¡¾ÌâÄ¿¡¿¸´ºÏÑõ»¯×ê¿ÉÓÃÓÚÉú²úºÏ½ð´ÅÐÔ²ÄÁÏ¡¢´ß»¯¼ÁµÈ¡£Îª³ä·ÖÀûÓÃ×ÊÔ´£¬ÔÚʵÑéÊÒÖÐ̽¾¿²ÉÓú¬îÜ·ÏÔü£¨º¬µÈ£©À´ÖƱ¸¸´ºÏÑõ»¯îÜ£¬¾ßÌåÁ÷³ÌÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©µÄµç×ÓʽΪ_______________________¡£

£¨2£©¡°¼îÖóˮϴ¡±µÄÄ¿µÄΪ______________________________________________¡£

£¨3£©¡°Ëá½þ¡±Ê±£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________________________£»ÂËÔü1ÖÐÖ÷Òª³É·ÖµÄ»¯Ñ§Ê½Îª_______________________¡£

£¨4£©¼ìÑ顰ˮϴҺ¡±Öк¬ÓеIJÙ×÷ºÍÏÖÏóΪ______________________________________________¡£ÉÏÊöÁ÷³ÌÖжà´ÎÓ¦ÓùýÂ˲Ù×÷¸Ã²Ù×÷ËùÓÃÖ÷Òª²£Á§ÒÇÆ÷ÓÐ_____________________¡£

£¨5£©ÏÂÁÐ×°ÖÃÖУ¬ÊʺϽøÐС°±ºÉÕ¡±²Ù×÷µÄÊÇ_____________________£¨ÌîÑ¡Ïî×Öĸ£©¡£

A. B. C. D.

£¨6£©³ÆÁ¿£¬³ä·Ö¡°±ºÉÕ¡±ºóµÃ¸´ºÏÑõ»¯îÜ£¬Ôò¸´ºÏÑõ»¯×êµÄ»¯Ñ§Ê½Îª________________£»¡°±ºÉÕ¡±¹ý³ÌÖÐÉú³ÉµÄÆøÌåÄܲÎÓë´óÆøÑ­»·£¬Ôò¸Ã¹ý³Ì_______________£¨Ìî¡°ÓС±»ò¡°ÎÞ¡±£©ÑõÆø²Î¼Ó·´Ó¦¡£

¡¾´ð°¸¡¿ ³ýÈ¥º¬îÜ·ÏÔü±íÃæµÄÓÍÎÛ Co2O3+4H++=2Co2+++2H2O CaSO4 Óýྻ²¬Ë¿ÕºÈ¡´ý¼ìÒºÔھƾ«µÆ»ðÑæÉÏ×ÆÉÕ£¬Í¸¹ýÀ¶É«îܲ£Á§¹Û²ì»ðÑæ³Ê×ÏÉ« ÉÕ±­¡¢Â©¶·¡¢²£Á§°ô C Co2O7 ÓÐ

¡¾½âÎö¡¿

¸ù¾ÝÁ÷³ÌʾÒâͼ·ÖÎö¿ÉÖª£¬º¬îÜ·ÏÔüÓÃÈȵĴ¿¼îÈÜҺϴȥ±íÃæµÄÓÍÎÛ£¬¼ÓË®K2OÈܽ⣬ͬˮϴҺÁ÷³ö£¬Ï´µÓºó¼ÓÈëÁòËáÈÜÒººÍNa2SO3½øÐÐËá½þ£¬Ëá½þʱ·¢Éú·´Ó¦Co2O3+4H++=2Co2+++2H2O£¬Í¬Ê±CaOºÍH2SO4·´Ó¦Éú³ÉCaSO4£¬µÃµ½º¬ÓÐCaSO4µÄÂËÔü1ºÍÖ÷Òªº¬CoSO4µÄÂËÒº1£¬ÔÙ¼ÓÈëNH4HCO3·¢Éú·´Ó¦CoSO4+NH4HCO3=CoCO3+NH4HSO4µÃµ½CoCO3µÄ³Áµí£¬³Áµí¾­±ºÉÕºóµÃµ½¸´ºÏÑõ»¯îÜ(CoxOy)£¬¾Ý´Ë·ÖÎö½â´ð¡£

(1)K2OÊÇÀë×Ó»¯ºÏÎK+ºÍO2-ÐγÉÀë×Ó¼ü£¬Æäµç×ÓʽΪ£¬¹Ê´ð°¸Îª£º£»

(2)ÈȵĴ¿¼îÈÜÒº¾ßÓнÏÇ¿µÄ¼îÐÔ£¬¿ÉʹÉúÎïÓÍÎÛË®½âΪ¿ÉÈÜÓÚË®µÄÎïÖÊ£¬±ãÓÚˮϴ³ýÈ¥£¬¹Ê´ð°¸Îª£º³ýÈ¥º¬îÜ·ÏÔü±íÃæµÄÓÍÎÛ£»

(3)¡°Ëá½þ¡±Ê±£¬Co2O3ÔÚËáÐÔÌõ¼þϱ»»¹Ô­£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪCo2O3+4H++=2Co2+++2H2O£¬Í¬Ê±£¬CaOºÍH2SO4·´Ó¦Éú³ÉÈܽâ¶ÈСµÄCaSO4£¬¹Ê´ð°¸Îª£ºCo2O3+4H++=2Co2+++2H2O£»CaSO4£»

(4)¼ìÑéK+¿ÉÓÃÑæÉ«·´Ó¦£¬²Ù×÷ºÍÏÖÏóΪÓýྻ²¬Ë¿ÕºÈ¡´ý¼ìÒºÔھƾ«µÆ»ðÑæÉÏ×ÆÉÕ£¬Í¸¹ýÀ¶É«îܲ£Á§¹Û²ì»ðÑæ³Ê×ÏÉ«£¬¹ýÂËËùÓõÄÖ÷Òª²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢Â©¶·ºÍ²£Á§°ô£¬¹Ê´ð°¸Îª£ºÓýྻ²¬Ë¿ÕºÈ¡´ý¼ìÒºÔھƾ«µÆ»ðÑæÉÏ×ÆÉÕ£¬Í¸¹ýÀ¶É«îܲ£Á§¹Û²ì»ðÑæ³Ê×ÏÉ«£»ÉÕ±­¡¢Â©¶·¡¢²£Á§°ô£»

(5)¡°±ºÉÕ¡±²Ù×÷Ó¦ÔÚÛáÛöÖнøÐУ¬CÑ¡ÏîÕýÈ·£¬¹Ê´ð°¸Îª£ºC£»

(6)5.95gCoCO3µÄÎïÖʵÄÁ¿Îª£¬Ôò¸´ºÏÑõ»¯îÜÖк¬îÜÔªËصÄÖÊÁ¿Îª0.05mol¡Á59g¡¤mol-1=2.95g£¬ÔòÑõÔªËصÄÖÊÁ¿Îª4.07g-2.95g=1.12g£¬¼´0.07mol£¬Òò´Ë¸´ºÏÑõ»¯îÜÖÐCoºÍOµÄ¸öÊý±ÈΪ5£º7£¬Ôò¸´ºÏÑõ»¯îܵĻ¯Ñ§Ê½ÎªCo2O7£¬CoCO3ת»¯ÎªCo5O7ʱCoÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬ÔòÓÐÑõÆø²Î¼Ó·´Ó¦£¬¹Ê´ð°¸Îª£ºCo2O7£»ÓС£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿A¡¢B·Ö±ðΪµÚ3¡¢4ÖÜÆÚͬһÖ÷×åµÄ²»Í¬ÔªËصÄÔ­×Ó£¬ËüÃÇÔ­×ÓºËÄÚµÄÖÊ×ÓÊýµÈÓÚÖÐ×ÓÊý¡£¢ÙÈôAΪµÚ¢òA×åÔªËØ£¬ÆäÖÊÁ¿ÊýΪx£¬ÔòBµÄÖÊ×ÓÊýΪy¡£¢ÚÈôAΪµÚ¢ôA×åÔªËØ£¬ÆäÖÊ×ÓÊýΪm£¬ÔòBµÄÖÊÁ¿ÊýΪn£¬ÔòyºÍnµÄÖµ·Ö±ðÊÇ£¨¡¡¡¡£©

A.£¨£«18£©¡¢£¨2m£«18£©

B.£¨£«8£©¡¢£¨2m£«18£©

C.£¨£«8£©¡¢£¨2m£«36£©

D.£¨£«18£©¡¢£¨2m£«36£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÈçͼÊÇʵÑéÊÒÑо¿Ç±Ë®Í§Öй©ÑõÌåϵ·´Ó¦»úÀíµÄ×°ÖÃͼ(¼Ð³ÖÒÇÆ÷ÂÔ)¡£

£¨1£©A×°ÖÃΪCO2µÄ·¢Éú×°Ö㬷´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________¡£

£¨2£©B×°ÖÿɳýÈ¥A×°ÖÃÖпÉÄܻӷ¢³öµÄ___________£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________¡£

£¨3£©C×°ÖÃΪO2µÄ·¢Éú×°Ö㬷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________¡¢________________¡£

£¨4£©D×°ÖÿɳýÈ¥C×°ÖÃÖÐδ·´Ó¦µÄ__________£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________¡£

£¨5£©E×°ÖÃΪÅÅË®·¨ÊÕ¼¯O2µÄ×°Ö㬼ìÑéËùÊÕ¼¯µÄÆøÌåΪO2µÄ·½·¨Îª_____________________¡£

£¨6£©C×°ÖÃÖйÌÌåÓɵ­»ÆÉ«ÍêÈ«±äΪ°×É«£¬¼ìÑé¹ÌÌå³É·ÖµÄʵÑé·½°¸ÎªÈ¡ÉÙÁ¿C×°ÖÃÖз´Ó¦ºóµÄ¹ÌÌåÈÜÓÚË®£¬ÏòÈÜÒºÖеÎÈë¹ýÁ¿___ÈÜÒº£¬ÈôÓа×É«³ÁµíÉú³É£¬ÔòÖ¤Ã÷¹ÌÌåÖк¬ÓÐ___£»¹ýÂË£¬ÏòÂËÒºÖеÎÈ뼸µÎ·Ó̪ÈÜÒº£¬Èô__ÇÒ²»ÍÊÉ«£¬ÔòÖ¤Ã÷¹ÌÌåÖк¬ÓÐ__¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿25 ¡æ£¬101 k Paʱ£¬Ç¿ËáÓëÇ¿¼îµÄÏ¡ÈÜÒº·¢ÉúÖкͷ´Ó¦µÄÖкÍÈÈΪ57.3 kJ/mol£¬ÐÁÍéµÄȼÉÕÈÈΪ5518 kJ/mol¡£ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÊéдÕýÈ·µÄÊÇ£¨¡¡¡¡£©

A. 2H+(aq) +(aq)+(aq)+2OH(aq)=BaSO4(s)+2HO(l);H=57.3 kJ/mol

B. KOH(aq)+HSO4(aq)=KSO4(aq)+HO(l);H=57.3kJ/mol

C. C8H18(l)+O(g)=8CO(g)+ 9HO;H=5518 kJ/mol

D. 2C8H18(g)+25O(g)=16CO(g)+18HO(l);H=5518 kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿I.ÓлúÎï¼ä´æÔÚÈçͼת»¯¹Øϵ£¬CÔں˴Ź²ÕñÇâÆ×ÖгöÏÖ4×é·å£¬Æä·åÃæ»ýÖ®±ÈΪ6:2:1:1¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)CÎïÖʵÄͬϵÎïÖУ¬ÓÐÒ»ÖÖÎïÖʵÄŨ¶ÈΪµÄÈÜÒº£¬ÄÜ´ïµ½Á¼ºÃµÄÏû¶¾¡¢É±¾ú×÷Ó㬸ÃÎïÖʵÄÃû³ÆÊÇ________¡£

(2)DµÄ¹ÙÄÜÍŵÄÃû³ÆÊÇ________¡£

(3)ÁùÖÖÓлúÎïÖÐÄܸú·´Ó¦µÄÊÇ_______________(Ìî×Öĸ)¡£

(4)ÓÉAÉú³ÉB¡¢CµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ____¡£

(5)ÒÑÖªFÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£¬GÊÇFͬϵÎïÖÐ×î¼òµ¥µÄÎïÖÊ£¬Íê³ÉÓÃGºÍ¶Ô±½¶þ¼×ËáΪԭÁϾ­3²½·´Ó¦ÖƱ¸¶Ô±½¶þ¼×ËáÒÒ¶þ´¼õ¥µÄºÏ³É·Ïß(ÓлúÎïд½á¹¹¼òʽ£¬ÆäËüÊÔ¼ÁÈÎÑ¡)_________

¢ò.ÎïÖÊJÊǾßÓÐÇ¿ÁÒõ¹åÏãÆøµÄÏãÁÏ£¬¿ÉÓÉÏÂÁз´Ó¦Â·ÏߺϳÉ(²¿·Ö·´Ó¦Ìõ¼þÒÑÂÔ)

(1)AÎïÖʵÄÀà±ðÊÇ_______¡£·´Ó¦¢ÚÊǼӳɷ´Ó¦£¬ÔòDÎïÖʵĽṹ¼òʽÊÇ_______¡£

(2)¢ÛµÄ·´Ó¦ÀàÐÍΪ_______¡£

(3)GµÄͬ·ÖÒì¹¹ÌåLÓöÈÜÒºÏÔ×ÏÉ«£¬Óë×ãÁ¿±¥ºÍäåË®·´Ó¦Î´¼û°×É«³Áµí²úÉú£¬ÔòLµÄ½á¹¹¼òʽΪ_______¡£(ֻдһÖÖ)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿°ÑζÈΪ20 ¡æ£¬Å¨¶ÈΪ1.0 mol/LµÄH2SO4ÈÜÒººÍ2.2 mol/LµÄ¼îÈÜÒº¸÷50 mL»ìºÏ[ÈÜÒºÃܶȾùΪ1g/mL£¬±ÈÈÈÈÝΪ4.18 kJ/(kg¡¤¡æ)]ÇáÇá½Á¶¯£¬²âµÃËá¼î»ìºÏÒºµÄζȱ仯Êý¾ÝÈçÏ£º

·´Ó¦Îï

ÆðʼζÈt1 ¡æ

ÖÕֹζÈt2 ¡æ

H2SO4£«NaOH

20

33.6

H2SO4£«NH3¡¤H2O

20

32.6

£¨1£©·´Ó¦NH3¡¤H2O(aq)NH4+ (aq)£«OH£­(aq)µÄìʱäÔ¼____¡£

£¨2£©¼ÆËãÉÏÊöÁ½×éʵÑé²â³öµÄÖкÍÈÈ£º¦¤H1£½___kJ/mol£»¦¤H2£½__kJ/mol¡£

£¨3£©Óɱ¾Ìâ½áÂÛ¿ÉÔ¤²â½«µÚ1×éÖеÄ1 mol/LµÄH2SO4ÈÜÒº»»³É2mol/LµÄCH3COOHÈÜÒº½øÐÐʵÑ飬²âµÃµÄÖкÍÈÈÊýÖµ__(Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)56.848¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑ֪þÄÜÔÚO2¡¢N2¡¢CO2ÖÐȼÉÕÉú³ÉÏàÓ¦µÄ»¯ºÏÎï¡£ÊԻشð£º

(1)þÔÚ¿ÕÆøÖÐȼÉÕ£¬³ý·¢Éú·´Ó¦N2£«3MgMg3N2Í⣬»¹ÄÜ·¢ÉúÆäËû·´Ó¦£¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________¡£

(2)þÌõÔÚÑõÆøÖÐȼÉÕʱ¿´µ½µÄÏÖÏóÊÇ________¡£

(3)Èç¹ûµÈÖÊÁ¿µÄþ·Ö±ðÔÚ×ãÁ¿µÄÑõÆø¡¢µªÆø¡¢¶þÑõ»¯Ì¼ÖÐȼÉÕ£¬È¼ÉÕºó¹ÌÌåµÄÖÊÁ¿·Ö±ðΪm1¡¢m2¡¢m3£¬ÔòËüÃǵĴóС˳ÐòΪ___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³Ñ§Éú¿ÎÍâ»î¶¯Ð¡×éÀûÓÃÏÂͼËùʾװÖ÷ֱð×öÈçÏÂʵÑ飺

£¨1£©ÔÚÊÔ¹ÜÖÐ×¢ÈëijºìÉ«ÈÜÒº£¬¼ÓÈÈÊԹܣ¬ÈÜÒºÑÕÉ«Öð½¥±ädz£¬ÀäÈ´ºó»Ö¸´ºìÉ«£¬ÔòÔ­ÈÜÒº¿ÉÄÜÊÇ________ÈÜÒº£»¼ÓÈÈʱÈÜÒºÓɺìÉ«Öð½¥±ädzµÄÔ­ÒòÊÇ________________¡£

£¨2£©ÔÚÊÔ¹ÜÖÐ×¢ÈëijÎÞÉ«ÈÜÒº£¬¼ÓÈÈÊԹܣ¬ÈÜÒº±äΪºìÉ«£¬ÀäÈ´ºó»Ö¸´ÎÞÉ«£¬Ôò´ËÈÜÒº¿ÉÄÜÊÇ________ÈÜÒº£»¼ÓÈÈʱÈÜÒºÓÉÎÞÉ«±äΪºìÉ«µÄÔ­ÒòÊÇ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½«1molI2ºÍ2molH2(g)ÖÃÓÚij2LÃܱÕÈÝÆ÷ÖУ¬ÔÚ¡ª¶¨Î¶ÈÏ·¢Éú·´Ó¦£ºI2(g)+H2(g)2HI(g) ¡÷H<0£¬²¢´ïƽºâ£¬HIµÄÌå»ý·ÖÊý¦Ø(HI)Ëæʱ¼ä±ä»¯ÈçͼÇúÏß

(1)´ïƽºâʱ£¬I2(g)µÄÎïÖʵÄÁ¿Å¨¶ÈΪ _______¡£H2(g)µÄƽºâת»¯ÂÊΪ___________¡£

ÔÚ´ËζÈÏ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK__________(±£ÁôһλСÊý£©¡£

(2)Èô¸Ä±ä·´Ó¦Ìõ¼þÏ£¬ÔÚ¼×Ìõ¼þϦØ(HI)µÄ±ä»¯ÈçͼÇúÏß(I)Ëùʾ£¬ÔÚÒÒÌõ¼þϦØ(HI)µÄ±ä»¯ÈçͼÇúÏß(III)Ëùʾ¡£Ôò¼×Ìõ¼þ¿ÉÄÜÊÇ______£¨ÌîÈëÏÂÁÐÌõ¼þµÄÐòºÅ¡£ÏÂͬ£©£¬ÒÒÌõ¼þ¿ÉÄÜÊÇ________¡£

¢ÙºãÈÝÌõ¼þÏ£¬Éý¸ßζÈ

¢ÚºãÈÝÌõ¼þÏ£¬½µµÍζÈ

¢ÛºãÎÂÌõ¼þÏ£¬ËõС·´Ó¦ÈÝÆ÷Ìå»ý

¢ÜºãÎÂÌõ¼þÏ£¬À©´ó·´Ó¦ÈÝÆ÷Ìå»ý

¢ÝºãκãÈÝÌõ¼þÏ£¬¼ÓÈëÊʵ±´ß»¯¼Á

(3)Èô±£³ÖζȲ»±ä£¬ÔÚÁíÒ»ÏàͬµÄ2LÃܱÕÈÝÆ÷ÖмÓÈëa mol I2(g)¡¢b mol H2ºÍc mol HI£¨a¡¢b¡¢ c¾ù´óÓÚ0£©£¬·¢Éú·´Ó¦£¬´ïƽºâʱ£¬HIµÄÌå»ý·ÖÊýÈÔΪ0.60£¬Ôòa¡¢b¡¢cµÄÓ¦Âú×ãµÄ¹ØϵÊÇ_______________ (Óú¬Ò»¸öa¡¢b¡¢cµÄ´úÊýʽ±íʾ)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸