«ÆøÌåCl2£¬ÔÚA¡¢B¡¢CÖзֱð·¢Éú·´Ó¦£ºCl2£«2KI===2KCl£«I2£¬Cl2£«2NaBr===2NaCl£«Br2£¬Cl2£«2NaBr===2NaCl£«Br2£¬ÓÉÓÚB¡¢CÖÐÉú³ÉÁËBr2¶øʹÈÜÒº±äΪ»ÆÉ«£¬´ò¿ª»îÈûb£¬CÖÐÉú³ÉµÄBr2ÔÚDÖз¢Éú·´Ó¦£ºBr2£«2KI===2KBr£«I2¡£¹ý³Ì¢óʵÑ飬µ±BÖлÆÉ«ÈÜÒº¼ÌÐøͨÈë¹ýÁ¿Cl2ʱ£¬ÈÜÒº±äΪºì×ØÉ«£¬ÒÔ´ËΪ¶ÔÕÕ£¬ËµÃ÷CÖлÆÉ«ÈÜÒºÎÞCl2£¬´Ó¶øÅųýCl2¶ÔäåÖû»µâʵÑéµÄ¸ÉÈÅ¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


       ÇàʯÃÞÊÇÒ»ÖÖÖ°©ÎïÖÊ£¬ÊÇ¡¶Â¹Ìص¤¹«Ô¼¡·ÖÐÊÜÏÞÖƵÄ46ÖÖ»¯Ñ§Æ·Ö®Ò»£¬Æ仯ѧʽΪNa2Fe5Si8O22(OH)2¡£ÇàʯÃÞÓÃÏ¡ÏõËáÈÜÒº´¦Àíʱ£¬»¹Ô­²úÎïÖ»ÓÐNO£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ                                                                                                          (¡¡¡¡)

A£®ÇàʯÃÞÊÇÒ»ÖÖ¹èËáÑβÄÁÏ

B£®ÇàʯÃÞÖк¬ÓÐÒ»¶¨Á¿µÄʯӢ¾§Ìå

C£®ÇàʯÃ޵Ļ¯Ñ§×é³É¿É±íʾΪNa2O¡¤3FeO¡¤Fe2O3¡¤8SiO2¡¤H2O

D£®1 molÇàʯÃÞÄÜʹ1 mol HNO3±»»¹Ô­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ÏòCa(HCO3)2ÈÜÒºÖмÓÈë¹ýÁ¿NaOHÈÜÒº£º

Ca2£«£«HCO£«OH£­===CaCO3¡ý£«H2O

B£®ÓÃʯīµç¼«µç½â±¥ºÍʳÑÎË®£º

2H£«£«2Cl£­Cl2¡ü£«H2¡ü

C£®AlCl3ÈÜÒººÍ¹ýÁ¿°±Ë®·´Ó¦£º

Al3£«£«3OH£­===Al(OH)3¡ý

D£®Ïòµí·Ûµâ»¯¼ØÈÜÒºÖеμÓÏ¡ÁòËᣬÔÚ¿ÕÆøÖзÅÖÃÒ»¶Îʱ¼äºó±äÀ¶£º4H£«£«4I£­£«O2===2I2£«2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


       ÂÈË®ÖдæÔÚ¶àÖÖ·Ö×ÓºÍÀë×Ó£¬ËüÃÇÔÚ²»Í¬µÄ·´Ó¦ÖбíÏÖ³ö²»Í¬µÄÐÔÖÊ¡£ÏÂÁнáÂÛÕýÈ·µÄÊÇ                                                                                                                    (¡¡¡¡)

A£®¼ÓÈëÓÐÉ«²¼Ìõ£¬Æ¬¿ÌºóÓÐÉ«²¼ÌõÍÊÉ«£¬ËµÃ÷ÓÐCl2´æÔÚ

B£®ÈÜÒº³Êdz»ÆÂÌÉ«£¬ÇÒÓд̼¤ÐÔÆøζ£¬ËµÃ÷ÓÐCl2´æÔÚ

C£®ÏȼÓÈëÑÎËáËữ£¬ÔÙ¼ÓÈëAgNO3ÈÜÒº£¬Éú³É°×É«³Áµí£¬ËµÃ÷ÓÐCl£­´æÔÚ

D£®¼ÓÈëNaOHÈÜÒº£¬ÂÈË®µÄdz»ÆÂÌÉ«Ïûʧ£¬ËµÃ÷ÓÐHClO´æÔÚ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


       ÅжÏÕýÎó£¬ÕýÈ·µÄ»®¡°¡Ì¡±£¬´íÎóµÄ»®¡°¡Á¡±

(1)º£Ë®ÌáäåµÄ¹ý³ÌÖв»·¢ÉúÑõ»¯»¹Ô­·´Ó¦                                                       (¡¡¡¡)

(2)¿ÉÒÔÓõí·ÛÈÜÒº¼ìÑé¼ÓµâʳÑÎÖеÄKIO3                                                                                                             (¡¡¡¡)

(3)ÓÃCCl4ÝÍÈ¡µâË®Öеĵ⣬ÏÈÕñµ´£¬ºó¾²Öã¬ÒºÌå·Ö²ã£¬Ï²ã³ÊÎÞÉ«             (¡¡¡¡)

(4)µâÒ×Éý»ª£¬¿ÉÓüÓÈÈÉý»ª·¨³ýÈ¥NH4ClÖлìÓеÄI2                                                                            (¡¡¡¡)

(5)ÄÜʹʪÈóµÄµí·ÛKIÊÔÖ½±äÀ¶µÄÆøÌåÒ»¶¨ÊÇCl2                                                                                           (¡¡¡¡)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


       ÅжÏÕýÎó£¬ÕýÈ·µÄ»®¡°¡Ì¡±£¬´íÎóµÄ»®¡°¡Á¡±

(1)(2013¡¤Õã½­Àí×Û£¬8A)ʵÑéÊÒ´Óº£´øÌáÈ¡µ¥ÖʵâµÄ·½·¨ÊÇ£ºÈ¡Ñù¡ú×ÆÉÕ¡úÈܽâ¡ú¹ýÂË¡úÝÍÈ¡                                                                                                               (¡¡¡¡)

(2)´Óº£Ë®ÖÐÌáÈ¡ÎïÖʶ¼±ØÐëͨ¹ý»¯Ñ§·´Ó¦²ÅÄÜʵÏÖ                                         (¡¡¡¡)

(2013¡¤¸£½¨Àí×Û£¬6B)

(3)´ÎÂÈËáÄÆÈÜÒº¿ÉÓÃÓÚ»·¾³µÄÏû¶¾É±¾ú                                                           (¡¡¡¡)

(2013¡¤ËÄ´¨Àí×Û£¬1D)

(4)ÔÚ¡°»ð²ñÍ·ÖÐÂÈÔªËصļìÑ顱ʵÑéÖУ¬ÕªÏ¼¸¸ùδȼ¹ýµÄ»ð²ñÍ·£¬½«Æä½þÓÚË®ÖУ¬ÉÔºóÈ¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼ÓÏõËáÒøÈÜÒººÍÏ¡ÏõËáºó£¬¼´¿ÉÅжÏÂÈÔªËصĴæÔÚ(¡¡¡¡)

(2012¡¤Õã½­Àí×Û£¬8B)

(5)¸ù¾Ý½ÏÇ¿Ëá¿ÉÒÔÖÆÈ¡½ÏÈõËáµÄ¹æÂÉ£¬ÍƳöCO2ͨÈëNaClO ÈÜÒºÖÐÄÜÉú³ÉHClO(¡¡¡¡)

(2011¡¤°²»ÕÀí×Û£¬11D)

(6)¢ÙʪÈóºìÖ½Ìõ£»¢Ú±¥ºÍÂÈË®¡£¢ÙÖкìÖ½ÌõÍÊÉ«                              (¡¡¡¡)

(2012¡¤±±¾©Àí×Û£¬10D)

(7)¢Ù·Ó̪ÈÜÒº£»¢ÚŨÑÎËá¡£¢ÙÖÐÎÞÃ÷ÏԱ仯                                     (¡¡¡¡)

(2012¡¤±±¾©Àí×Û£¬10B)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


       ijͬѧÓÃÏÂÁÐ×°ÖÃÖƱ¸²¢¼ìÑéCl2µÄÐÔÖÊ¡£

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ                                                                                           (¡¡¡¡)

A£®¢ñͼÖУºÈç¹ûMnO2¹ýÁ¿£¬Å¨ÑÎËá¾Í¿ÉÈ«²¿±»ÏûºÄ

B£®¢òͼÖУºÁ¿Í²Öз¢ÉúÁ˼ӳɷ´Ó¦

C£®¢óͼÖУº·¢ÉúµÄ·´Ó¦²»ÊÇȼÉÕ·´Ó¦

D£®¢ôͼÖУºÊªÈóµÄÓÐÉ«²¼ÌõÄÜÍÊÉ«£¬½«ÁòËáÈÜÒºµÎÈëÉÕ±­ÖУ¬ÖÁÈÜÒºÏÔËáÐÔ£¬½á¹ûÓÐCl2Éú³É

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


25¡æʱ£¬Âú×ãc(CH3COOH)+c(CH3COO-)=0£®1mol¡¤L-1µÄ´×ËáºÍ´×ËáÄÆ»ìºÏÈÜÒºÖУ¬c(CH3COO-)ÓëpHµÄ¹ØϵÈçͼËùʾ¡£ÏÂÁÐÐðÊö´íÎóµÄÊÇ

    A£®¸ÃζÈÏ´×ËáµÄµçÀë³£ÊýKaΪl0-4.75mol¡¤L-l

    B£®MµãËù±íʾµÄÈÜÒºÖУº

      c(Na+)+c(H+)+c(CH3COOH)=0£®1 mol¡¤L-1

    C£®NµãËù±íʾµÄÈÜÒºÖУº

       c(CH3COO-)>c(CH3COOH)>c(H+)>c(OH-)

    D£®QµãËù±íʾµÄÈÜÒºÖмÓÈëµÈÌå»ýµÄ0£®05mol¡¤L-1NaOHÈÜÒº³ä·Ö·´Ó¦ºópH>7

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÖÓз´Ó¦aA(g)£«bB(g) pC(g)£¬´ïµ½Æ½ºâºó£¬µ±Éý¸ßζÈʱ£¬BµÄת»¯Âʱä´ó£»µ±¼õСѹǿʱ£¬»ìºÏÌåϵÖÐCµÄÖÊÁ¿·ÖÊýÒ²¼õС£¬Ôò£º

(1)¸Ã·´Ó¦µÄÄæ·´Ó¦ÊÇ________ÈÈ·´Ó¦£¬ÇÒa£«b________p(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)¡£

(2)¼õѹʱ£¬AµÄÖÊÁ¿·ÖÊý________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£¬ÏÂͬ)£¬Õý·´Ó¦ËÙÂÊ________¡£

(3)Èô¼ÓÈëB(Ìå»ý²»±ä)£¬ÔòAµÄת»¯ÂÊ________£¬BµÄת»¯ÂÊ________¡£

(4)ÈôÉý¸ßζȣ¬Ôòƽºâʱ£¬B¡¢CµÄŨ¶ÈÖ®±È½«________¡£

(5)Èô¼ÓÈë´ß»¯¼Á£¬Æ½ºâʱÆøÌå»ìºÏÎïµÄ×ÜÎïÖʵÄÁ¿________¡£

(6)ÈôBÊÇÓÐÉ«ÎïÖÊ,A¡¢C¾ùΪÎÞÉ«ÎïÖÊ,Ôò¼ÓÈëC(Ìå»ý²»±ä)ʱ»ìºÏÎïµÄÑÕÉ«________£¬¶øά³ÖÈÝÆ÷ÄÚÆøÌåµÄѹǿ²»±ä,³äÈëÄÊÆøʱ,»ìºÏÎïµÄÑÕÉ«________¡£(Ìî¡°±ädz¡±¡°±äÉ»ò¡°²»±ä¡±)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸