ijÊÐÄâͶ×ʽ¨ÉèÒ»¸ö¹¤Òµ¾Æ¾«³§£¬Ä¿µÄÊÇÓù¤Òµ¾Æ¾«ÓëÆûÓÍ»ìºÏÖƳɡ°ÒÒ´¼ÆûÓÍ¡±£¬ÒÔ½ÚʡʯÓÍ×ÊÔ´¡£ÒÑÖªÖƾƾ«µÄ·½·¨ÓÐÈýÖÖ£º

¢ÙÔÚ´ß»¯¼Á×÷ÓÃÏÂÒÒÏ©ÓëË®·´Ó¦£»

¢ÚCH3CH2Br£«H2OCH3CH2OH£«HBr£»

¢Û(C6H10O5)n(µí·Û)£«nH2OnC6H12O6(ÆÏÌÑÌÇ)£»

C6H12O6(ÆÏÌÑÌÇ)2C2H5OH£«2CO2¡ü¡£

(1)·½·¨¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ________________________________________¡£

(2)·½·¨¢ÚµÄ»¯Ñ§·´Ó¦ÀàÐÍÊÇ____________¡£

(3)Ϊ»º½âʯÓͶÌȱ´øÀ´µÄÄÜԴΣ»ú£¬ÄãÈÏΪ¸ÃÊÐӦѡÓÃÄÄÒ»ÖÖ·½·¨Éú²ú¹¤Òµ¾Æ¾«£¿Çë¼òÊöÀíÓÉ

________________________________________________________________________

________________________________________________________________________

________________________________________________________________________¡£

(4)Èç¹û´ÓÂÌÉ«»¯Ñ§(¡°Ô­×ÓÀûÓÃÂÊ¡±×î´ó»¯)µÄ½Ç¶È¿´£¬Öƾƾ«×îºÃµÄÒ»×é·½·¨ÊÇ________¡£

A£®¢Ù                           B£®¢Û

C£®¢Ù¢Û                         D£®¢Ù¢Ú¢Û


´ð°¸£º(1)CH2===CH2£«H2OCH3CH2OH

(2)È¡´ú·´Ó¦

(3)Ó¦¸ÃÓ÷¨¢Û£¬ÒòΪµí·ÛÊÇ¿ÉÔÙÉú×ÊÔ´£¬¶øÒÒÏ©¡¢äåÒÒÍ鶼ÊÇÀ´×ÔÓÚÒÔʯÓÍΪԭÁÏÖƵõIJúÆ·£¬ÊDz»¿ÉÔÙÉú×ÊÔ´£¬ÓÃËüÃÇÖƳɾƾ«»¹²»ÈçÖ±½ÓÀûÓÃʯÓÍ¡¡(4)A


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)

    A£®Ô­×Ó×îÍâ²ãµç×ÓÊý´óÓÚ3(СÓÚ8)µÄÔªËØÒ»¶¨ÊǷǽðÊôÔªËØ

    B£®Ô­×Ó×îÍâ²ãÖ»ÓÐ1¸öµç×ÓµÄÔªËØÒ»¶¨ÊǽðÊôÔªËØ

    C£®×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ãµç×ÓÊý¶àµÄÔªËØÒ»¶¨Î»ÓÚµÚ2ÖÜÆÚ

    D£®Ä³ÔªËصÄÀë×ÓµÄ×îÍâ²ãÓë´ÎÍâ²ãµç×ÓÊýÏàͬ£¬¸ÃÔªËØÒ»¶¨Î»ÓÚµÚ3ÖÜÆÚ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÖÓÐAlCl3ºÍMgSO4»ìºÏÈÜÒº£¬ÏòÆäÖв»¶Ï¼ÓÈëNaOHÈÜÒº£¬µÃµ½µÄ³ÁµíµÄÁ¿Óë¼ÓÈëNaOHÈÜÒºµÄÌå»ýÈçͼËùʾ£¬ÔòÔ­ÈÜÒºÖÐCl£­¡¢SOµÄÎïÖʵÄÁ¿Ö®±ÈΪ(¡¡¡¡)

A£®1:1                                  B£®2:3

C£®3:2                                  D£®6:1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


´×µÄÖ÷Òª³É·ÖÊÇ´×Ëᣬ´×ÔÚÈÕ³£Éú»îÖÐÓкܶàÃîÓá£

(1)ÅëÓãʱ¼ÓÈëÉÙÁ¿Ê³´×ºÍÁϾƿÉÒÔʹÅëÖƵÄÓã¾ßÓÐÌØÊâµÄÏã棬ÕâÖÖÏãζÀ´×ÔÓÚ________¡£

A£®Ê³ÑÎ

B£®Ê³´×ÖеÄÒÒËá

C£®ÁϾÆÖеÄÒÒ´¼

D£®ÁϾÆÖеÄÒÒ´¼Óëʳ´×ÖеÄÒÒËá·´Ó¦Éú³ÉµÄÒÒËáÒÒõ¥

(2)ÈÈË®ºøÓþÃÁË»áÔÚÄÚ±Ú²úÉúÒ»²ãË®¹¸£¬Ë®¹¸µÄÖ÷Òª³É·ÖÊÇ̼Ëá¸ÆºÍÇâÑõ»¯Ã¾£¬ÔÚ¼ÓÈÈʱÏòºøÖе¹Èë°×´×£¬¼´¿É³ýȥˮ¹¸£¬Ð´³öÕâÒ»¹ý³ÌÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

________________________________________________________________________

________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚµí·ÛµÄ˵·¨Öв»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®µí·ÛÒ×ÈÜÓÚË®

B£®µí·ÛÊôÓڸ߷Ö×Ó»¯ºÏÎï

C£®µí·ÛÓëµâË®×÷ÓóÊÏÖÀ¶É«

D£®µí·ÛÔÚÈËÌåÄÚÄܹ»Ë®½â£¬Éú³ÉÆÏÌÑÌÇ£¬¾­³¦±ÚÎüÊÕ½øÈëѪҺ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÎïÖÊÖУ¬¿ÉÐγÉËáÓêµÄÊÇ(¡¡¡¡)

A£®¶þÑõ»¯Áò                            B£®·úÂÈ´úÌþ

C£®¶þÑõ»¯Ì¼                            D£®¼×Íé

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


º¬ÓÐ6¸ö̼ԭ×ÓµÄÍéÌþ£¬¾­´ß»¯ÁÑ»¯¿ÉÉú³ÉµÄÍéÌþ×î¶àÓÐ(¡¡¡¡)

A£®3ÖÖ                                B£®4ÖÖ

C£®5ÖÖ                                D£®6ÖÖ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ʵÑéÊÒÒªÓùÌÌåÂÈ»¯ÄÆ׼ȷÅäÖÆ0.5 L 0.2 mol¡¤L£­1µÄNaClÈÜÒº£¬ÏÂÁÐÄÄÖÖÒÇÆ÷²»ÊDZØÐëʹÓõÄ(¡¡¡¡)

A£®ÊԹܠ                                B£®½ºÍ·µÎ¹Ü

C£®500 mLµÄÈÝÁ¿Æ¿                       D£®ÍÐÅÌÌìƽ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÈçͼËùʾ£¬a¡¢bÊÇÁ½¸ùʯī°ô¡£ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ(¡¡¡¡)

A£®aÊÇÕý¼«£¬·¢Éú»¹Ô­·´Ó¦

B£®bÊÇÑô¼«£¬·¢ÉúÑõ»¯·´Ó¦

C£®Ï¡ÁòËáÖÐÁòËá¸ùÀë×ÓµÄÎïÖʵÄÁ¿²»±ä

D£®ÍùÂËÖ½ÉϵμӷÓ̪ÊÔÒº£¬a¼«¸½½üÑÕÉ«±äºì

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸