ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÂÁз½·¨ÖпÉÒÔ˵Ã÷2HI(g)H2(g)+I2(g)ÒѴﵽƽºâµÄÊÇ£º¢Ùµ¥Î»Ê±¼äÄÚÉú³Én mol H2µÄͬʱÉú³Én mol HI£»¢ÚÒ»¸öH¨CH¼ü¶ÏÁѵÄͬʱÓÐÁ½¸öH¨CI¼ü¶ÏÁÑ£»¢Û°Ù·Ö×é³É¦Ø(HI)=¦Ø(I2)£»¢Ü·´Ó¦ËÙÂʦÔ(H2)=¦Ô(I2)=1/2¦Ô(HI)ʱ£»¢Ý c(HI):c(H2):c(I2)=2:1:1ʱ£»¢ÞζȺÍÌå»ýÒ»¶¨Ê±£¬ÈÝÆ÷ÄÚѹǿ²»Ôٱ仯£»¢ßζȺÍÌå»ýÒ»¶¨Ê±£¬Ä³Ò»Éú³ÉÎïŨ¶È²»Ôٱ仯£»¢àÌõ¼þÒ»¶¨£¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»Ôٱ仯£»¢áζȺÍÌå»ýÒ»¶¨Ê±£¬»ìºÏÆøÌåµÄÑÕÉ«²»Ôٱ仯£»¢âζȺÍѹǿһ¶¨Ê±£¬»ìºÏÆøÌåµÄÃܶȲ»Ôٱ仯¡££¨ £©

A. ¢Ú¢Û¢Ý B. ¢Ù¢Ü¢ß C. ¢Ú¢ß¢á D. ¢à¢á¢â

¡¾´ð°¸¡¿C

¡¾½âÎö¡¿

ÊÔÌâÅжϿÉÄæ·´Ó¦´ïƽºâµÄ±êÖ¾£ºÒ»ÊÇ¿´ÕýÄæ·´Ó¦ËÙÂÊÊÇ·ñÏàµÈ£»¶þÊÇ¿´¸÷ÎïÖʵÄŨ¶ÈÊÇ·ñ¸Ä±ä£¨»ò¸÷×é·ÖµÄ°Ù·Öº¬Á¿ÊÇ·ñ¸Ä±ä£©¡£¢Ú±íʾv£¨Õý£©=v£¨Ä棩¢ßŨ¶È²»±ä¢áÑÕÉ«²»Ôٱ仯¼´Å¨¶ÈÒ»¶¨¡£ÓÉÓڸ÷´Ó¦Êǵķ´Ó¦£¬ÇÒÆøÌåÖÊÁ¿²»·¢Éú±ä»¯£¬»ìºÏÆøÌåѹǿ¡¢Æ½¾ùÏà¶Ô·Ö×ÓÖÊÁ¿¡¢ÃܶȾù²»ÄÜ×÷ΪÅжÏƽºâ״̬µÄ±êÖ¾£¬¹ÊCÕýÈ·¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¡°µÍ̼ѭ»·¡±ÒÑÒýÆð¸÷¹ú¼ÒµÄ¸ß¶ÈÖØÊÓ£¬¶øÈçºÎ½µµÍ´óÆøÖÐCO2µÄº¬Á¿ºÍÓÐЧµØ¿ª·¢ÀûÓÃCO2Õý³ÉΪ»¯Ñ§¼ÒÑо¿µÄÖ÷Òª¿ÎÌâ¡£

(1)½«²»Í¬Á¿µÄCO(g)ºÍH2O(g)·Ö±ðͨÈëµ½Ìå»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬½øÐз´Ó¦CO(g)£«H2O(g)CO2(g)£«H2(g)£¬µÃµ½ÈçÏÂÈý×éÊý¾Ý£º

¢ÙʵÑé2Ìõ¼þÏÂƽºâ³£ÊýK=______________¡£

¢ÚʵÑé3ÖУ¬Èôƽºâʱ£¬COµÄת»¯ÂÊ´óÓÚË®ÕôÆø£¬Ôòa/b µÄÖµ_______(Ìî¾ßÌåÖµ»òÈ¡Öµ·¶Î§)¡£

¢ÛʵÑé4£¬Èô900¡æʱ£¬ÔÚ´ËÈÝÆ÷ÖмÓÈëCO¡¢H2O¡¢CO2¡¢H2¾ùΪ1mol£¬Ôò´ËʱvÕý_________vÄæ(Ìî¡°<¡±£¬¡°>¡±£¬¡°=¡±)¡£

(2)ÒÑÖªÔÚ³£Î³£Ñ¹Ï£º

¢Ù 2CH3OH(l) £« 3O2(g) £½ 2CO2(g) £« 4H2O(g) ¦¤H£½£­1275.6 kJ£¯mol

¢Ú2CO (g)+ O2(g) £½ 2CO2(g) ¦¤H£½£­566.0 kJ£¯mol

¢Û H2O(g) £½ H2O(l) ¦¤H£½£­44.0 kJ£¯mol

д³ö¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º____________

(3)ÒÑÖª²ÝËáÊÇÒ»ÖÖ¶þÔªÈõËᣬ²ÝËáÇâÄÆ(NaHC2O4)ÈÜÒºÏÔËáÐÔ¡£³£ÎÂÏ£¬Ïò10 mL 0.01 mol¡¤L-1H2C2O4ÈÜÒºÖеμÓ10mL 0.01mol¡¤L-1NaOHÈÜҺʱ£¬±È½ÏÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈµÄ´óС¹Øϵ_____________£»

(4)CO2ÔÚ×ÔÈ»½çÑ­»·Ê±¿ÉÓëCaCO3·´Ó¦£¬CaCO3ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÆäKsp=2.8¡Á10-9¡£CaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ¿ÉÐγÉCaCO3³Áµí£¬ÏÖ½«µÈÌå»ýµÄCaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ£¬ÈôNa2CO3ÈÜÒºµÄŨ¶ÈΪ2¡Á10-4mol/L £¬ÔòÉú³É³ÁµíËùÐèCaCl2ÈÜÒºµÄ×îСŨ¶ÈΪ_______________mol/L¡£

(5)ÒÔ¶þ¼×ÃÑ(CH3OCH3)¡¢¿ÕÆø¡¢H2SO4ΪԭÁÏ£¬²¬Îªµç¼«¿É¹¹³ÉȼÁϵç³Ø£¬Æ乤×÷Ô­ÀíÓë¼×ÍéȼÁϵç³ØµÄÔ­ÀíÏàËÆ¡£Çëд³ö¸Ãµç³Ø¸º¼«Éϵĵ缫·´Ó¦Ê½£º__________________¡£

¡¾ÌâÄ¿¡¿50mL0.5mol/LµÄÑÎËáÓë50mL0.55moL/L µÄ NaOH ÈÜÒºÔÚÈçͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ¡£

»Ø´ðÎÊÌ⣺

(1)´ÓʵÑé×°ÖÃÉÏ¿´£¬Í¼ÖÐÉÐȱÉÙÒ»ÖÖ²£Á§ÒÇÆ÷£¬ÕâÖÖ²£Á§ÒÇÆ÷µÄÃû³ÆÊÇ_____£»ÄÜ·ñ¸ÄÓÃÌúË¿________ (Ìî¡°ÄÜ¡¢·ñ¡±)£¬Ô­ÒòÊÇ____£»

(2)ÉÕ±­¼äÌîÂúËéÅÝÄ­ËÜÁϵÄ×÷ÓÃÊÇ_____£»ÊµÑéʱÇâÑõ»¯ÄÆÈÜÒºµÄŨ¶ÈÒªÓà 0.55 mol/L µÄÔ­ÒòÊÇ£º_____________¡£

(3)ʵÑéÖиÄÓà 60mL0.50moL/L ÑÎËá¸ú 50mL0.55moL/L µÄ NaOH ÈÜÒº½øÐз´Ó¦£¬ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿____ (Ìî¡°Æ«´ó¡±¡°ÏàµÈ¡±»ò¡°Æ«Ð¡¡±)£¬ËùÇóÖкÍÈÈ_______(Ìî¡°ÏàµÈ¡±»ò¡°²»ÏàµÈ¡±)¡£

(4) ÈôÈý´ÎƽÐвÙ×÷²âµÃÊý¾ÝÖÐÆðʼʱÑÎËáÓëÉÕ¼îÈÜҺƽ¾ùζÈÏàͬ, ¶øÖÕֹζÈÓëÆðʼζȲî·Ö±ðΪ¢Ù3.1¡æ¢Ú3.2¡æ ¢Û2.7¡æ,Ôò×îÖÕ´úÈë¼ÆËãʽµÄβî¾ùֵΪ______ ¡æ (±£Áô 2 λСÊý)

(5)½üËƵØÈÏΪ 0.55mol/L NaOH ÈÜÒººÍ 0.5mol/L ÑÎËáÃܶȶ¼ÊÇ1 g/cm3£¬ÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝ c£½4.18 J/(g¡¤¡æ)£¬ ÔòÖкÍÈȦ¤H£½_____kJ/moL (ȡСÊýµãºóһλ)

(6)ÉÏÊöʵÑéÊýÖµ½á¹ûÓë 57.3 kJ/mol Ïà±ÈƫС£¬²úÉúÆ«²îµÄÔ­Òò¿ÉÄÜÊÇ(Ìî×Öĸ)___________

a.ʵÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î

b.ÅäÖÆ 0.55 mol/L NaOH ÈÜÒº¶¨ÈÝʱ¸©Êӿ̶ÈÏ߶ÁÊý

c.·Ö¶à´Î°Ñ NaOH ÈÜÒºµ¹ÈëÊ¢ÓÐÑÎËáµÄСÉÕ±­ÖÐ

d.ÓÃζȼƲⶨ NaOH ÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨ÑÎËáÈÜÒºµÄζÈ

e.ÓÃÁ¿Í²Á¿È¡ÑÎËáÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý

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

¾«Ó¢¼Ò½ÌÍø