ÌâÄ¿ÄÚÈÝ

»ð¼ýµÄȼÁÏ·ÖΪ¹ÌÌåȼÁϺÍÒºÌåȼÁϵȣ¬ÒÔÂÁ·ÛÓë¸ßÂÈËá淋ĻìºÏÎïΪ¹ÌÌåȼÁÏ£¬ÆäÖиßÂÈËá淋ķ´Ó¦Îª£º2NH4ClO4N2¡ü+Cl2¡ü+2O2¡ü+4H2O¡£ÏÂÁÐÓйØÐðÊö²»ÕýÈ·µÄÊÇ£¨    £©

A.¸Ã·´Ó¦ÊôÓڷֽⷴӦ¡¢Ñõ»¯»¹Ô­·´Ó¦

B.ÉÏÊö·´Ó¦Ë²¼äÄܲúÉú´óÁ¿¸ßÎÂÆøÌ壬ÕâÊÇÍÆ¶¯·É´¬·ÉÐеÄÖ÷ÒªÒòËØ

C.ÂÁ·ÛµÄ×÷ÓÃÊǵãȼʱÑõ»¯·ÅÈÈÒý·¢¸ßÂÈËáï§·´Ó¦

D.ÔÚ·´Ó¦ÖÐNH4ClO4½öÆðµ½Ñõ»¯¼Á×÷ÓÃ

˼·½âÎö£ºNH4ClO4¼ÈÊÇÑõ»¯¼ÁÓÖÊÇ»¹Ô­¼Á¡£

´ð°¸£ºD

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(¢ñ) ÒÑÖªÔÚ25 ¡æµÄË®ÈÜÒºÖУ¬AgX¡¢AgY¡¢AgZ¾ùÄÑÈÜÓÚË®£¬ÇÒKsp(AgX)=1.8¡Á10-10£¬Ksp(AgY)=1.0¡Á10-12£¬Ksp(AgZ)=8.7¡Á10-17¡£

(1)¸ù¾ÝÒÔÉÏÐÅÏ¢£¬ÅжÏAgX¡¢AgY¡¢AgZÈýÕßµÄÈܽâ¶È(ÓÃÒѱ»ÈܽâµÄÈÜÖʵÄÎïÖʵÄÁ¿/1 LÈÜÒº±íʾ)S(AgX)¡¢S(AgY)¡¢S(AgZ)µÄ´óС˳ÐòΪ_________________¡£

(2)ÈôÏòAgYµÄ±¥ºÍÈÜÒºÖмÓÈëÉÙÁ¿µÄAgX¹ÌÌ壬Ôòc(Y-)_______________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)¡£

(3)ÔÚ25 ¡æÊ±£¬ÈôÈ¡0.188 gµÄAgY(Ïà¶Ô·Ö×ÓÖÊÁ¿188)¹ÌÌå·ÅÈë100 mLË®ÖÐ(ºöÂÔÈÜÒºÌå»ýµÄ±ä»¯)£¬ÔòÈÜÒºÖÐY-µÄÎïÖʵÄÁ¿Å¨¶ÈΪ____________¡£

(4)ÓÉÉÏÊöKspÅжϣ¬ÔÚÉÏÊö(3)µÄÌåϵÖУ¬___________(Ìî¡°ÄÜ¡±»ò¡°·ñ¡±)ʵÏÖAgYÏòAgZµÄת»¯£¬¼òÊöÀíÓÉ£º_______________________________________________________¡£

(¢ò) ¡°æÏ¶ðÒ»ºÅ¡±³É¹¦·¢É䣬ʵÏÖÁËÖйúÈ˵ġ°±¼Ô¡±ÃÎÏë¡£

(1)·¢Éä¡°æÏ¶ðÒ»ºÅ¡±µÄ³¤Õ÷ÈýºÅ¼×»ð¼ýµÄµÚÈý¼¶Ê¹ÓõÄÍÆ½ø¼ÁÊÇÒºÇâºÍÒºÑõ£¬ÏÂÁÐÊÇ298 Kʱ£¬ÇâÆø(H2)¡¢Ì¼(C)¡¢ÐÁÍé(C8H18)¡¢¼×Íé(CH4)ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£ºH2(g)+O2(g)H2O(l)£»¦¤H=-285.8 kJ¡¤mol-1  C(s)+O2(g)CO2(g)£»¦¤H=-393.5 kJ¡¤mol-1  C8H18(l) +O2(g)8CO2(g)+9H2O(l)£»¦¤H=-5 518 kJ¡¤mol-1  CH4(g)+2O2(g)CO2(g)+2H2O(l)£»¦¤H=-890.3 kJ¡¤mol-1

ͨ¹ý¼ÆËã˵Ã÷µÈÖÊÁ¿µÄH2¡¢C¡¢C8H18¡¢CH4ÍêȫȼÉÕʱ£¬·Å³öÈÈÁ¿×î¶àµÄÊÇ_____________£¬·¢Éä¡°æÏ¶ðÒ»ºÅ¡±µÄ³¤Õ÷ÈýºÅ¼×»ð¼ýµÄµÚÈý¼¶Ê¹ÓÃÒºÇâºÍÒºÑõÕâÖÖÍÆ½ø¼ÁµÄÓŵãÊÇ__________

__________________________£»____________________________________¡£(ÇëдÁ½Ìõ)

(2)ÒÑÖª£ºH2(g)H2(l)£»¦¤H=-0.92 kJ¡¤mol-1

O2(g)O2(l)£»¦¤H=-6.84 kJ¡¤mol-1

H2O(l)H2O(g)£»¦¤H=44.0 kJ¡¤mol-1

Çëд³öÒºÇâºÍÒºÑõÉú³ÉÆøÌ¬Ë®µÄÈÈ»¯Ñ§·½³Ìʽ______________________________________¡£

Èç¹û´Ë´Î·¢Éä¡°æÏ¶ðÒ»ºÅ¡±µÄ³¤Õ÷ÈýºÅ¼×»ð¼ýËùЯ´øµÄȼÁÏΪ45¶Ö£¬ÒºÇâ¡¢ÒºÑõÇ¡ºÃÍêÈ«·´Ó¦Éú³ÉÆøÌ¬Ë®£¬×ܹ²ÊÍ·ÅÄÜÁ¿________kJ(±£Áô3λÓÐЧÊý×Ö)¡£

(3)ÇâÆø¡¢ÑõÆø²»½öȼÉÕʱÄÜÊÍ·ÅÈÈÄÜ£¬¶þÕßÐγɵÄÔ­µç³Ø»¹ÄÜÌṩµçÄÜ£¬ÃÀ¹úµÄ̽Ô·ɴ¬¡°°¢²¨Â޺š±Ê¹ÓõľÍÊÇÇâÑõȼÁÏµç³Ø£¬µç½âҺΪKOHÈÜÒº£¬Æäµç³Ø·´Ó¦Ê½Îª£º¸º¼«£º____________£»Õý¼«£º____________£»×Ü·´Ó¦Ê½£º________________________¡£

¼×´¼È¼ÁÏ·ÖΪ¼×´¼ÆûÓͺͼ״¼²ñÓÍ¡£¹¤ÒµÉϺϳɼ״¼µÄ·½·¨ºÜ¶à¡£

£¨1£©Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º

CO2(g) +3H2(g) £½CH3OH(g)+H2O(g)? ¡÷H1

2CO (g) +O2(g) £½2CO2(g)?? ¡÷H2

2H2(g)+O2(g) £½2H2O(g)??? ¡÷H3

Ôò¡¡CO(g) + 2H2(g) CH3OH(g)¡¡µÄ¡÷H£½??????????????? ¡£

£¨2£©ÔÚÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£ºCO(g)+2H2(g)CH3OH(g) £¬ÆäËûÌõ¼þ²»±ä£¬ÔÚ300¡æºÍ500¡æÊ±£¬ÎïÖʵÄÁ¿n(CH3OH) Ó뷴Ӧʱ¼ätµÄ±ä»¯ÇúÏßÈçͼËùʾ¡£¸Ã·´Ó¦µÄ¡÷H???? 0 £¨Ìî>¡¢<»ò=£©¡£

£¨3£©ÈôÒªÌá¸ß¼×´¼µÄ²úÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ____________£¨Ìî×Öĸ£©¡£

A£®ËõСÈÝÆ÷Ìå»ý

B£®½µµÍζÈ

C£®Éý¸ßζÈ

D£®Ê¹ÓúÏÊʵĴ߻¯¼Á

E£®½«¼×´¼´Ó»ìºÏÌåϵÖзÖÀë³öÀ´

£¨4£©CH4ºÍH2OÔÚ´ß»¯¼Á±íÃæ·¢Éú·´Ó¦CH4+H2OCO+3H2£¬T¡æÊ±£¬Ïò1 LÃܱÕÈÝÆ÷ÖÐͶÈë1 mol CH4ºÍ1 mol H2O(g)£¬5Сʱºó²âµÃ·´Ó¦Ìåϵ´ïµ½Æ½ºâ״̬£¬´ËʱCH4µÄת»¯ÂÊΪ50% £¬¼ÆËã¸ÃζÈÏÂµÄÆ½ºâ³£Êý????? £¨½á¹û±£ÁôСÊýµãºóÁ½Î»Êý×Ö£©¡£

£¨5£©ÒÔ¼×´¼ÎªÈ¼ÁϵÄÐÂÐÍµç³Ø£¬Æä³É±¾´ó´óµÍÓÚÒÔÇâΪȼÁ쵀 ´«Í³È¼ÁÏµç³Ø£¬Ä¿Ç°µÃµ½¹ã·ºµÄÑо¿£¬ÈçͼÊÇĿǰÑо¿½Ï¶àµÄÒ»Àà¹ÌÌåÑõ»¯ÎïȼÁÏµç³Ø¹¤×÷Ô­ÀíʾÒâͼ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙB¼«µÄµç¼«·´Ó¦Ê½Îª????????????????????? ¡£

¢ÚÈôÓøÃȼÁÏµç³Ø×öµçÔ´£¬ÓÃʯī×öµç¼«µç½âÁòËáÍ­ÈÜÒº£¬µ±µç·ÖÐ×ªÒÆ1mole- ʱ£¬Êµ¼ÊÉÏÏûºÄµÄ¼×´¼µÄÖÊÁ¿±ÈÀíÂÛÉϴ󣬿ÉÄÜÔ­ÒòÊÇ????????????????????? ¡£

£¨6£©25¡æÊ±£¬²ÝËá¸ÆµÄKsp=4.0¡Á10-8,̼Ëá¸ÆµÄKsp=2.5¡Á10-9¡£Ïò20ml̼Ëá¸ÆµÄ±¥ºÍÈÜÒºÖÐÖðµÎ¼ÓÈë8.0¡Á10-4 mol¡¤L-1µÄ²ÝËá¼ØÈÜÒº20ml£¬ÄÜ·ñ²úÉú³Áµí?????? £¨Ìî¡°ÄÜ¡±»ò¡°·ñ¡±£©¡£

 

 

¼×´¼È¼ÁÏ·ÖΪ¼×´¼ÆûÓͺͼ״¼²ñÓÍ¡£¹¤ÒµÉϺϳɼ״¼µÄ·½·¨ºÜ¶à¡£

£¨1£©Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º

¢ÙCO­2£¨g£© +3H2£¨g£© £½CH­­3OH£¨g£©+H2O£¨g£© ? ¡÷H1????

¢Ú2CO­£¨g£© +O2£¨g£© £½2CO­2£¨g£©?? ¡÷H2

¢Û2H2£¨g£©+O2£¨g£© £½2H2O£¨g£©????????? ¡÷H3

ÔòCO£¨g£© + 2H2£¨g£© CH3OH£¨g£©¡¡µÄ¡÷H£½??????????????? ¡£

£¨2£©ÔÚÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£º CO£¨g£©+2H2£¨g£©CH3OH£¨g£© £¬ÆäËûÌõ¼þ²»±ä£¬ÔÚ300¡æºÍ500¡æÊ±£¬ÎïÖʵÄÁ¿n£¨CH3OH£© Ó뷴Ӧʱ¼ätµÄ±ä»¯ÇúÏßÈçͼËùʾ¡£¸Ã·´Ó¦µÄ¡÷H???? 0 £¨Ìî>¡¢<»ò=£©¡£

£¨3£©ÈôÒªÌá¸ß¼×´¼µÄ²úÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ____________£¨Ìî×Öĸ£©¡£

A£®ËõСÈÝÆ÷Ìå»ý

B£®½µµÍζÈ

C£®Éý¸ßζÈ

D£®Ê¹ÓúÏÊʵĴ߻¯¼Á

E£®½«¼×´¼´Ó»ìºÏÌåϵÖзÖÀë³öÀ´

£¨4£©CH4ºÍH2OÔÚ´ß»¯¼Á±íÃæ·¢Éú·´Ó¦CH4+H2OCO+3H2£¬T¡æÊ±£¬Ïò1 LÃܱÕÈÝÆ÷ÖÐͶÈë1 mol CH4ºÍ1 mol H2O£¨g£©£¬5Сʱºó²âµÃ·´Ó¦Ìåϵ´ïµ½Æ½ºâ״̬£¬´ËʱCH4µÄת»¯ÂÊΪ50% £¬¼ÆËã¸ÃζÈÏÂµÄÆ½ºâ³£Êý???? £¨½á¹û±£ÁôСÊýµãºóÁ½Î»Êý×Ö£©¡£

£¨5£©ÒÔ¼×´¼ÎªÈ¼ÁϵÄÐÂÐÍµç³Ø£¬Æä³É±¾´ó´óµÍÓÚÒÔÇâΪȼÁϵĴ«Í³È¼ÁÏµç³Ø£¬Ä¿Ç°µÃµ½¹ã·ºµÄÑо¿£¬ÈçͼÊÇĿǰÑо¿½Ï¶àµÄÒ»Àà¹ÌÌåÑõ»¯ÎïȼÁÏµç³Ø¹¤×÷Ô­ÀíʾÒâͼ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙB¼«µÄµç¼«·´Ó¦Ê½Îª????????????????????? ¡£

¢ÚÈôÓøÃȼÁÏµç³Ø×öµçÔ´£¬ÓÃʯī×öµç¼«µç½âÁòËáÍ­ÈÜÒº£¬µ±µç·ÖÐ×ªÒÆ1mole- ʱ£¬Êµ¼ÊÉÏÏûºÄµÄ¼×´¼µÄÖÊÁ¿±ÈÀíÂÛÉϴ󣬿ÉÄÜÔ­ÒòÊÇ????????????????????? ¡£

£¨6£©25¡æÊ±£¬²ÝËá¸ÆµÄKsp=4.0¡Á10-8,̼Ëá¸ÆµÄKsp=2.5¡Á10-9¡£Ïò20ml̼Ëá¸ÆµÄ±¥ºÍÈÜÒºÖÐÖðµÎ¼ÓÈë8.0¡Á10-4 mol¡¤L-1µÄ²ÝËá¼ØÈÜÒº20ml£¬ÄÜ·ñ²úÉú³Áµí?????? £¨Ìî¡°ÄÜ¡±»ò¡°·ñ¡±£©¡£

 

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø