ôÈËáAºÍ´¼B£¨A¡¢BÖоùÖ»ÓÐÒ»¸ö¹ÙÄÜÍÅ£©ÐγɵÄõ¥C£¬ÒÑÖª£º¢Ù2.2g CȼÉÕ¿ÉÉú³É4.4gCO2ºÍ1.8g H2O£»¢Ú1.76g CÓë50mL 0.5 mol¡¤ L¡ª1µÄNaOHÈÜÒº¹²ÈȺó£¬ÎªÁËÖкÍÊ£ÓàµÄ¼îÒº£¬ºÄÈ¥0.2 mol¡¤ L¡ª1µÄÑÎËá25mL£»¢ÛÈ¡CË®½âºóËùµÃµÄ´¼B 0.64g£¬¸ú×ãÁ¿µÄÄÆ·´Ó¦Ê±£¬±ê¿öÏÂÊÕ¼¯µ½224mL H2¡£Çó£º
£¨1£©CµÄÏà¶Ô·Ö×ÓÖÊÁ¿¡¢·Ö×Óʽ¡¢½á¹¹¼òʽ£»
£¨2£©Ð´³öCÓë NaOH¹²ÈÈ·´Ó¦µÄ»¯Ñ§·½³Ìʽ¡£
£¨1£©88£»C4H8O2£»CH3CH2COOCH3
£¨2£©CH3CH2COOCH3 + NaOH¡úCH3CH2COONa + CH3OH
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨13·Ö£©A¡¢B¡¢C¡¢D¡¢E¡¢F¡¢G¾ùΪÓлúÎÆäÖÐA ³£ÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½µÄ±êÖ¾ÐÔÎïÖÊ£¬ËüÃÇÖ®¼äÓÐÈçÏÂת»¯¹Øϵ¡£

H2

 
¢Ý
 

ÒÑÖª£º£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©B¡¢DÖйÙÄÜÍŵÄÃû³Æ£ºB  ____________  C  ____________ 
£¨2£©Ö¸³öÏÂÁбàºÅ¶ÔÓ¦·´Ó¦µÄ·´Ó¦ÀàÐÍ£º
¢Ù  ___________  ¢Þ____________
£¨3£©ÔÚFµÄͬϵÎïÖÐ×î¼òµ¥µÄÓлúÎïµÄ¿Õ¼ä¹¹ÐÍΪ____________£¬µç×ÓʽΪ  ____________  ¡£
£¨4£©Ð´³öÓëF»¥ÎªÍ¬ÏµÎïµÄº¬5¸ö̼ԭ×ÓµÄËùÓÐͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º____________
£¨5£©Ð´³öÏÂÁбàºÅ¶ÔÓ¦·´Ó¦µÄ»¯Ñ§·´Ó¦·½³Ìʽ£º
¢Ú ____________________________________________________                                                              
¢Ü  ____________________________________________________                                                                      

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

£¨12·Ö£©Ä³»¯Ñ§ÐËȤС×éµÄͬѧ¶ÔʵÑéÊÒÒÒËáÒÒõ¥µÄÖÆÈ¡ºÍ·ÖÀë½øÐÐÁËʵÑé̽¾¿¡£
¡¾ÖƱ¸¡¿ÏÂÁÐÊǸÃС×éͬѧÉè¼ÆµÄʵÑé×°Ö᣻شðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÇÆ÷AµÄÃû³Æ     ¡ø    £¬×°ÖÃÖ㤵¼¹ÜµÄÖ÷Òª×÷ÓÃÊÇ     ¡ø      £»
£¨2£©Ö¤Ã÷׶ÐÎÆ¿ÖÐÊÕ¼¯µ½ÒÒËáÒÒõ¥µÄ¼òµ¥·½·¨ÊÇ          ¡ø              £»
¡¾·ÖÀ롿׶ÐÎÆ¿Öеõ½µÄ²úÎïÊÇ»ìºÏÎΪÁË·ÖÀë¸Ã»ìºÏÎÉè¼ÆÁËÈçÏÂÁ÷³Ì£º

£¨3£©aÊÔ¼Á×îºÃÑ¡Óà    ¡ø     £»
£¨4£©²Ù×÷¢ñ¡¢²Ù×÷¢ò·Ö±ðÊÇ        ¡ø           
A£®¹ýÂË¡¢·ÖÒºB£®ÝÍÈ¡¡¢ÕôÁóC£®·ÖÒº¡¢ÕôÁóD£®¹ýÂË¡¢Õô·¢
¡¾ÌÖÂÛ¡¿Ñо¿±íÃ÷ÖÊ×ÓËáÀë×ÓÒºÌåÒ²¿ÉÓÃ×÷õ¥»¯·´Ó¦µÄ´ß»¯¼Á¡£Í¨¹ý¶Ô±ÈʵÑé¿ÉÒÔÑо¿²»Í¬´ß»¯¼ÁµÄ´ß»¯Ð§ÂÊ£¬ÊµÑéÖгýÁËÐè¿ØÖÆ·´Ó¦ÎïÒÒËá¡¢ÒÒ´¼µÄÓÃÁ¿ÏàͬÍ⣬»¹Ðè¿ØÖƵÄʵÑéÌõ¼þÊÇ        ¡ø        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

£¨16·Ö£©¡°¾ÆÊdzµÄÏ㡱£¬ÊÇÒòΪ¾ÆÔÚ´¢´æ¹ý³ÌÖÐÉú³ÉÁËÓÐÏãζµÄÒÒËáÒÒõ¥¡£ÔÚʵÑéÊÒÎÒÃÇ¿ÉÒÔÓÃÈçͼËùʾµÄ×°ÖÃÀ´ÖÆÈ¡ÒÒËáÒÒõ¥¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öÊÔ¹ÜaÖз¢ÉúµÄ»¯Ñ§·½³Ìʽ£º
                                                              ¡£
£¨2£©ÊÔ¹ÜbÖÐËùÊ¢µÄÈÜҺΪ              £¬ÆäÖ÷Òª×÷ÓÃÊÇ                                       ¡£
£¨3£©ÊÔ¹ÜbÖеĵ¼¹ÜÒªÔÚÒºÃæµÄÉÔÉÏ·½£¬²»ÄܲåÈëÒºÃæÒÔÏ£¬ÆäÄ¿µÄÊÇ                    ¡£
£¨4£©¸ÃʵÑéÖУ¬ÈôÓÃ3molÒÒ´¼ºÍ1molÒÒËáÔÚŨÁòËá×÷ÓÃϼÓÈÈ£¬³ä·Ö·´Ó¦ºó£¬ÄÜ·ñÉú³É1molÒÒËáÒÒõ¥£¿        £¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©Ô­ÒòÊÇ            
                        ¡££¨²»¿¼ÂÇÔ­ÁÏËðºÄ£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

Ò»ÖÖÓÃÓÚÖÎÁƸßѪ֬µÄÐÂÒ©ÃðÖ¬ÁéÊÇ°´ÈçÏ·ÏߺϳɵģºÒÑÖª£º

ÒÑÖªGµÄ·Ö×ÓʽΪC10H22O3,ÊԻشð£º
(1)д³ö½á¹¹¼òʽ£ºB_________£¬G _______ £»
(2)ÉÏÊö·´Ó¦ÖÐÊôÓÚÈ¡´ú·´Ó¦µÄÊÇ___________(ÌîÐòºÅ)£»
(3)д³ö·´Ó¦·½³Ìʽ:
I¡¢·´Ó¦¢Ú___________________________________________________£»
¢ò¡¢·´Ó¦¢Ý___________________________________________________
¢ó¡¢FÓëÒø°±ÈÜÒº·´Ó¦£º _______________________________________________ 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

Óɱ¥ºÍÒ»ÔªËáÓë±¥ºÍÒ»Ôª´¼ÐγɵÄõ¥Í¬±¥ºÍһԪȩ×é³ÉµÄ»ìºÏÎï¹²xg£¬²âµÃÆäÖк¬Ñõyg£¬ÔòÆäÖÐ̼µÄÖÊÁ¿·ÖÊýΪ
A£®(x-y) B£®1-y/xC£®6/7(x-y)D£®6/7(1-y/x)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÓлúÎïA¸úNaOHÏ¡ÈÜÒº¹²ÈȺó£¬Éú³ÉBºÍC£»BµÄ¹ÌÌåÓë¼îʯ»Ò¹²ÈÈÉú³ÉÆøÌåD(ÌìÈ»ÆøÖ÷Òª³É·Ö)£»CÔÚ¼ÓÈȲ¢Óд߻¯¼Á´æÔÚµÄÌõ¼þϱ»¿ÕÆøÑõ»¯³ÉE£»E¸úÐÂÖƵÄCu(OH)2ÔÚ¼ÓÈÈʱÉú³ÉF£¬FÓëNaOHÈÜÒº×÷ÓÃÉú³ÉB£¬ÔòAÊÇ                          (    )                                        
A£®HCOOC2H5B£®CH3COOCH2CH2CH3C£®CH3CH2ClD£®CH3COOC2H5

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÓлúÎïÖв»ÈÜÓÚË®ÇÒ±ÈË®ÇáµÄÊÇ£¨  £©
A£®ÒÒ´¼B£®ÒÒËáC£®ÒÒËáÒÒõ¥D£®äå±½

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

5¿ËijһԪ´¼A £¬ÓëÒÒËá·¢Éúõ¥»¯·´Ó¦£¬µÃµ½ÒÒËáijõ¥6.5¿Ë£¬»ØÊÕδ·´Ó¦ÍêµÄA 0.6¿Ë£¬ÔòAµÄʽÁ¿Îª                                              £¨   £©
A£®98B£®116C£®126D£®88

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸