ËÄ´¨Ê¢²úÎå±¶×Ó¡£ÒÔÎå±¶×ÓΪԭÁÏ¿ÉÖÆµÃ»¯ºÏÎïA¡£AµÄ½á¹¹¼òʽÈçͼËùʾ£º

                   

Çë½â´ðÏÂÁи÷Ì⣺

£¨1£©AµÄ·Ö×ÓʽÊÇ                ¡£

£¨2£©Óлú»¯ºÏÎïBÔÚÁòËá´ß»¯Ìõ¼þϼÓÈÈ·¢Éúõ¥»¯·´Ó¦¿ÉµÃµ½A¡£

     Çëд³öBµÄ½á¹¹¼òʽ£º                               

£¨3£©Çëд³öAÓë¹ýÁ¿NaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

                                                   

£¨4£©Óлú»¯ºÏÎïCÊǺϳÉÖÎÁÆÇÝÁ÷¸ÐÒ©ÎïµÄÔ­ÁÏÖ®Ò»¡£C¿ÉÒÔ¿´³ÉÊÇBÓëÇâÆø°´ÎïÖʵÄÁ¿Ö®±È1¡Ã2·¢Éú¼Ó³É·´Ó¦µÃµ½µÄ²úÎï¡£C·Ö×ÓÖÐÎÞôÇ»ùÓë̼̼˫¼üÖ±½ÓÏàÁ¬µÄ½á¹¹£¬ËüÄÜÓëäåË®·´Ó¦Ê¹äåË®ÍÊÉ«¡£

    Çëд³öCÓëäåË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

                                                     


¡¾´ð°¸¡¿

£¨1£©C14H10O9

£¨2£©

£¨3£©

£¨4£©

¡¾½âÎö¡¿B·¢Éúõ¥»¯·´Ó¦¿ÉµÃµ½A£¬¼´A£ºË®½âºó£¨ÐéÏß¿ò²¿·Ö¶ÏÁÑ£©Éú³ÉB£¬È·¶¨BµÄ½á¹¹¼òʽ¡£BÓëH2°´1£º2¼Ó³É£¬²úÎïµÄÁùÔª»·ÉÏ»¹ÓÐÒ»¸öC£½C£¬ÓÖÒªÇóôÇ»ù²»ÄÜÓëC£½CÖ±½ÓÏàÁ¬£¬Ôò²úÎï½á¹¹Ö»ÄÜΪ¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ºÏ³ÉP(Ò»ÖÖ¿¹Ñõ»¯¼Á)µÄ·ÏßÈçÏ£º

A(C4H10O)B(C4H8)

 

DE(C15H22O3)P(C19H30O3)

 

¢ÚAºÍF»¥ÎªÍ¬·ÖÒì¹¹Ì壬A·Ö×ÓÖÐÓÐÈý¸ö¼×»ù£¬F·Ö×ÓÖÐÖ»ÓÐÒ»¸ö¼×»ù¡£

(1)A¡úBµÄ·´Ó¦ÀàÐÍΪ________¡£B¾­´ß»¯¼ÓÇâÉú³ÉG(C4H10)£¬GµÄ»¯Ñ§Ãû³ÆÊÇ______________¡£

(2)AÓëŨHBrÈÜÒºÒ»Æð¹²ÈÈÉú³ÉH£¬HµÄ½á¹¹¼òʽΪ______________¡£

(3)ʵÑéÊÒÖмìÑéC¿ÉÑ¡ÔñÏÂÁÐÊÔ¼ÁÖеÄ__________¡£

  a£®ÑÎËá¡¡  b£®FeCl3ÈÜÒº¡¡  c£®NaHCO3ÈÜÒº¡¡  d£®Å¨äåË®

(4)PÓë×ãÁ¿NaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ________(ÓлúÎïÓýṹ¼òʽ±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÌìÈ»Æø»¯¹¤ÊÇÖØÇìÊеÄÖ§Öù²úÒµÖ®Ò»¡£ÒÔÌìÈ»ÆøÎªÔ­ÁϾ­ÏÂÁз´Ó¦Â·Ï߿ɵù¤³ÌËÜÁÏPBT¡£

ÒÑÖª£º£¨R¡¢R¡ä¡¢R¡å¡ªÌþ»ù»òH£©

£¨1£©B·Ö×ӽṹÖÐÖ»ÓÐÒ»ÖÖÇâ¡¢Ò»ÖÖÑõ¡¢Ò»ÖÖ̼£¬ÔòBµÄ½á¹¹¼òʽÊÇ_____________£»BµÄͬ·ÖÒì¹¹ÌåÖÐÓëÆÏÌÑÌǾßÓÐÀàËÆ½á¹¹µÄÊÇ___________________¡££¨Ð´½á¹¹¼òʽ£©

£¨2£©FµÄ½á¹¹¼òʽÊÇ_____________£»PBTÊôÓÚ__________ÀàÓлú¸ß·Ö×Ó»¯ºÏÎï¡£

£¨3£©ÓÉA¡¢DÉú³ÉEµÄ·´Ó¦·½³ÌʽΪ____________£¬Æä·´Ó¦ÀàÐÍΪ____________________¡£

£¨4£©EµÄͬ·ÖÒì¹¹ÌåC²»ÄÜ·¢ÉúÒø¾µ·´Ó¦¡¢ÄÜʹäåË®ÍÊÉ«¡¢ÄÜË®½âÇÒ²úÎïµÄ̼ԭ×ÓÊý²»µÈ£¬ÔòGÔÚNaOHÈÜÒºÖз¢ÉúË®½â·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


LÊÇHͨ¹ýëļüÁ¬½Ó¶ø³ÉµÄ¸ß¾ÛÎ¹ÊLµÄ½á¹¹¼òʽΪ£º

  »ò

£¨2007¡¤º£ÄÏ¾í£©¡¶Óлú»¯Ñ§»ù´¡¡·Ä£¿é (Ñ¡¿¼Ìâ)21.

¸ù¾Ýͼʾ»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öA¡¢E¡¢GµÄ½á¹¹¼òʽ£ºA              £¬E                 £¬G                 £»

£¨2£©·´Ó¦¢ÚµÄ»¯Ñ§·½³Ìʽ£¨°üÀ¨·´Ó¦Ìõ¼þ£©ÊÇ                             £¬

·´Ó¦¢Ü»¯Ñ§·½³Ìʽ£¨°üÀ¨·´Ó¦Ìõ¼þ£©ÊÇ                             £»

£¨3£©Ð´³ö¢Ù¡¢¢ÝµÄ·´Ó¦ÀàÐÍ£º¢Ù         ¡¢¢Ý           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijÓлú»¯ºÏÎïX£¨C7H8O£©ÓëÁíÒ»Óлú»¯ºÏÎïY·¢ÉúÈçÏ·´Ó¦Éú³É»¯ºÏÎïZ£¨C11H14O2£©£º

X£«YZ£«H2O

£¨1£©XÊÇÏÂÁл¯ºÏÎïÖ®Ò»£¬ÒÑÖªX²»ÄÜÓëFeCl3ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦¡£ÔòXÊÇ________£¨Ìî±êºÅ×Öĸ£©¡£

                £¨A£©             £¨B£©             £¨C£©            £¨D£©

£¨2£©YµÄ·Ö×ÓʽÊÇ_____________£¬¿ÉÄܵĽṹ¼òʽÊÇ£º_____________ºÍ_____________¡£

£¨3£©YÓжàÖÖͬ·ÖÒì¹¹Ì壬ÆäÖÐÒ»ÖÖͬ·ÖÒì¹¹ÌåE·¢ÉúÒø¾µ·´Ó¦ºó£¬Æä²úÎï¾­Ëữ¿ÉµÃµ½F£¨C4H8O3£©¡£F¿É·¢ÉúÈçÏ·´Ó¦£ºF£«H2O

¸Ã·´Ó¦µÄÀàÐÍÊÇ_____________£¬EµÄ½á¹¹¼òʽÊÇ_____________¡£

£¨4£©ÈôYÓëE¾ßÓÐÏàͬµÄ̼Á´£¬ÔòZµÄ½á¹¹¼òʽΪ£º_____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚÒºÂȺÍÂÈË®µÄÐðÊöÖÐÕýÈ·µÄÊÇ£¨£©
A£®ÒºÂÈÊÇ´¿¾»Î¶øÂÈË®ÊÇ»ìºÏÎï
B£®ÒºÂÈÎÞËáÐÔ£¬ÂÈË®ÓÐËáÐÔ
C£®ÒºÂȽÏÂÈË®µÄƯ°××÷ÓøüÇ¿
D£®ÒºÂÈÎÞÉ«£¬ÂÈË®³Ê»ÆÂÌÉ«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚÎïÖʵı£´æ²»ÕýÈ·µÄÊÇ£¨    £©¡£

    A£®AgNO3ÈÜÒºÓ¦±£´æÔÚ×ØÉ«Æ¿ÖР   B£®ÂÈË®±£´æÔÚ×ØÉ«Ï¸¿ÚÆ¿ÖÐ

    C£®ÒºÂÈ¿ÉÒÔ±£´æÔÚ¸ÉÔïµÄ¸ÖÆ¿ÖР   D£®Æ¯°×·Û¿É¶ÖÃÓÚ¿ÕÆøÖб£´æ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Äܹ»Ö±½Ó¼ø±ðBaCl2¡¢NaCl¡¢Na2CO3ÈýÖÖÈÜÒºµÄÊÔ¼ÁÊÇ(¡¡¡¡)

A£®AgNO3ÈÜÒº                        B£®Ï¡ÁòËá

C£®Ï¡ÑÎËá                          D£®Ï¡ÏõËá

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÕÓÆøµÄÖ÷Òª³É·ÖÊÇCH4¡£0.5 mol CH4ÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ£¬·Å³ö445kJÈÈÁ¿¡£ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÖУ¬ÕýÈ·µÄÊÇ£º£¨     £©

A£®2CH4(g)+4O2(g)£½2CO2(g)+4H2O(l) ¡÷H=+890kJ/mol

B£®CH4(g)+2O2(g)£½CO2(g)+2H2O(l) ¡÷H=+890kJ/mol

C£®CH4(g)+2O2(g)£½CO2(g)+2H2O(l) ¡÷H=-890kJ/mol

D£®1/2 CH4(g)+ O2(g)£½1/2CO2(g)+ H2O(g) ¡÷H=-890kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸