ijÐËȤС×éΪÑéÖ¤ÈÕ³£Éú»îÓõĻð²ñÍ·ÉϺ¬ÓÐKClO3¡¢MnO2¡¢S£¬Éè¼ÆÁËÒÔÏÂʵÑéÁ÷³Ìͼ£º

Çë»Ø´ðÒÔÏÂÎÊÌ⣺

(1)ΪÑéÖ¤ÆøÌåA£¬°´ÈçͼËùʾװÖýøÐÐʵÑ飺ÈôÄܹ۲쵽______µÄÏÖÏ󣬼´¿ÉÖ¤Ã÷»ð²ñÍ·ÉϺ¬ÓÐSÔªËØ¡£

(2)²½Öè¢ÚµÄʵÑé²Ù×÷×°ÖÃÈçͼËùʾ£¬¸Ã²Ù×÷µÄÃû³ÆÊÇ________£¬Æä¹¤×÷Ô­ÀíÊÇ____________________________¡£

(3)ÒªÖ¤Ã÷»ð²ñÍ·Öк¬ÓÐClÔªËØµÄºóÐøÊµÑé²½ÖèÊÇ

_____________________________________________¡£

(4)ÓÐѧÉúÌá³ö¼ìÑé»ð²ñÍ·ÉÏKClO3µÄÁíÒ»Ì×ʵÑé·½°¸£º

ÓйصÄÀë×Ó·½³ÌʽΪ________£¬

ÓÐÈËÌá³öÉÏÊö·½·¨ÖгöÏÖ°×É«³Áµí²¢²»Äܳä·Ö˵Ã÷»ð²ñÍ·ÉÏKClO3µÄ´æÔÚ£¬ÆäÀíÓÉÊÇ______________________________________________¡£

(5)¸ÃС×é²Â²âÂËÔüD¶ÔË«ÑõË®·Ö½âÖÆÑõÆøµÄËÙÂÊ»á²úÉúÒ»¶¨µÄÓ°Ï죬Éè¼Æ²¢½øÐÐÁËÒÔÏÂ5´ÎʵÑé¡£

ʵÑé´ÎÊý

H2O2ÈÜÒºÖÊÁ¿·ÖÊý/%

H2O2ÈÜÒºÓÃÁ¿/mL

ÎïÖÊDÓÃÁ¿/g

·´Ó¦Î¶È/¡æ

ÊÕ¼¯ÆøÌåÌå»ý/mL

ËùÐèʱ¼ä/s

¢Ù

30

5

0

85

2

3.8

¢Ú

15

2

0.1

20

2

2.8

¢Û

15

2

0.2

20

2

2.2

¢Ü

5

2

0.1

20

2

7.4

¢Ý

30

5

0

55

2

10.5

ÓÉÉϱí¿ÉÖª£¬ÊµÑé¢ÙºÍ¢ÝÄÜÖ¤Ã÷ζÈÔ½¸ß£¬»¯Ñ§·´Ó¦ËÙÂÊÔ½¿ì£¬ÊµÑé________ºÍ________ÄÜÖ¤Ã÷ÎïÖÊDµÄÓÃÁ¿Ô½´ó£¬·´Ó¦ËÙÂÊÔ½¿ì¡£

(6)д³ö²½Öè¢ÙÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_________________________________________________¡£


¡¾½âÎö¡¿¡¡(1)²úÉúµÄÆøÌåAΪ¶þÑõ»¯Áò£¬¶þÑõ»¯ÁòÄܱ»¸ßÃÌËá¼ØËáÐÔÈÜÒºÑõ»¯¡£(2)¹Û²ìͼʾ£¬ÆäÓÐÈý²¿·Ö×é³É£¬×ó±ßΪÎüÂËÆ¿£¬ÖмäΪ°²È«Æ¿£¬ÓÒ²àÊÇ³éÆø±Ã£¬ËùÒÔ¸Ã×°ÖÃΪ¼õѹ¹ýÂË(»ò³éÂË)×°Öá£(3)ȼÉÕºóÂÈËá¼Ø×ª»¯ÎªÂÈ»¯¼Ø£¬ÂËÒºÖдæÔÚÂÈÀë×Ó£¬¿ÉÒÔÓÃÏõËáÒøÈÜÒº¼ìÑéÂÈÀë×Ó¡£(4)ClO¾ßÓÐÑõ»¯ÐÔ£¬¶øNO¾ßÓл¹Ô­ÐÔ£¬¶þÕß·¢ÉúÑõ»¯»¹Ô­·´Ó¦¡£(5)ʵÑé¢Ú¡¢¢ÛζÈÏàͬ£¬ÆäÖÐÎïÖÊD(¶þÑõ»¯ÃÌ)µÄÓÃÁ¿²»Í¬£¬ÊµÑé¢Û¼ÓÈëµÄD¶àÓÚ¢Ú£¬¶øÇÒÊÕ¼¯ÏàͬÌå»ýµÄÆøÌåʱ£¬ÊµÑé¢ÛËùÐèµÄʱ¼ä¶Ì¡£(6)²½Öè¢Ù·¢ÉúÁËÈý¸ö»¯Ñ§·´Ó¦£¬¼´ÂÈËá¼ØµÄ·Ö½â¡¢ÁòµÄȼÉÕ¡¢»ð²ñ¹£ÖÐ̼µÄȼÉյȡ£

¡¾´ð°¸¡¿¡¡(1)KMnO4ËáÐÔÈÜÒºÍÊÉ«

(2)¼õѹ¹ýÂË(»ò³éÂË)¡¡µ±´ò¿ª×ÔÀ´Ë®Áúͷʱ£¬×°ÖÃÄÚ²¿µÄ¿ÕÆøËæ×ÔÀ´Ë®±»´ø×ߣ¬µ¼ÖÂ×°ÖÃÄÚ²¿Ñ¹Ç¿¼õС£¬Ê¹¹ýÂËËٶȼӿ죬µÃµ½½Ï¸ÉÔïµÄ¹ÌÌåÎïÖÊ

(3)È¡ÂËÒºC£¬¼ÓÈëÏ¡ÏõËáºÍAgNO3ÈÜÒº£¬Èô¹Û²ìµ½Óа×É«³Áµí²úÉú£¬¼´¿ÉÖ¤Ã÷»ð²ñÍ·Öк¬ÓÐÂÈÔªËØ

(4)ClO£«3NO£«Ag£«AgCl¡ý£«3NO

AgNO2ÓëAgCl¾ùΪ²»ÈÜÓÚË®µÄ°×É«³Áµí

(5)¢Ú¡¡¢Û¡¡(6)2KClO32KCl£«3O2¡ü£¬S£«O2SO2£¬C£«O2CO2


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÓйØÎïÖʵÄÐÔÖʺ͸ÃÐÔÖʵÄÓ¦ÓþùÕýÈ·µÄÊÇ

A.³£ÎÂÏÂŨÁòËáÄÜÊÇÂÁ·¢Éú¶Û»¯£¬¿ÉÔÚ³£ÎÂÏÂÓÃÂÁÖÆÖü²ØÖüÔËŨÁòËá

B.¶þÑõ»¯¹è²»ÓëÈκÎËá·´Ó¦£¬¿ÉÓÃÊ¯Ó¢ÖÆÔìÄÍËáÈÝÆ÷

C.¶þÑõ»¯ÂȾßÓл¹Ô­ÐÔ£¬¿ÉÓÃÓÚ×ÔÀ´Ë®µÄɱ¾úÏû¶¾

D.Í­µÄ½ðÊô»îÆÃÐÔ±ÈÌúµÄ²î£¬¿ÉÔÚº£ÂÖÍâ¿ÇÉÏ×°Èô¸ÉÍ­¿éÒÔ¼õ»ºÆä¸¯Ê´

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÑÌÆøÖÐNOxÊÇNOºÍNO2µÄ»ìºÏÎï(²»º¬N2O4)¡£

(1)¸ù¾Ý·ÏÆøÅŷűê×¼£¬1 m3ÑÌÆø×î¸ßÔÊÐíº¬400 mg NOx¡£ÈôNOxÖÐNOÖÊÁ¿·ÖÊýΪ0.85£¬Ôò1 m3ÑÌÆøÖÐ×î¸ßÔÊÐíº¬NO___________________L(±ê×¼×´¿ö£¬±£Áô2λСÊý)¡£

(2)¹¤ÒµÉÏͨ³£ÓÃÈÜÖÊÖÊÁ¿·ÖÊýΪ0.150µÄNa2CO3Ë®ÈÜÒº(ÃܶÈ1.16 g¡¤mL-1)×÷ΪNOxÎüÊÕ¼Á£¬Ôò̼ËáÄÆÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈΪ__________________________mol¡¤L-1(±£Áô2λСÊý)¡£

(3)ÒÑÖª£ºNO+NO2+Na2CO32NaNO2+CO2¢Ù

2NO2+Na2CO3NaNO2+NaNO3+CO2¢Ú

1 m3º¬2 000 mg NOxµÄÑÌÆøÓÃÖÊÁ¿·ÖÊýΪ0.150µÄ̼ËáÄÆÈÜÒºÎüÊÕ¡£ÈôÎüÊÕÂÊΪ80%£¬ÎüÊÕºóµÄÑÌÆø_______________Åŷűê×¼(Ìî¡°·ûºÏ¡±»ò¡°²»·ûºÏ¡±)£¬ÀíÓÉ£º_________________¡£

(4)¼ÓÈëÏõËá¿É¸Ä±äÑÌÆøÖÐNOºÍNO2µÄ±È£¬·´Ó¦Îª£ºNO+2HNO33NO2+H2O

µ±ÑÌÆøÖÐn(NO)¡Ãn(NO2)=2¡Ã3ʱ£¬ÎüÊÕÂÊ×î¸ß¡£

1 m3ÑÌÆøº¬2 000 mg NOx£¬ÆäÖÐn(NO)¡Ãn(NO2)=9¡Ã1¡£

¼ÆË㣺(¢¡)ΪÁË´ïµ½×î¸ßÎüÊÕÂÊ£¬1 m3ÑÌÆøÐèÓÃÏõËáµÄÎïÖʵÄÁ¿(±£Áô3λСÊý)¡£

(¢¢)1 m3ÑÌÆø´ïµ½×î¸ßÎüÊÕÂÊ90%ʱ£¬ÎüÊÕºóÉú³ÉNaNO2µÄÖÊÁ¿(¼ÙÉèÉÏÊöÎüÊÕ·´Ó¦ÖУ¬·´Ó¦¢Ù±È·´Ó¦¢ÚѸËÙ¡£¼ÆËã½á¹û±£Áô1λСÊý)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁйØÓÚÒÇÆ÷¡°0¡±¿Ì¶ÈλÖÃÐðÊöÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ÔÚÁ¿Í²µÄÉ϶Ë

B£®Ôڵζ¨¹ÜÉ϶Ë

C£®ÔÚÍÐÅÌÌìÆ½¿Ì¶È³ßµÄÕýÖÐ

D£®ÔÚÍÐÅÌÌìÆ½¿Ì¶È³ßµÄÓÒ±ß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÑÎËáµÎ¶¨Î´ÖªÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜҺʱ£¬ÏÂÁвÙ×÷Öв»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºó£¬Ö±½Ó¼ÓÈëÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÑÎËá

B£®×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºó£¬Ö±½Ó¼ÓÈëÒ»¶¨Ìå»ýµÄδ֪ÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº

C£®µÎ¶¨Ç°£¬ÒªÅųýµÎ¶¨¹Ü¼â×ì´¦µÄÆøÅÝ

D£®¶ÁÊýʱ£¬ÊÓÏßÓëµÎ¶¨¹ÜÄÚÒºÌå°¼ÒºÃæ×îµÍ´¦±£³ÖÒ»ÖÂ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚʯ»ÒÒ¤ÖÐÉÕÖÆÉúʯ»Ò£¬1 mol CaCO3ÍêÈ«·Ö½âËùÐèÒªµÄÄÜÁ¿£¬¿ÉȼÉÕ0.453 mol̼À´Ìṩ¡£Éè¿ÕÆøÖÐO2Ìå»ý·ÖÊýΪ0.21£¬N2Ϊ0.79£¬Ôòʯ»ÒÒ¤²úÉúµÄÆøÌåÖÐCO2µÄÌå»ý·ÖÊý¿ÉÄÜÊÇ£¨    £©

A.0.43        B.0.46      C.0.49          D..0.52

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


.Ŀǰ£¬¹ØÓÚ¶þÑõ»¯Ì¼ÊÇ·ñΪ´óÆøÎÛȾÎïÓв»Í¬µÄ¹Ûµã¡£ÈÏΪ¡°¶þÑõ»¯Ì¼²»ÊÇ´óÆøÎÛȾÎµÄÀíÓÉÊÇ(   )

¢Ù¶þÑõ»¯Ì¼ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ

¢Ú¶þÑõ»¯Ì¼ÊÇÖ²Îï¹âºÏ×÷ÓõıØÐèÔ­ÁÏ

¢Û¶þÑõ»¯Ì¼ÊÇÎÞÉ«¡¢ÎÞζ¡¢ÎÞ¶¾µÄÆøÌå

¢Ü³ý¶þÑõ»¯Ì¼Í⣬¼×Íé¡¢Ò»Ñõ»¯¶þµªÒ²ÊÇÎÂÊÒÆøÌå

A.¢Ù¢Ú              B.¢Ú¢Û               C.¢Û¢Ü             D.¢Ù¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


U¡¢V¡¢W¡¢X¡¢Y¡¢ZÊÇÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄÁùÖÖ³£¼ûÔªËØ¡£YµÄµ¥ÖÊÔÚW2ÖÐȼÉյIJúÎï¿ÉʹƷºìÈÜÒºÍÊÉ«¡£ZºÍWÔªËØÐγɵϝºÏÎïZ3W4¾ßÓдÅÐÔ¡£UµÄµ¥ÖÊÔÚW2ÖÐȼÉÕ¿ÉÉú³ÉUWºÍUW2Á½ÖÖÆøÌå¡£XµÄµ¥ÖÊÊÇÒ»ÖÖ½ðÊô£¬¸Ã½ðÊôÔÚUW2ÖоçÁÒȼÉÕÉú³ÉºÚ¡¢°×Á½ÖÖ¹ÌÌå¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©VµÄµ¥ÖÊ·Ö×ӵĽṹʽΪ                              £»XWµÄµç×ÓʽΪ                             £»ZÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ                                ¡£

£¨2£©UÔªËØÐγɵÄÍ¬ËØÒìÐÎÌåµÄ¾§ÌåÀàÐÍ¿ÉÄÜÊÇ£¨ÌîÐòºÅ£©                                  ¡£

¢ÙÔ­×Ó¾§Ìå    ¢ÚÀë×Ó¾§Ìå    ¢Û·Ö×Ó¾§Ìå    ¢Ü½ðÊô¾§Ìå

£¨3£©U¡¢V¡¢WÐγɵÄ10µç×ÓÇ⻯ÎïÖУ¬U¡¢VµÄÇ⻯Îï·Ðµã½ÏµÍµÄÊÇ£¨Ð´»¯Ñ§Ê½£©
                                   £»V¡¢WµÄÇ⻯Îï·Ö×Ó½áºÏH+ÄÜÁ¦½ÏÇ¿µÄÊÇ£¨Ð´»¯Ñ§Ê½£©                                £»ÓÃÒ»¸öÀë×Ó·½³Ìʽ¼ÓÒÔÖ¤Ã÷                                                      ¡£

£¨4£©YW2ÆøÌåͨÈëBaCl2ºÍHNO3µÄ»ìºÏÈÜÒº£¬Éú³É°×É«³ÁµíºÍÎÞÉ«ÆøÌåVW£¬Óйط´Ó¦µÄÀë×Ó·½³ÌʽΪ                                                  £¬ÓÉ´Ë¿ÉÖªVWºÍYW2»¹Ô­ÐÔ½ÏÇ¿µÄÊÇ£¨Ð´»¯Ñ§Ê½£©                                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


.ÒÑÖª£º4NH3+5O24NO+6H2O    4NO+3O2+2H2O¡ú4HNO3

    Éè¿ÕÆøÖÐÑõÆøµÄÌå»ý·ÖÊýΪ0.20£¬µªÆøÌå»ý·ÖÊýΪ0.80£¬ÇëÍê³ÉÏÂÁÐÌî¿Õ¼°¼ÆËã

(1)a mol NOÍêȫת»¯ÎªHNO3ÐèÒªÑõÆø_______ mol

(2)ΪʹNH3Ç¡ºÃÍêÈ«Ñõ»¯ÎªÒ»Ñõ»¯µª£¬°±£­¿ÕÆø»ìºÏÎïÖа±µÄÌå»ý·ÖÊýΪ________(±£Áô2λСÊý)¡£

(3)20.0 moLµÄNH3ÓÃ¿ÕÆøÑõ»¯£¬²úÉú»ìºÏÎïµÄ×é³ÉΪ£ºNO 18.0 mol¡¢O2 12.0 mol¡¢N2 150.0 molºÍÒ»¶¨Á¿µÄÏõËᣬÒÔ¼°ÆäËü³É·Ö¡£(¸ßÎÂÏÂNOºÍO2²»·´Ó¦)¼ÆË㰱ת»¯ÎªNOºÍHNO3µÄת»¯ÂÊ¡£

(4)20.0 moL µÄNH3ºÍÒ»¶¨Á¿¿ÕÆø³ä·Ö·´Ó¦ºó£¬ÔÙת»¯ÎªHNO3

   ¢ÙÔÚÏÂͼÖл­³öHNO3µÄÎïÖʵÄÁ¿n(A)ºÍ¿ÕÆøµÄÎïÖʵÄÁ¿n(B)¹ØÏµµÄÀíÂÛÇúÏß¡£

       

   ¢Úд³öµ±125¡Ün(B)¡Ü200ʱ£¬n(A)ºÍn(B)µÄ¹ØÏµÊ½______________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸