ÏÂͼ¼×Êǿα¾ÖÐÑé֤ͭºÍŨÏõËá·´Ó¦µÄ×°Öã¬ÒÒ¡¢±ûÊÇʦÉú¶ÔÑÝʾʵÑé¸Ä½øºóµÄ×°Öãº

(1)¼×¡¢ÒÒ¡¢±ûÈý¸ö×°ÖÃÖй²Í¬·¢ÉúµÄ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______________________  

(2)ºÍ¼××°ÖÃÏà±È£¬ÒÒ×°ÖõÄÓŵãÊÇ¢Ù____________________£»¢Ú                  

(3)ΪÁ˽øÒ»²½ÑéÖ¤NO2ºÍË®µÄ·´Ó¦£¬Ä³Ñ§ÉúÉè¼ÆÁ˱û×°Öá£Ïȹرյ¯»É¼Ð________£¬ÔÙ´ò¿ªµ¯»É¼Ð________£¬²ÅÄÜʹNO2ÆøÌå³äÂúÊԹܢڡ£

(4)µ±ÆøÌå³äÂúÊԹܢں󣬽«Í­Ë¿ÌáÆðÓëÈÜÒºÍÑÀ룬ÓûʹÉÕ±­ÖеÄË®½øÈëÊԹܢÚÓ¦ÈçºÎ²Ù×÷

                                                                                                                                                      

 

¡¾´ð°¸¡¿

£¨1£©Cu+4HNO3==Cu(NO3)2£«2NO2¡ü£«2H2

   £¨2£©¿ÉËæÓÃËæÖÆ£»¶þÑõ»¯µª±»ÎüÊÕ£¬·ÀÖ¹ÎÛȾ»·¾³

   £¨3£©C £»a£¬b

   £¨4£©¹Ø±Õab´ò¿ªcÓÃÈÈë½íÎæÊÔ¹Ü

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂͼ¼×Êǿα¾ÖÐÑé֤ͭºÍŨÏõËá·´Ó¦µÄ×°Öã¬ÒÒ¡¢±ûÊÇʦÉú¶ÔÑÝʾʵÑé¸Ä½øºóµÄ×°Öãº
£¨1£©Ð´³öÍ­ºÍŨÏõËá·´Ó¦µÄÀë×Ó·½³Ìʽ
Cu+4H++2NO3_=Cu2++2NO2¡ü+2H2O
Cu+4H++2NO3_=Cu2++2NO2¡ü+2H2O
£®
£¨2£©ºÍ¼××°ÖÃÏà±È£¬ÒÒ×°ÖõÄÓŵãÓÐ
¢Ù¿ÉÒÔ¿ØÖÆ·´Ó¦¢ÚÎüÊÕNO2ÆøÌ壬·ÀÖ¹ÎÛȾ»·¾³£»
¢Ù¿ÉÒÔ¿ØÖÆ·´Ó¦¢ÚÎüÊÕNO2ÆøÌ壬·ÀÖ¹ÎÛȾ»·¾³£»

£¨3£©ÎªÁ˽øÒ»²½ÑéÖ¤NO2ºÍË®µÄ·´Ó¦£¬Ä³Ñ§ÉúÉè¼ÆÁ˱û×°Öã¬×öʵÑéʱÏȹرյ¯»É¼Ð
c
c
£¬ÔÙ´ò¿ªµ¯»É¼Ð
a¡¢b
a¡¢b
£¬²ÅÄÜʹNO2ÆøÌå³äÂú¢ÚÊԹܣ®
£¨4£©µ±ÆøÌå³äÂú¢ÚÊԹܺ󣬽«Í­Ë¿ÌáÆðÓëÈÜÒºÍÑÀ룬ÓûʹÉÕ±­ÖеÄË®½øÈë¢ÚÊÔ¹ÜÓ¦ÈçºÎ²Ù×÷
ÏȹرÕb£¬ÔٹرÕa£¬È»ºó´ò¿ªc£¬ÓÃÊÖÎæס£¨ÈÈË®¡¢ÈÈë½í¡¢¼ÓÈÈ£©ÊԹܢÚ
ÏȹرÕb£¬ÔٹرÕa£¬È»ºó´ò¿ªc£¬ÓÃÊÖÎæס£¨ÈÈË®¡¢ÈÈë½í¡¢¼ÓÈÈ£©ÊԹܢÚ
£®
£¨5£©¢ÚÊÔ¹ÜÖеÄNO2ºÍË®³ä·Ö·´Ó¦ºó£¬ËùµÃÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈµÄ×î´óÖµÊÇ
1/22.4mol?L-1»ò0.045mol?L-1
1/22.4mol?L-1»ò0.045mol?L-1
£¨ÆøÌåÌå»ý°´±ê×¼×´¿ö¼ÆË㣩£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂͼ¼×Êǿα¾ÖÐÑé֤ͭºÍŨÏõËá·´Ó¦µÄ×°Öã¬ÒÒ¡¢±ûÊÇʦÉú¶ÔÑÝʾʵÑé¸Ä½øºóµÄ×°Öãº

(1)¼×¡¢ÒÒ¡¢±ûÈý¸ö×°ÖÃÖй²Í¬·¢ÉúµÄ»¯Ñ§·½³ÌʽÊÇ________________¡£?

(2)ºÍ¼××°ÖÃÏà±È£¬ÒÒ×°ÖõÄÓŵãÊÇ¢Ù________________£¬¢Ú________________¡£?

(3)ΪÁ˽øÒ»²½ÑéÖ¤NO2ºÍË®µÄ·´Ó¦£¬Ä³Ñ§ÉúÉè¼ÆÁ˱û×°Öá£ÊµÑéʱÏȹرջîÈû________£¬ÔÙ´ò¿ª»îÈû_______£¬²ÅÄÜʹNO2ÆøÌå³äÂú¢ÚÊԹܣ»µ±ÆøÌå³äÂú¢ÚÊԹܺ󣬽«Í­Ë¿ÌáÆðÓëÈÜÒºÍÑÀë¡£ÓûʹÉÕ±­ÖеÄË®½øÈë¢ÚÊԹܣ¬Ó¦ÈçºÎ²Ù×÷?________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(10·Ö) ÏÂͼ¼×Êǿα¾ÖÐÑé֤ͭºÍŨÏõËá·´Ó¦µÄ×°Öã¬ÒÒ¡¢±ûÊÇʦÉú¶ÔÑÝʾʵÑé¸Ä½øºóµÄ×°Öãº

  

£¨1£©       д³öÍ­ºÍŨÏõËá·´Ó¦µÄÀë×Ó·½³Ìʽ                                 ¡£

£¨2£©ºÍ¼××°ÖÃÏà±È£¬ÒÒ×°ÖõÄÓŵãÓР                                        

£¨3£©ÎªÁ˽øÒ»²½ÑéÖ¤NO2ºÍË®µÄ·´Ó¦£¬Ä³Ñ§ÉúÉè¼ÆÁ˱û×°ÖÃ,×öʵÑéʱÏȹرյ¯»É¼Ð

        £¬ÔÙ´ò¿ªµ¯»É¼Ð            £¬²ÅÄÜʹNO2ÆøÌå³äÂú¢ÚÊԹܡ£

£¨4£©µ±ÆøÌå³äÂú¢ÚÊԹܺ󣬽«Í­Ë¿ÌáÆðÓëÈÜÒºÍÑÀ룬ÓûʹÉÕ±­ÖеÄË®½øÈë¢ÚÊÔ¹ÜÓ¦ÈçºÎ²Ù

×÷                                                                           ¡£

£¨5£©¢ÚÊÔ¹ÜÖеÄNO2ºÍË®³ä·Ö·´Ó¦ºó£¬ËùµÃÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈµÄ×î´óÖµÊÇ

             £¨ÆøÌåÌå»ý°´±ê×¼×´¿ö¼ÆË㣩¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º09¡ª10Äê³É¶¼Ê¯ÊÒÖÐѧ¸ßÒ»ÏÂÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§¾í ÌâÐÍ£ºÊµÑéÌâ

(10·Ö)ÏÂͼ¼×Êǿα¾ÖÐÑé֤ͭºÍŨÏõËá·´Ó¦µÄ×°Öã¬ÒÒ¡¢±ûÊÇʦÉú¶ÔÑÝʾʵÑé¸Ä½øºóµÄ×°Öãº

£¨1£©      Ð´³öÍ­ºÍŨÏõËá·´Ó¦µÄÀë×Ó·½³Ìʽ                                 ¡£
£¨2£©ºÍ¼××°ÖÃÏà±È£¬ÒÒ×°ÖõÄÓŵãÓР                                        
£¨3£©ÎªÁ˽øÒ»²½ÑéÖ¤NO2ºÍË®µÄ·´Ó¦£¬Ä³Ñ§ÉúÉè¼ÆÁ˱û×°ÖÃ,×öʵÑéʱÏȹرյ¯»É¼Ð
        £¬ÔÙ´ò¿ªµ¯»É¼Ð            £¬²ÅÄÜʹNO2ÆøÌå³äÂú¢ÚÊԹܡ£ 
£¨4£©µ±ÆøÌå³äÂú¢ÚÊԹܺ󣬽«Í­Ë¿ÌáÆðÓëÈÜÒºÍÑÀ룬ÓûʹÉÕ±­ÖеÄË®½øÈë¢ÚÊÔ¹ÜÓ¦ÈçºÎ²Ù
×÷                                                                           ¡£
£¨5£©¢ÚÊÔ¹ÜÖеÄNO2ºÍË®³ä·Ö·´Ó¦ºó£¬ËùµÃÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈµÄ×î´óÖµÊÇ
             £¨ÆøÌåÌå»ý°´±ê×¼×´¿ö¼ÆË㣩¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ì¸£½¨Ê¡¸ßÒ»ÏÂѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ

£¨8·Ö£©ÏÂͼ¼×Êǿα¾ÖÐÑé֤ͭºÍŨÏõËá·´Ó¦µÄ×°Öã¬ÒÒ¡¢±ûÊÇʦÉú¸Ä½øºóµÄ×°Öãº

£¨1£©Ð´³öÍ­ºÍŨÏõËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ                               ¡£

£¨2£©ºÍ¼××°ÖÃÏà±È£¬ÒÒ×°ÖõÄÓŵ㠠                              ¡£

£¨3£©ÎªÁ˽øÒ»²½ÑéÖ¤NO2ºÍË®µÄ·´Ó¦£¬Ä³Ñ§ÉúÉè¼ÆÁ˱û×°Öá£Ïȹرյ¯»É¼Ð         £¬ÔÙ´ò¿ªµ¯»É¼Ð           £¬²ÅÄÜʹNO2ÆøÌå³äÂú¢ÚÊԹܡ£

£¨4£©µ±ÆøÌå³äÂú¢ÚÊԹܺó£¬ÓûʹÉÕ±­ÖеÄË®½øÈë¢ÚÊÔ¹ÜÓ¦ÈçºÎ²Ù×÷             ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸