KMnO4ÈÜÒºÓëH2C2O4ÈÜÒº¿É·¢ÉúÈçÏ·´Ó¦£º
2KMnO4+5H2C2O4+3H2SO4=K2SO4+2MnSO4+10CO2¡ü+8H2O
£¨1£©¸Ã·´Ó¦ËÙÂÊ¿ªÊ¼Ê®·Ö»ºÂý£¬Ò»¶Îʱ¼äºóͻȻ¼Ó¿ì£¬ÕâÊÇÒòΪ________(Ìѧʽ)¶Ô¸Ã·´Ó¦¾ßÓд߻¯×÷Óá£
£¨2£©¾Ý´ËÔ­Àí£¬¿ÉÒÔÀûÓÃKMnO4ÈÜÒºÀ´²â¶¨H2C2O4ÈÜÒºµÄŨ¶È£¬¾ßÌå×ö·¨ÈçÏ£º
¢Ù׼ȷÅäÖÆ0.10mol/LµÄKMnO4ÈÜÒº
¢Ú½«KMnO4ÈÜҺʢ·ÅÔÚ________µÎ¶¨¹ÜÖУ¨Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©
¢Û׼ȷÁ¿È¡25.00mL H2C2O4ÈÜÒºÓÚ׶ÐÎÆ¿ÖÐ
¢Ü½øÐеζ¨
µÎ¶¨ÖÕµãÓÐʲôÏÖÏó________,ÊÇ·ñÐèҪָʾ¼Á________(Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)
£¨3£©ÔÚÏÂÁвÙ×÷ÖУ¬»áʹ²â¶¨µÄH2C2O4ÈÜҺŨ¶ÈÆ«´óµÄÊÇ________¡£
¢ÙÊ¢×°KMnO4ÈÜÒºµÄµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºóδÓÃKMnO4ÈÜÒºÈóÏ´
¢Ú׶ÐÎÆ¿ÖÐÊ¢ÓÐÉÙÁ¿ÕôÁóË®£¬ÔÙ¼Ó´ý²âÒº
¢ÛÊ¢×°H2C2O4ÈÜÒºµÄµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»ºó£¬Î´ÓÃH2C2O4ÈÜÒºÈóÏ´
¢ÜµÎ¶¨ºó¹Û²ìµÎ¶¨¹Ü¶ÁÊýʱ£¬ÊÓÏ߸ßÓڿ̶ÈÏß
£¨4£©µÎ¶¨Ê±ËùµÃµÄʵÑéÊý¾ÝÈçÏ£¬ÊÔ¼ÆËãËù²âH2C2O4ÈÜÒºµÄŨ¶ÈΪ________mol/L
ʵÑé´ÎÊý±àºÅ´ý²âÒºÌå»ýmLµÎÈëµÄ±ê×¼ÒºÌå»ý£¨mL£©
125.0028.95
225.0025.05
325.0024.95

MnSO4    Ëáʽ    ÈÜÒºÓÉÎÞÉ«±äΪ×ϺìÉ«    ·ñ    ¢Ù    0.25
£¨1£©¸ù¾Ý·½³Ìʽ¿ÉÖª£¬ÐÂÉú³ÉµÄÎïÖÊÊÇÁòËáÃ̺ÍCO2£¬µ«CO2ÊÇÆøÌ壬ËùÒÔÆð´ß»¯×÷ÓõÄÊÇÁòËáÃÌ¡£
£¨2£©¸ßÃÌËá¼ØÈÜÒº¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܸ¯Ê´Ï𽺣¬ËùÒÔÓ¦¸ÃÓÃËáʽµÎ¶¨¹ÜÊ¢·Å¸ßÃÌËá¼ØÈÜÒº¡£ÓÉÓÚËáÐÔ¸ßÃÌËá¼ØÈÜÒºÊÇÏÔ×ϺìÉ«µÄ£¬ËùÒÔ²»ÐèÒªÁí¼Óָʾ¼ÁÖÕµãʱµÄÏÖÏóÊÇÈÜÒºÓÉÎÞÉ«±äΪ×ϺìÉ«¡£
£¨3£©¢ÙÏ൱ÓÚÏ¡ÊÍÁ˸ßÃÌËá¼ØÈÜÒºµÄŨ¶È£¬ËùÒÔÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýÆ«´ó£¬²â¶¨½á¹ûÆ«¸ß¡£×¶ÐÎÆ¿²»ÄÜÓñê×¼ÒºÈóÏ´£¬ËùÒÔ¢Ú²»Ó°Ïì¡£¢ÛÒ²ÊÇÏ൱ÓÚÏ¡ÊͲÝËáµÄŨ¶È£¬ËùÒÔÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýÆ«ÉÙ£¬²â¶¨½á¹ûÆ«µÍ¡£µÎ¶¨ºó¹Û²ìµÎ¶¨¹Ü¶ÁÊýʱ£¬ÊÓÏ߸ßÓڿ̶ÈÏߣ¬ËµÃ÷¶ÁÊýƫС£¬ËùÒԲⶨ½á¹ûÆ«µÍ£¬´ð°¸Ñ¡¢Ù¡£
£¨4£©¸ù¾ÝʵÑéÊý¾Ý¿ÉÖª£¬µÚÒ»´ÎʵÑéÎó²îÌ«´ó£¬²»ÄÜÓá£ÊµÑé¸ù¾ÝºóÁ½´ÎµÄÊý¾Ý¿ÉÖª£¬ÏûºÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ýµÄƽ¾ùÖµÊÇ25.00ml£¬ËùÒÔ¸ù¾Ý·½³Ìʽ¿ÉÖª£¬²ÝËáÈÜÒºµÄŨ¶ÈÊÇ0.10mol/L¡Á2.5£½0.25mol/L¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º½­Î÷Ê¡°×ðØÖÞÖÐѧ2012½ì¸ßÈýÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£º013

ÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ»òÀë×Ó·½³ÌʽÖУ¬ÕýÈ·µÄÊÇ£º

[¡¡¡¡]

A£®

¼×ÍéµÄ±ê׼ȼÉÕÈȦ¤HΪ£­890.3 kJ¡¤mol£­1£¬Ôò¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ£ºCH4(g)£«2O2(g)£½CO2(g)£«2H2O(g)£»¦¤H£½£­890.3 kJ¡¤mol£­1

B£®

500¡æ¡¢30 MPaÏ£¬½«0.5 mol¡¡N2(g)ºÍ1.5 mol¡¡H2(g)ÖÃÓÚÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦Éú³ÉNH3(g)£¬·ÅÈÈ19.3 kJ£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ£ºN2(g)£«3H2(g)2NH3(g)£»¦¤H£½£­38.6 kJ¡¤mol£­1

C£®

ÏòÃ÷·¯ÈÜÒºÖмÓÈë¹ýÁ¿µÄÇâÑõ»¯±µÈÜÒº£ºAl3+£«2SO£«2Ba2+£«4OH£­£½2BaSO4¡ý£«AlO£«2H2O

D£®

ÓÃŨÑÎËáËữµÄKMnO4ÈÜÒºÓëH2O2·´Ó¦£¬Ö¤Ã÷H2O2¾ßÓл¹Ô­ÐÔ£º

2MnO£«6H+£«5H2O2£½2Mn2+£«5O2¡ü£«8H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ºÉ½¶«Ê¡ÇàÖÝÒ»ÖÐ2012½ì¸ßÈý»¯Ñ§Ò»ÂÖ¸ú×ÙѵÁ·£ºµÚ11Õ¡¡µÚ4½²¡¡È©¡¡ôÈËá¡¡õ¥(³¿Æ°æ) ÌâÐÍ£º013

´ÓÌð³ÈµÄ·¼ÏãÓÍÖпɷÖÀëµÃµ½ÈçϽṹµÄ»¯ºÏÎ

ÏÖÓÐÊÔ¼Á£º

¢ÙËáÐÔKMnO4ÈÜÒº£»

¢ÚH2/Ni£»

¢ÛAg(NH3)2OH£»

¢ÜÐÂÖÆCu(OH)2Ðü×ÇÒº£¬

ÄÜÓë¸Ã»¯ºÏÎïÖÐËùÓйÙÄÜÍŶ¼·¢Éú·´Ó¦µÄÊÔ¼ÁÓÐ

[¡¡¡¡]

A£®

¢Ù¢Ú

B£®

¢Ú¢Û

C£®

¢Û¢Ü

D£®

¢Ù¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ººÓ±±Ê¡Õý¶¨ÖÐѧ2012½ì¸ßÈýµÚ¶þ´Î×ۺϿ¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£º013

ÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ»òÀë×Ó·½³ÌʽÖУ¬ÕýÈ·µÄÊÇ

[¡¡¡¡]

A£®

¼×ÍéµÄȼÉÕÈÈΪ890.3kJ¡¤mol£­1£¬Ôò¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ£º

CH4(g)£«2O2(g)£½CO2(g)£«2H2O(g)£»¦¤H£½£­890.3kJ¡¤mol£­1

B£®

500¡æ¡¢30MPaÏ£¬½«0.5 mol¡¡N2ºÍ1.5 mol¡¡H2ÖÃÓÚÃܱյÄÈÝÆ÷Öгä·Ö·´Ó¦Éú³ÉNH3(g)£¬·ÅÈÈ19.3 kJ£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ£º¦¤H£½£­38.6 kJ¡¤mol£­1

C£®

ÏòÃ÷·¯ÈÜÒºÖмÓÈë¹ýÁ¿µÄÇâÑõ»¯±µÈÜÒº£ºAl3+£«2SO42£­£«2Ba2+£«4OH£­£½2BaSO4¡üAlO2£­£«2H2O

D£®

ÓÃŨÑÎËáËữµÄKMnO4ÈÜÒºÓëH2O2·´Ó¦£¬Ö¤Ã÷H2O2¾ßÓл¹Ô­ÐÔ£º2MnO4£­£«6H+£«5H2O2£½2Mn2+£«5O2¡ü£«8H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º½­Î÷Ê¡ÆÚÖÐÌâ ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ»òÀë×Ó·½³ÌʽÖУ¬ÕýÈ·µÄÊÇ
[     ]
A£®¼×ÍéµÄ±ê׼ȼÉÕÈÈ¡÷HΪ-890.3kJ/mol£¬Ôò¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ£ºCH4(g)+2O2(g)=CO2(g)+2H2O(g)   ¡÷H=£­890.3kJ/mol
B£®500¡æ¡¢30MPaÏ£¬½«0£®5molN2£¨g£©ºÍ1£®5mol H2£¨g£©ÖÃÓÚÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦Éú³ÉNH3£¨g£©£¬·ÅÈÈ19£®3kJ£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ£ºN2£¨g£©+3H2£¨g£©2NH3£¨g£©   ¡÷H= -38£®6kJ/mol
C. ÏòÃ÷·¯ÈÜÒºÖмÓÈë¹ýÁ¿µÄÇâÑõ»¯±µÈÜÒº£º
 
D. ÓÃŨÑÎËáËữµÄKMnO4ÈÜÒºÓëH2O2·´Ó¦£¬Ö¤Ã÷H2O2¾ßÓл¹Ô­ÐÔ£º2MnO42- + 6H+ 5H2O2 £½ 2Mn2+ + 5O2¡ü + 8H2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

´ÓÌð³ÈµÄ·¼ÏãÓÍÖпɷÖÀëµÃµ½ÈçÓҽṹµÄ»¯ºÏÎï

ÏÖÔÚÊÔ¼Á£º¢ÙËáÐÔKMnO4ÈÜÒº£»¢ÚH2/Ni£»¢ÛAg(NH3)2OHÈÜÒº£»¢ÜÐÂÖÆCu(OH)2Ðü×ÇÒº£¬ÄÜÓë¸Ã»¯ºÏÎïÖÐËùÓйÙÄÜÍŶ¼·¢Éú·´Ó¦µÄÊÔ¼ÁÓÐ

A£®¢Ù¢Ú¡¡¡¡¡¡B£®¢Ú¢Û¡¡¡¡¡¡C£®¢Û¢Ü¡¡¡¡¡¡¡¡¡¡D£®¢Ù¢Ü

¸ß¿¼×ÊÔ´Íø( www.ks5u.com)£¬Öйú×î´óµÄ¸ß¿¼ÍøÕ¾£¬ÄúÉí±ßµÄ¸ß¿¼×¨¼Ò¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸