¡¾ÌâÄ¿¡¿¡¾ÌâÄ¿¡¿ÏÂÁÐÃèÊöÖÐÕýÈ·µÄÊÇ ( )

A. ´¿¼îÈÜÓÚÈÈË®ºóÈ¥ÎÛЧ¹ûÔöÇ¿£¬ËµÃ÷´¿¼îµÄË®½â·´Ó¦ÊÇÎüÈÈ·´Ó¦

B. NH3(g) + HCl(g) = NH4Cl(s) ÔڽϵÍζÈÏÂÄÜ×Ô·¢½øÐУ¬ËµÃ÷¸Ã·´Ó¦µÄ¦¤H>0

C. 500¡æ¡¢30 MPaÏ£¬½«7 g N2ºÍ3 g H2ÖÃÓÚÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦Éú³ÉNH3(g)£¬·ÅÈÈ19.3 kJ£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪN2(g)£«3H2(g)2NH3(g) ¦¤H£½£­38.6 kJ¡¤mol£­1

D. ¼×ÍéµÄȼÉÕÈÈ£¨¦¤H£©Îª£­890.3 kJ¡¤mol£­1£¬Ôò¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪCH4(g)£«2O2(g)£½ CO2(g)£«2H2O(g)¡¡¦¤H£½£­890.3 kJ¡¤mol£­1

¡¾´ð°¸¡¿A

¡¾½âÎö¡¿A. ´¿¼îÈÜÓÚÈÈË®ºóÈ¥ÎÛЧ¹ûÔöÇ¿£¬ËµÃ÷Éý¸ßζȴٽøÁËË®½â£¬Ê¹Ë®½âƽºâÕýÏòÒƶ¯£¬ËùÒÔ¿ÉÒÔ˵Ã÷´¿¼îµÄË®½âÊÇÎüÈÈ·´Ó¦£¬¹ÊAÕýÈ·£»B. ¸ù¾Ý¡÷G£½¡÷H£­T¡÷S¿ÉÖª£¬µ±¡÷GСÓÚ0ʱ·´Ó¦×Ô·¢½øÐУ¬ÓÉÓÚ·´Ó¦NH3(g) + HCl(g) = NH4Cl(s)ÊÇìØÖµ¼õСµÄ·´Ó¦£¬ËùÒÔÈç¹û¸Ã·´Ó¦ÔڽϵÍζÈÏÂÄÜ×Ô·¢½øÐУ¬Ôò˵Ã÷¸Ã·´Ó¦µÄ¡÷H£¼0£¬¹ÊB´íÎó£»C.ÒòºÏ³É°±µÄ·´Ó¦ÊÇ¿ÉÄæ·´Ó¦£¬ËùÒÔÎÞ·¨È·¶¨½«7 g N2ºÍ3 g H2ÖÃÓÚÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦ºó£¬Éú³ÉÁ˶àÉÙNH3(g)£¬¹ÊÎÞ·¨¼ÆËã¸Ã·´Ó¦µÄìʱ䣬¹ÊC´íÎó£»D. ȼÉÕÈÈÖ¸µÄÊÇ1mol´¿ÎïÖÊÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯ÎïʱËù·Å³öµÄÄÜÁ¿£¬ËùÒÔÉú³ÉµÄˮӦΪҺ̬£¬¼´¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽӦ±íʾΪCH4(g)£«2O2(g)£½ CO2(g)£«2H2O(l)¡¡¦¤H£½£­890.3 kJ¡¤mol£­1£¬¹ÊD´íÎó£»´ð°¸Ñ¡A¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÀûÓÃÈçͼËùʾװÖòⶨÖкÍÈȵÄʵÑé²½ÖèÈçÏ£º

¢ÙÓÃÁ¿¼òÁ¿È¡30mL0.50 molL-1ÁòËáµ¹ÈëСÉÕ±­ÖУ¬²â³öÁòËáζȣ»

¢ÚÓÃÁíÒ»Á¿Í²Á¿È¡50mL0.50molL-1NaOHÈÜÒº£¬²¢ÓÃͬһζȼƲâ³öÆäζȣ»

¢Û½«NaOHÈÜÒºµ¹ÈëСÉÕ±­ÖУ¬É跨ʹ֮»ìºÏ¾ùÔÈ£¬²â³ö»ìºÏÒº×î¸ßζȡ£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Èç¹ûÓÃ30mL0.50mol/LÁòËáÓë70mL0.50mol/LNaOHÈÜÒº½øÐз´Ó¦£¬ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿___________£¨Ìî¡°ÏàµÈ¡±¡¢¡°²»ÏàµÈ¡±£©£¬ËùÇóÖкÍÈÈ___________£¨Ìî¡°ÏàµÈ¡±¡¢¡°²»ÏàµÈ¡±£©£¬¼òÊöÀíÓÉ_____________________¡£

£¨2£©Ä³ÊµÑéС×éÅäÖÆ0.50mol/LNaOHÈÜÒº£¬ÈôʵÑéÖдóԼҪʹÓÃ245mLNaOHÈÜÒº£¬ÖÁÉÙÐèÒª³ÆÁ¿NaOH¹ÌÌå_____________________g¡£

£¨3£©ÈôÓÃKOH´úÌæNaOH£¬¶Ô²â¶¨½á¹û___________(Ìî¡°ÓС±»ò¡°ÎÞ¡±)Ó°Ï죻ÈôÓô×Ëá´úÌæHCl×öʵÑ飬¶Ô²â¶¨½á¹û_____________(Ìî¡°ÓС±»ò¡°ÎÞ¡±)Ó°Ïì¡£

£¨4£©¢Ù½üËÆÈÏΪ0.50 molL-1ÁòËáºÍ0.50 molL-1NaOHÈÜÒºÃܶȶ¼ÊÇ1g¡¤cm3£¬ÖкͺóÈÜҺζÈÉÏÉýÁË4¡æ£¬Éú³ÉÈÜÒºµÄ±ÈÈÈÈÝc=4.18J¡¤(g¡¤¡æ)-1£¬ÔòÖкÍÈÈ¡÷H=______£¨Ð¡Êýµãºó±£ÁôһλÓÐЧÊý×Ö£©¡£

¢ÚÉÏÊöʵÑé½á¹ûÓë-57.3kJ¡¤mol-1ÓÐÆ«²î£¬²úÉúÆ«²îµÄÔ­Òò¿ÉÄÜÊÇ____________£¨Ìî×Öĸ£©¡£

A.ʵÑé×°Öñ£Î¸ôÈÈЧ¹û²î

B.Á¿È¡NaOHÈÜÒºÌå»ýʱÑöÊÓ¶ÁÊý

C.·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÑÎËáµÄСÉÕ±­ÖÐ

D.ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨ÑÎËáµÄζÈ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿A¡¢B¡¢C¡¢D¡¢E¡¢F ÊÇÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÖ÷×åÔªËØ¡£E ÊÇͬÖÜÆÚ½ðÊôÐÔ×îÇ¿µÄÔªËØ¡£¼×¡¢ÒÒ¡¢±û¡¢¶¡¡¢ÎìÊÇÓÉÉÏÊö²¿·ÖÔªËØÖеÄÁ½ÖÖ»ò¼¸ÖÖ×é³ÉµÄ»¯ºÏÎËüÃÇÖ®¼äµÄת»¯¹ØϵÈçͼËùʾ¡£ÆäÖм×ÊÇÉú»îÖеĵ÷ζƷ£¬¶¡Êǵ­»ÆÉ«¹ÌÌå¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ

A. ÔªËØBµÄÇ⻯Îï¿ÉÄܾßÓÐÕýËÄÃæÌå½á¹¹

B. ÔªËØÉϵÄÑõ»¯Îï¶ÔӦˮÎﻯµÄËáÐÔÒ»¶¨±ÈÁòËáµÄËáÐÔÇ¿

C. Ô­×Ӱ뾶:r(E)>r(F)>r(C)>r(D)

D. »¯ºÏÎïA4BC2D¼È¿ÉÄÜÊǹ²¼Û»¯ºÏÎҲ¿ÉÊÇÀë×Ó»¯ºÏÎï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁл¯Ñ§ÓÃÓïÕýÈ·µÄÊÇ £¨ £©

A. NaHCO3µÄË®½â£ºHCO3£­£«H2OH3O£«£«CO32£­

B. CaCO3µÄµçÀ룺CaCO3Ca2+£«CO32-

C. ̼Ëá¸ÆµÄÈܽâƽºâ£ºCaCO3(s)Ca2£«(aq)£«CO32£­(aq)

D. Na2SÈÜÒºµÄË®½â£ºS2-£«2H2OH2S£«2OH-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿È«·°Ñõ»¯»¹Ô­µç³ØÊÇÒ»ÖÖÐÂÐͿɳäµç³Ø£¬²»Í¬¼Û̬µÄº¬·°Àë×Ó×÷ΪÕý¼«ºÍ¸º¼«µÄ»îÐÔÎïÖÊ£¬·Ö±ð´¢´æÔÚ¸÷×ÔµÄËáÐÔµç½âÒº´¢¹ÞÖС£Æä½á¹¹Ô­ÀíÈçͼËùʾ£¬¸Ãµç³Ø·Åµçʱ£¬ÓÒ²ÛÖеĵ缫·´Ó¦Îª£ºV2+-e-=V3+£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. ·Åµçʱ£¬ÓÒ²Û·¢Éú»¹Ô­·´Ó¦

B. ·Åµçʱ£¬×ó²ÛµÄµç¼«·´Ó¦Ê½£ºVO2++2H++e-=VO2++H2O

C. ³äµçʱ£¬Ã¿×ªÒÆ1molµç×Ó£¬n(H+)µÄ±ä»¯Á¿Îª1mol

D. ³äµçʱ£¬Òõ¼«µç½âÒºpHÉý¸ß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¹ØÓÚÒÒÏ©ºÍÒÒÍéµÄ˵·¨ÖУ¬²»ÕýÈ·µÄÊÇ

A.ÒÒÏ©ÊôÓÚ²»±¥ºÍÌþ£¬ÒÒÍéÊôÓÚ±¥ºÍÌþ

B.ÒÒÏ©ÖÐËùÓÐÔ­×Ó´¦ÓÚͬһƽÃ棬ÒÒÍéÖÐÔ­×Ó²»ÔÚͬһƽÃæ

C.ÒÒÏ©·Ö×ÓÖеÄ̼̼˫¼ü±ÈÒÒÍéÖÐ̼̼µ¥¼ü¸üÎȶ¨£¬²»Ò×·¢Éú»¯Ñ§·´Ó¦

D.¶¼¿ÉÒÔʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚÓлú»¯ºÏÎïµÄ˵·¨ÕýÈ·µÄÊÇ

A. ·Ö×ÓʽΪC3H6Cl2µÄÓлúÎïÓÐ4ÖÖͬ·ÖÒì¹¹Ìå(²»¿¼ÂÇÁ¢ÌåÒì¹¹)

B. ºÍ»¥ÎªÍ¬ÏµÎï

C. ÒÒȲºÍäåµÄËÄÂÈ»¯Ì¼ÈÜÒº·´Ó¦Éú³É1,2-¶þäåÒÒÍé

D. ¼×±½·Ö×ÓÖÐËùÓÐÔ­×Ó¶¼ÔÚͬһƽÃæÉÏ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚµç½âÁ¶ÂÁ¹ý³ÌÖмÓÈë±ù¾§Ê¯(Óá°A¡±´úÌæ)£¬¿ÉÆðµ½½µµÍAl2O3ÈÛµãµÄ×÷Óᣱù¾§Ê¯µÄÉú²úÔ­ÀíΪ2Al(OH)3£«12HF£«3Na2CO3=2A£«3CO2¡ü£«9H2O¡£¸ù¾ÝÌâÒâÍê³ÉÏÂÁÐÌî¿Õ£º

£¨1£©±ù¾§Ê¯µÄ»¯Ñ§Ê½Îª____________£¬ º¬ÓÐÀë×Ó¼ü¡¢____________µÈ»¯Ñ§¼ü¡£

£¨2£©Éú³ÉÎïÖк¬ÓÐ10¸öµç×ӵķÖ×ÓÊÇ________(д·Ö×Óʽ)£¬¸Ã·Ö×ӵĿռ乹ÐÍ_______£¬ÖÐÐÄÔ­×ÓµÄÔÓ»¯·½Ê½Îª___________________¡£

£¨3£©·´Ó¦ÎïÖе縺ÐÔ×î´óµÄÔªËØΪ________(ÌîÔªËØ·ûºÅ)£¬Ð´³öÆäÔ­×Ó×îÍâ²ãµÄµç×ÓÅŲ¼Í¼£º_______________¡£

£¨4£©±ù¾§Ê¯ÓÉÁ½ÖÖ΢Á£¹¹³É£¬±ù¾§Ê¯µÄ¾§°û½á¹¹Èçͼ¼×Ëùʾ£¬¡ñλÓÚ´óÁ¢·½ÌåµÄ¶¥µãºÍÃæÐÄ£¬¡ðλÓÚ´óÁ¢·½ÌåµÄ12ÌõÀâµÄÖеãºÍ8¸öСÁ¢·½ÌåµÄÌåÐÄ£¬ÄÇô´óÁ¢·½ÌåµÄÌåÐÄ´¦Ëù´ú±íµÄ΢Á£ÊÇ__________(Ìî΢Á£·ûºÅ)¡£

£¨5£©Alµ¥Öʵľ§ÌåÖÐÔ­×ӵĶѻý·½Ê½ÈçͼÒÒËùʾ£¬Æ侧°ûÌØÕ÷Èçͼ±ûËùʾ£¬Ô­×ÓÖ®¼äÏ໥λÖùØϵµÄƽÃæͼÈçͼ¶¡Ëùʾ£º

ÈôÒÑÖªAlµÄÔ­×Ӱ뾶Ϊd£¬NA´ú±í°¢·ü¼ÓµÂÂÞ³£Êý£¬AlµÄÏà¶ÔÔ­×ÓÖÊÁ¿ÎªM£¬ÔòÒ»¸ö¾§°ûÖÐAlÔ­×ÓµÄÊýĿΪ___________¸ö£» Al¾§ÌåµÄÃܶÈΪ________(ÓÃ×Öĸ±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª£ºHCN(aq)ÓëNaOH(aq)·´Ó¦µÄ¡÷H=£­12.1kJ /mol£»HCl(aq)ÓëNaOH(aq)·´Ó¦µÄ¡÷H =£­55.6kJ/ mol¡£ÔòHCNÔÚË®ÈÜÒºÖеçÀëµÄ¦¤HµÈÓÚ

A. £­67.7 kJ /mol B. £­43.5kJ /mol C. +43.5 kJ/ mol D. +67.7 kJ/ mol

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸