¿Æѧ¼Ò·¢ÏÖijҩÎïMÄÜÖÎÁÆÐÄѪ¹Ü¼²²¡ÊÇÒòΪËüÔÚÈËÌåÄÚÄÜÊÍ·ÅÒ»ÖÖ¡°ÐÅʹ·Ö×Ó¡±D£¬²¢²ûÃ÷ÁËDÔÚÈËÌåÄÚµÄ×÷ÓÃÔ­Àí¡£Îª´ËËûÃÇÈÙ»ñÁË1998Äêŵ±´¶ûÉúÀíѧ»òҽѧ½±¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÑÖªMµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª227£¬ÓÉC¡¢H¡¢O¡¢NËÄÖÖÔªËØ×é³É£¬C¡¢H¡¢NµÄÖÊÁ¿·ÖÊýÒÀ´ÎΪ15.86%¡¢2.20%ºÍ18.5%¡£ÔòMµÄ·Ö×ÓʽÊÇ________________¡£DÊÇË«Ô­×Ó·Ö×Ó£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª30£¬ÔòDµÄ·Ö×ÓʽΪ________________________¡£

£¨2£©ÓÍõ¥A¾­ÏÂÁÐ;¾¶¿ÉµÃµ½M¡£

ͼ4-6

ͼÖТڵÄÌáʾ£º

C2H5OH+HO¡ªNO2C2H5O¡ªNO2+H2O

                  ÏõËá                         ÏõËáÒÒõ¥

·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ_____________________________________¡£

·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ_____________________________________¡£

£¨3£©CÊÇBºÍÒÒËáÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉµÄ»¯ºÏÎÏà¶Ô·Ö×ÓÖÊÁ¿Îª134£¬Ð´³öCËùÓпÉÄܵĽṹ¼òʽ£º_____________________________________¡£

£¨4£©Èô½«0.1 mol BÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦£¬ÔòÐèÏûºÄ__________ g½ðÊôÄÆ¡£

˼·½âÎö£º¸ù¾ÝMµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÎªÔªËØÖÊÁ¿·ÖÊý¾Í¿ÉÒÔ¼ÆËã³öËüµÄ»¯Ñ§Ê½³öÀ´¡£MÖÐÖ»ÓÐC¡¢H¡¢O¡¢NËÄÖÖÔªËØ£¬¿ÉÒÔ·Ö½â²úÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª30µÄÖ»ÓÐNO»òC2H6¡£

¸ù¾ÝÌáʾ¿ÉÖª£¬BΪһÖÖ´¼£¬Ëü±»Ïõ»¯ºóµÄ²úÎïΪM£¬¼´C3H5N3O9£¬´ÓMµÄ»¯Ñ§Ê½¿ÉÖªÊÇÈýÏõ»¯²úÎÔòBÖÐÓÐÈý¸öôÇ»ù£¬º¬Èý¸öôÇ»ùµÄÈý̼ÓлúÎïÊDZûÈý´¼£¬0.1 mol±ûÈý´¼±»ÄÆ»¹Ô­£¬ÏûºÄ0.3 molÄÆ£¬ÖÊÁ¿Îª6.9 g¡£

´ð°¸£º£¨1£©C3H5N3O9  NO

£¨2£©CH2£¨OOCR£©CH£¨OOCR£©CH2£¨OOCR£©+H2OCH2OHCHOHCH2OH+3RCOOH

CH2OHCHOHCH2OH+3HONO2CH2£¨ONO2£©CH£¨ONO2£©CH2£¨ONO2£©+3H2O

£¨3£©CH2OHCH£¨OOCR£©CH2OH£¬CH2£¨OOCR£©CHOHCH2OH 

£¨4£©6.9


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¿Æѧ¼Ò·¢ÏÖijҩÎïMÄÜÖÎÁÆÐÄѪ¹Ü¼²²¡ÊÇÒòΪËüÔÚÈËÌåÄÚÄÜÊͷųöÒ»ÖÖ¡°ÐÅʹ·Ö×Ó¡±D£¬²¢²ûÃ÷ÁËDÔÚÈËÌåÄÚµÄ×÷ÓÃÔ­Àí¡£Îª´ËËûÃÇÈÙ»ñÁË1998Äêŵ±´¶ûÉúÀíѧ»òҽѧ½±¡£

ÇëÍê³ÉÏÂÁÐÎÊÌ⣺

(1)ÒÑÖªMµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª227£¬ÓÉC¡¢H¡¢O¡¢NËÄÖÖÔªËØ×é³É£¬C¡¢H¡¢NµÄÖÊÁ¿·ÖÊýÒÀ´ÎΪ15.86%¡¢2.20%ºÍ18.50%¡£ÔòMµÄ·Ö×ÓʽÊÇ__________________________¡£DÊÇË«Ô­×Ó·Ö×Ó£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª30£¬ÔòDµÄ·Ö×ÓʽΪ_______________________________¡£

(2)ÓÍÖ¬A¾­Í¼2-3Ëùʾ;¾¶¿ÉµÃµ½M¡£

              

                           ͼ2-3

ͼÖТڵÄÌáʾ£º

C2H5OH+HO¡ªNO2C2H5O¡ªNO2+H2O

                  ÏõËá                              ÏõËáÒÒõ¥

·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ_____________________________________________________¡£

·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ_____________________________________________________¡£

(3)CÊÇBºÍÒÒËáÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉµÄ»¯ºÏÎÏà¶Ô·Ö×ÓÖÊÁ¿Îª134£¬Ð´³öCËùÓпÉÄܵĽṹ¼òʽ¡£___________________________________________________________

(4)Èô½«0.1 mol BÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦£¬ÔòÐèÏûºÄ___________________g½ðÊôÄÆ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¿Æѧ¼Ò·¢ÏÖijҩÎïMÄÜÖÎÁÆÐÄѪ¹Ü¼²²¡ÊÇÒòΪËüÔÚÈËÌåÄÚÄÜÊͷųöÒ»ÖÖ¡°ÐÅʹ·Ö×Ó¡±D£¬²¢²ûÃ÷ÁËDÔÚÈËÌåÄÚµÄ×÷ÓÃÔ­Àí¡£Îª´ËËûÃÇÈÙ»ñÁË1998Äêŵ±´¶ûÉúÀíѧ»òҽѧ½±¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÒÑÖªMµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª227£¬ÓÉC¡¢H¡¢O¡¢NËÄÖÖÔªËØ×é³É£¬C¡¢H¡¢NµÄÖÊÁ¿·ÖÊýÒÀ´ÎΪ15.86%¡¢2.20%ºÍ18.50%¡£ÔòMµÄ·Ö×ÓʽÊÇ______________¡£DÊÇË«Ô­×Ó·Ö×Ó£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª30£¬ÔòDµÄ·Ö×ÓʽΪ______________¡£

(2)ÓÍÖ¬A¾­ÏÂÁÐ;¾¶¿ÉµÃµ½M¡£

ͼÖТڵÄÌáʾ£º

·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ_____________________________¡£

·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ_____________________________¡£

(3)CÊÇBºÍÒÒËáÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉµÄ»¯ºÏÎ·Ö×ÓÁ¿Îª134£¬Ð´³öCËùÓпÉÄܵĽṹ¼òʽ£º________________________________¡£

(4)Èô½«0.1 mol BÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦£¬ÔòÐèÏûºÄ___________g½ðÊôÄÆ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¿Æѧ¼Ò·¢ÏÖijҩÎïMÄÜÖÎÁÆÐÄѪ¹Ü¼²²¡ÊÇÒòΪËüÔÚÈËÌåÄÚÄÜÊͷųöÒ»ÖÖ¡°ÐÅʹ·Ö×Ó¡±D£¬²¢²ûÃ÷ÁËDÔÚÈËÌåÄÚµÄ×÷ÓÃÔ­Àí¡£Îª´ËËûÃÇÈÙ»ñÁË1998Äêŵ±´¶ûÉúÀíѧ»òҽѧ½±¡£

ÇëÍê³ÉÏÂÁÐÎÊÌ⣺

(1)ÒÑÖªMµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª227£¬ÓÉC¡¢H¡¢O¡¢NËÄÖÖÔªËØ×é³É£¬C¡¢H¡¢NµÄÖÊÁ¿·ÖÊýÒÀ´ÎΪ15.86%¡¢2.20%ºÍ18.50%¡£ÔòMµÄ·Ö×ÓʽÊÇ__________¡£DÊÇË«Ô­×Ó·Ö×Ó£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª30£¬ÔòDµÄ·Ö×ÓʽΪ__________¡£

(2)ÓÍÖ¬A¾­ÏÂÁÐ;¾¶¿ÉµÃµ½M¡£

ͼÖТڵÄÌáʾ£º

C2H5OH+HO¡ªNO2C2H5O¡ªNO2+H2O

                    ÏõËá                            ÏõËáÒÒõ¥

·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ_________________________________£»

·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ_________________________________¡£

(3)CÊÇBºÍÒÒËáÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉµÄ»¯ºÏÎÏà¶Ô·Ö×ÓÖÊÁ¿Îª134£¬Ð´³öCËùÓпÉÄܵĽṹ¼òʽ__________________________________________________________________¡£

(4)Èô½«0.1 mol BÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦£¬ÔòÐèÏûºÄ___________g½ðÊôÄÆ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¿Æѧ¼Ò·¢ÏÖijҩÎïMÄÜÖÎÁÆÐÄѪ¹Ü¼²²¡ÊÇÒòΪËüÔÚÈËÌåÄÚÄÜÊͷųöÒ»ÖÖ¡°ÐÅʹ·Ö×Ó¡±D£¬²¢²ûÃ÷ÁËDÔÚÈËÌåÄÚµÄ×÷ÓÃÔ­Àí¡£Îª´ËËûÃÇÈÙ»ñÁË1998Äêŵ±´¶ûÉúÀíѧ»òҽѧ½±¡£ÇëÍê³ÉÏÂÁÐÎÊÌ⣺

(1)ÒÑÖªMµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª227£¬ÓÉC¡¢H¡¢O¡¢NËÄÖÖÔªËØ×é³É£¬C¡¢H¡¢NµÄÖÊÁ¿·ÖÊýÒÀ´ÎΪ15.86%¡¢2.20%ºÍ18.50%£¬ÔòMµÄ·Ö×ÓʽÊÇ________¡£DÊÇË«Ô­×Ó·Ö×Ó£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª30£¬ÔòDµÄ·Ö×ÓʽΪ________¡£

(2)ÓÍÖ¬A¾­ÏÂÁÐ;¾¶¿ÉµÃµ½M¡£

ͼ8-6

ͼÖТڵÄÌáʾ£º

C2H5OH+HO¡ªNO2C2H5O¡ªNO2+H2O

·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ_______________________________________¡£

·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ_______________________________________¡£

(3)CÊÇBºÍÒÒËáÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉµÄ»¯ºÏÎÏà¶Ô·Ö×ÓÖÊÁ¿Îª134£¬Ð´³öCËùÓпÉÄܵĽṹ¼òʽ_______________________________¡£

(4)Èô½«0.1 mol BÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦£¬ÔòÐèÏûºÄ________g½ðÊôÄÆ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¿Æѧ¼Ò·¢ÏÖijҩÎïMÄÜÖÎÁÆÐÄѪ¹Ü¼²²¡ÊÇÒòΪËüÔÚÈËÌåÄÚÄÜÊͷųöÒ»ÖÖ¡°ÐÅʹ·Ö×Ó¡±D£¬²¢²ûÃ÷ÁËDÔÚÈËÌåÄÚµÄ×÷ÓÃÔ­Àí¡£Îª´ËËûÃÇÈÙ»ñÁË1998Äêŵ±´¶ûÉúÀíѧºÍҽѧ½±¡£ÇëÍê³ÉÏÂÁÐÎÊÌâ¡£

(1)ÒÑÖªMµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª227£¬ÓÉC¡¢H¡¢O¡¢NËÄÖÖÔªËØ×é³É£¬C¡¢H¡¢NµÄÖÊÁ¿·ÖÊýÒÀ´ÎΪ15.86%¡¢2.20%ºÍ18.50%¡£ÔòMµÄ·Ö×ÓʽÊÇ________________¡£DÊÇË«Ô­×Ó·Ö×Ó£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª30£¬ÔòDµÄ·Ö×ÓʽΪ________________¡£

(2)ÓÍÖ¬A¾­ÏÂÁÐ;¾¶¿ÉµÃµ½M¡£

ͼÖТڵÄÌáʾ£º

C2H5OH£«HO¡ªNO2C2H5O¡ªNO2£«H2O

                   ÏõËá                               ÏõËáÒÒõ¥

·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽÊÇ__________________________________________________¡£

·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ__________________________________________________¡£

(3)CÊÇBºÍÒÒËáÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉµÄ»¯ºÏÎÏà¶Ô·Ö×ÓÖÊÁ¿Îª134£¬Ð´³öCËùÓпÉÄܵĽṹ¼òʽ£º________________________________________________________________¡£

(4)Èô½«0.1 mol BÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦£¬ÔòÐèÏûºÄ_______________g½ðÊôÄÆ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸