¡¾ÌâÄ¿¡¿Ä³100mL»ìºÏÒºÖУ¬HNO3ºÍH2SO4µÄÎïÖʵÄÁ¿Å¨¶È·Ö±ðΪ0.1mol/LºÍ0.4mol/L¡£Ïò¸Ã»ìºÏÒºÖмÓÈë1.92gÍ­·Û£¬¼ÓÈÈʹ·´Ó¦·¢ÉúÍêÈ«¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(ºöÂÔ·´Ó¦Ç°ºóÈÜÒºÌå»ý±ä»¯)£¨ £©

A.ËùµÃÈÜÒºÖÐc(Cu2+)=0.225mol/L

B.ËùµÃÈÜÒºÖÐc(H+)=0.5mol/L

C.ËùµÃÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ0.448L

D.·´Ó¦ÖÐתÒÆ0.06molµÄµç×Ó

¡¾´ð°¸¡¿B

¡¾½âÎö¡¿

n(Cu)==0.03mol£»ÈÜÒºÖÐn(H+)=0.1mol/L¡Á0.1L+0.4mol/L¡Á0.1L¡Á2=0.09mol£¬n(NO3)=0.1mol/L¡Á0.1L=0.01mol£¬Ïò»ìºÏÈÜÒº¼ÓÈëÍ­·Ûºó·¢Éú·´Ó¦£º3Cu+8H++2NO3-=3Cu2++2NO¡ü+4H2O£¬¸ù¾ÝÀë×Ó·´Ó¦·½³Ìʽ¿ÉÖªNO3µÄÁ¿²»×㣬Ôò·´Ó¦µÄÍ­µÄÎïÖʵÄÁ¿Îª0.015mol£¬ÏûºÄµÄÇâÀë×ÓµÄÎïÖʵÄÁ¿Îª0.04mol¡£

A. ·´Ó¦µÄÍ­µÄÎïÖʵÄÁ¿Îª0.15mol£¬ËùÒÔc(Cu2+)==0.15mol/L£¬¹ÊA´íÎó£»

B. ·´Ó¦ÏûºÄµÄÇâÀë×ÓµÄÎïÖʵÄÁ¿Îª0.04mol£¬ÔòÊ£ÓàµÄn(H+)=0.09mol-0.04mol=0.05mol£¬ÈÜÒºÌå»ýΪ100mL£¬ËùÒÔŨ¶ÈΪ0.5mol/L£¬¹ÊBÕýÈ·£»

C. ¸ù¾ÝÀë×Ó·½³Ìʽ¿ÉÖªÉú³ÉµÄÆøÌån(NO)= n(NO3)=0.01mol£¬±ê¿öÏÂÌå»ýΪ0.224L£¬¹ÊC´íÎó£»

D. ·´Ó¦¹ý³ÌÖÐ0.01mol NO3±»»¹Ô­³ÉNO£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª0.03mol£¬¹ÊD´íÎó£»

¹Ê´ð°¸ÎªB¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ

A. ÏòËÄôÇ»ùºÏÂÁËáÄÆÈÜÒºÖеμӹýÁ¿µÄ̼ËáÇâÄÆÈÜÒº£º[Al(OH)4]¡ª+4H£«=Al3++2H2O

B. ½«ÉÙÁ¿SO2ÆøÌåͨÈë×ãÁ¿µÄNaClOÈÜÒºÖУºSO2£«2ClO£­£«H2O=SO32-£«2HClO

C. NaHSO4ÈÜÒºÓëBa(OH)2ÈÜÒº·´Ó¦ÖÁÖÐÐÔ£º2H£«+SO42-+Ba2++2OH¡ª=BaSO4¡ý+2H2O

D. Ïò·ÐË®Öеμӱ¥ºÍÂÈ»¯ÌúÈÜÒº£ºFe3++3H2O=Fe(OH)3¡ý+3H£«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©25 ¡æʱ£¬ÖƱ¸ÑÇÏõõ£ÂÈËùÉæ¼°µÄÈÈ»¯Ñ§·½³ÌʽºÍƽºâ³£ÊýÈç±í£º

ÈÈ»¯Ñ§·½³Ìʽ

ƽºâ³£Êý

¢Ù

2NO2(g)+NaCl(s)NaNO3(s)+NOCl(g) ¦¤H1=a kJmol-1

K1

¢Ú

4NO2(g)+2NaCl(s)2NaNO3(s)+ 2NO(g)+Cl2(g) ¦¤H2=b kJmol-1

K2

¢Û

2NO(g)+Cl2(g)2NOCl(g) ¦¤H3

K3

Ôò¸ÃζÈÏ£¬¦¤H3=_______________kJmol-1£»K3=_____________£¨ÓÃK1ºÍK2±íʾ£©¡£

£¨2£©25¡æʱ£¬ÔÚÌå»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖÐͨÈë0.08 mol NOºÍ0.04 molCl2·¢ÉúÉÏÊö·´Ó¦¢Û£¬Èô·´Ó¦¿ªÊ¼Óë½áÊøʱζÈÏàͬ£¬Êý×ÖѹǿÒÇÏÔʾ·´Ó¦¹ý³ÌÖÐѹǿ(p)Ëæʱ¼ä(t)µÄ±ä»¯Èçͼ¢ñʵÏßËùʾ£¬Ôò¦¤H3 ___£¨Ìî¡°>¡±¡°<¡±»ò¡°=¡±£©0£»ÈôÆäËûÌõ¼þÏàͬ£¬½ö¸Ä±äijһÌõ¼þ£¬²âµÃÆäѹǿËæʱ¼äµÄ±ä»¯Èçͼ¢ñÐéÏßËùʾ£¬Ôò¸Ä±äµÄÌõ¼þÊÇ_____________£»ÔÚ5 minʱ£¬ÔÙ³äÈë0.08 mol NOºÍ0.04 molCl2£¬Ôò»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿½«_____________£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©¡£Í¼¢òÊǼס¢ÒÒÁ½Í¬Ñ§Ãè»æÉÏÊö·´Ó¦¢ÛµÄƽºâ³£ÊýµÄ¶ÔÊýÖµ£¨lgK£©Óëζȵı仯¹Øϵͼ£¬ÆäÖÐÕýÈ·µÄÇúÏßÊÇ______£¨Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©£¬aֵΪ__________¡£25 ¡æʱ²âµÃ·´Ó¦¢ÛÔÚijʱ¿Ì£¬NO(g)¡¢Cl2(g)¡¢NOCl(g)µÄŨ¶È·Ö±ðΪ0.8¡¢0.1¡¢0.3£¬Ôò´ËʱvÕý_________vÄ棨Ìî¡°>¡±¡°£¼¡±»ò¡°=¡±£©

(3)ÔÚ300 ¡æ¡¢8 MPaÏ£¬½«CO2ºÍH2°´ÎïÖʵÄÁ¿Ö®±È1¡Ã3 ͨÈëÒ»ÃܱÕÈÝÆ÷Öз¢ÉúCO2(g)£«3H2(g)CH3OH(g)£«H2O(g)Öз´Ó¦£¬´ïµ½Æ½ºâʱ£¬²âµÃCO2µÄƽºâת»¯ÂÊΪ50%£¬Ôò¸Ã·´Ó¦Ìõ¼þϵÄƽºâ³£ÊýΪKp£½_____(ÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È¼ÆË㣬·Öѹ£½×Üѹ¡ÁÎïÖʵÄÁ¿·ÖÊý)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª·´Ó¦N2(g)+3H2(g)2NH3(g)£¬Ä³Î¶ÈÏ£¬ÏòÌå»ý¹Ì¶¨µÄ1LÃܱÕÈÝÆ÷ÖгäÈë1mol N2(g)ºÍ3mol H2(g)£¬²âµÃ²»Í¬Ê±¿Ì·´Ó¦Ç°ºóµÄѹǿ¹ØϵÈçϱíËùʾ£º

ʱ¼ä/min

5

10

15

20

25

30

ѹǿ±ÈÖµPºó/PÇ°

0.98

0.88

0.80

0.75

0.75

0.75

(1)0~15minÄÚ,ÓÃH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪv(H2) =___________________mol¡¤L-1¡¤min-1¡£

(2)´ïµ½Æ½ºâʱN2µÄת»¯ÂÊΪ________£¬¸ÃζÈϵÄƽºâ³£ÊýΪ___________(±£ÁôÁ½Î»Ð¡Êý)¡£

(3)ÒÑÖª¸Ã·´Ó¦Îª·ÅÈÈ·´Ó¦£¬ÏÂͼΪ²»Í¬Ìõ¼þÏ·´Ó¦ËÙÂÊËæʱ¼äµÄ±ä»¯Çé¿ö(ÿ´Î½ö¸Ä±äÒ»¸öÌõ¼þ)£ºaʱ¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ____________£»bʱ¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ_______________¡£

(4)Ò»¶¨Ìõ¼þϵÄÃܱÕÈÝÆ÷ÖУ¬¸Ã·´Ó¦´ïµ½Æ½ºâºóÒªÌá¸ßH2µÄת»¯ÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÓУ¨_________£©

A£®µÍεÍѹ B£®¼ÓÈë´ß»¯¼Á C£®Ôö¼ÓN2µÄŨ¶È D£®Ôö¼ÓH2µÄŨ¶È E£®·ÖÀë³öNH3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ïØ£¨31Ga£©ÊÇÒ»ÖÖÖØÒª½ðÊôÔªËØ£¬ïؼ°Æ仯ºÏÎïÔÚµç×Ó¹¤Òµ¡¢¹âµç×Ó¹¤Òµ¡¢¹ú·À¹¤ÒµºÍ³¬µ¼²ÄÁϵÈÁìÓòÓÐ׏㷺µÄÓ¦Ó᣻شðÏÂÁÐÎÊÌ⣺

£¨1£©»ù̬GaÔ­×ÓÕ¼¾Ý×î¸ßÄܼ¶µç×ӵĵç×ÓÔÆÂÖÀªÍ¼ÐÎ״Ϊ__________£¬Î´³É¶Ôµç×ÓÊýΪ________________¡£

£¨2£©Ga(NO3)3ÖÐÒõÀë×ÓµÄÁ¢Ìå¹¹ÐÍÊÇ_____________£¬Ð´³öÒ»¸öÓë¸ÃÒõÀë×ÓµÄÁ¢Ìå¹¹ÐÍÏàͬµÄ·Ö×ӵĻ¯Ñ§Ê½___________¡£

£¨3£©2-¼×»ù-8-ôÇ»ùà­ßøïØ£¨Èçͼ£©Ó¦ÓÃÓÚ·Ö×ÓÓ¡¼£¼¼Êõ£¬2-¼×»ù-8-ôÇ»ùà­ßøïØÖÐÎåÖÖÔªËص縺ÐÔÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ____________________________£¨ÌîÔªËØ·ûºÅ)£¬Ìṩ¹Âµç×ӶԵijɼüÔ­×ÓÊÇ_____________¡£

£¨4£©Ò»ÖÖ¹èïØ°ëµ¼Ìå²ÄÁϵľ§°û½á¹¹ÈçͼËùʾÓÉÁò¡¢ïØ¡¢ÒøÐγɵĻ¯ºÏÎïµÄ¾§°ûÊǵ×ÃæΪÕý·½Ðεij¤·½Ì壬½á¹¹ÈçÏÂͼËùʾ£¬Ôò¸Ã¾§ÌåÖÐÁòµÄÅäλÊýΪ___________£¬¾§°ûµ×ÃæµÄ±ß³¤a=5.75 nm£¬¸ßh=10.30nm£¬¸Ã¾§ÌåÃܶÈΪ__________________g¡¤cm-3£¨Áгö¼ÆËãʽ¼´¿É£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ã÷·¯Ê¯ÊÇÖÆÈ¡¼Ø·ÊºÍÇâÑõ»¯ÂÁµÄÖØÒªÔ­ÁÏ¡£Ã÷·¯Ê¯µÄ×é³ÉºÍÃ÷·¯ÏàËÆ£¬´ËÍ⻹º¬ÓÐÑõ»¯ÂÁºÍÉÙÁ¿Ñõ»¯ÌúÔÓÖÊ¡£¾ßÌåʵÑé²½ÖèÈçͼËùʾ£º

¸ù¾ÝÉÏÊöͼʾ£¬Íê³ÉÏÂÁÐÌî¿Õ¡£

£¨1£©Ã÷·¯Ê¯±ºÉÕºóÓÃÏ¡°±Ë®½þ³ö¡£ÊµÑéÐèÒª500mLÏ¡°±Ë®(ÿÉýº¬ÓÐ19.60g°±)ÐèҪȡŨ°±Ë®(ÿÉýº¬ÓÐ250.28g°±)___mL£¬Óùæ¸ñΪ___mLÁ¿Í²Á¿È¡¡£

£¨2£©Ð´³ö³ÁµíÎïÖÐËùÓÐÎïÖʵĻ¯Ñ§Ê½£º___¡£

£¨3£©²Ù×÷¢ñµÄÃû³ÆÊÇ___£¬ËùÓõIJ£Á§ÒÇÆ÷ÓÐ___¡£

£¨4£©Îª²â¶¨»ìºÏ·ÊÁÏK2SO4¡¢(NH4)2SO4Öмصĺ¬Á¿(ÒÔK2O¼Æ)£¬ÍêÉÆÏÂÁв½Ö裺

¢Ù³ÆÈ¡¼Øµª·ÊÊÔÑù²¢ÈÜÓÚË®£¬¼ÓÈë×ãÁ¿BaCl2ÈÜÒº£¬²úÉú___¡£

¢Ú___¡¢___¡¢___(ÒÀ´ÎÌîдʵÑé²Ù×÷Ãû³Æ)¡£

¢ÛÀäÈ´¡¢³ÆÖØ¡£

¢ÜÈôÊÔÑùΪmg£¬³ÁµíµÄÎïÖʵÄÁ¿Îªnmol£¬ÔòÊÔÑùÖмصĺ¬Á¿(ÒÔK2O¼Æ)Ϊ___%£¨ÖÊÁ¿·ÖÊý£©(Óú¬m¡¢nµÄ´úÊýʽ±íʾ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿A¡¢B¡¢C¡¢D¡¢E¡¢FΪԪËØÖÜÆÚ±íÖÐÇ°ËÄÖÜÆÚÔªËØ£¬ÇÒÔ­×ÓÐòÊýÒÀ´ÎÔö´ó£¬AÓëÆäÓàÎåÖÖÔªËؼȲ»Í¬ÖÜÆÚÒ²²»Í¬Ö÷×壬BµÄÒ»ÖÖºËËØÔÚ¿¼¹Åʱ³£ÓÃÀ´¼ø¶¨Ò»Ð©ÎÄÎïµÄÄê´ú£¬CµÄÑõ»¯ÎïÊǵ¼ÖÂËáÓêµÄÖ÷ÒªÎïÖÊÖ®Ò»£¬DÔ­×ÓºËÍâµç×ÓÓÐ8ÖÖ²»Í¬µÄÔ˶¯×´Ì¬£¬EµÄ»ù̬ԭ×ÓÔÚÇ°ËÄÖÜÆÚÔªËصĻù̬ԭ×ÓÖе¥µç×ÓÊý×î¶à£¬FÔªËصĻù̬ԭ×Ó×îÍâÄܲãÖ»ÓÐÒ»¸öµç×Ó£¬ÆäËûÄܲã¾ùÒѳäÂúµç×Ó¡£

£¨1£©Ð´³ö»ù̬EÔ­×ӵļ۵ç×ÓÅŲ¼Ê½_______¡£

£¨2£©AÓëC¿ÉÐγÉCA3·Ö×Ó£¬¸Ã·Ö×ÓÖÐCÔ­×ÓµÄÔÓ»¯ÀàÐÍΪ______£¬¸Ã·Ö×ÓµÄÁ¢Ìå½á¹¹Îª_____£»CµÄµ¥ÖÊÓëBD»¯ºÏÎïÊǵȵç×ÓÌ壬¾ÝµÈµç×ÓÌåµÄÔ­Àí£¬Ð´³öBD»¯ºÏÎïµÄµç×Óʽ______£»A2DÓÉҺ̬Ðγɾ§ÌåʱÃܶȼõС£¬ÆäÖ÷ÒªÔ­ÒòÊÇ__________£¨ÓÃÎÄ×ÖÐðÊö£©¡£

£¨3£©ÒÑÖªD¡¢FÄÜÐγÉÒ»ÖÖ»¯ºÏÎÆ侧°ûµÄ½á¹¹ÈçͼËùʾ£¬Ôò¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½Îª___£»ÈôÏàÁÚDÔ­×ÓºÍFÔ­×Ó¼äµÄ¾àÀëΪa cm£¬°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪNA£¬Ôò¸Ã¾§ÌåµÄÃܶÈΪ______g¡¤cm£­3£¨Óú¬a¡¢NAµÄʽ×Ó±íʾ£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÈÝ»ýÒ»¶¨µÄÃܱÕÈÝÆ÷ÖУ¬ÖÃÈëÒ»¶¨Á¿µÄ NO(g)ºÍ×ãÁ¿C(s)£¬·¢Éú·´Ó¦ C(s)£«2NO(g) CO2(g)£«N2(g)£¬Æ½ºâ״̬ʱ NO(g)µÄÎïÖʵÄÁ¿Å¨¶È c(NO)ÓëÎÂ¶È T µÄ¹ØϵÈçͼËùʾ¡£ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ( )

A.¸Ã·´Ó¦µÄ ¦¤H>0B.Èô¸Ã·´Ó¦ÔÚ T1¡¢T2 ʱµÄƽºâ³£Êý·Ö±ðΪ K1¡¢K2£¬Ôò K1<K2

C.ÔÚ T3 ʱ£¬Èô»ìºÏÆøÌåµÄÃܶȲ»Ôٱ仯£¬Ôò¿ÉÒÔÅжϷ´Ó¦´ïµ½Æ½ºâ״̬ CD.ÔÚ T2 ʱ£¬Èô·´Ó¦Ìåϵ´¦ÓÚ״̬D£¬Ôò´Ëʱһ¶¨ÓÐ v Õý<v Äæ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿H2AÊǶþÔªÈõËᣬ25¡æʱ£¬ÅäÖÆÒ»×éc(H2A)£«c(HA£­)£«c(A2£­)£½0.1mol¡¤L-1µÄH2AºÍNaOH»ìºÏÈÜÒº£¬ÈÜÒºÖÐH2A¡¢HA£­ºÍA2£­ËùÕ¼ÈýÖÖÁ£×Ó×ÜÊýµÄÎïÖʵÄÁ¿·ÖÊý£¨¦Á£©ËæÈÜÒºpH±ä»¯µÄ¹ØϵÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A.c(Na£«)£½0.1mol¡¤L-1µÄÈÜÒºÖУºc(H£«)£½c(A2£­)£«c(OH£­)£­c(H2A)

B.c (HA£­)£½c(A2£­)µÄÈÜÒºÖУºc(Na£«)£¾3c(A2£­)

C.c (HA£­)£½0.5mol¡¤L-1µÄÈÜÒºÖУº2c(H2A)£«c(H£«)£½c(OH£­)£«1.5mol¡¤L-1

D.pH£½2µÄÈÜÒºÖУºc(HA£­)£«2c(A2£­)£¼0.1

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸