-+
ÀÁ¶èÊÇºÜÆæ¹ÖµÄ¶«Î÷£¬ËüʹÄãÒÔΪÄÇÊǰ²ÒÝ£¬ÊÇÐÝÏ¢£¬ÊǸ£Æø£»µ«Êµ¼ÊÉÏËüËù¸øÄãµÄÊÇÎÞÁÄ£¬ÊǾ뵡£¬ÊÇÏû³Á;Ëü°þ¶áÄã¶Ôǰ;µÄÏ£Íû£¬¸î¶ÏÄãºÍ±ðÈËÖ®¼äµÄÓÑÇ飬ʹÄãÐÄÐØÈÕ½¥ÏÁÕ£¬¶ÔÈËÉúÒ²Ô½À´Ô½»³ÒÉ¡£
¡ªÂÞÀ¼
µÚ¶þ½Ú ȼÉÕÈÈ ÄÜÔ´
ѧϰĿ±ê£º
Àí½âȼÉÕÈȵĸÅÄÈÏʶÄÜÔ´ÊÇÈËÀàÉú´æºÍ·¢Õ¹µÄÖØÒª»ù´¡£¬Á˽⻯ѧÔÚ½â¾öÄÜԴΣ»úÖеÄÖØÒª×÷Óá£ÖªµÀ½ÚÔ¼ÄÜÔ´¡¢Ìá¸ßÄÜÁ¿ÀûÓÃЧÂʵÄʵ¼ÊÒâÒå¡£ Ñ§Ï°ÖØµã¡¢Äѵ㣺
ȼÉÕÈȵĸÅÄî ѧϰ¹ý³Ì£º
¡¾ÎÂϰ¾ÉÖª¡¿¡¤
»ØÒä·´Ó¦ÈÈ ¡¢ìʱäµÄ֪ʶ£¬½â´ð£º
1¡¢ÔÚ25 ¡æ¡¢101 kPaÏ£¬1 g¼×´¼È¼ÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22£®68 kJ,ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ
3A£® CH3OH£¨l£©£«O2(g) CO2(g)+2H2O(l) £»¦¤H=+725£®8 kJ¡¤mol£1
2B£® 2CH3OH£¨l£©+3O2(g) 2CO2(g)+4H2O(l) £»¦¤H=£1 452 kJ¡¤mol£1 C£® 2CH3OH(l)+3O2(g) 2CO2(g)+4H2O(l) £»¦¤H=£725£®8 kJ¡¤mol£1 D£® 2CH3OH£¨l£©+3O2(g) 2CO2(g)+4H2O(l) £» ¦¤H=+1 452 kJ¡¤mol£1 2¡¢ÈçÓÒͼËùʾ£¬°ÑÊԹܷÅÈëÊ¢ÓÐ25 ¡æÊ±±¥ºÍʯ»ÒË®µÄÉÕ±ÖУ¬ÊÔ¹ÜÖпªÊ¼·ÅÈ뼸С¿éþƬ£¬ÔÙÓõιܵÎÈë5 mLÑÎËáÓÚÊÔ¹ÜÖС£ÊԻشðÏÂÁÐÎÊÌ⣺
ÑÎËá (1)ʵÑéÖй۲쵽µÄÏÖÏóÊÇ___________________¡£
(2)²úÉúÉÏÊöÏÖÏóµÄÔÒòÊÇ____ __¡£ (3)д³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ£º____________ _¡£
þƬ (4)ÓÉʵÑéÍÆÖª£¬MgCl2ÈÜÒººÍH2µÄ×ÜÄÜÁ¿______£¨Ìî
¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©Ã¾Æ¬ºÍÑÎËáµÄ×ÜÄÜÁ¿¡£ ±¥ºÍʯ»ÒË® ¡¾Ñ§Ï°ÐÂÖª¡¿ Ò»¡¢È¼ÉÕÈÈ
ÔĶÁ½Ì²Ä£¬ÕÒ³öȼÉÕÈȵĸÅÄ²¢°ÑËüÌîдÔÚÏÂÃæµÄ¿Õ¸ñÖС£
1¡¢¶¨Ò壺 ×¢Ò⣺
£¨1£©Ìõ¼þ£º £¨2£©È¼ÉÕµÄÎïÖÊÒ»¶¨Îª mol £¨3£©Éú³ÉµÄÎïÖÊÒ»¶¨ÒªÎȶ¨ a¡¢×´Ì¬ÒªÎȶ¨£»
b¡¢ÒªÍêȫȼÉÕ£¨Éú³ÉÎï²»ÄÜÔÙȼÉÕ£©¡£ Ò»°ãÖ¸£º
C -- CO2 £¨g£© N¨C N2 £¨g£©
S ¨C SO2 £¨g£© H¨C H2O£¨l£© 2.±íʾµÄÒâÒ壺
Èç: CH4µÄȼÉÕÈÈΪ890.3KJ/mol.
º¬Òå: 3.±íʾȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽµÄÊéд£º
Ó¦ÒÔ molÎïÖʵıê×¼À´Å䯽ÆäÓàÎïÖʵĻ¯Ñ§¼ÆÁ¿Êý(³£³öÏÖ·ÖÊý) 4.Ò»¶¨Á¿¿ÉȼÎïÍêȫȼÉÕʱ·Å³öµÄÈÈÁ¿ Q·Å=n(¿ÉȼÎï)¡Á¡÷H
5.ȼÉÕÈÈÓëÖкÍÈȵÄÇø±ðÓëÁªÏµ Ïà ͬ µã ÄÜÁ¿±ä»¯ ¦¤H ·´Ó¦ÎïµÄÁ¿ Éú³ÉÎïµÄÁ¿ ²» ͬ µã ·´Ó¦ÈÈ µÄº¬Òå 1mol¿ÉȼÎï ²»ÏÞÁ¿ ȼÉÕÈÈ ·ÅÈÈ·´Ó¦ ¦¤H<0 , µ¥Î» kJ/mol ¿ÉÄÜÊÇ1molÒ²¿ÉÒÔÊÇ0.5mol(²»ÏÞ) H2O 1mol ÖкÍÈÈ 1mol·´Ó¦ÎïÍêȫȼÉÕʱ·Å³öËá¼îÖкÍÉú³É1molH2Oʱ·Å³öµÄµÄÈÈÁ¿;²»Í¬µÄÎïÖÊȼÉÕÈȲ»ÈÈÁ¿,Ç¿ËáÇ¿¼î¼äµÄÖкͷ´Ó¦ÖкÍͬ ÈÈ´óÖÂÏàͬ,¾ùԼΪ57.3kJ/mol ¡¾Ë¼¿¼Óë½»Á÷¡¿ 1¡¢Ñ¡ÔñȼÁϵıê×¼ÊÇʲô?
2¡¢Ãº×÷ΪȼÁϵÄÀû±×?Ìá¸ßúȼÉÕЧÂʵĴëÊ©ÓÐÄÄЩ? ¡¾ËæÌÃÁ·Ï°¡¿ 1¡¢ÒÑÖª£º
2H2( g ) + O2 ( g) = 2H2O ( l ) ¦¤H=-571.6 kJ/mol H2( g ) +1/2 O2 ( g) = H2O ( g ) ¦¤H=-241.8 kJ/mol
ÇóÇâÆøµÄȼÉÕÈÈ¡£
2¡¢Ê¯Ä«Óë½ð¸ÕʯµÄȼÉÕÈÈÊÇ·ñÏàͬ£¬ÎªÊ²Ã´£¿ Çë·Ö±ðд³öʯīºÍ½ð¸ÕʯȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¡£
3.ÏÂÁи÷×éÎïÖʵÄȼÉÕÈÈÏàµÈµÄÊÇ£º£¨ £©
A.̼ºÍÒ»Ñõ»¯Ì¼ B.1moL̼ºÍ2moL̼ C.1moLÒÒȲºÍ2moL̼ D.µí·ÛºÍÏËÎ¬ËØ
4.ÒÑÖªÈÈ»¯Ñ§·½³Ìʽ£º
H2(g)+1/2O2(g)=H2O(g)£»¡÷H=-241.8 kJ/mol 2H2(g)+O2(g)=2H2O(g)£» ¡÷H=-483.6 kJ/mol H2(g)+1/2O2(g)=H2O(l)£» ¡÷H = -285.8 kJ/mol 2H2(g)+O2(g)=2H2O(l)£» ¡÷H=-571.6 kJ/mol
ÔòÇâÆøµÄȼÉÕÈÈΪ£º ¶þ¡¢ÄÜÔ´
ÔĶÁ¿ÎÎÄ£¬ÌÖÂÛÏÂÁÐÎÊÌ⣺
ÄÜÔ´¾ÍÊÇÄÜÌṩÄÜÁ¿µÄ×ÔÈ»×ÊÔ´£¬Ëü°üÀ¨»¯Ê¯È¼ÁÏ£¨Ãº¡¢Ê¯ÓÍ¡¢ÌìÈ»Æø£©¡¢Ñô¹â¡¢·çÁ¦¡¢Á÷Ë®¡¢³±Ï«ÒÔ¼°²ñ²ÝµÈ¡£
1£®ÄÜÔ´¾ÍÊÇÄÜÌṩ µÄ×ÔÈ»×ÊÔ´£¬
°üÀ¨ µÈ¡£
2.ÎÒ¹úĿǰʹÓõÄÖ÷ÒªÄÜÔ´ÊÇ £¬ÊDz»ÄÜ ¡£ 3.½â¾öÄÜÔ´µÄ°ì·¨ÊÇ £¬¼´¿ª·¢ ºÍ½ÚÔ¼ £¬Ìá¸ßÄÜÔ´µÄ ¡£
4.ÏÖÔÚ̽Ë÷¿ª·¢µÄÐÂÄÜÔ´ÓÐ µÈ£¬ÐÂÄÜÔ´µÄÖ÷ÒªÓÅÊÆÊÇ ¡£ 5¡¢ÄÜÔ´¡¢Ò»¼¶ÄÜÔ´¡¢¶þ¼¶ÄÜÔ´
ÏÂÁÐÊôÓÚÒ»¼¶ÄÜÔ´µÄÊÇ ÊôÓÚ¶þ¼¶ÄÜÔ´µÄÊÇ A ·çÄÜ BµçÄÜ C ÇâÄÜ D Ñô¹â E ÌìÈ»Æø 6¡¢ÎÒ¹úµÄÄÜÔ´×´¿öÈçºÎ£¿
£¨1£©Ä¿Ç°Ê¹ÓõÄÖ÷ÒªÄÜÔ´ÊǺÎÖÖÄÜÔ´£¿ £¨2£©ÎÒ¹úµÄÄÜÔ´´¢Á¿¡£
£¨3£©ÎÒ¹úµÄÈ˾ùÄÜÔ´ÓµÓÐÁ¿¡£
£¨4£©½üÄêÀ´ÎÒ¹úÄÜÔ´µÄ×ÜÏû·ÑÁ¿ÓëÈ˾ùÏû·ÑÁ¿Çé¿ö¡£ 3¡¢Á˽âÁËÎÒ¹úµÄÄÜÔ´ÀûÓÃÂÊ£¬ÄãÓкθÐÏ룿
£ÛÒéÒ»Òé£Ý1.³ÇÊмÒÓÃȼÁÏʹÓÃÆøÌåȼÁϱÈʹÓÃú¡¢ÃºÇòµÈ¹ÌÌåȼÁÏÓÐʲôºÃ´¦£¿
£ÛÒéÒ»Òé£Ý2.ÍÆ¹ãʹÓÃÌìÈ»Æø£¨Ïà¶ÔÓڹܵÀÃºÆøºÍÒº»¯Æø£©ÎªÊ²Ã´ÊdzÇÊÐÈ¼ÆøµÄ·¢Õ¹·½Ïò£¿
ÌÖÂÛ£ºÏÂÃæÁгöµÄÊÇÓйØÃº×÷ȼÁÏÀû±×ÎÊÌâµÄһЩÖ÷ÒªÂ۵㣬Çë²Î¿¼ÕâЩÂ۵㣬²¢×¼±¸ÓйزÄÁÏ£¬¼òÒªÂÛÊöÄã¶Ô¸ÃÎÊÌâµÄ¿´·¨¡£ ¢ÙúÊÇÎÒ¹ú´¢Á¿×î¶àµÄÄÜÔ´×ÊÔ´£¬Ó¦³ä·ÖÀûÓøÃ×ÊԴΪÎÒ¹úµÄÉç»áÖ÷Ò彨Éè·þÎñ¡£
¾ÝÓÐÈ˹À¼Æ£¬ÎÒ¹úµÄú̿´¢Á¿×㹻ʹÓü¸°ÙÄê¡£ ¢ÚúµÄÐγÉÐè¾¹ýÊýÒÚÄêµÄʱ¼ä£¬ÓÃһЩ¾ÍÉÙһЩ£¬²»¿ÉÔÙÉú¡£ ¢Û úÊÇÖØÒªµÄ»¯¹¤ÔÁÏ£¬°Ñú×÷ȼÁϼòµ¥ÉÕµôÌ«¿Éϧ£¬Ó¦¸Ã×ÛºÏÀûÓᣠ¢Ü úÊÇ·¢ÈÈÁ¿ºÜ¸ßµÄ¹ÌÌåȼÁÏ£¬ÎÒ¹úú̿×ÊÔ´Ïà¶Ô±È½Ï¼¯ÖУ¬¿ª²É³É±¾½ÏµÍ£¬ÓÃú×÷ȼÁϺÏËã¡£
ú×÷Ϊ¹ÌÌåȼÁÏ£¬È¼ÉÕ·´Ó¦ËÙÂÊС£¬ÈÈÀûÓÃЧÂʵͣ¬ÇÒÔËÊä²»·½±ã¡£ ¢Ýúֱ½ÓȼÉÕʱ²úÉúSO2µÈÓж¾ÆøÌåºÍÑ̳¾£¬¶Ô»·¾³Ôì³ÉÑÏÖØÎÛȾ¡£ ¢Þ¿ÉÒÔͨ¹ýÇå½àú¼¼Êõ£¬ÈçúµÄÒº»¯ºÍÆø»¯£¬ÒÔ¼°ÊµÐÐÑÌÆø¾»»¯ÍÑÁòµÈ£¬´ó´ó¼õÉÙȼú¶Ô»·¾³Ôì³ÉµÄÎÛȾ£¬Ìá¸ßúȼÉÕµÄÈÈÀûÓÃÂÊ¡£ ¢ßú´óÁ¿¿ª²Éʱ£¬»áÔì³ÉµØÃæËúÏÝ¡£
¡¾×÷Òµ¡¿
1.ÏÂÁз´Ó¦¼ÈÊôÓÚÑõ»¯»¹Ô·´Ó¦,ÓÖÊÇÎüÈÈ·´Ó¦µÄÊÇ( ) A.ÂÁƬÓëÏ¡ÑÎËáµÄ·´Ó¦. B.Ba(OH)2¡¤8H2OÓëNH4ClµÄ·´Ó¦. C.ׯÈȵÄ̼ÓëCO2µÄ·´Ó¦ D.¼×ÍéÔÚÑõÆøÖеÄȼÉÕ·´Ó¦
2.ÏÂÁÐȼÁÏÖÐ,²»ÊôÓÚ»¯Ê¯È¼ÁϵÄÊÇ( )
A.ú B.ʯÓÍ C.ÌìÈ»Æø D.Ë®ÃºÆø 3.ËáÓêΣº¦¿É°üÀ¨( ) ¢Ù¶ÔÈËÌåµÄÖ±½ÓΣº¦,ÒýÆð·Î²¿¼²²¡¶øÊ¹ÈËÖÂËÀ,¢ÚÒýÆðºÓÁ÷,ºþ²´µÄË®ÌåËữ,ÑÏÖØÓ°ÏìË®Éú¶¯¸ËÎïµÄÉú³¤,¢ÛÆÆ»µÍÁÈÀ,Ö²±»,ÉÁÖ ¢Ü¸¯Ê´½ðÊô,ÓÍÆá,Ƥ¸ï,·Ä֯Ʒ¼°½¨Öþ²ÄÁϵÈ,¢ÝÉøÈëµØÏÂ,¿ÉÄÜÒýÆðµØÏÂË®Ëữ. A.¢Ù¢Û¢Ý B.¢Ù¢Ú¢Û¢Ü C.¢Ù¢Û¢Ü¢Ý D.¢Ù¢Ú¢Û¢Ü¢Ý
4.ÎÒ¹ú·¢ÉäµÄ¨DÉñÖÛÎåºÅ¡¬ÔØÈË·É´¬µÄȼÁÏÊÇÂÁ·ÛÓë¸ßÂÈËá淋ĻìºÏÎï¡£µãȼʱ£¬ÂÁ·ÛÑõ»¯·ÅÈÈÒý·¢¸ßÂÈËáï§·´Ó¦2NH4ClO4=N2¡ü+4H2O+Cl2¡ü+O2¡ü£»¡÷H<0¡£¹ØÓڸ÷´Ó¦ÐðÊö²»ÕýÈ·µÄÊÇ £¨ £©
A£®¸Ã·´Ó¦ÊôÓڷֽⷴӦ¡¢Ñõ»¯»¹Ô·´Ó¦¡¢·ÅÈÈ·´Ó¦ B£®¸Ã·´Ó¦Ë²¼äÄܲúÉú´óÁ¿¸ßÎÂÆøÌå£¬ÍÆ¶¯·É´¬·ÉÐÐ C£®´ÓÄÜÁ¿±ä»¯ÉÏ¿´£¬¸Ã·´Ó¦ÊÇ»¯Ñ§ÄÜת±äΪÈÈÄܺͶ¯ÄÜ D£®·´Ó¦ÖÐNH4ClO4Ö»Æðµ½Ñõ»¯¼ÁµÄ×÷ÓÃ
5.ͨ³£×´¿öÏ£¬½«1g ÇâÆøÔÚÂÈÆøÖÐÍêȫȼÉÕ£¬·Å³ö92£®4KJÈÈÁ¿£¬ÏÂÁÐËùʾµÄÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ ( ) A.H2(g)£«Cl2(g)£½1/2HCl(g)£»¦¤£È£½92.4KJ/mol B.H2(g)£«Cl2(g)£½1/2HCl(g)£»¦¤£È£½£92.4KJ/mol C.H2(g)£«Cl2(g)£½2HCl(g)£» ¦¤£È£½£184.8KJ/mol D.H2(g)£«Cl2(g)£½2HCl(l)£» ¦¤£È£½184.8/mol
6.ÇâÆø£¨H2£©¡¢Ò»Ñõ»¯Ì¼£¨CO£©¡¢ÐÁÍ飨C8H18£©¡¢¼×Í飨CH4£©µÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£º
H2(g) + 1/2O2(g) = H2O(l)£» ¡÷H= ¡ª285.8 kJ/mol CO(g) + 1/2O2(g) = CO2(g)£»¡÷H= ¡ª283.0 kJ/mol
C8H18(l) + 25/2O2(g) = 8CO2(g) + 9 H2O(l)£» ¡÷H= ¡ª5518 kJ/mol CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(l)£» ¡÷H= ¡ª890.3 kJ/mol
ÏàͬÎïÖʵÄÁ¿µÄH2¡¢CO¡¢C8H18¡¢CH4ÍêȫȼÉÕʱ£¬·Å³öÈÈÁ¿×îÉÙµÄÊÇ( ) A£®H2(g) B£®CO(g) C£®C8H18(l) D£®CH4(g)
7.ÒÑÖª³ä·ÖȼÉÕa g ÒÒÈ²ÆøÌåʱÉú³É1mol¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮ£¬²¢·Å³öÈÈÁ¿bkJ£¬ÔòÒÒȲȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ
A.2C2H2(g) + 5O2(g) = 4CO2(g) + 2H2O(l)£»¡÷H = £4b KJ/mol B.C2H2(g) + 5/2O2(g) = 2CO2(g) + H2O(l)£»¡÷H = 2b KJ/mol C.2C2H2(g) + 5O2(g) = 4CO2(g) + 2H2O(l)£»¡÷H = £2b KJ/mol D2C2H2(g) + 5O2(g) = 4CO2(g) + 2H2O(l)£»¡÷H = b KJ/mol
8.ÒÑÖª1mol°×Á×±ä³É1molºìÁ׷ųö18.39KJÈÈÁ¿¡£ÏÂÁÐÁ½¸ö·´Ó¦ 4P(°×¡¢s)£«5O2£½2P2O5(s)£»¡÷H1=kJ/mol 4P(ºì¡¢s)£« 5O2£½2P2O5(s)£»¡÷H2=kJ/mol Ôò¦¤H1Ó릤H2¹ØÏµÕýÈ·µÄÊÇ ( )
A£®¦¤H1£½¦¤H2 B£®¦¤H1<¦¤H2 C£®¦¤H1>¦¤H2 D£®ÎÞ·¨È·¶¨ 9.ÏÂÁйØÓÚȼÉÕÈȵÄ˵·¨ÖÐÕýÈ·µÄÊÇ ( ) A.1molÎïÖÊȼÉÕËù·Å³öµÄÈÈÁ¿ B.³£ÎÂÏ£¬¿ÉȼÎïȼÉշųöµÄÈÈÁ¿ C.ÔÚ25¡æ¡¢1.01¡Á105Paʱ£¬1molÎïÖÊȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯ÎïʱËù·Å³öµÄÈÈÁ¿ D.ȼÉÕÈÈËæ»¯Ñ§·½³ÌʽǰµÄ»¯Ñ§¼ÆÁ¿ÊýµÄ¸Ä±ä¶ø¸Ä±ä 10.ÔÚ25¡æ¡¢1.01¡Á105Paʱ£¬1gCH4ȼÉÕʱÉú³ÉCO2ÓëҺ̬H2O£¬·Å³ö55.6kJµÄÈÈÁ¿£¬ÔòCH4µÄȼÉÕÈÈΪ ( )
A£®55.6kJ/mol B.889.6kJ/mol C.£889.6kJ/mol D.444.8kJ/mol 11.¹Ø»¯Ê¯È¼ÁϵÄ˵·¨ÕýÈ·µÄÊÇ ( )
A£®»¯Ê¯È¼ÁÏÊÇ¿ÉÔÙÉúµÄ£¬Òò´ËÔÚµØÇòÉϵÄÔ̲ØÁ¿Ò²ÊÇÎÞÏÞµÄ
B£®»¯Ê¯È¼ÁÏËäÈ»ÔÚµØÇòÉϵÄÔ̲ØÁ¿ÓÐÏÞ£¬µ«Ðγɻ¯Ê¯È¼ÁϵÄËÙÂÊÏ൱¿ì£¬ËùÒÔ»¯Ê¯È¼ÁÏÏ൱ÓÚÎÞÏÞµÄ
C£®»¯Ê¯È¼ÁϵÄÐγÉÊǷdz£¸´Ôӵģ¬ËùÐèʱ¼äºÜ³¤£¬µ«»¯Ê¯È¼ÁÏÔÚµØÇòÉϵÄÔ̲ØÁ¿ÊÇÎÞÏÞµÄ
D£®»¯Ê¯È¼ÁÏÔÚµØÇòÉϵÄÔ̲ØÁ¿ÊÇÓÐÏÞµÄ,¶øÇÒÓÖ¶¼ÊǾ¹ýÒÚÍòÄê²ÅÄÜÐγɵķÇÔÙÉú×ÊÔ´
12. 1998Äê³öÏÖµÄÈ«Çòζȴó·ù¶ÈÉý¸ß£¬²úÉúÁËÏÔÖøµÄ¨D¶ò¶ûÄáŵ¡¬ÏÖÏ󣮸ɺµºÍ±©ÓêÔÖÄÑ£¬Î£º¦ÁËÐí¶àµØÇø¡£ÎªÁË·ÀÖ¹Æøºò±äůµÄ½øÒ»²½¶ñ»¯£¬ÁªºÏ¹ú»·¾³±£»¤×éÖ¯ÓÚ1998Äêͨ¹ý´ó»áÒªÇó¸÷¹ú¼õÉÙ¹¤ÒµÅÅ·ÅÁ¿µÄÆøÌåÊÇ( ) A. ¶þÑõ»¯Áò B. ¶þÑõ»¯Ì¼ C. µªµÄÑõ»¯Îï D. ̼Ç⻯ºÏÎï
13.ÎÒ¹úÈýÏ¿¹¤³ÌËùÌṩµÄÇå½à¡¢Á®¼ÛÇ¿¾¢¡¢¿ÉÔÙÉúµÄË®µç£¬Ï൱ÓÚÿÄêȼÉÕ3¡Á106tÔúµÄ»ðÁ¦·¢µç³§²úÉúµÄµçÄÜ¡£Òò´Ë£¬ÈýÏ¿¹¤³ÌÓÐÖúÓÚ¿ØÖÆ ( ) A. ÎÂÊÒЧӦ B. ËáÓê C. °×É«ÎÛȾ D. ÆÆ»µ´óÆø³ôÑõ²ã 14.ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ ( )
A. »¯Ê¯È¼ÁÏÔÚÈκÎÌõ¼þ϶¼Äܳä·ÖȼÉÕ
B. »¯Ê¯È¼ÁÏÔÚȼÉÕ¹ý³ÌÖÐÄܲúÉúÎÛȾ»·¾³µÄCO¡¢SO2µÈÓк¦ÆøÌå C. Ö±½ÓȼÉÕú²»È罫ú½øÐÐÉî¼Ó¹¤ºóÔÙȼÉÕµÄЧ¹ûºÃ D. ¹ÌÌåú±äÎªÆøÌåȼÁϺó£¬È¼ÉÕЧÂʽ«Ìá¸ß
15.ΪÁ˼õÉÙ´óÆøÎÛȾ£¬Ðí¶à³ÇÊÐÕý´óÁ¦ÍƹãÆû³µÇå½àȼÁÏ¡£Ä¿Ç°Ê¹ÓõÄÇå½àȼÁÏÖ÷ÒªÓÐÁ½À࣬һÀàÊÇѹËõÌìÈ»Æø(CNG)£¬ÁíÒ»ÀàÊÇÒº»¯Ê¯ÓÍÆø(LPG)¡£Õâ
Á½ÀàȼÁϵÄÖ÷Òª³É·Ö¾ùΪ £¨ £© A£®Ì¼Ç⻯ºÏÎï B£®Ì¼Ë®»¯ºÏÎï C£®ÇâÆø D£®´¼Àà
16.ʵÑé²âµÃ25¡æ¡¢101kPaʱ1molCÍêȫȼÉշųö393.5 kJµÄÈÈÁ¿£¬Ð´³öCȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º ¡£ 17.ʵÑé²âµÃ25¡æ¡¢101kPaʱ1molH2ÍêȫȼÉշųö285.8 kJµÄÈÈÁ¿£¬Ð´³öH2ȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º ¡£
5
18.ÔÚ1.01¡Á10Paʱ£¬4gÇâÆøÔÚO2ÖÐÍêȫȼÉÕÉú³ÉҺ̬ˮ£¬·Å³ö572KJµÄÈÈÁ¿£¬ÔòH2µÄȼÉÕÈÈΪ £»±íʾÇâÆøÈ¼ÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ ¡£ 19.ÔÚ1.01¡Á105Paʱ£¬16g SÔÚ×ãÁ¿µÄÑõÆøÖгä·ÖȼÉÕÉú³É¶þÑõ»¯Áò£¬·Å³ö148.5KJµÄÈÈÁ¿£¬ÔòSµÄȼÉÕÈÈΪ £¬SȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ ¡£