ÌâÄ¿ÄÚÈÝ

(8·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º

(1)ÒÑ֪ϡÈÜÒºÖУ¬1 mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Ê±£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ

________________________________________________________________________

________________________________________________________________________¡£

(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾­²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ

________________________________________________________________________¡£

(3)ÈçͼÊÇ101 kPaʱÇâÆøÔÚÂÈÆøÖеãȼÉú³ÉÂÈ»¯ÇâÆøÌåµÄÄÜÁ¿±ä»¯Ê¾Òâͼ£º

д³ö´Ë·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_________________________________________________¡£

(4)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º

¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l)

¦¤H1£½£­870.3 kJ¡¤mol£­1

¢ÚC(s)£«O2(g)===CO2(g)¡¡¦¤H2£½£­393.5 kJ¡¤mol£­1

¢ÛH2(g)£«O2(g)===H2O(l)¡¡¦¤H3£½£­285.8 kJ¡¤mol£­1

д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ________________________________________________________________________¡£

 

¡¾´ð°¸¡¿

(1)H2SO4(aq)£«NaOH(aq)===Na2SO4(aq)£«H2O(l)¡¡¦¤H£½£­57.3 kJ¡¤mol£­1

(2)C4H10(g)£«O2(g)===4CO2(g)£«5H2O(l)

¦¤H£½£­16QkJ¡¤mol£­1

(3)H2(g)£«Cl2(g)===2HCl(g)¡¡¦¤H£½£­183 kJ¡¤mol£­1

(4)2C(s)£«2H2(g)£«O2(g)===CH3COOH(l)

¦¤H£½£­488.3 kJ¡¤mol£­1

¡¾½âÎö¡¿(1)ÓÉÖкÍÈȵĶ¨Òå¿ÉÖª£¬1 mol NaOHÓë mol H2SO4Éú³É1 molˮʱ·ÅÈÈ57.3 kJ¡£

(2)CO2ͨÈë³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬¼´n(CO2)£½0.25 mol£¬Ôòn(C4H10)£½mol,1 mol¶¡ÍéÍêȫȼÉշųöÈÈÁ¿16QkJ¡£

(3)1 mol H2Óë1 mol Cl2·´Ó¦Éú³É2 mol HClµÄÈÈЧӦÊÇ£º

¦¤H£½436 kJ¡¤mol£­1£«243 kJ¡¤mol£­1£­2¡Á431 kJ¡¤mol£­1£½£­183 kJ¡¤mol£­1¡£

(4)ºÏ³ÉCH3COOHµÄ·´Ó¦Îª£º

2C(s)£«2H2(g)£«O2(g)===CH3COOH(l)

¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ú¡Á2£«¢Û¡Á2£­¢Ù¼´µÃ£¬

¦¤H£½(£­393.5 kJ¡¤mol£­1)¡Á2£«(£­285.8 kJ¡¤mol£­1)¡Á2£­(£­870.3 kJ¡¤mol£­1)£½£­488.3 kJ¡¤mol£­1¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2010?ÌÆɽÈýÄ££©ÈçͼËùʾÊÇÓÃÓÚÆøÌåÖƱ¸¡¢¸ÉÔï¡¢ÐÔÖÊÑéÖ¤¡¢Î²Æø´¦ÀíµÄ²¿·ÖÒÇÆ÷×°Ö㨼ÓÈȼ°¼Ð³Ö¹Ì¶¨×°ÖþùÒÑÂÔÈ¥£¬¸÷×°ÖÿÉÖظ´Ê¹Óã©£®Çë¸ù¾ÝÏÂÁÐÒªÇó»Ø´ðÎÊÌ⣮

£¨1£©Èô׶ÐÎÆ¿ÖÐʢװƯ°×·Û¹ÌÌ壬·ÖҺ©¶·ÖÐʢװŨÑÎËᣬÔò£º
µ±ÒÇÆ÷Á¬½Ó˳ÐòΪA¡úC¡úC¡úB¡úDʱ£¬B×°ÖÃÖÐΪÂÁ·Û£¬ÖƱ¸ÉÙÁ¿ÎÞË®ÈýÂÈ»¯ÂÁ£®AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
4HCl+Ca£¨ClO£©2=CaCl2+2Cl2¡ü+2H2O
4HCl+Ca£¨ClO£©2=CaCl2+2Cl2¡ü+2H2O
£¬µÚ¶þ¸öC×°ÖÃÖÐÊ¢×°µÄÊÔ¼ÁΪ
ŨÁòËá
ŨÁòËá
£¬D×°ÖõÄ×÷ÓÃÊÇ
³ýȥβÆøÂÈÆø£¬²¢·ÀÖ¹¿ÕÆøÖеÄË®½øÈëB×°ÖÃ
³ýȥβÆøÂÈÆø£¬²¢·ÀÖ¹¿ÕÆøÖеÄË®½øÈëB×°ÖÃ
£®
£¨2£©Èô׶ÐÎÆ¿ÖÐÊ¢×°×ãÁ¿µÄNa2O2·ÛÄ©£¬·ÖҺ©¶·ÖÐʢװŨ°±Ë®£¬ÂýÂý´ò¿ª·ÖҺ©¶·µÄ»îÈû£¬Ôò²úÉúµÄÆøÌåͨ¹ýºìÈȵIJ¬·Û£¬¸÷ÒÇÆ÷×°Öð´ÆøÁ÷·½Ïò´Ó×óµ½ÓÒÁ¬½Ó˳ÐòÊÇ£¨Ìî×Öĸ£©
A
A
¡ú
D
D
¡ú
B
B
¡ú
D
D
£»×°ÖÃBÖпÉÄܹ۲쵽µÄÏÖÏóÊÇ
B×°ÖÃÖÐÓкì×ØÉ«ÆøÌå³öÏÖ
B×°ÖÃÖÐÓкì×ØÉ«ÆøÌå³öÏÖ
£®
£¨3£©Èô׶ÐÎÆ¿ÖÐΪ´óÀíʯ£¬·ÖҺ©¶·ÖÐΪϡÑÎËᣬB×°ÖÃÖÐΪþ·Û£¬ÎªÑé֤þÓë¶þÑõ»¯Ì¼µÄ·´Ó¦£¬ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪA¡úC¡úC¡úB£¬µÚÒ»¸ö×°ÖÃÖеÄÊÔ¼ÁÃû³ÆΪ
±¥ºÍNaHCO3ÈÜÒº
±¥ºÍNaHCO3ÈÜÒº
£¬BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2Mg+CO2
  ¡÷  
.
 
2MgO+C
2Mg+CO2
  ¡÷  
.
 
2MgO+C
£¬ÔÚ¼ÓÈÈB×°ÖÃ֮ǰ£¬Ó¦¸Ã
´ò¿ª·ÖҺ©¶·µÄ»îÈû£¬ÏÈͨһ¶Îʱ¼äµÄ¶þÑõ»¯Ì¼
´ò¿ª·ÖҺ©¶·µÄ»îÈû£¬ÏÈͨһ¶Îʱ¼äµÄ¶þÑõ»¯Ì¼
£®
£¨4£©Èô׶ÐÎÆ¿ÖÐΪ¶þÑõ»¯Ã̹ÌÌ壬·ÖҺ©¶·ÖÐΪ˫ÑõË®£¬B×°ÖÃÖÐΪµçʯ£¨Ö»º¬ÔÓÖÊÁò»¯¸Æ£©£¬ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪA¡úC¡úB¡úC¡úC£¬µÚ¶þ¸öC×°ÖÃÖÐΪƷºìÈÜÒº£¬µÚÈý¸öC×°ÖÃÖÐΪ³ÎÇåʯ»ÒË®£¬ÇÒʵÑéºóÆ·ºìÈÜÒººìÉ«Ã÷ÏÔ±ädz£¬³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ð´³öBÖз¢ÉúµÄ¿ÉÄÜ·´Ó¦
2CaC2+5O2
  ¡÷  
.
 
2CaO+4CO2
2CaC2+5O2
  ¡÷  
.
 
2CaO+4CO2
£¬
2CaS+3O2
  ¡÷  
.
 
2CaO+2SO2
2CaS+3O2
  ¡÷  
.
 
2CaO+2SO2
£®Èô·´Ó¦Ç°BÖÐÑùÆ·ÖÊÁ¿Îª8.40g£¬·´Ó¦ºóBÖвÐÁô¹ÌÌåÖÊÁ¿Îª7.28g£¬ÔòµçʯÖÐ̼»¯¸ÆµÄÖÊÁ¿·ÖÊýΪ
91.4%
91.4%
£®£¨¼ÙÉè¹ÌÌå·´Ó¦ÍêÈ«£¬½á¹û±£ÁôÈýλÓÐЧÊý×Ö£©£®

ÈçͼËùʾÊÇÓÃÓÚÆøÌåÖƱ¸¡¢¸ÉÔï¡¢ÐÔÖÊÑéÖ¤¡¢Î²Æø´¦ÀíµÄ²¿·ÖÒÇÆ÷×°Ö㨼ÓÈȼ°¼Ð³Ö¹Ì¶¨×°ÖþùÒÑÂÔÈ¥£¬¸÷×°ÖÿÉÖظ´Ê¹Óã©£®Çë¸ù¾ÝÏÂÁÐÒªÇó»Ø´ðÎÊÌ⣮

£¨1£©Èô׶ÐÎÆ¿ÖÐʢװƯ°×·Û¹ÌÌ壬·ÖҺ©¶·ÖÐʢװŨÑÎËᣬÔò£º
µ±ÒÇÆ÷Á¬½Ó˳ÐòΪA¡úC¡úC¡úB¡úDʱ£¬B×°ÖÃÖÐΪÂÁ·Û£¬ÖƱ¸ÉÙÁ¿ÎÞË®ÈýÂÈ»¯ÂÁ£®AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ______£¬µÚ¶þ¸öC×°ÖÃÖÐÊ¢×°µÄÊÔ¼ÁΪ______£¬D×°ÖõÄ×÷ÓÃÊÇ______£®
£¨2£©Èô׶ÐÎÆ¿ÖÐÊ¢×°×ãÁ¿µÄNa2O2·ÛÄ©£¬·ÖҺ©¶·ÖÐʢװŨ°±Ë®£¬ÂýÂý´ò¿ª·ÖҺ©¶·µÄ»îÈû£¬Ôò²úÉúµÄÆøÌåͨ¹ýºìÈȵIJ¬·Û£¬¸÷ÒÇÆ÷×°Öð´ÆøÁ÷·½Ïò´Ó×óµ½ÓÒÁ¬½Ó˳ÐòÊÇ£¨Ìî×Öĸ£©______¡ú______¡ú______¡ú______£»×°ÖÃBÖпÉÄܹ۲쵽µÄÏÖÏóÊÇ______£®
£¨3£©Èô׶ÐÎÆ¿ÖÐΪ´óÀíʯ£¬·ÖҺ©¶·ÖÐΪϡÑÎËᣬB×°ÖÃÖÐΪþ·Û£¬ÎªÑé֤þÓë¶þÑõ»¯Ì¼µÄ·´Ó¦£¬ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪA¡úC¡úC¡úB£¬µÚÒ»¸ö×°ÖÃÖеÄÊÔ¼ÁÃû³ÆΪ______£¬BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______£¬ÔÚ¼ÓÈÈB×°ÖÃ֮ǰ£¬Ó¦¸Ã______£®
£¨4£©Èô׶ÐÎÆ¿ÖÐΪ¶þÑõ»¯Ã̹ÌÌ壬·ÖҺ©¶·ÖÐΪ˫ÑõË®£¬B×°ÖÃÖÐΪµçʯ£¨Ö»º¬ÔÓÖÊÁò»¯¸Æ£©£¬ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪA¡úC¡úB¡úC¡úC£¬µÚ¶þ¸öC×°ÖÃÖÐΪƷºìÈÜÒº£¬µÚÈý¸öC×°ÖÃÖÐΪ³ÎÇåʯ»ÒË®£¬ÇÒʵÑéºóÆ·ºìÈÜÒººìÉ«Ã÷ÏÔ±ädz£¬³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ð´³öBÖз¢ÉúµÄ¿ÉÄÜ·´Ó¦______£¬______£®Èô·´Ó¦Ç°BÖÐÑùÆ·ÖÊÁ¿Îª8.40g£¬·´Ó¦ºóBÖвÐÁô¹ÌÌåÖÊÁ¿Îª7.28g£¬ÔòµçʯÖÐ̼»¯¸ÆµÄÖÊÁ¿·ÖÊýΪ______£®£¨¼ÙÉè¹ÌÌå·´Ó¦ÍêÈ«£¬½á¹û±£ÁôÈýλÓÐЧÊý×Ö£©£®

ÈçͼËùʾÊÇÓÃÓÚÆøÌåÖƱ¸¡¢¸ÉÔï¡¢ÐÔÖÊÑéÖ¤¡¢Î²Æø´¦ÀíµÄ²¿·ÖÒÇÆ÷×°Ö㨼ÓÈȼ°¼Ð³Ö¹Ì¶¨×°ÖþùÒÑÂÔÈ¥£¬¸÷×°ÖÿÉÖظ´Ê¹Óã©£®Çë¸ù¾ÝÏÂÁÐÒªÇó»Ø´ðÎÊÌ⣮

¾«Ó¢¼Ò½ÌÍø

£¨1£©Èô׶ÐÎÆ¿ÖÐʢװƯ°×·Û¹ÌÌ壬·ÖҺ©¶·ÖÐʢװŨÑÎËᣬÔò£º
µ±ÒÇÆ÷Á¬½Ó˳ÐòΪA¡úC¡úC¡úB¡úDʱ£¬B×°ÖÃÖÐΪÂÁ·Û£¬ÖƱ¸ÉÙÁ¿ÎÞË®ÈýÂÈ»¯ÂÁ£®AÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ______£¬µÚ¶þ¸öC×°ÖÃÖÐÊ¢×°µÄÊÔ¼ÁΪ______£¬D×°ÖõÄ×÷ÓÃÊÇ______£®
£¨2£©Èô׶ÐÎÆ¿ÖÐÊ¢×°×ãÁ¿µÄNa2O2·ÛÄ©£¬·ÖҺ©¶·ÖÐʢװŨ°±Ë®£¬ÂýÂý´ò¿ª·ÖҺ©¶·µÄ»îÈû£¬Ôò²úÉúµÄÆøÌåͨ¹ýºìÈȵIJ¬·Û£¬¸÷ÒÇÆ÷×°Öð´ÆøÁ÷·½Ïò´Ó×óµ½ÓÒÁ¬½Ó˳ÐòÊÇ£¨Ìî×Öĸ£©______¡ú______¡ú______¡ú______£»×°ÖÃBÖпÉÄܹ۲쵽µÄÏÖÏóÊÇ______£®
£¨3£©Èô׶ÐÎÆ¿ÖÐΪ´óÀíʯ£¬·ÖҺ©¶·ÖÐΪϡÑÎËᣬB×°ÖÃÖÐΪþ·Û£¬ÎªÑé֤þÓë¶þÑõ»¯Ì¼µÄ·´Ó¦£¬ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪA¡úC¡úC¡úB£¬µÚÒ»¸ö×°ÖÃÖеÄÊÔ¼ÁÃû³ÆΪ______£¬BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______£¬ÔÚ¼ÓÈÈB×°ÖÃ֮ǰ£¬Ó¦¸Ã______£®
£¨4£©Èô׶ÐÎÆ¿ÖÐΪ¶þÑõ»¯Ã̹ÌÌ壬·ÖҺ©¶·ÖÐΪ˫ÑõË®£¬B×°ÖÃÖÐΪµçʯ£¨Ö»º¬ÔÓÖÊÁò»¯¸Æ£©£¬ÒÇÆ÷µÄÁ¬½Ó˳ÐòΪA¡úC¡úB¡úC¡úC£¬µÚ¶þ¸öC×°ÖÃÖÐΪƷºìÈÜÒº£¬µÚÈý¸öC×°ÖÃÖÐΪ³ÎÇåʯ»ÒË®£¬ÇÒʵÑéºóÆ·ºìÈÜÒººìÉ«Ã÷ÏÔ±ädz£¬³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ð´³öBÖз¢ÉúµÄ¿ÉÄÜ·´Ó¦______£¬______£®Èô·´Ó¦Ç°BÖÐÑùÆ·ÖÊÁ¿Îª8.40g£¬·´Ó¦ºóBÖвÐÁô¹ÌÌåÖÊÁ¿Îª7.28g£¬ÔòµçʯÖÐ̼»¯¸ÆµÄÖÊÁ¿·ÖÊýΪ______£®£¨¼ÙÉè¹ÌÌå·´Ó¦ÍêÈ«£¬½á¹û±£ÁôÈýλÓÐЧÊý×Ö£©£®

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

¾«Ó¢¼Ò½ÌÍø