ÏÂͼËùʾװÖÃÖУ¬¼×¡¢ÒÒ¡¢±ûÈý¸öÉÕ±­ÒÀ´Î·Ö±ðÊ¢·Å109g5.51£¥µÄNaOHÈÜÒº¡¢×ãÁ¿µÄCuSO4ÈÜÒººÍ200g10.00£¥µÄK2SO4ÈÜÒº£®µç¼«¾ùΪʯīµç¼«¡£

½ÓͨµçÔ´£¬¾­¹ýÒ»¶Îʱ¼äºó£¬²âµÃ±ûÖÐK2SO4Ũ¶ÈΪ10.47£¥£¬ÒÒÖÐcµç¼«ÖÊÁ¿Ôö¼Ó¡£¾Ý´Ë»Ø´ðÎÊÌ⣺
£¨1£©µç¼«bÉÏ·¢ÉúµÄµç¼«·´Ó¦Îª___________________________________¡£
£¨2£©µç¼«bÉÏÉú³ÉµÄÆøÌåÔÚ±ê×´¿öϵÄÌå»ýΪ__________________£¬´Ëʱ¼×ÉÕ±­ÖÐNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ£¨ÉèÈÜÒºµÄÃܶÈΪ1g/cm3£©_______________¡£
£¨3£©µç¼«cµÄÖÊÁ¿±ä»¯ÊÇ___________g£¬Óûʹµç½âºóÒÒÖеĵç½âÒº»Ö¸´µ½Æðʼ״̬£¬Ó¦¸ÃÏòÈÜÒºÖмÓÈëÊÊÁ¿µÄ___________£¨Ìî×Öĸ±àºÅ£©¡£

A£®Cu(OH)2B£®Cu2OC£®CuCO3D£®Cu2(OH)2CO3
£¨4£©ÆäËûÌõ¼þ²»±ä£¬Èç¹û°ÑÒÒ×°ÖøÄΪµç½â¾«Á¶Í­£¬Ôòcµç¼«µÄ²ÄÁÏΪ___________£¬dµç¼«µÄ²ÄÁÏΪ______¡£

£¨1£© 4OH£­£­ 4e£­£½ 2H2O + O2¡ü£¨2·Ö£©
£¨2£©5.6 L£¨2·Ö£© 1.5mol/L£¨2·Ö£©
(3)32£¨2·Ö£© C£¨1·Ö£©
(4)¾«Í­»ò´¿Í­£¨1·Ö£© ´ÖÍ­£¨1·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝÒÒÖÐcµç¼«ÖÊÁ¿Ôö¼Ó¿ÉÖªcµç½âΪÒõ¼«£¬ÔòMµç¼«ÎªµçÔ´µÄ¸º¼«£¬Nµç¼«ÎªÕý¼«£¬bµç¼«ÎªÑô¼«£¬·¢ÉúOH?ʧȥµç×ӵķ´Ó¦£º4OH£­£­ 4e£­£½ 2H2O + O2¡ü¡£
£¨2£©µç½âºó200g10.00£¥µÄK2SO4ÈÜҺŨ¶È±äΪ10.47£¥£¬Ôòµç½âµÄH2OµÄÖÊÁ¿Îª200g -200g¡Á10.00£¥¡Â10.47£¥=9g£¬µç½âH2OʱÓëתÒƵç×ӵĶÔÓ¦¹ØϵΪ£ºH2O~2e?£¬Ôòn(e?)=2¡Á9g¡Â18g/mol=1mol£¬¸ù¾Ýµç¼«b·¢Éú·´Ó¦£º4OH£­£­ 4e£­£½ 2H2O + O2¡ü£¬¿ÉÇó³öÉú³ÉµÄÆøÌ壺V(O2)=1mol¡Â4¡Á22.4L/mol=5.6L£»¼×ÉÕ±­ÖÐNaOHÈÜÒºÒ²µç½âÁË9gH2O£¬ÈÜÒºÌå»ýΪ£º£¨109g-9g£©¡Â1000g/L=0.1L£¬c(NaOH)=109g¡Á5.51%¡Â40g/mol¡Â0.1L=1.5mol?L?1¡£
£¨3£©µç¼«cΪÒõ¼«£¬·¢Éú·´Ó¦£ºCu2++2e?=Cu£¬Éú³ÉµÄCuÖÊÁ¿Îªm(Cu)=1mol¡Â2¡Á64g/mol=32g£»µç½âÉú³ÉH2SO4¡¢O2¡¢Cu£¬¸ù¾Ý³öÀ´Ê²Ã´¼ÓʲôµÄÔ­Ôò£¬¼ÓÈëCuCO3ºó¿Éʹµç½âºóÒÒÖеĵç½âÒº»Ö¸´µ½Æðʼ״̬¡£
£¨4£©µç½â¾«Á¶Í­£¬´¿Í­×÷Òõ¼«£¬µç¼«´ÖÍ­×÷Ñô¼«£¬¹Êcµç¼«µÄ²ÄÁÏΪ´¿Í­£¬dµç¼«µÄ²ÄÁÏΪ´ÖÍ­¡£
¿¼µã£º±¾Ì⿼²éµç¼«·½³ÌʽµÄÊéд¡¢µç½â²úÎïµÄ¼ÆËã¡¢µç¼«²ÄÁϵÄÑ¡Ôñ¡¢µç½âÖÊÈÜÒº»Ø¸´¼ÓÈëÎïÖʵÄÑ¡Ôñ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¢ñ.ºÏÀíÒûʳºÍÕýÈ·ÓÃÒ©£¬ÊÇÈËÌ彡¿µµÄÖØÒª±£Ö¤¡£
ÏÖÓÐÏÂÁÐÎåÖÖÎïÖÊAʳÑΣ»BСËÕ´ò£»CÆ»¹ûÖ­£»DÆÏÌÑÌÇ£»EÇàùËØ£¬Çë°´ÏÂÁÐÒªÇóÌî¿Õ(ÌîÐòºÅ)¡£
¸»º¬Î¬ÉúËØCµÄÊÇ          £¬¿ÉÖ±½Ó½øÈëѪҺ£¬²¹³äÄÜÁ¿µÄÊÇ           £¬Ó¦ÓÃ×î¹ã·ºµÄ¿¹ÉúËØÖ®Ò»µÄÊÇ          £¬¼È¿É×÷ΪÊèËɼÁ£¬ÓÖ¿ÉÖÎÁÆθËá¹ý¶àµÄÊÇ          £¬Ê³Óùý¶à»áÒýÆðѪѹÉý¸ß¡¢ÉöÔàÊÜËðµÄ          ¡£
¢ò.¸ÖÌúµÄÉú²úÓëʹÓÃÊÇÈËÀàÎÄÃ÷ºÍÉú»î½ø²½µÄÒ»¸öÖØÒª±êÖ¾¡£
£¨1£©Ð´³ö¹¤ÒµÉÏÓóàÌú¿óÁ¶ÌúµÄÖ÷Òª»¯Ñ§·´Ó¦·½³Ìʽ£º                        ¡£
£¨2£©³´¹ý²ËµÄÌú¹øδ¼°Ê±Ï´¾»£¨²ÐÒºÖк¬NaCl£©,µÚ¶þÌì±ã»áÒò¸¯Ê´³öÏÖºìºÖÉ«Ðâ°ß¡£ÊԻشð£º
¢ÙÌú¹øµÄ¸¯Ê´Ö÷ÒªÊÇÓÉ          £¨Ìѧ»òµç»¯Ñ§£©¸¯Ê´Ôì³ÉµÄ¡£ÐγɵÄÌúÐâµÄÖ÷Òª³É·ÖÊÇ       ¡£
¢ÚΪ·ÀÖ¹ÂÖ´¬µÄ´¬ÌåÔÚº£Ë®Öи¯Ê´£¬Ò»°ãÔÚ´¬ÉíÁ¬½Ó           £¨Ìп¿é¡±»ò¡°Í­¿é¡±£©¡£
¢ó.²ÄÁÏÊÇÈËÀàÀµÒÔÉú´æºÍ·¢Õ¹µÄÖØÒªÎïÖÊ»ù´¡£¬ºÏÀíʹÓòÄÁÏ¿ÉÒÔ¸ÄÉÆÎÒÃǵÄÉú»î¡£
£¨1£©ÌåÓý³¡¹Ý½¨ÉèÐè´óÁ¿½¨Öþ²ÄÁÏ¡£ÏÂÁвÄÁϲ»ÊôÓÚ¹èËáÑβÄÁϵÄÊÇ          £¨Ìî×Öĸ£©¡£
A£®Ê¯»Òʯ           B£®Ë®Äà               C£®²£Á§
£¨2£©ÔÚÏÂÁвÄÁÏÖУ¬ÊôÓÚÎÞ»ú·Ç½ðÊô²ÄÁϵÄÊÇ    (Ìî×Öĸ)¡£ÊôÓÚËÜÁÏÖÆÆ·µÄÊÇ     ¡£
A£®µÓÂÚ¡¡¡¡¡¡¡¡  B£®¾ÛÂÈÒÒÏ©ËÜÁÏ¡¡¡¡C£®µª»¯¹èÌÕ´É         D£®²£Á§¸Ö
£¨3£©ÏÂÁÐÓйغϽðÐÔÖʵÄ˵·¨ÕýÈ·µÄÊÇ          £¨Ìî×Öĸ£©¡£
A£®ºÏ½ðµÄÈÛµãÒ»°ã±ÈËüµÄ³É·Ö½ðÊô¸ß
B£®ºÏ½ðµÄÓ²¶ÈÒ»°ã±ÈËüµÄ³É·Ö½ðÊôµÍ
C£®×é³ÉºÏ½ðµÄÔªËØÖÖÀàÏàͬ£¬ºÏ½ðµÄÐÔÄܾÍÒ»¶¨Ïàͬ
D£®ºÏ½ðÓë¸÷³É·Ö½ðÊôÏà±È£¬¾ßÓÐÐí¶àÓÅÁ¼µÄÎïÀí¡¢»¯Ñ§»ò»úеÐÔÄÜ
£¨4£©ºÏ³ÉËÜÁÏ¡¢ºÏ³ÉÏ𽺺͠         Êdz£ËµµÄÈý´óºÏ³É²ÄÁÏ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ijʵÑéС×éÒÀ¾Ý¼×´¼È¼Éյķ´Ó¦Ô­Àí£¬Éè¼ÆÈçͼËùʾµÄ×°Öá£ÒÑÖª¼×³ØµÄ×Ü·´Ó¦Ê½Îª£º2CH3OH+3O2+4KOH=2K2CO3+6H2O¡£Çë»Ø´ð£º

¢ÙͨÈëO2µÄµç¼«Ãû³ÆÊÇ   £¬Bµç¼«µÄÃû³ÆÊÇ    ¡£
¢ÚͨÈëCH3OHµÄµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ        £¬Aµç¼«µÄµç¼«·´Ó¦Ê½Îª                        ¡£
¢ÛÒÒ³ØÌå»ýΪ1L£¬ÔÚAgNO3×ãÁ¿µÄÇé¿öϵç½âÒ»¶Îʱ¼äºóÈÜÒºµÄPH±äΪ1£¬ÔòÔÚÕâ¶Îʱ¼äÄÚתÒƵĵç×ÓÊýĿΪ          

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ðîµç³ØÊÇÒ»ÖÖ¿ÉÒÔ·´¸´³äµç¡¢·ÅµçµÄ×°Öá£ÓÐÒ»ÖÖÐîµç³ØÔÚ³äµçºÍ·Åµçʱ·¢ÉúµÄ·´Ó¦ÎªNiO2+Fe+2H2OFe(OH)2+Ni(OH)2
£¨1£©¸ÃÐîµç³Ø·Åµçʱ£¬·¢Éú»¹Ô­·´Ó¦µÄÎïÖÊÊÇ       £¨Ìî×Öĸ£¬ÏÂͬ£©¡£
A£®NiO2             B£®Fe             C£®Fe(OH)2            D£®Ni(OH)2
£¨2£©ÏÂÁÐÓйظõç³ØµÄ˵·¨ÖÐÕýÈ·µÄÊÇ        
A£®·Åµçʱµç½âÖÊÈÜÒºÏÔÇ¿ËáÐÔ
B£®·Åµçʱ5.6g FeÈ«²¿×ª»¯ÎªFe(OH)2ʱ£¬Íâµç·ÖÐͨ¹ýÁË0.2 molµç×Ó
C£®³äµçʱÑô¼«·´Ó¦ÎªNi(OH)2+2OH£­?2e£­==NiO2+2H2O
D£®³äµçʱÒõ¼«¸½½üÈÜÒºµÄ¼îÐÔ±£³Ö²»±ä
£¨3£©ÓôËÐîµç³Øµç½âº¬ÓÐ0.01 mol CuSO4ºÍ0.01 mol NaClµÄ»ìºÏÈÜÒº100 mL£¬µç½â³ØµÄµç¼«¾ùΪ¶èÐԵ缫¡£µ±ÈÜÒºÖеÄCu2+ È«²¿×ª»¯³ÉCuʱ£¬Ñô¼«²úÉúµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ         £»½«µç½âºóµÄÈÜÒº¼ÓˮϡÊÍÖÁ1L£¬´ËʱÈÜÒºµÄpH=      ¡£
£¨4£©ÓôËÐîµç³Ø½øÐеç½â£¬ÇÒµç½â³ØµÄµç¼«¾ùΪͭµç¼«£¬µç½âÖÊÈÜҺΪŨ¼îÒºÓëNaClÈÜÒºµÄ»ìºÏÒº£¬µç½âÒ»¶Îʱ¼äºó£¬Í¬Ñ§ÃǾªÆæµØ·¢ÏÖ£¬Ñô¼«¸½½ü²»ÊÇÉú³ÉÀ¶É«³Áµí£¬¶øÊÇÉú³ÉºìÉ«³Áµí£¬²éÔÄ×ÊÁϵÃÖª¸ÃºìÉ«³ÁµíÊÇCu2O¡£Ð´³ö¸ÃÑô¼«Éϵĵ缫·´Ó¦Ê½£º                                           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÀ¾ÝÑõ»¯»¹Ô­·´Ó¦£º2Ag+(aq)+Cu(s)=Cu2+(aq)+2Ag(s)Éè¼ÆµÄÔ­µç³ØÈçÏÂͼ£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©µç¼«XµÄ²ÄÁÏÊÇ    £¬µç½âÖÊÈÜÒºYÊÇ           ¡£
£¨2£©Òøµç¼«Îªµç³ØµÄ     ¼«¡£
£¨3£©ÑÎÇÅÖÐÑôÀë×ÓÏò      Òƶ¯£¨Ìî¡°×ó ¡±»ò¡°ÓÒ¡±£©¡£
£¨4£©Íâµç·Öеç×ÓÊÇ´Ó      £¨Ìîµç¼«²ÄÁÏÃû³Æ£¬ÏÂͬ£©µç¼«Á÷Ïò        µç¼«¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

°´ÒªÇóÌî¿Õ£º

A                    B
£¨1£©ÔÚAͼÖУ¬Ï¡ÁòËáΪµç½âÖÊÈÜÒº£¬Óõ¼ÏßÁ¬½Óºó£¬Í­Æ¬µç¼«·´Ó¦Ê½        ¡£
£¨2£©ÔÚBͼÖÐÍâ½ÓÖ±Á÷µçÔ´£¬ÈôÒªÔÚa¼«¶ÆÍ­£¬¼ÓÒÔ±ØÒªµÄÁ¬½Óºó£¬¸Ã×°ÖýР       £¬b¼«µç¼«·´Ó¦Ê½                                    ¡£
£¨3£©ÔÚBͼÖÐÍâ½ÓÖ±Á÷µçÔ´£¬Èôµç¼«Îª¶èÐԵ缫£¬µç½âÖÊÈÜÒºÊÇCuSO4ÈÜÒº£¨×ãÁ¿£©£¬µç½â×Ü·´Ó¦Àë×Ó·½³ÌʽΪ                        £¬Òõ¼«ÔöÖØ3.2 g£¬ÔòÑô¼«ÉϷųöµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýÊÇ_____L£¬¼ÓÈëÒ»¶¨Á¿µÄ     ºó£¨Ìѧʽ£©£¬ÈÜÒºÄָܻ´ÖÁÓëµç½âÇ°ÍêÈ«Ò»Ö¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

º£Ë®ÊǷḻµÄ×ÊÔ´±¦¿â£¬´Óº£Ë®ÖÐÌáÈ¡ÔªËØÊÇ»¯Ñ§¹¤ÒµµÄÖØÒª×é³É²¿·Ö¡£
£¨1£©´ÖÑξ«ÖƾÍÊdzýÈ¥ÆäÖеÄCa2+¡¢Fe3+¡¢SO42-¼°ÄàɳµÈÔÓÖÊ£¬Ðè¼ÓÈëµÄÊÔ¼ÁÓУº¢ÙNa2CO3ÈÜÒº ¢ÚHCl£¨ÑÎËᣩ ¢ÛBa£¨OH£©2ÈÜÒº£¬ÕâÈýÖÖÊÔ¼ÁµÄÌí¼Ó˳ÐòÊÇ_________£¨ÌîÐòºÅ£©¡£
£¨2£©ÓÃÂÈÆø½øÐС°º£Ë®Ìáä塱ÖÐÖÆÈ¡äåµ¥ÖÊ·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º__________¡£
£¨3£©Ä³Í¬Ñ§Éè¼ÆÁËÈçͼװÖýøÐÐÒÔϵ绯ѧʵÑé¡£

¢Ùµ±¿ª¹ØKÓëaÁ¬½Óʱ£¬Á½¼«¾ùÓÐÆøÅݲúÉú£¬ÔòÒõ¼«Îª_______¼«¡£
¢ÚÒ»¶Îʱ¼äºó£¬Ê¹¿ª¹ØKÓëa¶Ï¿ª¡¢ÓëbÁ¬½Óʱ£¬ÐéÏß¿òÄÚµÄ×°ÖÿɳÆΪ__________¡£Çëд³ö´ËʱFeµç¼«Éϵĵ缫·´Ó¦Ê½_________________¡£
£¨4£©Ä³¹«³§ÏòÊ¢ÓÐCaSO4Ðü×ÇÒºµÄ·´Ó¦³ØÖÐͨÈë°±ÆøÓûÖÆÈ¡µª·Ê£¨NH4£©2SO4£¬Ð§¹û²»ºÃ¡£ÔÙͨÈëCO2£¬ÔòÖð½¥²úÉú´óÁ¿£¨NH4£©2SO4¡£Çë·ÖÎöÆäÔ­Òò¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÈçÏÂͼװÖÃÖУ¬bµç¼«ÓýðÊô MÖƳɣ¬a¡¢c¡¢dΪʯīµç¼«£¬½ÓͨµçÔ´£¬½ðÊôM³Á»ýÓÚb¼«£¬Í¬Ê±a¡¢dµç¼«ÉϲúÉúÆøÅÝ¡£ÊԻشð£º

£¨1£©aΪ        ¼«£¬c¼«µÄµç¼«·´Ó¦Ê½Îª                             ¡£
£¨2£©µç½â½øÐÐÒ»¶Îʱ¼äºó£¬ÕÖÔÚc¼«ÉϵÄÊÔ¹ÜÖÐÒ²ÄÜÊÕ¼¯µ½µÄÆøÌ壬´Ëʱc¼«Éϵĵ缫·´Ó¦Ê½Îª          ¡£
£¨3£©µ±d¼«ÉÏÊÕ¼¯µ½44.8mLÆøÌ壨±ê×¼×´¿ö£©Ê±Í£Ö¹µç½â£¬a¼«ÉϷųöÁËÆøÌåµÄÎïÖʵÄÁ¿Îª       £¬Èôbµç¼«ÉϳÁ»ý½ðÊôMµÄÖÊÁ¿Îª0.432g£¬Ôò´Ë½ðÊôµÄĦ¶ûÖÊÁ¿Îª                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

ÓÃÁâÃÌ¿ó£¨MnCO3£©³£º¬ÓÐFe2O3¡¢FeO¡¢HgCO3¡¤2HgOµÈÔÓÖÊ£¬¹¤Òµ³£ÓÃÁâÃÌ¿óÖÆÈ¡ÃÌ£¬¹¤ÒÕÁ÷³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ïò´ÖÒº1ÖмÓÈëµÄË®×îºóÐèÒª          ·½·¨²ÅÄÜ´ïµ½¼¼ÊõÒªÇó¡£
£¨2£©Á÷³ÌÖÐÓõĿÕÆøÊÇÓÃĤ·ÖÀë·¨ÖƱ¸µÄ¸»Ñõ¿ÕÆø£¬¸Ã·½·¨µÄÔ­ÀíÊÇ       ¡£
£¨3£©¾»»¯¼ÁÖ÷Òª³É·ÖΪ(NH4)2S£¬´ÖÒº2Öз¢ÉúÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ      ¡£
£¨4£©Ð´³öÑô¼«µÄµç¼«·´Ó¦Ê½       ¡£ËµÃ÷µç½âҺѭ»·µÄÔ­Òò              ¡£
£¨5£©Ð´³öÂÁÈÈ·¨Á¶Ã̵Ļ¯Ñ§·½³Ìʽ                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸