¡¾ÌâÄ¿¡¿Ä³ÊµÑéС×éÓÃ0.50 mol/L NaOHÈÜÒººÍ0.50 mol/LÁòËáÈÜÒº½øÐÐÖкÍÈȵIJⶨ¡£

¢ñ£®ÅäÖÆ0.50 mol/L NaOHÈÜÒº

ÈôʵÑéÖдóԼҪʹÓÃ245 mL NaOHÈÜÒº£¬ÖÁÉÙÐèÒª³ÆÁ¿NaOH¹ÌÌå_____g¡£

¢ò£®²â¶¨Ï¡ÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÖкÍÈȵÄʵÑé×°ÖÃÈçͼËùʾ¡£

£¨1£©Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£¨ÖкÍÈÈΪ57.3 kJ/mol£©£º_____________¡£

£¨2£©È¡50 mL NaOHÈÜÒººÍ30 mLÁòËáÈÜÒº½øÐÐʵÑ飬ʵÑéÊý¾ÝÈçÏÂ±í¡£

¢ÙÇëÌîдϱíÖеĿհףº

ÆðʼζÈt1/¡æ

ÖÕֹζÈ

t2/¡æ

ζȲîƽ¾ùÖµ

£¨t2-t1£©/¡æ

H2SO4

NaOH

ƽ¾ùÖµ

1

26.2

26.0

26.1

30.1

______

2

27.0

27.4

27.2

33.3

3

25.9

25.9

25.9

29.8

4

26.4

26.2

26.3

¢Ú½üËÆÈÏΪ0.50 mol/L NaOHÈÜÒººÍ0.50 mol/LÁòËáÈÜÒºµÄÃܶȶ¼ÊÇ1 g/cm3£¬ÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc=4.18 J/(g¡¤¡æ£©¡£ÔòÖкÍÈÈ¡÷H=______________ȡСÊýµãºóһ룩¡£

¢ÛÉÏÊöʵÑéÊýÖµ½á¹ûÓë57.3 kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔ­Òò¿ÉÄÜÊÇ£º______¡£

A£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î

B£®Á¿È¡NaOHÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý

C£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±­ÖÐ

D£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζÈ

¡¾´ð°¸¡¿5.01/2H2SO4(aq£©£«NaOH(aq£©£½1/2Na2SO4(aq£©£«H2O(l£© ¡÷H£½£­57.3kJ/mol4.0-53.5kJ/molacd

¡¾½âÎö¡¿

±¾Ì⿼²éÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÅäÖƵÄÓйؼÆË㣬ÈÈ»¯Ñ§·½³ÌʽµÄÊéдÒÔ¼°ÖкÍÈȵļÆË㣬Îó²î·ÖÎöµÈ£¬ÌâÄ¿ÄѶȲ»´ó£¬×¢ÒâÀí½âÖкÍÈȵĸÅÄî¡¢°ÑÎÕÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨£¬ÒÔ¼°²â¶¨·´Ó¦ÈȵÄÎó²îµÈÎÊÌâ¡£

I¡¢£¨1£©ÓÉÓÚʵÑéÊÒÖÐûÓÐ245mLÈÝÁ¿Æ¿£¬Ó¦Ñ¡ÓÃ250mLµÄÈÝÁ¿Æ¿À´ÅäÖÆ250mLµÄÈÜÒº£¬ËùÐèµÄÇâÑõ»¯ÄƹÌÌåµÄÖÊÁ¿m=cVM=0.50mol/L¡Á0.25L¡Á40g/mol=5.0g¡£

II¡¢£¨1£©¸ù¾ÝÖкÍÈȵĶ¨Ò壬ϡǿËᡢϡǿ¼î·´Ó¦Éú³É1molˮʱ·Å³ö57.3kJµÄÈÈÁ¿£¬Ï¡ÁòËáºÍÇâÑõ»¯ÄÆÏ¡ÈÜÒº·Ö±ðÊÇÇ¿Ëᡢǿ¼î£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ1/2H2SO4(aq£©£«NaOH(aq£©£½1/2Na2SO4(aq£©£«H2O(l£© ¡÷H£½£­57.3kJ/mol¡£

£¨2£©¢Ù¸ù¾Ý±í¸ñÌṩµÄÐÅÏ¢£¬µÚ2×éÊý¾ÝÎó²î½Ï´ó£¬ÉáÈ¥£¬ÔòζȲîƽ¾ùÖµ=£¨4.0+3.9+4.1£©¡Â3=4.0¡ãC¡£¢Ú50mL0.50mol/LÇâÑõ»¯ÄÆÓë30mL0.50mol/LÁòËáÈÜÒº½øÐÐÖкͷ´Ó¦Éú³ÉË®µÄÎïÖʵÄÁ¿Îª0.05L¡Á0.50mol/L=0.025mol£¬ÈÜÒºµÄÖÊÁ¿Îª£º80mL¡Á1g/mL=80g£¬Î¶ȱ仯µÄֵΪ¡÷T=4¡æ£¬ÔòÉú³É0.025molË®·Å³öµÄÈÈÁ¿ÎªQ=mc¡÷T=80g¡Á4.18J/£¨g¡æ£©¡Á4.0¡æ=1337.6J£¬¼´1.3376kJ£¬ËùÒÔʵÑé²âµÃµÄÖкÍÈÈ¡÷H=-1.3376kJ¡Â0.025mol=-53.5 kJ/mol¡£

¢Ûa£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û±ØÐëºÃ£¬·ñÔò»áµ¼ÖÂÈÈÁ¿É¢Ê§£¬Ê¹²â¶¨½á¹ûƫС£¬aÏîÕýÈ·£»b£®Á¿È¡NaOHÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý£¬»áµ¼ÖÂËùÁ¿µÄÇâÑõ»¯ÄÆÌå»ýÆ«´ó£¬·Å³öµÄÈÈÁ¿Æ«¸ß£¬Ôò´óÓÚ57.3kJ/mol£¬bÏî´íÎó£»c£®¾¡Á¿Ò»´Î¿ìËÙ½«NaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±­ÖУ¬Èô·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±­ÖУ¬»áµ¼ÖÂÈÈÁ¿É¢Ê§£¬cÕýÈ·£»d£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺó£¬Òª½«Î¶ȼƻØÁãÔٲⶨH2SO4ÈÜÒºµÄζȣ¬·ñÔò»áÒýÆðβî¼õС£¬dÏîÕýÈ·¡£´ð°¸Ñ¡acd¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂͼËùʾװÖÃÖУ¬¼×¡¢ÒÒ¡¢±ûÈý¸öÉÕ±­ÒÀ´Î·Ö±ðÊ¢·Å100 g 5.00%µÄNaOHÈÜÒº¡¢×ãÁ¿µÄCuSO4ÈÜÒººÍ100 g 10.00%µÄK2SO4ÈÜÒº£¬µç¼«¾ùΪʯīµç¼«¡£

(1)½ÓͨµçÔ´£¬¾­¹ýÒ»¶Îʱ¼äºó£¬²âµÃ±ûÖÐK2SO4Ũ¶ÈΪ10.47%£¬ÒÒÖÐcµç¼«ÖÊÁ¿Ôö¼Ó¡£¾Ý´Ë»Ø´ðÎÊÌ⣺

¢ÙµçÔ´µÄN¶ËΪ____________________¼«£»

¢Úµç¼«bÉÏ·¢ÉúµÄµç¼«·´Ó¦Îª____________________£»

¢ÛÁÐʽ¼ÆËãµç¼«bÉÏÉú³ÉµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý£º________________£»

¢Üµç¼«cµÄÖÊÁ¿±ä»¯ÊÇ__________g£»

¢Ýµç½âÇ°ºó¸÷ÈÜÒºµÄËá¡¢¼îÐÔ´óСÊÇ·ñ·¢Éú±ä»¯£¬¼òÊöÆäÔ­Òò£º¼×ÈÜÒº______________________________£»ÒÒÈÜÒº______________________________£»±ûÈÜÒº______________________________£»

(2)Èç¹ûµç½â¹ý³ÌÖÐÍ­È«²¿Îö³ö£¬´Ëʱµç½âÄÜ·ñ¼ÌÐø½øÐУ¬ÎªÊ²Ã´£¿ ________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»¯Ñ§ÓëÉú»îÃÜÇÐÏà¹Ø£¬ÏÂÁÐÓйØ˵·¨´íÎóµÄÊÇ£¨ £©
A.ÓÃ×ÆÉյķ½·¨¿ÉÒÔÇø·Ö²ÏË¿ºÍÈËÔìÏËά
B.ʳÓÃÓÍ·´¸´¼ÓÈÈ»á²úÉú³í»··¼ÏãÌþµÈÓк¦ÎïÖÊ
C.¼ÓÈÈÄÜɱËÀÁ÷¸Ð²¡¶¾ÊÇÒòΪµ°°×ÖÊÊÜÈȱäÐÔ
D.Ò½ÓÃÏû¶¾¾Æ¾«ÖÐÒÒ´¼µÄŨ¶ÈΪ95%

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÖÒÑÖªO3·Ö×ÓΪV×ÖÐνṹ£¬O3ÔÚË®ÖеÄÈܽâ¶ÈºÍO2±È½Ï

A.O3ÔÚË®ÖеÄÈܽâ¶ÈºÍO2Ò»Ñù B.O3ÔÚË®ÖеÄÈܽâ¶È±ÈO2С

C.O3ÔÚË®ÖеÄÈܽâ¶È±ÈO2Òª´ó D.û°ì·¨±È½Ï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ò»¶¨Î¶ÈÏ£¬¶ÔÓÚ¿ÉÄæ·´Ó¦N2(g) +3H2(g) 2NH3(g)£¨Õý·´Ó¦·ÅÈÈ£©µÄÏÂÁÐÐðÊö£¬²»ÄÜ˵Ã÷·´Ó¦ÒѴﻯѧƽºâ״̬µÄÊÇ£¨ £©

A. ºãÈÝÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄÃܶȲ»Ôٱ仯

B. NH3µÄÉú³ÉËÙÂÊÓëH2µÄÉú³ÉËÙÂÊÖ®±ÈΪ2:3

C. ºãѹÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄ×ÜÎïÖʵÄÁ¿²»Ôٱ仯

D. µ¥Î»Ê±¼äÄÚ¶ÏÁÑa mol NÈýN¼ü£¬Í¬Ê±¶ÏÁÑ6amol N-H¼ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»·¾³¼à²âÏÔʾ£¬Ä³µØÊеÄÖ÷ÒªÆøÌåÎÛȾÎïΪSO2¡¢NOx¡¢COµÈ£¬ÆäÖ÷ÒªÀ´Ô´ÎªÈ¼Ãº¡¢»ú¶¯³µÎ²Æø¡£½øÐÐÈçÏÂÑо¿£º

£¨1£©Îª¼õÉÙȼú¶ÔSO2µÄÅÅ·Å£¬¿É½«Ãº×ª»¯ÎªÇå½àȼÁÏˮúÆø£¨COºÍH2£©¡£

ÒÑÖª£º ¦¤H£½241.8kJ¡¤mol£­1£¬

¦¤H£½£­110.5kJmol£­1

д³ö½¹Ì¿Óë1molË®ÕôÆø·´Ó¦Éú³ÉˮúÆøµÄÈÈ»¯Ñ§·½³Ìʽ£º________¡£

£¨2£©Æû³µÎ²ÆøÖÐNOÊÇÔÚ·¢¶¯»úÆû¸×ÖÐÉú³ÉµÄ£¬·´Ó¦ÎªN2£¨g£©£«O2£¨g£©2NO£¨g£© ¦¤H£¾0¡£

¢Ù½«º¬0.8molN2ºÍ0.2molO2£¨½üËÆ¿ÕÆø×é³É£©µÄ»ìºÏÆøÌå³äÈëijÃܱÕÈÝÆ÷ÖУ¬±£³Ö1300¡æ·´Ó¦´ïµ½Æ½ºâ£¬²âµÃÉú³É8¡Á10£­4molNO¡£¼ÆËã¸ÃζÈÏ´˷´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK£½________£¨Ìî½üËƼÆËã½á¹û£©¡£

¢ÚÆû³µÆô¶¯ºó£¬Æû¸×ÄÚζÈÔ½¸ß£¬µ¥Î»Ê±¼äÄÚNOÅÅ·ÅÁ¿Ô½´ó£¬Ô­ÒòÊÇ________£®

£¨3£©ÀûÓÃÈçͼËùʾװÖ㨵缫¾ùΪ¶èÐԵ缫£©¿ÉÎüÊÕSO2£¬²¢ÀûÓÃÒõ¼«ÅųöµÄÈÜÒºÎüÊÕNO2¡£

¢Ùµç¼«AµÄµç¼«·´Ó¦Ê½Îª________£»

µç¼«BµÄµç¼«·´Ó¦Ê½Îª________¡£

¢Ú¼îÐÔÌõ¼þÏ£¬ÓÃÒõ¼«ÅųöµÄÈÜÒºÎüÊÕNO2£¬Ê¹Æäת»¯ÎªÎÞº¦ÆøÌ壬ͬʱÉú³ÉSO32£­¡£¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊµÑéÊÒÄ£ÄâºÏ³ÉÁòËáµÄÁ÷³ÌÈçÏ£º

´ÓÏÂͼÖÐÑ¡ÔñÖÆÈ¡ÆøÌåµÄºÏÊÊ×°Öãº

£¨1£©×°ÖÃCµÄÃû³ÆΪ_________________£¬ÊµÑéÊÒͨ³£ÄÜÓÃ×°ÖÃCÖƱ¸_________¡£

A£®H2 B£®H2S C£®CO2 D£®O2

£¨2£©ÊµÑéÊÒÓÃ×°ÖÃDÖƱ¸NH3µÄ»¯Ñ§·½³ÌʽΪ_______________________________¡£

£¨3£©ÈôÓÃ×°ÖÃBÖƱ¸SO2£¬¿ÉÒÔÑ¡ÓÃÊÔ¼ÁΪ___________¡£

A¡¢Å¨ÁòËá¡¢ÑÇÁòËáÄƹÌÌå B£®Å¨ÁòËᡢͭƬ

C£®Ï¡ÁòËá¡¢ÑÇÁòËáÈÜÒº D£®Å¨ÁòËá¡¢Ìúм

£¨4£©SO2ºÍO2ͨ¹ý¼××°Ö㬼××°ÖõÄ×÷ÓóýÁË¿ÉÒÔ¿ØÖÆSO2¡¢O2µÄÁ÷ËÙÍ⣬»¹¿ÉÒÔ__________¡£

£¨5£©Èô½«Ò»¶¨Á¿µÄSO2(g)ºÍO2(g)·Ö±ðͨÈëµ½Ìå»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬ÔÚ²»Í¬Î¶ÈϽøÐз´Ó¦µÃµ½ÈçϱíÖеÄÁ½×éÊý¾Ý£º

ʵÑé±àºÅ

ζÈ/¡æ

ÆðʼÁ¿/mol

ƽºâÁ¿/mol

SO2

O2

SO2

O2

1

T1

4

2

x

0.8

2

T2

4

2

0.4

y

±íÖÐx=_______mol£¬y=_______mol£¬T1______T2£¨Ìî¡°>¡±¡¢¡°=¡±»ò¡°<¡±)

£¨6£©SO2βÆø¿ÉÓÃNaOHÈÜÒºÎüÊÕ£¬µÃµ½Na2SO3ºÍNaHSO3Á½ÖÖÑΡ£ÈôÒ»¶¨ÎïÖʵÄÁ¿µÄSO2ÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£¬ËùµÃÈÜÒºÖÐNa2SO3ºÍNaHSO3µÄÎïÖʵÄÁ¿Ö®±ÈΪ2:3£¬Ôò²Î¼Ó·´Ó¦µÄSO2ºÍNaOHµÄÎïÖʵÄÁ¿Ö®±ÈΪ________¡£

A. 5:7 B£®1:2 C£®9:4 D£®9:13

£¨7£©ÁòËáÓ백ˮ·´Ó¦Éú³ÉÁòËá炙òÁòËáÇâ泥¬ÏÖ³ÆÈ¡(NH4)2SO4ºÍNH4HSO4»ìºÏÎïÑùÆ·7.58 g£¬¼ÓÈ뺬0.1 molNaOHµÄÈÜÒº£¬·¢Éú·´Ó¦£º¢ÙOH-+H+¡úH2O, ¢ÚOH-+NH4+¡úNH3¡ü+H2O¡£³ä·Ö·´Ó¦£¬Éú³É1792 mL£¨±ê×¼×´¿ö£©¡£ÒÑÖª¢ÙÓÅÏÈ·´Ó¦£¬ÔòÑùÆ·ÖÐ(NH4)2SO4µÄÎïÖʵÄÁ¿Îª_______mol£¬NH4HSO4µÄÎïÖʵÄÁ¿Îª_______mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÓйص绯ѧԭÀíµÄ˵·¨´íÎóµÄÊÇ£¨ £©

A£®ÂÈ»¯ÂÁµÄÈÛµã±ÈÑõ»¯ÂÁµÍ£¬Òò´Ë¹¤ÒµÉÏ×îºÃ²ÉÓõç½âÈÛÈÚÂÈ»¯ÂÁÀ´ÖƱ¸µ¥ÖÊÂÁ

B£®µç¶Æʱ£¬Í¨³£°Ñ´ý¶ÆµÄ½ðÊôÖÆÆ·×÷Òõ¼«£¬°Ñ¶Æ²ã½ðÊô×÷Ñô¼«

C£®¶ÔÓÚÒ±Á¶ÏñÄÆ¡¢¸Æ¡¢Ã¾¡¢ÂÁµÈÕâÑù»îÆõĽðÊô£¬µç½â·¨¼¸ºõÊÇΨһ¿ÉÐеĹ¤Òµ·½·¨

D£®¶Ô´óÐÍ´¬²°µÄÍâ¿Ç½øÐеġ°ÎþÉüÑô¼«µÄÒõ¼«±£»¤·¨¡±£¬ÊÇÓ¦ÓÃÁËÔ­µç³ØÔ­Àí

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁи÷×éÈÈ»¯Ñ§·½³ÌʽÖУ¬»¯Ñ§·´Ó¦µÄ¦¤HÇ°Õß´óÓÚºóÕßµÄÊÇ( )

¢ÙC(s)£«O2(g)===CO2(g) ¦¤H1

C(s)£«O2(g)===CO(g) ¦¤H2

¢ÚS(s)£«O2(g)===SO2(g) ¦¤H3

S(g)£«O2(g)===SO2(g) ¦¤H4

¢ÛH2(g)£«O2(g)===H2O(l) ¦¤H5

2H2(g)£«O2(g)===2H2O(l) ¦¤H6

¢ÜCaCO3(s)===CaO(s)£«CO2(g) ¦¤H7

CaO(s)£«H2O(l)===Ca(OH)2(s) ¦¤H8

A£®¢Ù¢Ú B£®¢Ù¢Ü C£®¢Ú¢Û¢Ü D£®¢Ù¢Ú¢Û

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸