Ò»Ïî¿ÆÑ§Ñо¿³É¹û±íÃ÷£¬Í­ÃÌÑõ»¯Îï(CuMn2O4)ÄÜÔÚ³£ÎÂÏ´߻¯Ñõ»¯¿ÕÆøÖеÄÒ»Ñõ»¯Ì¼ºÍ¼×È©(HCHO)¡£

(1)ÏòÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄCu(NO3)2ºÍMn(NO3)2ÈÜÒºÖмÓÈëNa2CO3ÈÜÒº£¬ËùµÃ³Áµí¾­¸ßÎÂׯÉÕ£¬¿ÉÖÆµÃCuMn2O4¡£

¢ÙMn2£«»ù̬µÄµç×ÓÅŲ¼Ê½¿É±íʾΪ                                       ¡£

¢ÚNO3£­µÄ¿Õ¼ä¹¹ÐÍ                           (ÓÃÎÄ×ÖÃèÊö)¡£

(2)ÔÚÍ­ÃÌÑõ»¯ÎïµÄ´ß»¯Ï£¬CO±»Ñõ»¯³ÉCO2£¬HCHO±»Ñõ»¯³ÉCO2ºÍH2O¡£

¢Ù¸ù¾ÝµÈµç×ÓÔ­Àí£¬CO·Ö×ӵĽṹʽΪ                        ¡£

¢ÚH2O·Ö×ÓÖÐOÔ­×Ó¹ìµÀµÄÔÓ»¯ÀàÐÍΪ                        ¡£

¢Û1molCO2Öк¬ÓеĦҼüÊýĿΪ                        ¡£

(3)ÏòCuSO4ÈÜÒºÖмÓÈë¹ýÁ¿NaOHÈÜÒº¿ÉÉú³É[Cu(OH)4]2£­¡£²»¿¼Âǿռ乹ÐÍ£¬[Cu(OH)4]2£­µÄ½á¹¹¿ÉÓÃʾÒâͼ±íʾΪ                        ¡£


¡¾²Î¿¼´ð°¸¡¿

(1)¢Ù1s22s22p63s23p63d5(»ò[Ar]3d5)

 ¢ÚÆ½ÃæÈý½ÇÐÎ

(2)¢ÙC¡ÔO  ¢Úsp3     ¢Û2¡Á6.02¡Á1023¸ö(»ò2mol)

(3)


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÉ5mol Fe2O3¡¢4mol Fe3O4ºÍ3mol FeO×é³ÉµÄ»ìºÏÎ¼ÓÈë´¿Ìú1mol²¢ÔÚ¸ßÎÂϺÍFe2O3·´Ó¦¡£Èô´¿ÌúÍêÈ«·´Ó¦£¬Ôò·´Ó¦ºó»ìºÏÎïÖÐFeOÓëFe2O3µÄÎïÖʵÄÁ¿Ö®±È¿ÉÄÜÊÇ

A£®4:3    B£®3:2    C£®3:1    D£®2:l

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ʯīÔÚ²ÄÁÏÁìÓòÓÐÖØÒªÓ¦Óá£Ä³³õ¼¶Ê¯Ä«Öк¬SiO2(7.8%)¡¢Al2O3(5.1%)¡¢Fe2O3(3.1%)ºÍMgO(0.5%)µÈÔÓÖÊ¡£Éè¼ÆµÄÌá´¿Óë×ÛºÏÀûÓù¤ÒÕÈçÏ£º

(×¢£ºSiCl4µÄ·ÐµãΪ57.6 ¡æ£¬½ðÊôÂÈ»¯ÎïµÄ·Ðµã¾ù¸ßÓÚ150 ¡æ)

(1)Ïò·´Ó¦Æ÷ÖÐͨÈëCl2ǰ£¬Ðèͨһ¶Îʱ¼äN2£¬Ö÷ҪĿµÄÊÇ____________________¡£

(2)¸ßη´Ó¦ºó£¬Ê¯Ä«ÖÐÑõ»¯ÎïÔÓÖʾùת±äΪÏàÓ¦µÄÂÈ»¯Îï¡£ÆøÌå¢ñÖеÄ̼Ñõ»¯ÎïÖ÷ҪΪ________¡£ÓÉÆøÌå¢òÖÐijÎïµÃµ½Ë®²£Á§µÄ»¯Ñ§·´Ó¦·½³ÌʽΪ____________________________________________¡£

(3)²½Öè¢ÙΪ£º½Á°è¡¢________¡£ËùµÃÈÜÒº¢ôÖеÄÒõÀë×ÓÓÐ________¡£

(4)ÓÉÈÜÒº¢ôÉú³É³Áµí¢õµÄ×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________________________________£¬100 kg³õ¼¶Ê¯Ä«×î¶à¿ÉÄÜ»ñµÃ¢õµÄÖÊÁ¿Îª______kg¡£

(5)ʯī¿ÉÓÃÓÚ×ÔȻˮÌåÖÐÍ­¼þµÄµç»¯Ñ§·À¸¯£¬Íê³ÉÈçͼ·À¸¯Ê¾Òâͼ£¬²¢×÷ÏàÓ¦±ê×¢¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


FeCl3ÔÚÏÖ´ú¹¤ÒµÉú²úÖÐÓ¦Óù㷺¡£Ä³»¯Ñ§Ñо¿ÐÔѧϰС×éÄ£Ä⹤ҵÉú²úÁ÷³ÌÖÆ±¸ÎÞË®FeCl3£¬ÔÙÓø±²úÆ·FeCl3ÈÜÒºÎüÊÕÓж¾µÄH2S¡£

I.¾­²éÔÄ×ÊÁϵÃÖª£ºÎÞË®FeCl3ÔÚ¿ÕÆøÖÐÒ׳±½â£¬¼ÓÈÈÒ×Éý»ª¡£ËûÃÇÉè¼ÆÁËÖÆ±¸ÎÞË®FeCl3µÄʵÑé·½°¸£¬×°ÖÃʾÒâͼ£¨¼ÓÈȼ°¼Ð³Ö×°ÖÃÂÔÈ¥£©¼°²Ù×÷²½ÖèÈçÏ£º

¢Ù¼ì²é×°ÖÃµÄÆøÃÜÐÔ£»

¢ÚͨÈë¸ÉÔïµÄCl2£¬¸Ï¾¡×°ÖÃÖÐµÄ¿ÕÆø£»

¢ÛÓþƾ«µÆÔÚÌúмÏ·½¼ÓÈÈÖÁ·´Ó¦Íê³É

¢Ü¡­¡­

¢ÝÌåϵÀäÈ´ºó£¬Í£Ö¹Í¨ÈëCl2£¬²¢ÓøÉÔïµÄN2¸Ï¾¡Cl2£¬½«ÊÕ¼¯Æ÷ÃÜ·â

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

×°ÖÃAÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________________¡£

µÚ¢Û²½¼ÓÈȺó£¬Éú³ÉµÄÑÌ×´FeCl3´ó²¿·Ö½øÈëÊÕ¼¯Æ÷£¬ÉÙÁ¿³Á»ýÔÚ·´Ó¦¹ÜAµÄÓÒ¶Ë¡£ÒªÊ¹³Á»ýµÃFeCl3½øÈëÊÕ¼¯Æ÷£¬µÚ¢Ü²½²Ù×÷ÊÇ_______________________________________________¡£

²Ù×÷²½ÖèÖУ¬Îª·ÀÖ¹FeCl3³±½âËù²ÉÈ¡µÄ´ëÊ©ÓУ¨Ìî²½ÖèÐòºÅ£©_________________________¡£

×°ÖÃBÖеÄÀäˮԡµÄ×÷ÓÃΪ__________________£»×°ÖÃCµÄÃû³ÆÎª__________________£»×°ÖÃDÖÐFeCl2È«²¿·´Ó¦Íêºó£¬ÒòΪʧȥÎüÊÕCl2µÄ×÷ÓöøÊ§Ð§£¬Ð´³ö¼ìÑéFeCl2ÊÇ·ñʧЧµÄÊÔ¼Á£º___________¡£

ÔÚÐéÏß¿òÄÚ»­³öÎ²ÆøÎüÊÕ×°ÖÃE²¢×¢Ã÷ÊÔ¼Á¡£

II.¸Ã×éͬѧÓÃ×°ÖÃDÖеĸ±²úÆ·FeCl3ÈÜÒºÎüÊÕH2S£¬µÃµ½µ¥ÖÊÁò£»¹ýÂ˺ó£¬ÔÙÒÔʯīΪµç¼«£¬ÔÚÒ»¶¨Ìõ¼þϵç½âÂËÒº¡£

FeCl3ÓëH2S·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________________________________________¡£

µç½â³ØÖÐH+ÔÚÒõ¼«·Åµç²úÉúH2£¬Ñô¼«µÄµç¼«·´Ó¦Îª___________________________________¡£

×ۺϷÖÎöʵÑéIIµÄÁ½¸ö·´Ó¦£¬¿ÉÖª¸ÃʵÑéÓÐÁ½¸öÏÔÖøÓŵ㣺

¢ÙH2SµÄÔ­×ÓÀûÓÃÂÊ100%£»¢Ú____________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ñõ»¯Ã¾ÔÚÒ½Ò©¡¢½¨ÖþµÈÐÐÒµÓ¦Óù㷺¡£ÁòËáþ»¹Ô­ÈȽâÖÆ±¸¸ß´¿Ñõ»¯Ã¾ÊÇÒ»ÖÖеÄ̽Ë÷¡£ÒÔÁâþ¿ó(Ö÷Òª³É·ÖΪMgCO3£¬º¬ÉÙÁ¿FeCO3)ΪԭÁÏÖÆ±¸¸ß´¿Ñõ»¯Ã¾µÄʵÑéÁ÷³ÌÈçÏ£º

(1)MgCO3ÓëÏ¡ÁòËá·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                          ¡£

(2)¼ÓÈëH2O2Ñõ»¯Ê±£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                    ¡£

(3)ÂËÔü2µÄ³É·ÖÊÇ                  (Ìѧʽ)¡£

(4)ìÑÉÕ¹ý³Ì´æÔÚÒÔÏ·´Ó¦£º

2MgSO4£«C2MgO£«2SO2¡ü£«CO2¡ü

MgSO4£«CMgO£«SO2¡ü£«CO¡ü

MgSO4£«3CMgO£«S¡ü£«3CO¡ü

ÀûÓÃÓÒͼװÖöÔìÑÉÕ²úÉúµÄÆøÌå½øÐзֲ½ÎüÊÕ»òÊÕ¼¯¡£

¢ÙDÖÐÊÕ¼¯µÄÆøÌå¿ÉÒÔÊÇ                  (Ìѧʽ)¡£

¢ÚBÖÐÊ¢·ÅµÄÈÜÒº¿ÉÒÔÊÇ                  (Ìî×Öĸ)¡£

a.NaOH ÈÜÒº      b.Na2CO3ÈÜÒº      c.Ï¡ÏõËá      d.KMnO4ÈÜÒº

¢ÛAÖеõ½µÄµ­»ÆÉ«¹ÌÌåÓëÈȵÄNaOHÈÜÒº·´Ó¦£¬²úÎïÖÐÔªËØ×î¸ß¼Û̬Ϊ£«4£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º                                                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÂÈ»¯¼ØÑùÆ·Öк¬ÓÐÉÙÁ¿Ì¼Ëá¼Ø¡¢ÁòËá¼ØºÍ²»ÈÜÓÚË®µÄÔÓÖÊ¡£ÎªÁËÌá´¿ÂÈ»¯¼Ø£¬ÏȽ«ÑùÆ·ÈÜÓÚÊÊÁ¿Ë®ÖУ¬³ä·Ö½Á°èºó¹ýÂË£¬ÔÚ½«ÂËÒº°´ÏÂͼËùʾ²½Öè½øÐвÙ×÷¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

ÆðʼÂËÒºµÄpH_____________7£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£¬ÆäÔ­ÒòÊÇ_________________________________________________¡£

ÊÔ¼ÁIµÄ»¯Ñ§Ê½Îª______________________£¬¢ÙÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________________¡£

ÊÔ¼Á¢òµÄ»¯Ñ§Ê½Îª______________________£¬¢ÚÖмÓÈëÊÔ¼Á¢òµÄÄ¿µÄÊÇ__________________________________________________________________£»

ÊÔ¼Á¢óµÄÃû³ÆÊÇ______________________£¬¢ÛÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________________________________________£»

ijͬѧ³ÆÈ¡Ìá´¿µÄ²úÆ·0.7759g£¬Èܽâºó¶¨¶¨ÈÝÔÚ100mLÈÝÁ¿Æ¿ÖУ¬Ã¿´ÎÈ¡25.00mLÈÜÒº£¬ÓÃ0.1000mol¡¤L-1µÄÏõËáÒø±ê×¼ÈÜÒºµÎ¶¨£¬Èý´ÎµÎ¶¨ÏûºÄ±ê×¼ÈÜÒºµÄƽ¾ùÌå»ýΪ25.62mL£¬¸Ã²úÆ·µÄ´¿¶ÈΪ____________________________________________¡££¨ÁÐʽ²¢¼ÆËã½á¹û£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÏÖÏóÒò·¢Éú¼Ó³É·´Ó¦¶ø²úÉúµÄÊÇ(    )¡£

A£®ÒÒϩʹËáÐÔKMnO4ÈÜÒºÍÊÉ«          B£®ÒÒϩʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«

 C£®½«±½¼ÓÈëäåË®ÖУ¬Õñµ´ºóË®²ã½Ó½üÎÞÉ«  D£®¼×ÍéÓëÂÈÆø»ìºÏ£¬¹âÕÕÒ»¶Îʱ¼äºó»ÆÂÌÉ«Ïûʧ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


PHBËÜÁÏÊÇÒ»ÖÖ¿ÉÔÚ΢ÉúÎï×÷ÓÃϽµ½âµÄ»·±£ÐÍËÜÁÏ£¬Æä½á¹¹¼òʽΪ£¬ÏÂÃæÓйØPHB˵·¨²»ÕýÈ·µÄÊÇ                 (¡¡  ¡¡)

A£®PHBͨ¹ý¼Ó¾Û·´Ó¦ÖƵÃ.

B£®PHBµÄµ¥ÌåÊÇCH3CH2CH(OH)COOH

C£®PHBÔÚ΢ÉúÎï×÷ÓÃϵĽµ½â²úÎï¿ÉÄÜÓÐCO2ºÍH2O

D£®PHBÊÇÒ»ÖÖ¾Ûõ¥

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸