4£®Èç¹û1¸ö·´Ó¦¿ÉÒÔ·Ö¼¸²½½øÐУ¬Ôò¸÷·Ö²½·´Ó¦µÄ·´Ó¦ÈÈÖ®ºÍÓë¸Ã·´Ó¦Ò»²½Íê³ÉʱµÄ·´Ó¦ÈÈÊÇÏàͬµÄ£¬Õâ¸ö¹æÂɳÆΪ¸Ç˹¶¨ÂÉ£®¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©±±¾©°ÂÔ˻ᡰÏéÔÆ¡±»ð¾æȼÁÏÊDZûÍ飨C3H8£©£¬ÑÇÌØÀ¼´ó°ÂÔË»á»ð¾æȼÁÏÊDZûÏ©£¨C3H6£©£®±ûÍéÍÑÇâ¿ÉµÃ±ûÏ©£®
ÒÑÖª£º¢ÙCH3CH2CH3£¨g£©¡úCH4£¨g£©+HC¡ÔCH£¨g£©+H2£¨g£©£»¡÷H1=+156.6kJ•mol-1?
¢ÚCH3CH=CH2£¨g£©¡úCH4£¨g£©+HC¡ÔCH£¨g£©£»¡÷H2=+32.4kJ•mol-1?
ÔòÏàͬÌõ¼þÏ£¬±ûÍéÍÑÇâµÃ±ûÏ©µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºC3H8£¨g£©¡úCH3CH=CH2£¨g£©+H2£¨g£©¡÷H=+124.2KJ/mol£®
£¨2£©ÒÒÅðÍ飨B2H6£©ÊÇÒ»ÖÖÆø̬¸ßÄÜȼÁÏ£¬0.3molµÄÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5kJÈÈÁ¿£¬ÆäÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£»
ÈôH2O£¨l£©=H2O£¨g£©£»¡÷H=+44kJ/mol£¬Ôò5.6L£¨±ê¿ö£©ÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ£¬·Å³öµÄÈÈÁ¿ÊÇ508.25 kJ£®

·ÖÎö £¨1£©ÒÀ¾ÝÌâÖÐÈÈ»¯Ñ§·½³ÌʽºÍ¸Ç˹¶¨ÂÉд³ö±ûÍéÍÑÇâµÃ±ûÏ©µÄÈÈ»¯Ñ§·½³Ìʽ£»
£¨2£©¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨¿ÉÖª£¬ÎïÖʵÄÎïÖʵÄÁ¿Óë·´Ó¦·Å³öµÄÈÈÁ¿³ÉÕý±È£¬²¢×¢Òâ±êÃ÷¸÷ÎïÖʵľۼ¯×´Ì¬À´½â´ð£»
Ïȸù¾Ý¸Ç˹¶¨ÂÉд³ö·½³Ìʽ£¬È»ºó¸ù¾ÝÎïÖʵÄÎïÖʵÄÁ¿Óë·´Ó¦·Å³öµÄÈÈÁ¿³ÉÕý±ÈÀ´½â´ð£®

½â´ð ½â£º£¨1£©¢ÙC3H8£¨g£©¡úCH4£¨g£©+HC¡ÔCH£¨g£©+H2£¨g£©¡÷H1=+156.6kJ•mol-1
¢ÚCH3CH=CH2£¨g£©¡úCH4£¨g£©+HC¡ÔCH£¨g£©¡÷H2=+32.4kJ•mol-1
ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢Ù-¢Ú¿ÉµÃ£ºC3H8£¨g£©¡úCH3CH=CH2£¨g£©+H2£¨g£©¡÷H=+124.2KJ/mol£¬
¹Ê´ð°¸Îª£ºC3H8£¨g£©¡úCH3CH=CH2£¨g£©+H2£¨g£©¡÷H=+124.2KJ/mol£»
£¨2£©0.3molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5KJµÄÈÈÁ¿£¬Ôò1molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö2165KJµÄÈÈÁ¿£¬·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£¬
¸ù¾Ý£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£» ¢Ù
H2O£¨l£©¡úH2O£¨g£©£»¡÷H=+44kJ/moL£¬¢Ú
¢Ù+¢Ú¡Á3µÃ£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨g£©¡÷H=-2033kJ/mol£¬¹Ê5.6L£¨±ê×¼×´¿ö£©¼´0.25molÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇΪ2033kJ¡Á0.25=508.25kJ£»
¹Ê´ð°¸Îª£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£»508.25£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÈÈ»¯Ñ§·½³ÌʽµÄÊéд£¬¸Ç˹¶¨ÂɵÄÔËÓÃÒÔ¼°ÈÈÁ¿µÄ¼ÆË㣬ÐèҪעÒâµÄÊÇ·´Ó¦ÈȵÄÊýÖµÓ뻯ѧ·½³ÌʽǰÃæµÄϵÊý³ÉÕý±È£¬ÌâÄ¿ÄѶÈÖеȣ¬²àÖØÓÚ¿¼²éѧÉúµÄ·ÖÎöÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

14£®ÒÑÖªÒÒ´¼µÄ·ÐµãΪ78.3¡æ£¬ÓëË®ÒÔÈÎÒâ±È»¥ÈÜ£¬ÇÒÔÚ78.15¡æʱÓëË®¿ÉÐγÉ95.57%£¨Ìå»ý·ÖÊý£¬ÏÂͬ£©µÄºã·Ð»ìºÏÎ¼´Ë®ºÍ¾Æ¾«¿Éͬʱ»Ó·¢£©£®¾Æ¾«ÖеÄË®¿É¼ÓÈëÉúʯ»Ò³ýÈ¥£¬·´Ó¦ÈçÏ£ºCaO+H2O¨TCa£¨OH£©2£¬CaOºÍCa£¨OH£©2¾ù²»ÈÜÓھƾ«£®
ij»¯Ñ§»î¶¯Ð¡×éÒªÓÃ35%µÄ¾Æ¾«ÖÆÈ¡ÎÞË®¾Æ¾«£¬Éè¼ÆÁËÈçÏÂËÄÖÖ·½°¸£®
·½°¸Ò»£º¶Ô35%µÄ¾Æ¾«Ö±½ÓÕôÁóµÃÎÞË®¾Æ¾«£®
·½°¸¶þ£ºÏȶÔ35%µÄ¾Æ¾«ÕôÁóµÃ95.57%µÄ¾Æ¾«£¬ÔÙ¼ÓÈëÉúʯ»Ò£¬È»ºó½øÐÐÕôÁóµÃÎÞË®¾Æ¾«£®
·½°¸Èý£ºÓÃCCl4¶Ô35%µÄ¾Æ¾«½øÐÐÝÍÈ¡£¬È»ºó¶Ô¾Æ¾«µÄCCl4ÈÜÒº½øÐÐÕôÁóµÃÎÞË®¾Æ¾«£®
·½°¸ËÄ£ºÏò35%µÄ¾Æ¾«ÖÐÖ±½Ó¼ÓÈëÉúʯ»Ò£¬È»ºó½øÐÐÕôÁóµÃÎÞË®¾Æ¾«£®
Çë¶ÔËÄÖÖ·½°¸½øÐÐÆÀ¼Û£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×îºÃµÄ·½°¸ÊÇ·½°¸¶þ£¬´íÎóµÄ·½°¸ÊÇ·½°¸Ò»ºÍ·½°¸Èý£¬´íÎóµÄÔ­ÒòÊÇ·½°¸Ò»¾Æ¾«ÓëË®¿ÉÐγÉ95.57%µÄºã·Ð»ìºÏÎֱ½ÓÕôÁóÎÞ·¨µÃÎÞË®¾Æ¾«£»·½°¸Èý£¬¾Æ¾«ÓëË®»ìÈÜ£¬²»ÄÜÓÃCCl4ÝÍÈ¡£®
£¨2£©ÆäÓà·½°¸µÄ²»×ãÊÇ·½°¸Ëĵľƾ«ÖÐˮ̫¶à£¬ÒªÏûºÄ´óÁ¿µÄÉúʯ»Ò£®
£¨3£©ÓÃ×îºÃµÄ·½°¸½øÐÐʵÑéʱ£¬ËùÓõ½µÄ²£Á§ÒÇÆ÷ÓÐÕôÁóÉÕÆ¿¡¢ÀäÄý¹Ü¡¢Å£½Ç¹Ü¡¢×¶ÐÎÆ¿¡¢Î¶ȼƺ;ƾ«µÆ£®
£¨4£©ÕôÁó²Ù×÷ÖÐҪעÒâµÄÊÂÏîÊÇζȼÆË®ÒøÇòÒªÔÚÉÕÆ¿Ö§¹Ü¿Ú´¦£»ÀäÄýˮҪÏ¿ڽøÉÏ¿Ú³ö£»¼ÓÈÈʱҪÊʵ±µ÷Õû¾Æ¾«µÆµÄλÖã¬ÒÔ¿ØÖÆζȣ»ÏÈͨÀäÄýË®ºó¼ÓÈÈ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

15£®ÏÂÁÐÈÜÒº¼ÓÈÈÕô¸Éºó£¬²»ÄÜÎö³öÔ­ÈÜÖʹÌÌåµÄÊÇ£¨¡¡¡¡£©
A£®Al2£¨SO4£©3B£®Na2CO3C£®AlCl3D£®KNO3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®ÖÐÃÀÑо¿ÈËÔ±ÔÚÐÂÒ»ÆÚÃÀ¹ú£¨»·¾³¿ÆѧÓë¼¼Êõ£©ÔÓÖ¾Éϱ¨¸æ˵£¬»Æ·Û³æ¿ÉÒÔÍÌʳºÍÍêÈ«½µ½âËÜÁÏ£®ËûÃÇÒÑÔڻƷ۳æÌåÄÚ·ÖÀë³ö¿¿¾Û±½ÒÒÏ©Éú´æµÄϸ¾ú£¬²¢½«Æä±£´æ£®¾Û±½ÒÒÏ©ÊôÓÚ£¨¡¡¡¡£©
A£®ºÏ½ðB£®¹èËáÑβÄÁÏC£®Óлú¸ß·Ö×Ó²ÄÁÏD£®ÎÞ»ú·Ç½ðÊô²ÄÁÏ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

19£®¿Æѧ¼ÒÖÂÁ¦ÓÚ¶þÑõ»¯Ì¼µÄ¡°×éºÏת»¯¡¯¼¼ÊõµÄÑо¿£¬°Ñ¹ý¶à¶þÑõ»¯Ì¼×ª»¯ÎªÓÐÒæÓÚÈËÀàµÄÎïÖÊ£®
¢ÙÈç¹û½«CO2ºÍH2ÒÔ1£º4µÄ±ÈÀý»ìºÏ£¬Í¨ÈË·´Ó¦Æ÷£¬ÔÚÊʵ±µÄÌõ¼þÏ·´Ó¦£¬¿É»ñµÃÒ»ÖÖÖØÒªµÄÄÜÔ´£®ÇëÍê³ÉÒÔÏ»¯Ñ§·½³Ìʽ£ºCO2+4H2¡úCH4+2H20
¢ÚÈô½«CO2ºÍH2ÒÔ1£º3µÄ±ÈÀý»ìºÏ£¬Ê¹Ö®·¢Éú·´Ó¦Éú³ÉijÖÖÖØÒªµÄ»¯¹¤Ô­ÁϺÍË®£¬ÔòÉú³ÉµÄ¸ÃÖØÒª»¯¹¤Ô­ÁÏ¿ÉÄÜÊÇB £¨Ìî×Öĸ£©£®
A£®ÍéÌþ          B£®Ï©Ìþ          C£®È²Ìþ           D£®·¼ÏãÌþ
¢Û¶þÑõ»¯Ì¼¡°×éºÏת»¯¡±µÄijÍéÌþ̼¼Ü½á¹¹ÈçͼËùʾ£º£¬´ËÍéÌþµÄÒ»äå´úÎïÓÐ6 ÖÖ£¬Èô´ËÍéÌþΪȲÌþ¼ÓÇâÖƵã¬Ôò´ËȲÌþµÄ½á¹¹¼òʽΪ£¨CH3£©2CH£¨CH3£©CHC¡ÔCH£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

9£®£¨1£©ÓÉÇâÆøºÍÑõÆø·´Ó¦Éú³É1molË®ÕôÆø·ÅÈÈ241.8kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£ºH2£¨g£©+$\frac{1}{2}$O2£¨g£©¨TH2O£¨g£©¡÷H=-241.8kJ•mol-1£®ÒÑÖª18gҺ̬ˮת»¯³ÉË®ÕôÆøÐèÎüÈÈ44kJ£¬Ôò·´Ó¦2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©µÄ¡÷H=-571.6kJ•mol-1£¬ÇâÆøµÄ±ê׼ȼÉÕÈÈ¡÷H=-285.8kJ•mol-1£®
£¨2£©ÒÑÖªH2£¨g£©+Cl2£¨g£©¨T2HCl£¨g£©¡÷H=-184.6kJ•mol-1£®ÆäËüÏà¹ØÊý¾ÝÈçÏÂ±í£º
H2£¨g£©Cl2 £¨g£©HCl £¨g£©
1mol·Ö×ÓÖеĻ¯Ñ§¼ü¶ÏÁÑʱÐèÒªÎüÊÕµÄÄÜÁ¿/kJ436243a
Ôò±íÖÐaΪ431.8£®
£¨3£©ÒÑÖª£º2SO2£¨g£©+O2£¨g£©¨T2SO3£¨g£©¡÷H=-196.6kJ•mol-1
2NO£¨g£©+O2£¨g£©¨T2NO2£¨g£©¡÷H=-113.0kJ•mol-1
Ôò·´Ó¦NO2£¨g£©+SO2£¨g£©¨TSO3£¨g£©+NO£¨g£©µÄ¡÷H=-41.8kJ•mol-1£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

16£®ÓÐÒ»Õæ¿ÕÃܱÕÈÝÆ÷ÖУ¬Ê¢ÓÐamolPCl5£¬¼ÓÈȵ½200¡æʱ£¬·¢ÉúÈçÏ·´Ó¦PCl5£¨g£©?PCl3£¨g£©+Cl2£¨g£©£¬·´Ó¦´ïƽºâʱ£¬PCl5ËùÕ¼Ìå»ý·ÖÊýΪM%£¬ÈôÔÚͬһζÈÏ£¬Í¬Ò»ÈÝÆ÷ÖУ¬×î³õͶÈë2amolPCl5·´Ó¦´ïƽºâʱ£¬PCl5ËùÕ¼Ìå»ý·ÖÊýΪN%£¬ÔòMÓëNµÄ¹ØϵÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®M£¼NB£®2M=NC£®N£¼MD£®M=N

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®ÏÂÁйý³ÌÊôÓÚ»¹Ô­·´Ó¦µÄÊÇ£¨¡¡¡¡£©
A£®CO2¡úCOB£®Zn¡úZn2+C£®H2¡úH2OD£®CuO¡úCuCl2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®ÏÂÁÐÓйػ¯Ñ§ÓÃÓï±íʾÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ClÔ­×ӵĽṹʾÒâͼ£ºB£®´ÎÂÈËáµÄ½á¹¹Ê½£ºH-Cl-O
C£®NH4ClµÄµç×Óʽ£ºD£®¼äÏõ»ù¼×±½µÄ½á¹¹¼òʽ£º

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸