ijͬѧÉè¼ÆÁËÈçͼËùʾװÖ㨲¿·Ö¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©½øÐÐʵÑéÑо¿£®
Çë»Ø´ð£º
£¨1£©ÓÃÉÏÊö×°ÖÃ̽¾¿Ó°Ï컯ѧ·´Ó¦ËÙÂʵÄÒòËØ£®
¢ÙÔ²µ×ÉÕÆ¿Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Zn+2H+=Zn2++H2¡ü
Zn+2H+=Zn2++H2¡ü
£®
¢ÚÓÃÈçͼËùʾװÖýøÐÐʵÑ飬ÒÔÉú³É9.0mLÆøÌåΪ¼ÆʱÖյ㣬½á¹ûΪt1£¾t2£®
ÐòºÅ V£¨H2SO4£©/mL c£¨H2SO4£©/mol?L-1 t/s ¢ñ 40 1 t1 ¢ò 40 4 t2
±È½ÏʵÑé¢ñºÍ¢ò¿ÉÒԵóöµÄʵÑé½áÂÛÊÇ
ÔÚÆäËüÌõ¼þÒ»¶¨Ê±£¬»¯Ñ§·´Ó¦ËÙÂÊËæ·´Ó¦ÎïŨ¶ÈµÄÔö´ó¶øÔö´ó
ÔÚÆäËüÌõ¼þÒ»¶¨Ê±£¬»¯Ñ§·´Ó¦ËÙÂÊËæ·´Ó¦ÎïŨ¶ÈµÄÔö´ó¶øÔö´ó
£®
¢ÛÈô½«Ð¿Æ¬»»³Éº¬´ÖпƬ£¬ÇÒ¿ØÖÆÆäËûÌõ¼þʹÆäÓëÉÏÊöʵÑéÍêÈ«Ò»Ö£¬Ëù²âµÃµÄ·´Ó¦ËÙÂʾù´óÓÚÉÏÊöʵÑéµÄÊý¾Ý£®´ÖпƬÖÐËùº¬ÔÓÖÊ¿ÉÄÜÊÇ£¨ÌîÐòºÅ£©
abc
abc
£®          
a£®Ê¯Ä«     b£®Òø      c£®Í­      d£®É³Á££¨¶þÑõ»¯¹è£©
£¨2£©ÓÃÉÏÊö×°ÖÃÑéÖ¤ÉúÌúÔÚ³±Êª¿ÕÆøÖлᷢÉúÎüÑõ¸¯Ê´£®ÉúÌú·ÅÔÚÉÕÆ¿ÖУ¬Ôò
¢ÙÔ²µ×ÉÕÆ¿ÖеÄÊÔ¼Á¿ÉÑ¡Óã¨ÌîÐòºÅ£©
ac
ac
£®
a£®NaOHÈÜÒº     b£®C2H5OH   c£®NaClÈÜÒº   d£®Ï¡ÁòËá
¢ÚÄÜÖ¤Ã÷ÉúÌúÔÚ³±Êª¿ÕÆøÖлᷢÉúÎüÑõ¸¯Ê´µÄÏÖÏóÊÇ
Á¿Æø¹ÜÓÒ¶ËÒºÃæϽµ£¬×ó¶ËÒºÃæÉÏÉý
Á¿Æø¹ÜÓÒ¶ËÒºÃæϽµ£¬×ó¶ËÒºÃæÉÏÉý
£®
·ÖÎö£º£¨1£©¢ÙпÓëÏ¡ÁòËá·´Ó¦Éú³ÉÁòËáпºÍÇâÆø£»
¢Út1£¾t2£¬ËµÃ÷ʵÑé¢ò·´Ó¦ËÙÂʽϴ󣬽áºÏŨ¶ÈµÄÓ°ÏìÅжϣ»
¢ÛËùº¬ÔÓÖÊÄÜÓëпÐγÉÔ­µç³Ø·´Ó¦£»
£¨2£©¢Ù·¢ÉúÎüÑõ¸¯Ê´Ó¦ÔÚ¼îÐÔ»òÖÐÐÔÈÜÒºÖнøÐУ»
¢Ú·¢ÉúÎüÑõ¸¯Ê´Ê±£¬ÏûºÄÑõÆø£¬Á¿Æø¹ÜÓÒ¶ËÒºÃæϽµ£¬×ó¶ËÒºÃæÉÏÉý£®
½â´ð£º½â£º£¨1£©¢ÙпÓëÏ¡ÁòËá·´Ó¦Éú³ÉÁòËáпºÍÇâÆø£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪZn+2H+=Zn2++H2¡ü£¬
¹Ê´ð°¸Îª£ºZn+2H+=Zn2++H2¡ü£»
¢Út1£¾t2£¬ËµÃ÷ʵÑé¢ò·´Ó¦ËÙÂʽϴ󣬶þÕßŨ¶È²»Í¬£¬ËµÃ÷ÔÚÆäËüÌõ¼þÒ»¶¨Ê±£¬»¯Ñ§·´Ó¦ËÙÂÊËæ·´Ó¦ÎïŨ¶ÈµÄÔö´ó¶øÔö´ó£¬¶ÔÁ¿Æø¹Ü¶ÁÊýʱ£¬Òª×¢Òâµ÷Õû×óÓÒÁ½¹ÜµÄÒºÃæ¸ß¶ÈÏàƽ£¬ÇÒÊÓÏßÓëÒºÃæÏàƽ£¬
¹Ê´ð°¸Îª£ºÔÚÆäËüÌõ¼þÒ»¶¨Ê±£¬»¯Ñ§·´Ó¦ËÙÂÊËæ·´Ó¦ÎïŨ¶ÈµÄÔö´ó¶øÔö´ó£»
¢ÛÓÉ·´Ó¦ËÙÂÊ´óÓÚÉÏÊöʵÑé¶ÔÓ¦µÄÊý¾Ý¿ÉÖªËùº¬ÔÓÖʱØÄÜÓëпÐγÉÔ­µç³Ø£¬Ê¹·´Ó¦ËÙÂÊÔö´ó£¬Ëùº¬ÔÓÖÊ¿ÉÒÔÊÇʯī¡¢Òø¡¢Í­µÈ£¬
¹Ê´ð°¸Îª£ºabc£»
£¨2£©¢Ù¸ÖÌúÔÚÖÐÐÔ»òÈõ¼îÐÔ»·¾³ÖÐÒ×·¢ÉúÎüÑõ¸¯Ê´£¬
¹Ê´ð°¸Îª£ºac£»
¢ÚÎüÑõ¸¯Ê´ÏûºÄÑõÆø£¬¹ÊÁ¿Æø¹ÜÓÒ¶ËÒºÃæϽµ£¬×ó¶ËÒºÃæÉÏÉý£¬
¹Ê´ð°¸Îª£ºÁ¿Æø¹ÜÓÒ¶ËÒºÃæϽµ£¬×ó¶ËÒºÃæÉÏÉý£®
µãÆÀ£º±¾Ì⿼²é»¯Ñ§·´Ó¦ËÙÂÊÒÔ¼°½ðÊôµÄµç»¯Ñ§¸¯Ê´µÄʵÑé̽¾¿£¬´ðÌâʱעÒâʵÑéÊý¾ÝµÄ´¦Àí¡¢·ÖÎö£¬°ÑÎÕʵÑéÔ­Àí¡¢¸ù¾ÝʵÑéÏÖÏóµÃ³öʵÑé½áÂÛ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¹úÎñԺǿµ÷£¬¡°ÄÏË®±±µ÷¡±¹¤³Ì±ØÐë¼á³Ö½ÚË®¡¢ÖÎÎÛºÍÉú̬»·¾³±£»¤Ó빤³Ì½¨ÉèÏàЭµ÷£¬ÒÔË®×ÊÔ´ºÏÀíÅäÖÃΪÖ÷Ïߣ¬°Ñ½ÚË®¡¢ÖÎÎÛ¡¢Éú̬»·¾³±£»¤Óëµ÷Ë®×÷Ϊһ¸öÍêÕûµÄϵͳÀ´·ÖÎöµÄÔ­Ôò£®ÔÚµ÷Ë®¹¤³ÌÖУ¬ÑØ;¹¤ÒµÎÛË®µÄÈÎÒâÅÅ·ÅÊÇÔì³ÉË®Öʶñ»¯µÄ×î´óÒþ»¼£®¼ì²âij¹¤³§·ÏÒºÖУ¬º¬ÓдóÁ¿µÄMg2+¡¢Al3+¡¢Cu2+¡¢Ag+£®ÊÔ·ÖÎö»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸Ã·ÏÒºÖпÉÄÜ´óÁ¿´æÔÚµÄÒ»ÖÖÒõÀë×ÓÊÇ
D
D
£¨Ñ¡ÌîÐòºÅ£©?
A£®SO42-   B£®CO32-     C£®Cl-   D£®NO3-
£¨2£©¾­¼ì²â·ÏÒºÖÐÂÁÔªËصĺ¬Á¿½Ï¸ß£¬Ð轫ÆäÓëÆäËûÈýÖÖÀë×Ó·Ö¿ª£¬ÇëÑ¡ÓúÏÊʵÄÊÔ¼Á£¬Ð´³öÂÁÔªËØÓë¸ÃÊÔ¼Á·´Ó¦Ê±µÄÀë×Ó·½³Ìʽ£º
Al3++4OH-=AlO2-+2H2O
Al3++4OH-=AlO2-+2H2O
£®
£¨3£©ÎªÁË»ØÊÕ·ÏÒºÖеĽðÊôÒø£¬Ä³Í¬Ñ§Éè¼ÆÁËÈçͼËùʾµÄ·½°¸£®ÈôÒÀ¸Ã·½°¸»ñµÃ½ðÊôÒø10.80g£¬Îª±£Ö¤²»ÎÛȾ»·¾³ºÍCl2µÄÑ­»·ÀûÓã¬ÀíÂÛÉÏÓ¦Ìṩ±ê×¼×´¿öϵÄH2Ìå»ýΪ
1.12
1.12
L£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijͬѧÉè¼ÆÁËÈçͼËùʾװÖ㨼гÖÒÇÆ÷Ê¡ÂÔ£©½øÐÐϵÁÐʵÑ飬ʵÑéʱ½«Ò©Æ·AÖðµÎ¼ÓÈëµ½¹ÌÌåBÖУ¬Çë¸ù¾ÝÏÂÁÐʵÑé»Ø´ðÎÊÌ⣺
£¨1£©ÈôAΪˮ£¬BΪ¹ýÑõ»¯ÄÆ£¬CÖÐÊ¢ÓÐÑÎËáËữ¹ýµÄFeCl2ÈÜÒº£¬Ðý¿ª»îÈûEºó£¬BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2Na2O2+2H2O=4NaOH+O2
2Na2O2+2H2O=4NaOH+O2
£»CÖÐÏÖÏóΪ
±ä»Æ»ò¼ÓÉî
±ä»Æ»ò¼ÓÉî
£®
£¨2£©ÈôAΪÁòËᣬBΪÑÇÁòËáÄƹÌÌ壬CÖÐÊ¢ÓÐÇâÁòËáÈÜÒº£¬Ðý¿ª»îÈûEºó£¬BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽµÄΪ
H2SO4+Na2SO3=Na2SO4+H2O+SO2
H2SO4+Na2SO3=Na2SO4+H2O+SO2
£¬CÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽµÄΪ
2H2S+SO2=3S+2H2O
2H2S+SO2=3S+2H2O
£®
£¨3£©ÀûÓÃÉÏÊö×°Öû¹¿ÉÒÔÑéÖ¤ÎïÖʵÄÐÔÖÊ£¬ÈçÉè¼ÆÖ¤Ã÷Ñõ»¯ÐÔ£ºKClO3£¾Cl2£¾Br2£¬ÔòAÖмÓŨÑÎËᣬBÖмÓ
KClO3
KClO3
£¬CÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
2Br-+Cl2=2Cl-+Br2
2Br-+Cl2=2Cl-+Br2
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijͬѧÉè¼ÆÁËÈçͼËùʾµÄʵÑé×°ÖÃÀ´´ÖÂԵزⶨµçʯÖÐ̼»¯¸ÆµÄÖÊÁ¿·ÖÊý£®
£¨1£©ÉÕÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
CaC2+2H2O¡úCa£¨OH£©2+C2H2¡ü
CaC2+2H2O¡úCa£¨OH£©2+C2H2¡ü
£®×°ÖÃB¡¢CµÄ×÷ÓÃÊÇ
×°ÖÃBÓÃÓÚÖüË®ÒÔ±ãÉÕÆ¿ÖÐÉú³ÉµÄC2H2ÆøÌå½øÈëBʱÅųöÓëËüµÈÌå»ýµÄË®½øÈëÁ¿Í²CÀ´²â¶¨Éú³ÉÒÒȲµÄÁ¿£»
×°ÖÃBÓÃÓÚÖüË®ÒÔ±ãÉÕÆ¿ÖÐÉú³ÉµÄC2H2ÆøÌå½øÈëBʱÅųöÓëËüµÈÌå»ýµÄË®½øÈëÁ¿Í²CÀ´²â¶¨Éú³ÉÒÒȲµÄÁ¿£»
£®
£¨2£©·ÖҺ©¶·ÓëÉÕÆ¿¼äÓÃÈ齺¹ÜÁ¬½ÓµÄÄ¿µÄÊÇ
±ãÓÚÒºÌå˳ÀûµÎÏÂ
±ãÓÚÒºÌå˳ÀûµÎÏÂ
£®
£¨3£©ËùÓõçʯÖÊÁ¿²»ÄÜÌ«´ó£¬·ñÔò
Éú³ÉC2H2Ì«¶à³¬¹ýBµÄ¿Õ¼ä
Éú³ÉC2H2Ì«¶à³¬¹ýBµÄ¿Õ¼ä
£»Ò²²»ÄÜ̫С£¬·ñÔò
ÎÞ·¨²â¶¨ÆäÌå»ý²â¶¨Îó²î»á¸ü´ó
ÎÞ·¨²â¶¨ÆäÌå»ý²â¶¨Îó²î»á¸ü´ó
£»ÈôÈÝÆ÷BµÄÈÝ»ýΪ250mL£¬ÔòËùÓõçʯµÄÖÊÁ¿Ó¦ÔÚ
0.60
0.60
g×óÓÒ£¨´ÓºóÃæÊý¾ÝÖÐÑ¡Ì0.03¡¢0.60¡¢1.00¡¢1.50¡¢2.00£©£®
£¨4£©ÎªÁ˵õ½±È½ÏƽÎȵÄÒÒȲÆøÁ÷£¬ÊµÑéÊÒͨ³£ÓÃ
±¥ºÍʳÑÎË®
±¥ºÍʳÑÎË®
´úÌæË®½øÐÐʵÑ飮
£¨5£©ÊµÑéÖвâµÃÅÅÈëÁ¿Í²ÖÐË®µÄÌå»ýΪVL¡¢µçʯµÄÖÊÁ¿ÎªWg£®ÔòµçʯÖÐ̼»¯¸ÆµÄÖÊÁ¿·ÖÊýÊÇ
0.29V
W
%
0.29V
W
%
£¨²»¼ÆËãµ¼¹ÜÖвÐÁôµÄË®£¬ÆøÌåÖб¥ºÍµÄË®ÕôÆøµÈÒ²ºöÂÔ²»¼Æ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2010?ÁÙÒÊһģ£©»¯Ñ§ÊµÑéÖÐÒ»Ì××°ÖÃÍùÍù¿ÉÒÔÍê³É¶à¸öʵÑ飬AijͬѧÉè¼ÆÁËÈçͼËùʾװÖ㨼гÖÒÇÆ÷Ê¡ÂÔ£©½øÐÐϵÁÐʵÑ飮Çë»Ø´ð£º
£¨1£©Ö¸³öÏÂÁÐÒÇÆ÷µÄÃû³Æ£º
A
·ÖҺ©¶·
·ÖҺ©¶·
£»
D
¸ÉÔï¹Ü
¸ÉÔï¹Ü
£®
£¨2£©ÈôAÖÐΪŨ°±Ë®£¬BÖÐΪÉռCÖÐΪAICl3ÈÜÒº£¬ÊµÑéÖпɹ۲⵽BÖÐÓÐÆøÌåÉú³É£¬CÖÐÓа×É«³Áµí£®ÔòCÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
Al3++3NH3?H2O¨TAl£¨OH£©3+3NH4+
Al3++3NH3?H2O¨TAl£¨OH£©3+3NH4+
£¬ÒÇÆ÷DµÄ×÷ÓÃΪ
·ÀÖ¹µ¹Îü
·ÀÖ¹µ¹Îü
£®
£¨3£©ÈôAÖÐ×°ÓÐŨÑÎËᣬBÖÐ×°ÓйÌÌåKMn04£¬CÖÐÊ¢ÓÐKIµí·ÛÈÜÒº£¬ÊµÑéÖпɹ۲⵽BÖгöÏÖ»ÆÂÌÉ«ÆøÌ壬CÖÐÈÜÒº±äÀ¶£®ÒÀ¾ÝʵÑéÏÖÏó¿ÉÒԵõ½µÄ½áÂÛΪ£º
ɾ³ý´Ë¿Õ
ɾ³ý´Ë¿Õ
£»
Ñõ»¯ÐÔÓÉÇ¿µ½ÈõµÄ˳Ðò£º
KMnO4£¾Cl2£¾I2
KMnO4£¾Cl2£¾I2
£»
µ«¸Ã×°ÖõIJ»×ãÖ®´¦ÊÇ£º
ȱÉÙβÆøÎüÊÕ×°ÖÃ
ȱÉÙβÆøÎüÊÕ×°ÖÃ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»ÆÍ­¿óÊǹ¤ÒµÁ¶Í­µÄÖ÷ÒªÔ­ÁÏ£¬Ö÷Òª³É·ÖΪCuFeS2£¬º¬ÉÙÁ¿Âöʯ£®Îª²â¶¨¸Ã»ÆÍ­¿óµÄ´¿¶È£¬Ä³Í¬Ñ§Éè¼ÆÁËÈçͼËùʾʵÑ飺
¾«Ó¢¼Ò½ÌÍø
ÏÖÓõç×ÓÌìƽ³ÆÈ¡ÑÐϸµÄ»ÆÍ­¿óÑùÆ·1.150g£¬ÔÚ¿ÕÆø´æÔÚϽøÐÐìÑÉÕ£¬Éú³ÉCu¡¢Fe3O4ºÍSO2ÆøÌ壬ʵÑéºóÈ¡dÖÐÈÜÒºµÄ
110
ÖÃÓÚ׶ÐÎÆ¿ÖУ¬ÓÃ0.05mol?L-1±ê×¼µâÈÜÒº½øÐе樣¬ÏûºÄ±ê×¼µâÈÜÒº20.00mL£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©½«ÑùÆ·ÑÐϸºóÔÙ½øÐз´Ó¦£¬ÆäÄ¿µÄÊÇ
 
£»±ê×¼µâÈÜҺӦʢ·ÅÔÚ£¨Ìî¡°¼îʽ¡±»ò¡°Ëáʽ¡±£©
 
µÎ¶¨¹ÜÖУ®
£¨2£©×°ÖÃaµÄ×÷ÓÃÊÇ
 
£®
A£®³ýÈ¥¿ÕÆøÖеĶþÑõ»¯Ì¼       
B£®³ýÈ¥¿ÕÆøÖеÄË®ÕôÆø
C£®ÓÐÀûÓÚÆøÌå»ìºÏ             
D£®ÓÐÀûÓڹ۲졢¿ØÖÆ¿ÕÆøÁ÷ËÙ
£¨3£©ÈôÈ¥µôc×°Ö㬻áʹ²â¶¨½á¹û£¨Ìî¡°Æ«µÍ¡±¡¢¡°Æ«¸ß¡±»ò¡°ÎÞÓ°Ï족£©
 
£¬Ð´³öÓ°Ïì²â¶¨½á¹ûµÄ»¯Ñ§·½³Ìʽ
 
£®
£¨4£©ÉÏÊö·´Ó¦½áÊøºó£¬ÈÔÐèͨһ¶Îʱ¼äµÄ¿ÕÆø£¬ÆäÄ¿µÄÊÇ
 
£®
£¨5£©Í¨¹ý¼ÆËã¿ÉÖª£¬¸Ã»ÆÍ­¿óµÄ´¿¶ÈΪ
 
£®
£¨6£©¼ÙÉèʵÑé²Ù×÷¾ùÕýÈ·£¬²âµÃµÄ»ÆÍ­¿ó´¿¶ÈÈÔȻƫµÍ£¬¿ÉÄܵÄÔ­ÒòÖ÷ÒªÓÐ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸