Çë¸ù¾ÝÈçͼ×÷´ð£º
ÒÑÖª£ºÒ»¸ö̼ԭ×ÓÉÏÁ¬ÓÐÁ½¸öôÇ»ùʱ£¬Ò×·¢ÉúÏÂÁÐת»¯£º
£¨1£©EÖк¬ÓеĹÙÄÜÍÅÊÇ
 

£¨2£©ÒÑÖªBµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäÍêȫȼÉյIJúÎïÖÐn£¨CO2£©£ºn£¨H2O£©=2£º1£¬ÔòBµÄ·Ö×ÓʽΪ
 
£®
£¨3£©·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ
 
£®
£¨4£©FÊǸ߷Ö×Ó¹â×è¼ÁÉú²úÖеÄÖ÷ÒªÔ­ÁÏ£®F¾ßÓÐÈçÏÂÌص㣺¢ÙÄܸúFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£»¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ£®FÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨5£©»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦£®G¿ÉÄÜÓÐ
 
Öֽṹ£¬Ð´³öÆäÖв»º¬¼×»ùµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ
 
£®
¿¼µã£ºÓлúÎïµÄÍƶÏ
רÌ⣺ÓлúÎïµÄ»¯Ñ§ÐÔÖʼ°ÍƶÏ
·ÖÎö£ºA·¢ÉúË®½â·´Ó¦C¡¢D£¬DËữµÃµ½E£¬CÑõ»¯µÃµ½E£¬ÔòCµÄ½á¹¹¼òʽΪCH3CHO£¬DΪCH3COONa£¬EΪCH3COOH£¬BµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäÍêȫȼÉյIJúÎïÖÐn£¨CO2£©£ºn £¨H2O£©=2£º1£¬ÔòBÖÐ̼ÇâÔ­×Ó¸öÊýÖ®±ÈΪ1£º1£¬ËùÒÔa=b£¬½áºÏÆäÏà¶Ô·Ö×ÓÖÊÁ¿Öª£ºa=b=
162-32
13
=10£¬ËùÒÔBµÄ·Ö×ÓʽΪC10H10O2£¬BË®½âÉú³ÉE£¨ÒÒËᣩºÍF£¬FµÄ·Ö×ÓʽΪC10H10O2+H2O-C2H4O2=C8H8O£¬²»±¥ºÍ¶ÈΪ5£¬F¾ßÓÐÈçÏÂÌص㣺¢ÙÄܸúFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£¬¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ£¬ÔòFº¬ÓзÓôÇ»ù¡¢º¬ÓÐ1¸ö-CH=CH2£¬ÇÒ2¸öÈ¡´ú»ù´¦ÓÚ¶Ô룬ÔòFµÄ½á¹¹¼òʽΪ£º£¬BΪ£¬¾Ý´Ë½â´ð£®
½â´ð£º ½â£ºA·¢ÉúË®½â·´Ó¦C¡¢D£¬DËữµÃµ½E£¬CÑõ»¯µÃµ½E£¬ÔòCµÄ½á¹¹¼òʽΪCH3CHO£¬DΪCH3COONa£¬EΪCH3COOH£¬BµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª162£¬ÆäÍêȫȼÉյIJúÎïÖÐn£¨CO2£©£ºn £¨H2O£©=2£º1£¬ÔòBÖÐ̼ÇâÔ­×Ó¸öÊýÖ®±ÈΪ1£º1£¬ËùÒÔa=b£¬½áºÏÆäÏà¶Ô·Ö×ÓÖÊÁ¿Öª£ºa=b=
162-32
13
=10£¬ËùÒÔBµÄ·Ö×ÓʽΪC10H10O2£¬BË®½âÉú³ÉE£¨ÒÒËᣩºÍF£¬FµÄ·Ö×ÓʽΪC10H10O2+H2O-C2H4O2=C8H8O£¬²»±¥ºÍ¶ÈΪ5£¬F¾ßÓÐÈçÏÂÌص㣺¢ÙÄܸúFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£¬¢ÚÄÜ·¢Éú¼Ó¾Û·´Ó¦£»¢Û±½»·ÉϵÄÒ»ÂÈ´úÎïÖ»ÓÐÁ½ÖÖ£¬ÔòFº¬ÓзÓôÇ»ù¡¢º¬ÓÐ1¸ö-CH=CH2£¬ÇÒ2¸öÈ¡´ú»ù´¦ÓÚ¶Ô룬ÔòFµÄ½á¹¹¼òʽΪ£º£¬BΪ£¬
£¨1£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬EΪCH3COOH£¬º¬ÓеĹÙÄÜÍÅÊÇ£ºôÈ»ù£¬¹Ê´ð°¸Îª£ºôÈ»ù£»
£¨2£©ÓÉÉÏÊö·ÖÎö£¬¿ÉÖªBµÄ·Ö×ÓʽΪC10H10O2£¬¹Ê´ð°¸Îª£ºC10H10O2£»
£¨3£©·´Ó¦¢ÛÊÇÒÒÈ©´ß»¯Ñõ»¯Éú³ÉÒÒËᣬ·´Ó¦»¯Ñ§·½³ÌʽΪ£ºCH3CHO+2Cu£¨OH£©2+NaOH
¡÷
CH3COONa+Cu2O¡ý+3H2O£¬¹Ê´ð°¸Îª£ºCH3CHO+2Cu£¨OH£©2+NaOH
¡÷
CH3COONa+Cu2O¡ý+3H2O£»
£¨4£©FΪ£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú¼Ó¾Û·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º£¬
¹Ê´ð°¸Îª£º£»
£¨5£©»¯ºÏÎïGÊÇFµÄͬ·ÖÒì¹¹Ì壬ÊôÓÚ·¼Ïã×廯ºÏÎÄÜ·¢ÉúÒø¾µ·´Ó¦£¬GÖк¬ÓÐÈ©»ù£¬¿ÉÄܵÄͬ·ÖÒì¹¹ÌåÓС¢¡¢¡¢¹²4ÖÖ£¬ÆäÖв»º¬¼×»ùµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽΪ£¬¹Ê´ð°¸Îª£º4£»£®
µãÆÀ£º±¾Ì⿼²éÓлúÎïµÄÍƶϣ¬ÄѶÈÖеȣ¬×¢Òâ¼ÆËãÈ·¶¨BµÄ·Ö×Óʽ£¬¸ù¾ÝÓлúÎïµÄ¹ÙÄÜÍŵı仯ΪͻÆÆ¿Ú½øÐÐÍƶϣ¬ÐèҪѧÉúÊìÁ·ÕÆÎÕ¹ÙÄÜÍŵĽṹÓëÐÔÖÊ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢1.8g Ë®Ëùº¬Óеĵç×ÓÊýĿΪNA
B¡¢2g ÇâÆøËùº¬Ô­×ÓÊýΪNA
C¡¢³£Î³£Ñ¹ÏÂ11.2LÑõÆøËùº¬·Ö×ÓÊýĿΪ0.5NA
D¡¢200 mL0.5mol?L-1Na2SO4ÈÜÒºËùº¬Na+ÊýÄ¿0.1NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÇëÓÃÏàÓ¦µÄ»¯Ñ§Ê½¼°·½³ÌʽÌî¿Õ£º
£¨1£©´ÓH¡¢O¡¢S¡¢NaËÄÖÖÔªËØÖУ¬Ñ¡ÔñÊʵ±µÄÔªËØÌî¿Õ£¨Ã¿¿ÕÒ»ÖÖÎïÖÊ£©£®
¢Ù¼îÐÔÑõ»¯Îï
 
£»¢ÚËáÐÔÑõ»¯Îï
 
£»¢Ûº¬ÑõËá
 
£»¢Ü¼î
 
£»¢ÝËáʽÑÎ
 
£®
£¨2£©Ð´³öÀë×Ó·´Ó¦·½³ÌʽCO2+2OH-=CO32-+H2OËù¶ÔÓ¦µÄÒ»¸ö»¯Ñ§·½³Ìʽ£º
 
£®
£¨3£©Ä³Ò»·´Ó¦ÌåϵÖÐÓз´Ó¦ÎïºÍÉú³ÉÎï¹²5ÖÖÎïÖÊ£ºS¡¢H2S¡¢HNO3¡¢NO¡¢H2O£¬ÒÑ֪ˮÊÇÉú³ÉÎïÖ®Ò»£®
¢ÙÆäÖÐÊôÓÚµç½âÖʵÄÊÇ
 
£»ÊôÓڷǵç½âÖʵÄÊÇ
 
£®
¢Ú
 
ÊÇÑõ»¯¼Á£»
 
ÊÇÑõ»¯²úÎ
¢Ûд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ²¢Åäƽ£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª£ºAÊÇʯÓÍÁѽâÆøµÄÖ÷Òª³É·Ý£¬AµÄ²úÁ¿Í¨³£ÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤Ë®Æ½£®ÏÖÒÔAΪÖ÷ÒªÔ­ÁϺϳÉÒÒËáÒÒõ¥£¬ÆäºÏ³É·ÏßÈçͼ1Ëùʾ£®

£¨1£©AÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔ¾ÛºÏÉú³ÉÒ»ÖÖ³£¼ûËÜÁÏ£¬¸ÃËÜÁϵĽṹ¼òʽΪ
 
£®
£¨2£©·´Ó¦¢ÜµÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨3£©¢Ù¡¢¢ÜµÄ·´Ó¦ÀàÐÍ·Ö±ðΪ
 
¡¢
 
£®ÓлúÎïBÖйÙÄÜÍŵÄÃû³ÆΪ
 
£®
£¨4£©ÔÚʵÑéÊÒÖлñµÃµÄÒÒËáÒÒõ¥ÍùÍùº¬ÓÐB¡¢D£¬ÎªÌá´¿ÒÒËáÒÒõ¥£¬¼ÓÈëµÄÊÔ¼ÁÒÔ¼°·ÖÀë²Ù×÷·½·¨
 
¡¢
 
£®
£¨5£©ÒÒËáÒÒõ¥ÊÇÖØÒªµÄÓлúºÏ³ÉÖмäÌ壬¹ã·ºÓ¦ÓÃÓÚ»¯Ñ§¹¤Òµ£®ÊµÑéÊÒÀûÓÃͼ2µÄ×°ÖÃÖƱ¸ÒÒËáÒÒõ¥£®
¢ÙÓë½Ì²Ä²ÉÓõÄʵÑé×°Öò»Í¬£¬´Ë×°ÖÃÖвÉÓÃÁËÇòÐθÉÔï¹Ü£¬Æä×÷ÓÃÊÇ£º
 
£®
¢ÚΪÁËÖ¤Ã÷ŨÁòËáÔڸ÷´Ó¦ÖÐÆðµ½ÁË´ß»¯¼ÁºÍÎüË®¼ÁµÄ×÷Óã¬Ä³Í¬Ñ§ÀûÓÃÉÏͼËùʾװÖýøÐÐÁËÒÔÏÂ4¸öʵÑ飮ʵÑ鿪ʼÏÈÓþƾ«µÆ΢ÈÈ3min£¬ÔÙ¼ÓÈÈʹ֮΢΢·ÐÌÚ3min£®ÊµÑé½áÊøºó³ä·ÖÕñµ´Ð¡ÊԹܢò£¬ÔÙ²âÆäÖÐÓлú²ãµÄºñ¶È£¬ÊµÑé¼Ç¼ÈçÏ£º
ʵÑé±àºÅÊԹܢñÖÐÊÔ¼ÁÊԹܢòÖÐÓлú²ãµÄºñ¶È/cm
A2mLÒÒ´¼¡¢1mLÒÒËá¡¢1mL 18mol?L-1ŨÁòËá3.0
B2mLÒÒ´¼¡¢1mLÒÒËá0.1
C2mLÒÒ´¼¡¢1mLÒÒËá 6mL 3mol?L-1 H2SO40.6
D2mLÒÒ´¼¡¢1mLÒÒËá¡¢ÑÎËá0.6
ʵÑéDµÄÄ¿µÄÊÇÓëʵÑéCÏà¶ÔÕÕ£¬Ö¤Ã÷H+¶Ôõ¥»¯·´Ó¦¾ßÓд߻¯×÷Óã®ÊµÑéDÖÐÓ¦¼ÓÈëÑÎËáµÄÌå»ýºÍŨ¶È·Ö±ðÊÇ
 
mLºÍ
 
mol?L-1£®·ÖÎöʵÑéAºÍʵÑéCµÄÊý¾Ý£¬¿ÉÒÔÍƶϳöŨH2SO4µÄ
 
£¨Ìî¡°´ß»¯¡±»ò¡°ÎüË®¡±£©×÷ÓÃÌá¸ßÁËÒÒËáÒÒõ¥µÄ²úÂÊ£®
£¨6£©ÒÒ´¼ÔÚÒ»¶¨Ìõ¼þÏ¿Éת»¯ÎªÓлúÎïE£¬EµÄÕôÆûÓëÏàͬ״¿öÏÂͬÌå»ýÇâÆøµÄÖÊÁ¿±ÈΪ37£¬Æä·Ö×ÓÖÐ̼¡¢ÇâµÄÖÊÁ¿·ÖÊý·ÖΪ±ð64.9%¡¢13.5%£¬ÆäÓàΪÑõ£®ÇóEµÄ·Ö×Óʽ£¬Çëд³ö¼ÆËã¹ý³Ì£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij·¼ÏãÌþAÊÇÓлúºÏ³ÉÖÐÖØÒªµÄÔ­ÁÏ£¬Í¨¹ýÖÊÆ×·¨²âµÃÆäÏà¶Ô·Ö×ÓÖÊÁ¿Îª118£¬Æä±½»·ÉÏÖ»ÓÐÒ»¸öÈ¡´ú»ù£®ÒÔÏÂÊÇÒÔAΪԭÁϺϳɻ¯ºÏÎïFºÍ¸ß·Ö×Ó»¯ºÏÎïIµÄ·Ïßͼ£¨Èçͼ1£©£¬ÆäÖл¯ºÏÎïFÖк¬ÓÐÈý¸öÁùÔª»·£®ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©EÖйÙÄÜÍŵÄÃû³ÆΪ
 
£»
£¨2£©IµÄ½á¹¹¼òʽΪ
 
£»
£¨3£©Ð´³ö·´Ó¦¢ÛµÄ»¯Ñ§·½³Ìʽ
 
£»
£¨4£©Ð´³ö·´Ó¦¢ÝµÄ»¯Ñ§·½³Ìʽ
 
£»
£¨5£©Ð´³öËùÓзûºÏÏÂÁÐÒªÇóµÄDµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ
 
£»
a£®±½»·ÉÏÖ»ÓÐÒ»¸öÈ¡´ú»ù£»
b£®ÄÜÓëÐÂÖÆÇâÑõ»¯Í­Ðü×ÇÒº·´Ó¦£»
c£®·Ö×ÓÖÐ-COO-º¬Óнṹ£®
£¨6£©ÒÔ±½ºÍ±ûϩΪԭÁϿɺϳÉA£¬ÇëÉè¼ÆºÏ³É·Ïߣ¨ÎÞ»úÊÔ¼Á¼°ÈܼÁÈÎÑ¡£©
×¢£ººÏ³É·ÏßµÄÊéд¸ñʽ²ÎÕÕÈçͼ2µÄʵÀýÁ÷³Ìͼ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Á½ÖÖÆø̬ÌþÒÔµÈÌå»ý»ìºÏ£¬ÔÚ120¡æʱ1L¸Ã»ìºÏÌþÓë9LÑõÆø»ìºÏ£¬³ä·ÖȼÉÕºó»Ö¸´µ½Ô­×´Ì¬£¬ËùµÃÆøÌåÈÔΪ10L£®ÏÂÁи÷×é»ìºÏÌþÖв»·ûºÏ´ËÌõ¼þµÄÊÇ£¨¡¡¡¡£©
A¡¢CH4¡¢C2H4
B¡¢C2H2¡¢C3H8
C¡¢C2H4¡¢C3H4
D¡¢C2H2¡¢C3H6

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁз´Ó¦ÖУ¬ÓгÁµí²úÉúÇÒ²»»áÏûʧµÄÊÇ£¨¡¡¡¡£©
A¡¢ÏòNaOHÈÜÒºÖÐÖðµÎµÎÈëFe2£¨SO4£©3ÈÜÒºÖ±ÖÁ¹ýÁ¿
B¡¢½«NaOHÈÜÒºÖðµÎµÎÈëAlCl3ÈÜÒºÖУ¬Ö±ÖÁ¹ýÁ¿
C¡¢ÏòAlCl3ÈÜÒºÖÐÖðµÎµÎÈëÏ¡ÁòËá
D¡¢½«°±Ë®ÖðµÎµÎÈëÏõËáÒøÈÜÒºÖУ¬Ö±ÖÁ¹ýÁ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢Í¬ÎÂͬѹÏ£¬ÎïÖʵÄÎïÖʵÄÁ¿ÏàµÈ£¬ÆäÌå»ýÒ»¶¨Ïàͬ
B¡¢µÈÌå»ýµÄ¶þÑõ»¯Ì¼ºÍÒ»Ñõ»¯Ì¼Ëùº¬µÄ·Ö×ÓÊýÒ»¶¨ÏàµÈ
C¡¢1LµªÆøÒ»¶¨±È1LÑõÆøµÄÖÊÁ¿Ð¡
D¡¢ÏàͬÌõ¼þϵÄÒ»Ñõ»¯Ì¼ÆøÌåºÍµªÆø£¬ÈôÌå»ýÏàµÈ£¬ÔòÖÊÁ¿Ò»¶¨ÏàµÈ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚÈçͼËùʾװÖÃÖнøÐа±µÄ´ß»¯Ñõ»¯ÊµÑ飺ÍùÈý¾±Æ¿ÄÚµÄŨ°±Ë®Ö⻶ÏͨÈË¿ÕÆø£¬½«ºìÈȵIJ¬Ë¿²åÈËÆ¿Öв¢½Ó½üÒºÃ森·´Ó¦¹ý³ÌÖУ¬¿É¹Û²ìµ½Æ¿ÖÐÓкì×ØÉ«ÆøÌå²úÉú£¬²¬Ë¿Ê¼ÖÕ±£³ÖºìÈÈ£®ÏÂÁÐÓйØ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©
A¡¢·´Ó¦ºóÈÜÒºÖк¬ÓÐNO3-
B¡¢·´Ó¦ºóÈÜÒºÖÐc£¨H+£©Ôö´ó
C¡¢ÊµÑé¹ý³ÌÖÐÓл¯ºÏ·´Ó¦·¢Éú
D¡¢ÊµÑé¹ý³ÌÖÐNH3?H2OµÄµçÀë³£Êý²»¿ÉÄÜ·¢Éú±ä»¯

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸