ʱÖкÍÈÈΪ57.3 kJ¡¤mol-1£¬ÔòÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÊéдÕýÈ·µÄÊÇ( )
¢ÙC8H18(1)+
O2(g)====8CO2(g)+9H2O(1)£»¡÷H=+5 518.kJ¡¤mol-1
¢ÚC8H18(1)+
O2(g)====8CO2(g)+9H2O(1)£»¡÷H=-5 518kJ¡¤mol-1
¢ÛH++OH-====H2O£»¡÷H=-57.3 kJ¡¤mol-1
¢ÜNaOH(aq)+
H2SO4(aq)====
Na2SO4(aq)+H2O(1)£»¡÷H=+57.3kJ¡¤mol-1
A.¢Ù¢Û B.¢Ú¢Û C.¢Ú¢Ü D.Ö»ÓТÚ
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
| A¡¢Í¬ÎÂͬѹÏ£¬H2£¨g£©+Cl2£¨g£©¨T2HCl£¨g£©ÔÚ¹âÕպ͵ãȼÌõ¼þϵġ÷HÏàͬ | B¡¢Ç¦Ðîµç³Ø·ÅµçʱµÄ¸º¼«ºÍ³äµçʱµÄÒõ¼«¾ù·¢Éú»¹Ô·´Ó¦ | C¡¢ÒÑÖª£ºH2£¨g£©+I2£¨g£©?2HI£¨g£©£»¡÷H=-9.48 kJ/mol£¬Èô½«254g I2£¨g£©ºÍ2gH2£¨g£©³ä·Ö·´Ó¦¿É·Å³ö9.48 kJµÄÈÈÁ¿ | D¡¢ÒÑÖªÔÚ101 kPaʱ£¬2 g̼ȼÉÕÉú³ÉCO·Å³öÈÈÁ¿ÎªQ kJ£¬Ôò̼µÄȼÉÕÈÈΪ6Q kJ?mol-1 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(1)ëÂ(N2H4)ÓÖ³ÆÁª°±£¬ÊÇÒ»ÖÖ¿ÉȼÐÔµÄÒºÌ壬¿ÉÓÃ×÷»ð¼ýȼÁÏ¡£ÒÑÖªÔÚ101 kPaʱ£¬32.0 g N2H4ÔÚÑõÆøÖÐÍêȫȼÉÕÉú³ÉµªÆø£¬·Å³öÈÈÁ¿624 kJ(25¡æÊ±)£¬N2H4ÍêȫȼÉÕ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ____________________________________________________________________¡£
(2)ëÂ-¿ÕÆøÈ¼ÁÏµç³ØÊÇÒ»ÖÖ¼îÐÔȼÁÏµç³Ø£¬µç½âÖÊÊÇ20%¡ª30%µÄKOHÈÜÒº¡£Ð´³öëÂ-¿ÕÆøÈ¼ÁÏµç³Ø·ÅµçʱÕý¡¢¸º¼«µÄµç¼«·´Ó¦Ê½¡£
Õý¼«£º________________________________£¬
¸º¼«£º________________________________
(3)ͼ2-2-5ÊÇÒ»¸öµç»¯Ñ§¹ý³ÌʾÒâͼ¡£
![]()
ͼ2-2-5
¢ÙпƬÉÏ·¢ÉúµÄµç¼«·´Ó¦ÊÇ________________________________________________¡£
¢Ú¼ÙÉèʹÓÃëÂ?¿ÕÆøÈ¼ÁÏµç³Ø×÷Ϊ±¾¹ý³ÌÖеĵçÔ´¡¢ÍƬµÄÖÊÁ¿±ä»¯128 g£¬ÔòëÂ-¿ÕÆøÈ¼ÁÏµç³ØÀíÂÛÉÏÏûºÄ±ê±ê×¼×´¿öÏÂµÄ¿ÕÆø___________L(¼ÙÉè¿ÕÆøÖÐÑõÆøÌå»ýº¬Á¿Îª20%)
(4)´«Í³ÖƱ¸ëµķ½·¨£¬ÊÇÒÔNaClOÑõ»¯NH3£¬ÖƵÃëµÄÏ¡ÈÜÒº¡£¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÕã½ÁÙº£Êа×ÔÆ¸ßÖиßÒ»ÏÂѧÆÚÆÚĩģÄ⻯ѧÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ
£¨6·Ö£©¢ñ¡¢ÒÑÖªÔÚ101 kPaʱ£¬CH4ÍêȫȼÉÕÉú³É1molҺ̬ˮ£¬·Å³öµÄÈÈÁ¿ÎªQkJ£¬ÔòCH4ÍêȫȼÉÕ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ£º ¡£
¢ò¡¢ÔÚÍÆ¬¡¢Ð¿Æ¬ºÍ400 mLÏ¡ÁòËá×é³ÉµÄÔµç³ØÖУ¬Èôµç·ÖÐͨ¹ý0.2 molµç×Ó£¬H2SO4Ç¡ºÃ·´Ó¦Íê±Ï¡£ÊÔ¼ÆË㣺
£¨1£©Éú³ÉÆøÌåµÄÌå»ý£¨ÔÚ±ê×¼×´¿öÏ£©£»
£¨2£©Ô400 mLÏ¡ÁòËáµÄÎïÖʵÄÁ¿Å¨¶È£¨²»¿¼ÂÇÈÜÒºµÄÌå»ý±ä»¯£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì°²»ÕÊ¡¸ß¶þ3ÔÂÁª¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÒÑÖª£º¢Ù101 kPaʱ£¬2C(s)£«O2(g)===2CO(g)¡¡¦¤H£½£221 kJ/mol
¢ÚÏ¡ÈÜÒºÖУ¬H£«(aq)£«OH£(aq)===H2O(l)¡¡¦¤H£½£57.3 kJ/molÏÂÁнáÂÛÕýÈ·µÄÊÇ(¡¡¡¡)
A£®Ì¼µÄȼÉÕÈÈ´óÓÚ110.5 kJ/mol
B£®¢ÙµÄ·´Ó¦ÈÈΪ 221 kJ/mol
C£®Ï¡ÁòËáÓëÏ¡NaOHÈÜÒº·´Ó¦µÄÖкÍÈÈΪ£57.3 kJ/mol
D£®Å¨ÁòËáÓëÏ¡NaOHÈÜÒº·´Ó¦Éú³É1 molË®£¬·Å³ö57.3 kJÈÈÁ¿
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì¹óÖÝÊ¡¸ß¶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£º¼ÆËãÌâ
£¨11·Ö£©CO¡¢CH4¾ùΪ³£¼ûµÄ¿ÉȼÐÔÆøÌå¡£
(1)µÈÌå»ýµÄCOºÍCH4ÔÚÏàͬÌõ¼þÏ·ֱðÍêȫȼÉÕ£¬×ªÒƵĵç×ÓÊýÖ®±ÈÊÇ ¡£
(2)ÒÑÖªÔÚ101 kPaʱ£¬COµÄȼÉÕÈÈΪ283 kJ/mol¡£ÏàͬÌõ¼þÏ£¬Èô2 molCH4ÍêȫȼÉÕÉú³ÉҺ̬ˮ£¬Ëù·Å³öµÄÈÈÁ¿Îª1 mol COÍêȫȼÉշųöÈÈÁ¿µÄ6.30 ±¶£¬CH4ÍêȫȼÉÕ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ£º ¡£
(3)120¡æ¡¢101 kPaÏ£¬a mLÓÉCO¡¢CH4×é³ÉµÄ»ìºÏÆøÌåÔÚbmL O2ÖÐÍêȫȼÉպ󣬻ָ´µ½ÔζȺÍѹǿ¡£
¢Ù Èô»ìºÏÆøÌåÓëO2Ç¡ºÃÍêÈ«·´Ó¦£¬²úÉúb mL CO2£¬Ôò»ìºÏÆøÌåÖÐCH4µÄÌå»ý·ÖÊýΪ
£¨±£Áô2λСÊý£©¡£
¢Ú ÈôȼÉÕºóÆøÌåÌå»ýËõСÁËa/4 mL £¬ÔòaÓëb¹ØÏµµÄÊýѧ±íʾʽÊÇ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com