£¨15·Ö£©°¢Ã÷Âå·Ò£¨G£©ÊôÓÚ±½±ûËáÀ࿹Ñ×Ò©£¬ÕòʹЧ¹ûÓÅÓÚ²¼Âå·Ò¡£ÏÂͼÊÇ°¢Ã÷Âå·ÒµÄÒ»ÌõºÏ³É·Ïß¡£

£¨1£©EµÄ½á¹¹¼òʽΪ                  £»¢ÚµÄ·´Ó¦ÀàÐÍ£º                  ¡£
£¨2£©DÖк¬Ñõ¹ÙÄÜÍŵÄÃû³ÆΪ              ¡¢                  ¡£
£¨3£©·´Ó¦¢Û¿ÉÒÔ¿´³ÉÊÇÁ½²½·´Ó¦µÄ×Ü·´Ó¦£¬µÚÒ»²½ÊÇÇè»ù£¨£­CN£©µÄÍêÈ«Ë®½â·´Ó¦Éú³ÉôÈ»ù£¨£­COOH£©£¬Çëд³öµÚ¶þ²½·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                           ¡£
£¨4£©ÈÎÒâд³öÁ½ÖÖ·ûºÏÏÂÁÐÌõ¼þµÄEµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ          ¡¢         ¡£
¢ÙÄÜÓëFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦     ¢ÚÄÜ·¢ÉúË®½â·´Ó¦£¬µ«²»ÄÜ·¢ÉúÒø¾µ·´Ó¦
¢Û±½»·ÉϵÄÈ¡´ú»ùÖ»ÓÐ2ÖÖ£¬ÇÒ±½»·Éϵĺ˴Ź²ÕñÇâÆ×ÓÐÁ½¸öÎüÊÕ·å
£¨5£©ÏÂÁÐÓйذ¢Ã÷Âå·Ò£¨G£©µÄ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A£®·Ö×ÓʽΪC13H17NO2B£®ÊôÓÚ°±»ùËá
C£®ÄÜÓëÑÎËá·´Ó¦Éú³ÉÑÎD£®ÄÜ·¢ÉúÈ¡´ú¡¢¼Ó¾Û¡¢Ñõ»¯¡¢»¹Ô­·´Ó¦

£¨1£©£¬È¡´ú·´Ó¦£»£¨2£©Ïõ»ù£¬õ¥»ù£»
£¨3£©£¨4£©£»£»; ; ;;;µÈ¡£
£¨5£©A¡¢C¡¢D£®

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝD¡¢EµÄ·Ö×Óʽ¿ÉÖªD·¢Éú»¹Ô­·´Ó¦²úÉúE£¬D·Ö×ÓÖеÄÏõ»ù±»»¹Ô­Îª°±»ù£¬ EµÄ½á¹¹¼òʽΪ£»¸ù¾ÝB¡¢CµÄ·Ö×ӽṹ¿ÉÒÔ¿´³öB·¢ÉúÈ¡´ú·´Ó¦£¨Ò²½ÐÏõ»¯·´Ó¦£©²úÉúC£¬ËùÒԢڵķ´Ó¦ÀàÐÍÊÇÈ¡´ú·´Ó¦£»£¨2£©ÔÚDµÄ·Ö×ӽṹÖеĺ¬Ñõ¹ÙÄÜÍŵÄÃû³ÆΪÏõ»ù£¬õ¥»ù£»£¨3£©·´Ó¦¢Û¿ÉÒÔ¿´³ÉÊÇÁ½²½·´Ó¦µÄ×Ü·´Ó¦£¬µÚÒ»²½ÊÇÇè»ù£¨£­CN£©µÄÍêÈ«Ë®½â·´Ó¦Éú³ÉôÈ»ù£¨£­COOH£©£¬µÚ¶þ²½ÊÇË®½â²úÉúµÄôÈ»ùÓëÒÒ´¼·¢Éúõ¥»¯·´Ó¦²úÉúõ¥»ù£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º£»£¨4£©E½á¹¹¼òʽÊÇ£¬ÈôÆäͬ·ÖÒì¹¹ÌåÄÜÓëFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦Ôòº¬ÓзÓôÇ»ù£»¢ÚÄÜ·¢ÉúË®½â·´Ó¦Ôòº¬ÓÐõ¥»ù»òëļü£¬µ«²»ÄÜ·¢ÉúÒø¾µ·´Ó¦ËµÃ÷²»ÄÜÊǼ×ËáÐγɵÄõ¥£»¢Û±½»·ÉϵÄÈ¡´ú»ùÖ»ÓÐ2ÖÖ£¬ÇÒ±½»·Éϵĺ˴Ź²ÕñÇâÆ×ÓÐÁ½¸öÎüÊշ壬Ôò·ÓôÇ»ùÓëõ¥»ùÔÚ±½»·ÉÏÏà¶ÔµÄλÖã¬Ôò·ûºÏÌâÒâµÄEµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽµÄÆäÖеÄÁ½ÖÖ¿ÉÒÔÊÇ£»£»; ;;;;µÈ¡££¨5£©A£®ÓÉ°¢Ã÷Âå·Ò£¨G£©µÄ½á¹¹¼òʽ¿ÉÖª£ºÆä·Ö×ÓʽÊÇC13H17NO£¬ÕýÈ·£»B£®ÓÉÓڸ÷Ö×Ó²»º¬Óа±»ù£¬Ö»º¬ÓÐôÈ»ù£¬ËùÒÔ²»ÊôÓÚ°±»ùËᣬ´íÎó£» C£®ÓÉÓÚÔÚÑÇ°±»ùµÄNÔ­×ÓÉÏÓÐÒ»¶Ô¹Â¶Ôµç×Ó£¬ËùÒÔÄÜÓëÑÎËá·´Ó¦Éú³ÉÑΣ¬ÕýÈ·£» D£®ÓÉÓÚÔÚÎïÖʵķÖ×ÓÖк¬ÓÐôÈ»ù£¬ËùÒÔÄÜ·¢ÉúÈ¡´ú·´Ó¦£¬º¬ÓÐ̼̼˫¼ü£¬ËùÒÔ¿ÉÒÔ·¢Éú¼Ó¾Û·´Ó¦£»Ò²Äܱ»ÑõÆøÑõ»¯²úÉúȼÉÕ·´Ó¦£¬Ò²¿ÉÒÔÓëÇâÆø·¢Éú¼Ó³É·´Ó¦£¬ÓëÇâÆøµÄ¼Ó³É·´Ó¦¾ÍÊÇ·¢ÉúÁË»¹Ô­·´Ó¦£¬Òò´ËÑ¡ÏîÊÇA¡¢C¡¢D¡£
¿¼µã£º¿¼²éÓлúÎïµÄ½á¹¹¡¢ÐÔÖÊ¡¢Ï໥ת»¯¡¢»¯Ñ§·½³Ìʽ¼°Í¬·ÖÒì¹¹ÌåµÄÊéдµÄ֪ʶ¡£  

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªËáÐÔ£ºH2SO4 > >H2CO3> > HCO3¡ª£¬×ۺϿ¼ÂÇ·´Ó¦ÎïµÄת»¯ÂʺÍÔ­Áϳɱ¾µÈÒòËØ£¬½«×ª±äΪµÄ×î¼Ñ·½·¨ÊÇ( )

A£®ÓëÏ¡H2SO4¹²ÈȺ󣬼ÓÈë×ãÁ¿µÄNaOHÈÜÒº
B£®ÓëÏ¡H2SO4¹²ÈȺ󣬼ÓÈë×ãÁ¿µÄNa2CO3ÈÜÒº
C£®Óë×ãÁ¿µÄNaOHÈÜÒº¹²ÈȺó£¬ÔÙͨÈë×ãÁ¿CO2
D£®Óë×ãÁ¿µÄNaOHÈÜÒº¹²ÈȺó£¬ÔÙ¼ÓÈëÊÊÁ¿H2SO4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÌþX²úÁ¿ÊǺâÁ¿Ò»¸ö¹ú¼ÒʯÓÍ»¯¹¤Ë®Æ½¸ßµÍµÄ±êÖ¾£¬A¡¢E¾ùΪXµÄͬϵÎï¡£
ÒÑÖª£º¢Ù 
 (ͪÖÐR¡¢R¡¯´ú±íÌþ»ùÇÒͪ²»ÄÜÔÙ±»Ñõ»¯ÎªôÈËᣬÆäËüÎïÖÊÖÐR¡¢R¡¯´ú±íÌþ»ù»òÇâÔ­×Ó)
¢ÚµÍ¼¶õ¥Í¨³£¾ßÓÐË®¹ûÏãζ

ÉÏÊöͼ½âÖÐEΪֱÁ´½á¹¹£¬FÖ»Äܱ»Ñõ»¯ÎªÍª¡£Ôò
£¨1£©X¹ÙÄÜÍŽṹʽΪ               £¬ A¡úB·´Ó¦ÀàÐÍ            ¡£
£¨2£©E¿ÉÄܵĽṹ¼òʽΪ                    £»B¡¢GÖйÙÄÜÍÅÃû³Æ·Ö±ðÊÇ                           ¡£
£¨3£©AËùÐγɵĸ߷Ö×Ó»¯ºÏÎïÊdz£ÓÃʳƷ°ü×°´üµÄÔ­²ÄÁÏ£¬Ð´³ö´Ë¸ß¾ÛÎïµÄ½á¹¹¼òʽ       ¡£
£¨4£©Ð´³öDÓëF·´Ó¦Éú³ÉGµÄ»¯Ñ§·½³Ìʽ                                            ¡£
£¨5£©·ûºÏÒÔÏÂÌõ¼þµÄGµÄͬ·ÖÒì¹¹Ìå¹²ÓР           ÖÖ£¨²»°üÀ¨G£©¡£(¢Ù¾ßÓÐË®¹ûÏãζ¢ÚËáÐÔÌõ¼þÏÂË®½â²úÎïµÈÏà¶Ô·Ö×ÓÖÊÁ¿)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨18·Ö£©ßÅßáÎô·Ó (idoxifene)¿ÉÓÃÓÚÖÎÁƹÇÖÊÊèËÉÖ¢£¬ËüµÄºÏ³É·ÏßÈçͼ¡£

£¨1£©»¯ºÏÎïAµÄ·Ö×ÓʽÊÇ             £¬1molA×î¶àÄÜÓë        H2·¢Éú·´Ó¦
£¨2£©·´Ó¦ÀàÐÍ£ºB¡úC                £» D¡úE                      
£¨3£©DÖк¬Ñõ¹ÙÄÜÍÅÓУº                                               £¨Ð´Ãû³Æ£©¡£
£¨4£©»¯ºÏÎïEÄÜ·¢ÉúµÄ·´Ó¦ÀàÐÍÊÇ         £¨ÌîÈëÐòºÅ£©

A£®¼Ó³É·´Ó¦B£®õ¥»¯·´Ó¦C£®Ë®½â·´Ó¦D£®¼Ó¾Û·´Ó¦
£¨5£©E¡úFÖл¹ÓÐÒ»ÖÖ¸±²úÎïGÉú³É£¬GÓëF»¥ÎªÍ¬·ÖÒì¹¹Ì壬ÇÒº¬ÓÐÈý¸öÁùÔª»·£¬G ½á¹¹¼òʽΪ£º                                                                 ¡£
£¨6£©Âú×ãÏÂÁÐËĸöÌõ¼þµÄAµÄͬ·ÖÒì¹¹ÌåÊýÄ¿ÓР       ÖÖ¡£
¢Ù±½µÄÑÜÉúÎÇÒ±½»·ÉÏÖ»ÓÐÁ½¸ö»¥Îª¶ÔλµÄÈ¡´ú»ù£»¢ÚÄÜ·¢ÉúÒø¾µ·´Ó¦£»¢ÛÓëFeCl3ÈÜÒº×÷Óò»ÏÔÉ«£»¢Ü²»ÓëÇâÑõ»¯ÄÆË®ÈÜÒº·´Ó¦
£¨7£©2£¬2-¶þ¼×»ùÎìËá[CH3CH2CH2C(CH3)2COOH]ÊÇÓлúºÏ³ÉÖмäÌ壬ÇëÉè¼ÆºÏÀíµÄ·½°¸ÒÔ±ûͪ£¨CH3COCH3£©ÎªÎ¨Ò»ÓлúÔ­ÁϺϳÉ2£¬2-¶þ¼×»ùÎìËᣨÓúϳÉ·ÏßÁ÷³Ìͼ±íʾ£¬×¢Ã÷·´Ó¦Ìõ¼þ£©
Ìáʾ£º¢ÙºÏ³É¹ý³ÌÖÐÎÞ»úÊÔ¼ÁÈÎÑ¡£»¢Ú±ûͪ·Ö×Ó¼äÄÜ·¢ÉúÉÏͼºÏ³É·ÏßÖÐA¡úBµÄÀàËÆ·´Ó¦£»
¢ÛºÏ³É·ÏßÁ÷³ÌͼʾÀý£ºCH3CH2OHCH2=CH2CH2Br-CH2Br
                                                                                          

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(12·Ö)A¡ªG¶¼ÊÇÓлúÎËüÃǵÄת»¯¹ØϵÈçÏ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÑÖª£º6£®0g»¯ºÏÎïEÍêȫȼÉÕÉú³É8.8gCO2ºÍ3.6gH2O£¬EµÄÕôÆûÓëÇâÆøµÄÏà¶ÔÃܶÈΪ30£¬ÔòEµÄ·Ö×ÓʽΪ             ¡£
£¨2£©AΪһȡ´ú·¼Ìþ£¬BÖк¬ÓÐÒ»¸ö¼×»ù£¬ÓÉBÉú³ÉCµÄ»¯Ñ§·½³ÌʽΪ
£¨3£©Ð´³öÏÂÁÐת»¯µÄ»¯Ñ§·½³Ìʽ£º
B¡úD                                         
E+C¡úF                                         
£¨4£©ÉÏÊö¹ý³ÌÖÐת»¯µÃµ½µÄFÓжàÖÖͬ·ÖÒì¹¹Ì壬ͬʱ·ûºÏÏÂÁÐÌõ¼þµÄFµÄͬ·ÖÒì¹¹ÌåÓР            ÖÖ¡£
¢Ù±½»·ÉÏÖ»ÓÐÁ½¸öÁÚλȡ´ú»ù£¬ÆäÖÐÒ»¸öΪ¼×»ù ¢ÚÓëF¾ßÓÐÏàͬµÄ¹ÙÄÜÍÅ¡£
ÆäÖÐ1molij·ûºÏÉÏÊöÌõ¼þµÄÓлúÎÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Ê±ÏûºÄ2molÇâÑõ»¯ÄÆ£¬Æä·´Ó¦·½³Ìʽ£º                                     

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓлúÎïA¡¢B¡¢C¡¢D¡¢E¡¢F¡¢G¡¢HÏ໥ת»¯¹ØϵÈçÏÂͼËùʾ¡£5.2 g FÄÜÓë100 mL 1 mol/L NaOHÈÜҺǡºÃÍêÈ«Öкͣ¬0.1 mol F»¹ÄÜÓë×ãÁ¿NaHCO3·´Ó¦£¬ÔÚ±ê×¼×´¿öÏ·ųö4.48 L CO2¡£DµÄ·Ö×ÓʽΪC3H3O2Na£¬EµÄ·Ö×ÓÖк¬ÓÐôÈ»ù¡£

£¨1£©Ð´³öÎïÖÊCÖеĹÙÄÜÍŵÄÃû³Æ£º                             £»
£¨2£©Ð´³öÎïÖÊF¡¢HµÄ½á¹¹¼òʽ£»
F                     ¡¢H                      £»
£¨3£©Ð´³ö·´Ó¦¢Ù¡¢¢ÜµÄ»¯Ñ§·´Ó¦ÀàÐÍ£º¢Ù             ¡¢¢Ü             £»
£¨4£©Ð´³ö±ä»¯¢Ù¡¢¢ÛµÄ»¯Ñ§·½³Ìʽ£»
¢Ù                                                                 
¢Û                                                                 
£¨5£©Ð´³öÏà¶Ô·Ö×ÓÖÊÁ¿±ÈB´ó14£¬ÇÒÓëB¾ßÓÐÏàͬ¹ÙÄÜÍŵÄËùÓÐÎïÖʵĽṹʽ£¨²»¿¼ÂÇÁ¢ÌåÒì¹¹£©¡£
                                                                   ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨7·Ö£©À´×ÔʯÓ͵ÄÓлú»¯¹¤Ô­ÁÏA£¬Æä²úÁ¿ÒÑ×÷ΪºâÁ¿Ò»¸ö¹ú¼ÒʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½µÄ±êÖ¾£¬A¿ÉÒÔ·¢ÉúÈçÏÂת»¯£º

ÒÑÖª£ºEÊǾßÓйûÏãζµÄÓлúÎÆä·Ö×ÓʽΪC4H8O2£¬FÊÇÒ»Öָ߷Ö×Ó»¯ºÏÎï¡£
£¨1£©AµÄ·Ö×ÓʽÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¬CµÄÃû³ÆÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
£¨2£©D·Ö×ÓÖеĹÙÄÜÍÅÃû³ÆÊÇ¡¡¡¡¡¡¡¡£¬Ö¤Ã÷¸Ã¹ÙÄÜÍžßÓÐËáÐԵķ½·¨ÊÇ    ¡¡¡¡¡¡¡¡¡¡¡£
£¨3£©·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽÊÇ              £»·´Ó¦¢ÜµÄÀàÐÍÊÇ                ·´Ó¦¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÒÏ©ÊÇʯÓÍ»¯¹¤×îÖØÒªµÄ»ù´¡Ô­ÁÏ£¬Çë¸ù¾ÝÒÔÏ¿òͼ»Ø´ð£º

41.ÒÔʯÓÍΪԭÁϵÄϵÁл¯¹¤Éú²ú¹ý³ÌÖУ¬µÃµ½´óÁ¿ÒÒÏ©µÄÖ÷Òª·½·¨ÊÇ_______£¨Ñ¡ÌîÐòºÅ£©¡£
a. Ë®½â         b. ·ÖÁó       c. Áѽ⠠         d. ÁÑ»¯
42.ÓлúÎïAË׳Æ______________£¬º¬ÓеĹÙÄÜÍÅÃû³ÆÊÇ_________________.
43.BµÄ·Ö×ÓʽΪC2H4O2£¬Óë´¿¼î·´Ó¦ÄÜÉú³É¶þÑõ»¯Ì¼ÆøÌ壬д³ö·´Ó¦A£«B¡úCµÄ»¯Ñ§
·½³Ìʽ___________________________________________________________  £¨ÓлúÎïÓýṹ¼òʽ±íʾ£©£¬¸Ã·´Ó¦ÀàÐÍΪ______________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¸ß·Ö×ÓÏËάE¹ã·ºÓÃÓÚͨѶ¡¢µ¼µ¯¡¢ÓµÈÁìÓò£¬¿ÉÓ÷¼ÏãÌþAΪԭÁÏÀ´ºÏ³É£¬ºÏ³É·ÏßÈçÏ£º

Çë»Ø´ð£º
(1)д³ö½á¹¹¼òʽ£ºA__________£¬B____________£¬
C____________¡£
(2)д³ö·´Ó¦ÀàÐÍ£º¢Ù__________£¬¢Ú_______________________________¡£
(3)д³öD¡úE¾ÛºÏµÄ»¯Ñ§·½³Ìʽ___________________¡£
(4)ÏÂÁл¯ºÏÎïÖÐÄÜÓëF·¢Éú»¯Ñ§·´Ó¦µÄÊÇ________¡£
a£®HCl        b£®NaCl        c£®Na2CO3  d£®NaOH

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸