ij»¯Ñ§Ð¡×é²ÉÓÃÀàËÆÖÆÒÒËáÒÒõ¥µÄ×°Öã¨Èçͼ¢ñ)ÒÔ»·¼º´¼ÖƱ¸»·¼ºÏ©¡£

ÒÑÖª£º

Ãܶȣ¨g¡¤cm-3)

È۵㣨¡æ)·Ð

µã£¨¡æ)

ÈܽâÐÔ

»·¼º´¼

0.96

25

161

ÄÜÈÜÓÚË®

»·¼ºÏ©

0.81

-103

83

ÄÑÈÜÓÚË®

ͼ¢ñ

(1)ÖƱ¸´ÖÆ·

½«12.5 mL»·¼º´¼¼ÓÈëÊÔ¹ÜAÖУ¬ÔÙ¼ÓÈë1 mLŨÁòËᣬҡÔȺó·ÅÈëËé´ÉƬ£¬»ºÂý¼ÓÈÈÖÁ·´Ó¦ÍêÈ«£¬ÔÚÊÔ¹ÜCÄڵõ½»·¼ºÏ©´ÖÆ·¡£

¢ÙAÖÐËé´ÉƬµÄ×÷ÓÃÊÇ_____________£¬µ¼¹ÜB³ýÁ˵¼ÆøÍ⻹¾ßÓеÄ×÷ÓÃÊÇ_______________¡£

¢ÚÊÔ¹ÜCÖÃÓÚ±ùˮԡÖеÄÄ¿µÄÊÇ_________________________________________________¡£

£¨2£©ÖƱ¸¾«Æ·

¢Ù»·¼ºÏ©´ÖÆ·Öк¬Óл·¼º´¼ºÍÉÙÁ¿ËáÐÔÔÓÖʵȡ£Öкͱ¥ºÍʳÑÎË®£¬Õñµ´¡¢¾²Öᢷֲ㣬»·¼ºÏ©ÔÚ_______________²ã£¨Ìî¡°ÉÏ¡±»ò¡°Ï¡±£©£¬·ÖÒººóÓÃ_______________£¨ÌîÈë±àºÅ£©Ï´µÓ¡£

a.KMnO4ÈÜÒº

b.Ï¡H2SO4

c.Na2CO3ÈÜÒº

ͼ¢ò

¢ÚÔÙ½«»·¼ºÏ©°´Í¼¢ò×°ÖÃÕôÁó£¬ÀäÈ´Ë®´Ó____________¿Ú½øÈë¡£ÕôÁóʱҪ¼ÓÈëÉúʯ»Ò£¬Ä¿µÄÊÇ____________¡£

¢ÛÊÕ¼¯²úƷʱ£¬¿ØÖƵÄζÈÓ¦ÔÚ____________×óÓÒ£¬ÊµÑéÖƵõĻ·¼ºÏ©¾«Æ·ÖÊÁ¿µÍÓÚÀíÂÛ²úÁ¿£¬¿ÉÄܵÄÔ­ÒòÊÇ________________________¡£

a.ÕôÁóʱ´Ó70 ¡æ¿ªÊ¼ÊÕ¼¯²úÆ·

b.»·¼º´¼Êµ¼ÊÓÃÁ¿¶àÁË

c.ÖƱ¸´ÖƷʱ»·¼º´¼Ëæ²úÆ·Ò»ÆðÕô³ö

£¨3£©ÒÔÏÂÇø·Ö»·¼ºÏ©¾«Æ·ºÍ´ÖÆ·µÄ·½·¨£¬ºÏÀíµÄÊÇ____________¡£

a.ÓøßÃÌËá¼ØËáÐÔÈÜÒº           b.ÓýðÊôÄÆ               c.²â¶¨·Ðµã

£¨1£©¢Ù·À±©·Ð  ÀäÄý  ¢Ú·ÀÖ¹»·¼ºÏ©»Ó·¢

£¨2£©¢ÙÉÏ  c  ¢Úg  ÎüÊÕË®·Ö£¬±ãÓÚÕôÁó³ö¸ü´¿¾»µÄ²úÆ·  ¢Û83 ¡æ  c

(3)bc


½âÎö:

(1)¶ÔÒºÌå¼ÓÈÈʱ£¬ÎªÁË·ÀÖ¹±©·Ð³£³£Òª¼ÓÈë·Ðʯ»òËé´ÉƬ£¬´Ó±íÖÐÊý¾Ý²»ÄÑ¿´³ö£¬µ¼¹ÜB¾ßÓÐÀäÄý×÷Óã¬ÊÔ¹ÜCÖñùˮԡÖÐÊÇΪÁË·ÀÖ¹»·¼ºÏ©µÄ»Ó·¢¡£

£¨2£©Òò»·¼ºÏ©µÄÃܶÈСÓÚË®£¬Ó¦ÔÚÉϲ㣬·ÖÒººó¼ÓÈëNa2CO3¿ÉÒÔÖк͵ôÉÙÁ¿Ëá¡£¸ù¾Ý¡°ÄæÁ÷¡±Ô­ÀíÀäÄý¹ÜÓ¦¸Ã´Óg´¦½øË®£¬¼ÓÈëÉúʯ»ÒµÄÄ¿µÄÊÇΪÁËÎüÊÕË®·Ö±ãÓÚÕôÁó³ö¸ü´¿¾»µÄ²úÆ·¡£¸ù¾Ý±íÖÐÊý¾Ý»·¼ºÏ©·ÐµãΪ83 ¡æ£¬ËùÒÔζȿØÖÆÔÚ83 ¡æ×óÓÒ¡£¶øÖÊÁ¿µÍÓÚÀíÂÛ²úÁ¿µÄÔ­Òò£¬a¡¢c¶¼Ó¦Åųý£¬Òòa¡¢c²Ù×÷¶¼»áʹ²úÁ¿¸ßÓÚÀíÂÛ²úÁ¿£¬¹ÊÔ­ÒòΪb¡£

£¨3£©Çø·Ö»·¼ºÏ©ÊǾ«Æ·»¹ÊÇ´ÖÆ·²ÉÓòⶨ·ÐµãµÄÎïÀí·½·¨¸üºÏÀí¡¢·½±ã£¬Òà¿ÉÓýðÊôÄÆ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij»¯Ñ§Ð¡×é²ÉÓÃÀàËÆÖÆÒÒËáÒÒõ¥µÄ×°Öã¨Èçͼ1£©£¬ÒÔ»·¼º´¼ÖƱ¸»·¼ºÏ©
ÒÑÖª£º

Ãܶȣ¨g/cm3£© È۵㣨¡æ£© ·Ðµã£¨¡æ£© ÈܽâÐÔ »·¼º´¼ 0.96 25 161 ÄÜÈÜÓÚË® »·¼ºÏ© 0.81 -103 83 ÄÑÈÜÓÚË®
£¨1£©ÖƱ¸´ÖÆ·
½«12.5mL»·¼º´¼¼ÓÈëÊÔ¹ÜAÖУ¬ÔÙ¼ÓÈë1mLŨÁòËᣬҡÔȺó·ÅÈëËé´ÉƬ£¬»ºÂý¼ÓÈÈÖÁ·´Ó¦ÍêÈ«£¬ÔÚÊÔ¹ÜCÄڵõ½»·¼ºÏ©´ÖÆ·£®
¢Ùµ¼¹ÜB³ýÁ˵¼ÆøÍ⻹¾ßÓеÄ×÷ÓÃÊÇ
ÀäÄý»ØÁ÷
ÀäÄý»ØÁ÷
£®
¢ÚŨH2SO4µÄ×÷ÓÃ
´ß»¯¼ÁºÍÎüË®¼Á
´ß»¯¼ÁºÍÎüË®¼Á
£®
£¨2£©ÖƱ¸¾«Æ·
¢Ù»·¼ºÏ©´ÖÆ·Öк¬Óл·¼º´¼ºÍÉÙÁ¿ËáÐÔÔÓÖʵȣ®¼ÓÈë±¥ºÍʳÑÎË®£¬Õñµ´¡¢¾²Öᢷֲ㣬»·¼ºÏ©ÔÚ
ÉÏ
ÉÏ
²ã£¨ÌîÉÏ»òÏ£©£¬·ÖÒººóÓÃ
c
c
 £¨ÌîÈë±àºÅ£©Ï´µÓ£®
a£®KMnO4ÈÜÒº      b£®Ï¡H2SO4  c£®±¥ºÍNa2CO3ÈÜÒº
¢ÚÔÙ½«»·¼ºÏ©°´ÓÒͼ2×°ÖÃÕôÁó£¬ÀäÈ´Ë®´Ó
g
g
¿Ú½øÈ룮ÕôÁóʱҪ¼ÓÈëÉúʯ»Ò£¬Ä¿µÄÊÇ
³ýÈ¥Á˲ÐÁôµÄË®
³ýÈ¥Á˲ÐÁôµÄË®
£®
¢ÛÊÕ¼¯²úƷʱ£¬¿ØÖƵÄζÈÓ¦ÔÚ
83¡æ
83¡æ
×óÓÒ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ij»¯Ñ§Ð¡×é²ÉÓÃÀàËÆÖÆÒÒËáÒÒõ¥µÄ×°Öã¨Èçͼ1£©£¬ÒÔ»·¼º´¼ÖƱ¸»·¼ºÏ©
ÒÑÖª£º

Ãܶȣ¨g/cm3£© È۵㣨¡æ£© ·Ðµã£¨¡æ£© ÈܽâÐÔ »·¼º´¼ 0.96 25 161 ÄÜÈÜÓÚË® »·¼ºÏ© 0.81 -103 83 ÄÑÈÜÓÚË®
£¨1£©ÖƱ¸´ÖÆ·½«12.5mL»·¼º´¼¼ÓÈëÊÔ¹ÜAÖУ¬ÔÙ¼ÓÈë1mLŨÁòËᣬҡÔȺó·ÅÈëËé´ÉƬ£¬»ºÂý¼ÓÈÈÖÁ·´Ó¦ÍêÈ«£¬ÔÚÊÔ¹ÜCÄڵõ½»·¼ºÏ©´ÖÆ·£®
¢ÙAÖÐËé´ÉƬµÄ×÷ÓÃÊÇ
·ÀÖ¹±©·Ð
·ÀÖ¹±©·Ð
£¬µ¼¹ÜB³ýÁ˵¼ÆøÍ⻹¾ßÓеÄ×÷ÓÃÊÇ
ÀäÄý
ÀäÄý
£®
¢ÚÊÔ¹ÜCÖÃÓÚ±ùˮԡÖеÄÄ¿µÄÊÇ
½øÒ»²½ÀäÈ´£¬·ÀÖ¹»·¼ºÏ©»Ó·¢
½øÒ»²½ÀäÈ´£¬·ÀÖ¹»·¼ºÏ©»Ó·¢
£®
£¨2£©ÖƱ¸¾«Æ·
¢Ù»·¼ºÏ©´ÖÆ·Öк¬Óл·¼º´¼ºÍÉÙÁ¿ËáÐÔÔÓÖʵȣ®¼ÓÈë±¥ºÍʳÑÎË®£¬Õñµ´¡¢¾²Öᢷֲ㣬»·¼ºÏ©ÔÚ
Éϲã
Éϲã
²ã£¨ÌîÉÏ»òÏ£©£¬·ÖÒººóÓÃ
c
c
 £¨ÌîÈë±àºÅ£©Ï´µÓ£®
a£®KMnO4ÈÜÒº      b£®Ï¡H2SO4   c£®Na2CO3ÈÜÒº
¢ÚÔÙ½«»·¼ºÏ©°´Í¼2ËùʾװÖÃÕôÁó£¬ÀäÈ´Ë®´Ó
g
g
¿Ú½øÈ룬ĿµÄÊÇ
ÀäÈ´Ë®ÓëÆøÌåÐγÉÄæÁ÷£¬ÀäÄýЧ¹û¸üºÃ£»¸üÈÝÒ×½«ÀäÄý¹Ü³äÂúË®£»ÒÔ·ÀÕôÆûÈë¿Ú´¦ÖèÀäÖèÈÈʹÀäÄý¹ÜÆÆÁÑ
ÀäÈ´Ë®ÓëÆøÌåÐγÉÄæÁ÷£¬ÀäÄýЧ¹û¸üºÃ£»¸üÈÝÒ×½«ÀäÄý¹Ü³äÂúË®£»ÒÔ·ÀÕôÆûÈë¿Ú´¦ÖèÀäÖèÈÈʹÀäÄý¹ÜÆÆÁÑ
£®
¢ÛÊÕ¼¯²úƷʱ£¬¿ØÖƵÄζÈÓ¦ÔÚ
83¡æ
83¡æ
×óÓÒ£¬ÊµÑéÖƵõĻ·¼ºÏ©¾«Æ·ÖÊÁ¿µÍÓÚÀíÂÛ²úÁ¿£¬¿ÉÄܵÄÔ­ÒòÊÇ
b
b

a£®ÕôÁóʱ´Ó70¡æ¿ªÊ¼ÊÕ¼¯²úÆ·
b£®»·¼º´¼Êµ¼ÊÓÃÁ¿¶àÁË
c£®ÖƱ¸´ÖƷʱ»·¼º´¼Ëæ²úÆ·Ò»ÆðÕô³ö
£¨3£©ÒÔÏÂÇø·Ö»·¼ºÏ©¾«Æ·ºÍ´ÖÆ·µÄ·½·¨£¬ºÏÀíµÄÊÇ
c
c
£®
a£®ÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒº        b£®ÓýðÊôÄÆ    c£®²â¶¨·Ðµã£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêºÓ±±Ê¡ÌÆɽһÖи߶þÏÂѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÊµÑéÌâ

(11·Ö)ij»¯Ñ§Ð¡×é²ÉÓÃÀàËÆÖÆÒÒËáÒÒõ¥µÄ×°ÖÃ(Èçͼ)£¬ÒÔ»·¼º´¼ÖƱ¸»·¼ºÏ©¡£
ÒÑÖª£º

 
ÃܶÈ(g/cm3)
ÈÛµã(¡æ)
·Ðµã(¡æ)
ÈܽâÐÔ
»·ÒÑ´¼
0.96
25
161
ÄÜÈÜÓÚË®
»·ÒÑÏ©
0.81
£­103
83
ÄÑÈÜÓÚË®
(1)ÖƱ¸´ÖÆ·
½«12.5mL»·¼º´¼¼ÓÈëÊÔ¹ÜAÖУ¬ÔÙ¼ÓÈëlmLŨÁòËᣬҡÔȺó·ÅÈëËé´ÉƬ£¬»ºÂý¼ÓÈÈÖÁ·´Ó¦ÍêÈ«£¬ÔÚÊÔ¹ÜCÄڵõ½»·¼ºÏ©´ÖÆ·¡£
¢ÙAÖÐËé´ÉƬµÄ×÷ÓÃÊÇ                £¬µ¼¹ÜB³ýÁ˵¼ÆøÍ⻹¾ßÓеÄ×÷ÓÃÊÇ      ¡£
¢ÚÊÔ¹ÜCÖÃÓÚ±ùˮԡÖеÄÄ¿µÄÊÇ                       ¡£
(2)ÖƱ¸¾«Æ·
¢Ù»·¼ºÏ©´ÖÆ·Öк¬Óл·¼º´¼ºÍÉÙÁ¿ËáÐÔÔÓÖʵȡ£¼ÓÈë±¥ºÍʳÑÎË®£¬Õñµ´¡¢¾²Öᢷֲ㣬»·¼ºÏ©ÔÚ    ²ã(ÌîÉÏ»òÏÂ)£¬·ÖÒººóÓà      (ÌîÈë±àºÅ)Ï´µÓ¡£
a£®KMnO4ÈÜÒº    b£®Ï¡H2SO4    c£®Na2CO3ÈÜÒº
¢ÚÔÙ½«»·¼ºÏ©°´Í¼×°ÖÃÕôÁó£¬ÀäÈ´Ë®´Ó       (ÌîÈë±àºÅ)¿Ú½øÈë¡£ÕôÁóʱҪ¼ÓÈëÉúʯ»ÒµÄÄ¿µÄ                    ¡£

¢ÛÊÕ¼¯²úƷʱ£¬¿ØÖƵÄζÈÓ¦ÔÚ       ×óÓÒ£¬ÊµÑéÖƵõĻ·
¼ºÏ©¾«Æ·ÖÊÁ¿µÍÓÚÀíÂÛ²úÁ¿£¬¿ÉÄܵÄÔ­ÒòÊÇ      ¡£
a£®ÕôÁóʱ´Ó70¡æ¿ªÊ¼ÊÕ¼¯²úÆ·
b£®»·¼º´¼Êµ¼ÊÓÃÁ¿¶àÁË
c£®ÖƱ¸´ÖƷʱ»·¼º´¼Ëæ²úÆ·Ò»ÆðÕô³ö
(3)ÒÔÏÂÇø·Ö»·¼ºÏ©¾«Æ·ºÍ´ÖÆ·µÄ·½·¨£¬ºÏÀíµÄÊÇ       ¡£
a£®ÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒº    b£®ÓýðÊôÄÆ            c£®²â¶¨·Ðµã    

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê¹ã¶«Ê¡Ã·ÖÝÊÐÔøÏÜè÷ÖÐѧ¸ß¶þ5ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÊµÑéÌâ

£¨20·Ö£©Ä³»¯Ñ§Ð¡×é²ÉÓÃÀàËÆÖÆÒÒËáÒÒõ¥µÄ×°Öã¨Èçͼ£©£¬ÒÔ»·¼º´¼ÖƱ¸»·¼ºÏ©

£¨1£©ÖƱ¸´ÖÆ·
½«12.5mL»·¼º´¼¼ÓÈëÊÔ¹ÜAÖУ¬ÔÙ¼ÓÈëlmLŨÁòËᣬҡÔȺó·ÅÈëËé´ÉƬ£¬»ºÂý¼ÓÈÈÖÁ·´Ó¦ÍêÈ«£¬ÔÚÊÔ¹ÜCÄڵõ½»·¼ºÏ©´ÖÆ·¡£
¢ÙAÖÐËé´ÉƬµÄ×÷ÓÃÊÇ____________£¬µ¼¹ÜB³ýÁ˵¼ÆøÍ⻹¾ßÓеÄ×÷ÓÃÊÇ____ ¡£
¢ÚÊÔ¹ÜCÖÃÓÚ±ùˮԡÖеÄÄ¿µÄÊÇ_____________________________¡£
£¨2)ÖƱ¸¾«Æ·
¢Ù»·¼ºÏ©´ÖÆ·Öк¬Óл·¼º´¼ºÍÉÙÁ¿ËáÐÔÔÓÖʵȡ£¼ÓÈë±¥ºÍʳÑÎË®£¬Õñµ´¡¢¾²Öᢷֲ㣬»·¼ºÏ©ÔÚ______²ã(ÌîÉÏ»òÏÂ)£¬·ÖÒººóÓÃ_________ (ÌîÈë±àºÅ)Ï´µÓ¡£
A£®KMnO4ÈÜÒº     B£®Ï¡H2SO  C£®Na2CO3ÈÜÒº
¢ÚÔÙ½«»·¼ºÏ©°´ÏÂͼװÖÃÕôÁó£¬ÀäÈ´Ë®´Ó_________¿Ú½øÈë¡£ÕôÁóʱҪ¼ÓÈëÉúʯ»Ò£¬Ä¿µÄÊÇ______________   ____¡£

¢ÛÊÕ¼¯²úƷʱ£¬¿ØÖƵÄζÈÓ¦ÔÚ_________×óÓÒ£¬ÊµÑéÖƵõĻ·¼ºÏ©¾«Æ·ÖÊÁ¿µÍÓÚÀíÂÛ²úÁ¿£¬¿ÉÄܵÄÔ­ÒòÊÇ____________________
A£®ÕôÁóʱ´Ó70¡æ¿ªÊ¼ÊÕ¼¯²úÆ·      
B£®»·¼º´¼Êµ¼ÊÓÃÁ¿¶àÁË
C£®ÖƱ¸´ÖƷʱ»·¼º´¼Ëæ²úÆ·Ò»ÆðÕô³ö
£¨3£©ÒÔÏÂÇø·Ö»·¼ºÏ©¾«Æ·ºÍ´ÖÆ·µÄ·½·¨£¬ºÏÀíµÄÊÇ_________¡£
A£®ÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒº    B£®ÓýðÊôÄÆ   C£®²â¶¨·Ðµã

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸