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                        ¡£

£¨1£©Ñô£¨1·Ö£©£»   2I¡ªÒ»2e¡ª£½I2£¨2·Ö£©
£¨2£©40H¡ª¡ª4e¡ª£½2H2O+O2¡ü £¨2·Ö£©
£¨3£©0£®001 mol£¨1·Ö£©£»108g£¯mol  £¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£ºÓɵç½âÔ­Àí¿ÉµÃ£º½ðÊôM³Á»ýÓÚb¼«£¬ËµÃ÷bÊÇÒõ¼«£¬ÔòaÊÇÑô¼«£¬cÊÇÑô¼«£¬dÊÇÒõ¼«£¬£¨1£©ÒòaÊÇÑô¼«£¬ÈÜÒºÖеÄÒõÀë×ӷŵ磬¸ù¾ÝÀë×ӵķŵç˳Ðò£¬¿ÉÖªÊÇ2I--2e-=I2£»£¨2£©ÔÚBÉÕ±­ÖУ¬ cÊÇÑô¼«£¬ÈÜÒºÖеÄÒõÀë×ӷŵ磬¼´2I--2e-=I2£¬I2Óöµ½µí·ÛÄÜʹµí·Û±äÀ¶£¬I-·ÅµçÍê±Ïºó£¬½Ó×ÅÊÇOH-·Åµç£º4OH--4e=2H2O+O2¡ü£¬c¼«ÉϵÄÊÔ¹ÜÖÐÊÕ¼¯µ½µÄÆøÌåΪÑõÆø£¬£¨3£©dµç¼«ÉÏÊÕ¼¯µÄ44.8mlÆøÌ壨±ê×¼×´¿ö£©ÊÇÇâÆø£¬a¼«ÉÏÊÕ¼¯µ½µÄÆøÌåÊÇÑõÆø£¬¸ù¾ÝתÒƵç×ÓÊýÏàµÈÖª£¬ÑõÆøºÍÇâÆøµÄÌå»ýÖ®±ÈÊÇ1£º2£¬dµç¼«ÉÏÊÕ¼¯µÄ44.8mlÆøÌåÇâÆø£¬Ôòaµç¼«ÉÏÊÕ¼¯µ½22.4mLÑõÆø£»ÎïÖʵÄÁ¿Îª0.01mol£¬dµç¼«ÉÏÎö³öµÄÇâÆøµÄÎïÖʵÄÁ¿=£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿ÊÇ0.04mol£¬ÏõËáÑÎÖÐMÏÔ+1¼Û£¬ËùÒÔµ±×ªÒÆ0.04molµç×ÓʱÎö³ö0.04mol½ðÊôµ¥ÖÊ£¬M=
¿¼µã£ºÔ­µç³ØºÍµç½â³ØµÄ¹¤×÷Ô­Àí£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ij»¯Ñ§ÐËȤС×éΪÁË̽Ë÷ÂÁµç¼«ÔÚÔ­µç³ØÖеÄ×÷ÓÃ,Éè¼Æ²¢½øÐÐÁËÒÔÏÂһϵÁÐʵÑé,ʵÑé½á¹û¼Ç¼ÈçÏÂ:

񅧏
µç¼«²ÄÁÏ
µç½âÖÊÈÜÒº
µçÁ÷¼ÆÖ¸Õë
ƫת·½Ïò
1
Mg¡¡Al
Ï¡ÑÎËá
Æ«ÏòAl
2
Al¡¡Cu
Ï¡ÑÎËá
Æ«ÏòCu
3
Al¡¡Ê¯Ä«
Ï¡ÑÎËá
Æ«Ïòʯī
4
Mg¡¡Al
NaOHÈÜÒº
Æ«ÏòMg
5
Al¡¡Zn
ŨÏõËá
Æ«ÏòAl
 
¸ù¾ÝÉϱíÖеÄʵÑéÏÖÏóÍê³ÉÏÂÁÐÎÊÌâ:
£¨1£©ÊµÑé1¡¢2ÖÐAlËù×÷µÄµç¼«ÊÇ·ñÏàͬ?¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 
£¨2£©Ö¸³öʵÑé3ÖÐÂÁºÍʯīµÄµç¼«Ãû³Æ£¬²¢Ð´³öʵÑéÖеĵ缫·´Ó¦ºÍµç³Ø×Ü·´Ó¦¡£
ÂÁΪ(¡¡¡¡)¡¡ ¡¡ ¡¡ ¡¡ ¡¡£»  Ê¯Ä«Îª(¡¡¡¡)¡¡ ¡¡ ¡¡ ¡¡ ¡¡¡£ 
µç³Ø×Ü·´Ó¦:                                            ¡£ 
£¨3£©ÊµÑé4ÖеÄÂÁ×÷Õý¼«»¹ÊǸº¼«?¡¡¡¡¡¡,Ϊʲô?¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

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______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÐÂÐ͸ßЧµÄ¼×ÍéȼÁϵç³Ø²ÉÓò¬Îªµç¼«²ÄÁÏ£¬Á½µç¼«ÉÏ·Ö±ðͨÈëCH4ºÍO2£¬µç½âÖÊΪKOHÈÜÒº¡£Ä³Ñо¿Ð¡×齫¼×ÍéȼÁϵç³Ø×÷ΪµçÔ´½øÐÐÂÈ»¯Ã¾ÈÜÒº¹ê½âʵÑ飬µç½â×°ÖÃÈçͼËùʾ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼×ÍéȼÁϵç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½Îª£º                                  ¡£
£¨2£©±ÕºÏ¿ª¹ØKºó£¬a¡¢bµç¼«ÉϾùÓÐÆøÌå²úÉú£¬ÆäÖÐaµç¼«ÉϵÄÆøÌå¿ÉÓà         ¼ìÑ飬bµç¼«Éϵõ½µÄÆøÌåÊÇ         £¬µç½âÂÈ»¯Ã¾ÈÜÒºµÄÀë×Ó·½³ÌʽΪ            ¡£
£¨3£©Èô¼×ÍéͨÈëÁ¿Îª1.12 L£¨±ê×¼×´¿ö£©£¬ÇÒ·´Ó¦ÍêÈ«£¬ÔòÀíÂÛÉÏͨ¹ýµç½â³ØµÄµç×ÓµÄÎïÖʵÄÁ¿Îª           £¬²úÉúµÄÂÈÆøÌå»ýΪ           L£¨±ê×¼×´¿ö£©¡£
£¨4£©ÒÑÖª³£Î³£Ñ¹Ï£¬0.25molCH4ÍêȫȼÉÕÉú³ÉCO2ºÍH2Oʱ£¬·Å³ö222.5kJÈÈÁ¿£¬Çëд³öCH4ȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ                                          ¡£
ÒÑÖª£»¢ÙC£¨Ê¯Ä«£©+O2£¨g£©=CO2£¨g£©¡÷H1=-393£®5kJ/mol
¢Ú2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H2=-571£®6kJ/mol
¼ÆË㣺C£¨Ê¯Ä«£©ÓëH2£¨g£©·´Ó¦Éú³É1molCH4£¨g£©µÄ¡÷H=             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÈçÏÂͼËùʾ£¬Ä³Í¬Ñ§Éè¼ÆÁËÒ»¸öȼÁϵç³Ø²¢Ì½¾¿ÂȼҵԭÀíºÍ´ÖÍ­µÄ¾«Á¶Ô­Àí£¬ÆäÖÐÒÒ×°ÖÃÖÐXΪÑôÀë×Ó½»»»Ä¤¡£Çë°´ÒªÇó»Ø´ðÏà¹ØÎÊÌâ:

£¨1£©¼×ÍéȼÁϵç³Ø¸º¼«µç¼«·´Ó¦Ê½ÊÇ:                                        
£¨2£©Ê¯Ä«µç¼«(C)µÄµç¼«·´Ó¦Ê½Îª                                       
£¨3£©ÈôÔÚ±ê×¼×´¿öÏ£¬ÓÐ2£® 24 LÑõÆø²Î¼Ó·´Ó¦£¬ÔòÒÒ×°ÖÃÖÐÌúµç¼«ÉÏÉú³ÉµÄÆøÌåÌå»ýΪ_  L;±û×°ÖÃÖÐÒõ¼«Îö³öÍ­µÄÖÊÁ¿Îª      g
£¨4£©Ä³Í¬Ñ§Àû¼×ÍéȼÁϵç³ØÉè¼Æµç½â·¨ÖÆȡƯ°×Òº»òFe(OH)2µÄʵÑé×°ÖÃ(ÈçͼËùʾ)¡£ÈôÓÃÓÚÖÆƯ°×ҺʱaΪµç³Ø_  ¼«£¬µç½âÖÊÈÜÒº×îºÃÓÃ_        ¡£ÈôÓÃÓÚÖÆ   Fe(OH)2£¬Ê¹ÓÃÁòËáÄÆ×öµç½âÖÊÈÜÒº£¬Ñô¼«Ñ¡Óà     ×÷µç¼«¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¢ñ£®ÔÚÏÂͼµÄ×°ÖÃÖУ¬ÊôÓÚÔ­µç³ØµÄÊÇ            ¡£

¢ò£®£¨1£©ÈçͼËùʾ£¬ÉÕ±­ÖÐΪCuCl2ÈÜÒº£¬ÔÚͼÖл­³ö±ØÒªµÄÁ¬Ïß»ò×°Öã¬Ê¹Á¬½ÓºóµÄ×°ÖÃΪԭµç³Ø¡£µç¼«·´Ó¦·½³Ìʽ£º

Ìú°å£º                       £»Ì¼°ô£º                         ¡£
£¨2£©Í­Æ¬¡¢Ð¿Æ¬Á¬½Óºó½þÈëÏ¡ÁòËáÖй¹³ÉÔ­µç³Ø£¬µ±µ¼ÏßÉÏͨ¹ý3.01¡Á1022¸öµç×Óʱ£¬Ð¿Æ¬ÖÊÁ¿¼õÉÙ________g¡£Í­Æ¬±íÃæÎö³öÇâÆø_________L(±ê×¼×´¿ö)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

µç½âÔ­ÀíÔÚ»¯Ñ§¹¤ÒµÖÐÓй㷺ӦÓá£ÏÂͼ±íʾһ¸öµç½â³Ø£¬×°Óеç½âÒºa£»X¡¢YÊÇÁ½¿éµç¼«°å£¬Í¨¹ýµ¼ÏßÓëÖ±Á÷µçÔ´ÏàÁ¬¡£

Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©ÈôX¡¢Y¶¼ÊǶèÐԵ缫£¬aÊDZ¥ºÍNaClÈÜÒº£¬ÊµÑ鿪ʼʱ£¬Í¬Ê±ÔÚÁ½±ß¸÷µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬Ôò
¢Ùµç½â³ØÖÐX¼«Éϵĵ缫·´Ó¦Ê½Îª                                  ¡£
ÔÚX¼«¸½½ü¹Û²ìµ½µÄÏÖÏóÊÇ                                        ¡£¡¡
¢ÚYµç¼«Éϵĵ缫·´Ó¦Ê½Îª                                         £¬
¼ìÑé¸Ãµç¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ                                       ¡£
£¨2£©ÈçÒªÓõç½â·½·¨¾«Á¶´ÖÍ­£¬µç½âÒºaÑ¡ÓÃCuSO4ÈÜÒº£¬Ôò
¢ÙXµç¼«µÄ²ÄÁÏÊÇ          £¬µç¼«·´Ó¦Ê½ÊÇ                         ¡£
¢ÚYµç¼«µÄ²ÄÁÏÊÇ          £¬µç¼«·´Ó¦Ê½ÊÇ                        ¡£
£¨ËµÃ÷£ºÔÓÖÊ·¢ÉúµÄµç¼«·´Ó¦²»±Øд³ö£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÎÊ´ðÌâ

(14·Ö)2013Äê³õ£¬Îíö²ÌìÆø¶à´ÎËÁÅ°Ìì½ò¡¢±±¾©µÈµØÇø¡£ÆäÖУ¬È¼ÃººÍÆû³µÎ²ÆøÊÇ
Ôì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»¡£
(l)Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO(g) + 2CO(g)2CO2(g)+ N2(g)¡÷H <0
¢Ù¸Ã·´Ó¦Æ½ºâ³£Êý±í´ïʽ____________________________
¢ÚÈô¸Ã·´Ó¦ÔÚ¾øÈÈ¡¢ºãÈݵÄÃܱÕÌåϵÖнøÐУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦ÔÚ½øÐе½t1ʱ¿Ì´ïµ½Æ½ºâ״̬µÄÊÇ________________£¨Ìî´úºÅ£©¡£

£¨2£©Ö±½ÓÅÅ·ÅúȼÉÕ²úÉúµÄÑÌÆø»áÒýÆðÑÏÖصĻ·¾³ÎÊÌ⡣úȼÉÕ²úÉúµÄÑÌÆøº¬µªµÄÑõ»¯ÎÓÃCH4´ß»¯»¹Ô­NOx¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£
ÒÑÖª£ºCH4(g)+2NO2(g)£½N2(g)£«CO2(g)+2H2O(g)¡¡¡÷H£½£­867 kJ/mol        ¢Ù
2NO2(g)N2O4(g)  ¡÷H£½£­56.9 kJ/mol         ¢Ú
H2O(g)£½H2O(l)   ¡÷H£½£­44.0 kJ/mol            ¢Û
д³öCH4´ß»¯»¹Ô­N2O4(g)Éú³ÉN2ºÍH2O£¨1£©µÄÈÈ»¯Ñ§·½³Ìʽ£º_____________________¡£
£¨3£©¼×ÍéȼÁϵç³Ø¿ÉÒÔÌáÉýÄÜÁ¿ÀûÓÃÂÊ¡£ÏÂͼÊÇÀûÓü×ÍéȼÁϵç³Øµç½â100mLlmol/LʳÑÎË®£¬µç½âÒ»¶Îʱ¼äºó£¬ÊÕ¼¯µ½±ê×¼×´¿öϵÄÇâÆø2.24L£¨Éèµç½âºóÈÜÒºÌå»ý²»±ä£©£®

¢Ù¼×ÍéȼÁϵç³ØµÄ¸º¼«·´Ó¦Ê½£º______________________________________.
¢Úµç½âºóÈÜÒºµÄpH=____£¨ºöÂÔÂÈÆøÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£©
¢ÛÑô¼«²úÉúÆøÌåµÄÌå»ýÔÚ±ê×¼×´¿öÏÂÊÇ________L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑ֪̼̼µ¥¼ü¿ÉÈƼüÖá×ÔÓÉÐýת,ijÌþµÄ½á¹¹¼òʽÈçÏÂͼËùʾ,ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ

A£®¸ÃÎïÖÊËùÓÐÔ­×Ó¾ù¿É¹²Ãæ
B£®·Ö×ÓÖÐÖÁÉÙÓÐ10¸ö̼ԭ×Ó´¦ÓÚͬһƽÃæÉÏ
C£®·Ö×ÓÖÐÖÁÉÙÓÐ11¸ö̼ԭ×Ó´¦ÓÚͬһƽÃæÉÏ
D£®¸ÃÌþÓë±½»¥ÎªÍ¬ÏµÎï

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸