ÌâÄ¿ÄÚÈÝ

ÎÒ¹ú»¯¹¤×¨¼ÒºîµÂ°ñÏÈÉú·¢Ã÷µÄ¡°ºîÊÏÖƼ¡±µÄ»ù±¾Ô­ÀíÊÇ£ºÔÚŨ°±Ë®ÖÐͨÈë×ãÁ¿µÄ¶þÑõ»¯Ì¼Éú³ÉÒ»ÖÖÑΣ¬È»ºóÔÚ´ËÈÜÒºÖмÓÈëϸСµÄʳÑηÛÄ©£»ÓÉÓÚ̼ËáÇâÄÆÔÚ¸Ã״̬ÏÂÈܽâ¶ÈºÜС£¬³Ê¾§ÌåÎö³ö£¬Í¬Ê±ÓÉÓÚ̼ËáÇâÄƲ»Îȶ¨£¬¼ÓÈȺóÉú³É´¿¼î£®Ë®ºÍ¶þÑõ»¯Ì¼£»ÔÙÏòÎö³ö̼ËáÇâÄƺóµÄÂÈ»¯ï§ÈÜÒºÖмÓÈëʳÑΣ¬Ê¹ÆäÖеÄÂÈ»¯ï§µ¥¶À½á¾§³öÀ´£¬ÂÈ»¯ÄÆÔò¿ÉÑ­»·Ê¹Ó㮸ù¾ÝÒÔÉÏÐðÊö¼òÒª»Ø´ðÎÊÌ⣺
£¨1£©ÓÃÉÏÊö·½·¨½øÐÐʵÑéʱ£¬ËùÓõÄÆðʼԭÁÏ
 
£®£¨Ð´³öËüÃǵĻ¯Ñ§Ê½£¬ÏÂͬ£©£®
£¨2£©×îÖÕ²úÆ·ÊÇ
 
£®
£¨3£©ÓйصĻ¯Ñ§·½³Ìʽ
 
£»
 
£»
 
£®
£¨4£©ÓÐÈËÈÏΪºîÊÏÖƼµÄÓŵãÓÐÒÔÏÂËĵ㣺
A£®Éú²ú¹ý³ÌÖв¿·Ö²úÆ·¿É×÷ΪÆðʼԭÁÏʹÓã»
B£®¸±²úÆ·ÊÇÒ»ÖÖ¿ÉÀûÓõĵª·Ê£»
C£®·´Ó¦²»ÐèÒª¼ÓÈÈ£»
D£®¸±²úÆ·²»»áÔì³É»·¾³ÎÛȾ£®
ÄãÈÏΪÆäÖÐÕýÈ·µÄÊÇ£¨ÓôúºÅ»Ø´ð£©
 
£®
·ÖÎö£º
Ũ°±Ë®
¶þÑõ»¯Ì¼
?Éú³ÉÒ»ÖÖÑμ´Ì¼ËáÇâ泥¨NH4HCO3£©
NaCl
NH4Cl+NaHCO3£¬NaHCO3?Na2CO3+CO2¡ü+H2O£®
ÓÉÉÏÃæÒ»Á¬´®µÄ·ÖÎö¿ÉÃ÷È·¿´³ö£¬Ô­ÁϾÍÊÇŨ°±Ë®¡¢¶þÑõ»¯Ì¼ºÍNaCl£»×îÖÕ²úÎï¾ÍÊÇNH4ClºÍNa2CO3£»Èý¸ö»¯Ñ§·½³ÌʽҲ¿ÉºÜÈÝÒ×µÄд³öÀ´£®
½â´ð£º½â£º£¨1£©ÓÉ¡°ºîÊÏÖƼ¡±µÄ»ù±¾Ô­Àí¿ÉÖª£ºÊ×ÏÈÊÇŨ°±Ë®ºÍ¶þÑõ»¯Ì¼·´Ó¦Éú³ÉÒ»ÖÖÑΣ¨NH4HCO3£©£¬
ÔÚÑÎÈÜÒºÖмÓÈëϸСµÄʳÑηÛÄ©£¬Îö³ö̼ËáÇâÄÆ£¬Ì¼ËáÇâÄƲ»Îȶ¨£¬¼ÓÈȺóÉú³É´¿¼î£¨Na2CO3£©¡¢
Ë®ºÍ¶þÑõ»¯Ì¼£¬²¿·ÖÔ­ÁÏ¿ÉÒÔÑ­»·Ê¹Óã®Í¨¹ý·ÖÎö¿ÉÖª½øÐÐʵÑéʱ£¬ËùÓõÄÆðʼԭÁÏΪ£ºNH3¡¢CO2¡¢NaCl£®
£¨2£©ÓÉʵÑéÔ­Àí¿ÉÖª£º·¢ÉúµÄ·´Ó¦ÓÐ NH3+H2O+CO2¨TNH4HCO3£¬NH4HCO3+NaCl¨TNH4Cl+NaHCO3
2NaHCO3
  ¡÷  
.
 
Na2CO3+CO2¡ü+H2O  ËùÒÔ×îÖÕ²úÆ·ÊÇNa2CO3¡¢NH4Cl
£¨3£©ÓÉ£¨2£©µÄ·ÖÎö¿É֪ʵÑé¹ý³ÌÖз¢ÉúµÄ·´Ó¦ÓУº
NH3+H2O+CO2¨TNH4HCO3£¬NH4HCO3+NaCl¨TNH4Cl+NaHCO3£¬2NaHCO3
  ¡÷  
.
 
Na2CO3+CO2¡ü+H2O
£¨4£©ÓÉ£¨2£©µÄ·ÖÎö¿ÉÖª·´Ó¦Éú³ÉµÄCO2¿ÉÒÔÑ­»·×÷ΪÆðʼԭÁÏʹÓ㬹ÊAÕýÈ·£»
ÒòΪ¸±²úÆ·£¨NH4Cl£©º¬ÓеªÔªËØ£¨N£©ÊôÓÚµª·Ê£¬¹Ê BÕýÈ·£»
ÒòΪ2NaHCO3
  ¡÷  
.
 
Na2CO3+CO2¡ü+H2O Õâ¸ö·´Ó¦ÐèÒª¼ÓÈÈ£¬¹ÊC´í£»
ÒòΪ¸±²úÆ·ÊÇNH4Cl¡¢CO2¡¢H2O ¶¼²»ÎÛȾ»·¾³£¬¹ÊDÕýÈ·£»ËùÒÔӦѡABD
¹Ê´ð°¸Îª£º
£¨1£©NH3¡¢CO2¡¢NaCl£®
£¨2£©Na2CO3¡¢NH4Cl
£¨3£©NH3+H2O+CO2¨TNH4HCO3 £»NH4HCO3+NaCl¨TNH4Cl+NaHCO3 £»2NaHCO3
  ¡÷  
.
 
Na2CO3+CO2¡ü+H2O
£¨4£©ABD
µãÆÀ£º´ËÌ⿼²éµÄÊÇÓйغîÊÏÖƼµÄÓйØ֪ʶ£¬Ö»ÒªÈÏÕæ¿´Ì⣬ÕæÕýÀí½âÁ˺îÊÏÖƼµÄÔ­Àí£¬´ËÌâ¾Í»áÓ­Èжø½â£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÎÒ¹ú»¯¹¤×¨¼ÒºîµÂ°ñÔøΪÊÀ½çÖƼҵ×÷³öÁËÍ»³ö¹±Ï×£®¡°ºîÊÏÖƼ¡±ÊÇÒÔʳÑΡ¢°±Æø¡¢¶þÑõ»¯Ì¼µÈΪԭÁÏÏÈÖƵÃN£®HCq£¬½ø¶øÉú²ú³ö´¿¼î£®ÆäÉú²ú¹ý³ÌÓû¯Ñ§·½³Ìʽ¿É¼òÒª±íʾΪ£º
¢ÙNH3+H2O+CO2=NH4HCO3£»¢ÚNH4HCO3+NaCl=NaHCO3¡ý+NH4Cl£»¢Û2NaHCO3
  ¡÷  
.
 
Na2CO3+H2O+CO2¡ü£®
£¨1£©¡°ºîÊÏÖƼ¡±Éú²ú¹ý³ÌÖÐÉæ¼°µ½µÄ»ù±¾·´Ó¦ÀàÐÍÊÇ
 
£®
A£®Öû»·´Ó¦£»B£®»¯ºÏ·´Ó¦£»C£®·Ö½â·´Ó¦£»D£®¸´·Ö½â·´Ó¦
£¨2£©µÚ¢Ú²½ÖмÓÈëµÄÊÇĥϸµÄʳÑηۣ¬Ê³ÑÎĥϸµÄÄ¿µÄÊÇ
 
£®
ÈôµÚ¢Ú²½Öеõ½µÄNaHCO3Öк¬ÓÐÉÙÁ¿µÄNH4HCO3£¬µ«²»»áÓ°Ïì²úÆ·Na2CO3µÄ´¿¶È£¬Ô­ÒòÊÇ
 
£®
£¨3£©Èô´¿¼îÖк¬ÓÐNaHCO3ÔÓÖÊ£¬Îª²â¶¨ÊÔÑùÖд¿¼îµÄÖÊÁ¿·ÖÊý£¬ÓÃÓÒͼÖеÄ×°ÖýøÐÐʵÑ飮Ö÷ҪʵÑé²½ÖèÈçÏ£º¾«Ó¢¼Ò½ÌÍø
ÓÃÌìƽ׼ȷ³ÆÁ¿³ö¸ÉÔïÊԹܵÄÖÊÁ¿Îª20.2¿Ë£¬È»ºó³ÆÈ¡4.0¿ËÒÑÑÐËéµÄ´¿¼îÑùÆ·²¢·ÅÈë¸ÉÔïµÄÊÔ¹ÜÖУ®Óþƾ«µÆ¼ÓÈÈ£¬´ý³ä·Ö·´Ó¦²¢ÀäÈ´ºó£¬ÔÚÌìƽÉϳƳöÊԹܺÍ̼ËáÄƵÄÖÊÁ¿23.5¿Ë£®
¢ÙÍ£Ö¹¼ÓÈÈÇ°£¬Òª½øÐеIJÙ×÷ÊÇ
 
£®
¢Ú¸ÃÊÔÑùÖд¿¼îµÄÖÊÁ¿·ÖÊýΪ
 
£®
£¨2012?ÑïÖÐÊÐÄ£Ä⣩ÎÒ¹ú»¯¹¤×¨¼ÒºîµÂ°ñµÄ¡°ºîÊÏÖƼ¡±ÔøΪÊÀ½çÖƼҵ×ö³öÁËÍ»³ö¹±Ï×£®ËûÒÔNaCl¡¢NH3¡¢CO2µÈΪԭÁÏÏÈÖƵÃNaHCO3£¬½ø¶øÉú²ú³ö´¿¼î£®Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
NH3+CO2+H2O¨TNH4HCO3£»
NH4HCO3+NaCl¨TNaHCO3¡ý+NH4Cl£»
2NaHCO3
  ¡÷  
.
 
Na2CO3+CO2¡ü+H2O
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ì¼ËáÇâï§Óë±¥ºÍʳÑÎË®·´Ó¦£¬ÄÜÎö³ö̼ËáÇâÄƾ§ÌåµÄÔ­ÒòÊÇ
c
c
£¨Ìî×Öĸ±êºÅ£©£®
a£®Ì¼ËáÇâÄÆÄÑÈÜÓÚË®    b£®Ì¼ËáÇâÄÆÊÜÈÈÒ׷ֽ⠠  c£®Ì¼ËáÇâÄƵÄÈܽâ¶ÈÏà¶Ô½ÏС£¬ËùÒÔÔÚÈÜÒºÖÐÊ×ÏȽᾧÎö³ö
£¨2£©Ä³Ì½¾¿»î¶¯Ð¡×é¸ù¾ÝÉÏÊöÖƼîÔ­Àí£¬½øÐÐ̼ËáÇâÄƵÄÖƱ¸ÊµÑ飬ͬѧÃÇ°´¸÷×ÔÉè¼ÆµÄ·½°¸ÊµÑ飮
¢Ùһλͬѧ½«¶þÑõ»¯Ì¼ÆøÌåͨÈ뺬°±µÄ±¥ºÍʳÑÎË®ÖÐÖƱ¸Ì¼ËáÇâÄÆ£¬ÊµÑé×°ÖÃÈçÏÂͼËùʾ£¨Í¼Öмг֡¢¹Ì¶¨ÓõÄÒÇÆ÷δ»­³ö£©£®

ÊԻشðÏÂÁÐÓйØÎÊÌ⣺
£¨¢ñ£©ÒÒ×°ÖÃÖеÄÊÔ¼ÁÊÇ ±¥ºÍµÄ̼ËáÇâÄÆÈÜÒº£¬×÷ÓÃÊÇ
ÎüÊÕ¼××°ÖÿÉÄܻӷ¢³öµÄÂÈ»¯ÇâÆøÌå
ÎüÊÕ¼××°ÖÿÉÄܻӷ¢³öµÄÂÈ»¯ÇâÆøÌå
£»
£¨¢ò£©¶¡×°ÖÃÖÐÏ¡ÁòËáµÄ×÷ÓÃÊÇ
ÎüÊÕÄ©·´Ó¦µÄNH3
ÎüÊÕÄ©·´Ó¦µÄNH3
£»
£¨¢ó£©ÊµÑé½áÊøºó£¬·ÖÀë³öNaHCO3¾§ÌåµÄ²Ù×÷ÊÇ
¹ýÂË
¹ýÂË
£¨Ìî·ÖÀë²Ù×÷µÄÃû³Æ£©£®
¢ÚÁíһλͬѧÓÃͼÖÐÎì×°Öã¨ÆäËü×°ÖÃδ»­³ö£©½øÐÐʵÑ飮
£¨¢ñ£©ÊµÑéʱ£¬ÐëÏÈ´Ó
a
a
¹ÜͨÈë
°±Æø
°±Æø
ÆøÌ壻
£¨¢ò£©ÓÐͬѧ½¨ÒéÔÚÎì×°ÖõÄb¹Ü϶ËÁ¬½Ó¼º×°Öã¬ÀíÓÉÊÇ
Ôö´óÆøÌåÓëÈÜÒº½Ó´¥Ãæ»ý£¬Ìá¸ßCO2ÎüÊÕÂÊ
Ôö´óÆøÌåÓëÈÜÒº½Ó´¥Ãæ»ý£¬Ìá¸ßCO2ÎüÊÕÂÊ
£»
£¨3£©ÇëÄãÔÙд³öÒ»ÖÖʵÑéÊÒÖÆÈ¡ÉÙÁ¿Ì¼ËáÇâÄƵķ½·¨£º
ÓÃ̼ËáÇâï§ÓëÊÊÁ¿±¥ºÍʳÑÎË®·´Ó¦£®£¨»òÍùÉÕ¼îÈÜÒºÖÐͨÈë¹ýÁ¿CO2£»Íù±¥ºÍNa2CO3 ÈÜÒºÖÐͨÈë¹ýÁ¿CO2 µÈ£®ÆäËûºÏÀí·½·¨¾ù¿É£©
ÓÃ̼ËáÇâï§ÓëÊÊÁ¿±¥ºÍʳÑÎË®·´Ó¦£®£¨»òÍùÉÕ¼îÈÜÒºÖÐͨÈë¹ýÁ¿CO2£»Íù±¥ºÍNa2CO3 ÈÜÒºÖÐͨÈë¹ýÁ¿CO2 µÈ£®ÆäËûºÏÀí·½·¨¾ù¿É£©
£®
£¨4£©¡°´¿¼îÖг£³£»á»ìÓÐÉÙÁ¿ÂÈ»¯ÄÆ£®¡±Ä³Ñо¿ÐÔѧϰС×éÒÔÒ»°ü´¿¼î£¨Ö»¿¼ÂǺ¬ÂÈ»¯ÄÆ£©ÎªÑо¿¶ÔÏó£¬Ì½¾¿´¿¼îÑùÆ·ÖÐ̼ËáÄƵĺ¬Á¿£®
¡¾ÊµÑéÉè¼Æ¡¿
¼×·½°¸
¢ñ£®Éè¼Æ˼·£º¸ù¾ÝÑùÆ·ÓëÂÈ»¯¸ÆÈÜÒº·´Ó¦Éú³É³Áµí̼Ëá¸ÆµÄÖÊÁ¿£¬Çó³ö̼ËáÄƵÄÖÊÁ¿£¬ÔÙ¼ÆËãÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊý£®
¢ò£®²Ù×÷²½Ö裺³ÆÈ¡13.25g´¿¼îÑùÆ·£¬¼ÓÈë¹ýÁ¿µÄÂÈ»¯¸ÆÈÜÒº£¬³ä·Ö½Á°è£®¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔµÃµ½µÄ°×É«³Áµí10.00g£®
¢ó£®Êý¾Ý´¦Àí£ºÇë¸ù¾ÝÉÏÊöʵÑéÊý¾Ý£¬¼ÆËã¸ÃÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊý£®
¼ÆËã¹ý³Ì£º
½â£ºÉè´¿¼îÑùÆ·Öк¬Na2CO3µÄÖÊÁ¿Îªx
Na2CO3+CaCl2¨TCaCO3¡ý+2NaCl
106 100
x 10.00g
106
100
=
x
10.00g

x=10.6g
´¿¼îÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿·ÖÊýΪ
10.6g
13.25g
¡Á100%=80%
½â£ºÉè´¿¼îÑùÆ·Öк¬Na2CO3µÄÖÊÁ¿Îªx
Na2CO3+CaCl2¨TCaCO3¡ý+2NaCl
106 100
x 10.00g
106
100
=
x
10.00g

x=10.6g
´¿¼îÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿·ÖÊýΪ
10.6g
13.25g
¡Á100%=80%


ÒÒ·½°¸
I£®Éè¼Æ˼·£º¸ù¾ÝÑùÆ·£¨ÖÊÁ¿Îªa g£©ÓëÏ¡ÁòËáÍêÈ«·´Ó¦Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£¨ÖÊÁ¿Îªb g£©£¬Çó³ö̼ËáÄƵÄÖÊÁ¿£¬ÔÙ¼ÆËãÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊý£®
II£®ÊµÑé½áÂÛ£º
£¨1£©µ±a¡¢bµÄÖÊÁ¿¹ØϵÂú×ã
a
b
=
53
22
£¨»ò22a=53b£»»òa=
53
22
bµÈ£©
a
b
=
53
22
£¨»ò22a=53b£»»òa=
53
22
bµÈ£©
£¨Ìîдº¬a¡¢b×ÖĸµÄ±í´ïʽ£¬ÏÂͬ£©Ê±£¬´¿¼îÑùÆ·ÖÐÖ»º¬ÓÐ̼ËáÄÆ£¬ÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊýÊÇ100%£®
£¨2£©µ±a¡¢bµÄÖÊÁ¿¹ØϵÂú×ã
a
b
£¾
53
22
a
b
£¾
53
22
ʱ£¬´¿¼îÑùÆ·ÊÒÓÐ̼ËáÄƺÍÉÙÁ¿ÂÈ»¯ÄÆ×é³ÉµÄ»ìºÏÎÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊýÊÇ
53b
22a
¡Á100%
53b
22a
¡Á100%
£®
¡¾ÊµÑéÆÀ¼Û¡¿
Îó²î·ÖÎö£º
£¨Ò»£©¼×·½°¸ÖУ¬°×É«³ÁµíµÄÖÊÁ¿ÔÚ¹ýÂË¡¢¸ÉÔïµÈ²Ù×÷¹ý³ÌÖлáÓÐËùËðºÄ£¬Ôì³É¼ÆËã½á¹ûÓëʵ¼ÊÖµÏà±ÈƫС£®Èç¹û½«ÂÈ»¯¸ÆÈÜÒº»»³ÉÂÈ»¯±µÈÜÒº£¬Ôò¿ÉÒÔʹÎó²î¼õС£¬ÀíÓÉÊÇ
BaCl2±ÈCaCl2µÄÏà¶Ô·Ö×ÓÖÊÁ¿´ó£¬²úÉúµÄ³ÁµíÖÊÁ¿´ó£¬³ÆÁ¿Îó²îС
BaCl2±ÈCaCl2µÄÏà¶Ô·Ö×ÓÖÊÁ¿´ó£¬²úÉúµÄ³ÁµíÖÊÁ¿´ó£¬³ÆÁ¿Îó²îС
£®
£¨¶þ£©ÒÒ·½°¸ÖУ¬Óв¿·ÖÉú³ÉµÄ¶þÑõ»¯Ì¼ÆøÌåÒòÈܽâÓÚË®¶øûÓÐÈ«²¿Òݳö£¬Ôì³É¼ÆËã½á¹ûÓëʵ¼ÊÖµÏà±È
ƫС
ƫС
£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢»ò¡°²»±ä¡±£©£®
ÎÒ¹ú»¯¹¤×¨¼ÒºîµÂ°ñµÄ¡°ºîÊÏÖƼ¡±ÔøΪÊÀ½çÖƼҵ×ö³öÁËÍ»³ö¹±Ï×£®ËûÒÔNaCl¡¢NH3¡¢CO2µÈΪԭÁÏÏÈÖƵÃNaHCO3£¬½ø¶øÉú²ú³ö´¿¼î£®Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
NH3+CO2+H2O¨TNH4HCO3£»
NH4HCO3+NaCl¨TNaHCO3¡ý+NH4Cl£»
2NaHCO3Na2CO3+CO2¡ü+H2O
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ì¼ËáÇâï§Óë±¥ºÍʳÑÎË®·´Ó¦£¬ÄÜÎö³ö̼ËáÇâÄƾ§ÌåµÄÔ­ÒòÊÇ______£¨Ìî×Öĸ±êºÅ£©£®
a£®Ì¼ËáÇâÄÆÄÑÈÜÓÚË®    b£®Ì¼ËáÇâÄÆÊÜÈÈÒ׷ֽ⠠  c£®Ì¼ËáÇâÄƵÄÈܽâ¶ÈÏà¶Ô½ÏС£¬ËùÒÔÔÚÈÜÒºÖÐÊ×ÏȽᾧÎö³ö
£¨2£©Ä³Ì½¾¿»î¶¯Ð¡×é¸ù¾ÝÉÏÊöÖƼîÔ­Àí£¬½øÐÐ̼ËáÇâÄƵÄÖƱ¸ÊµÑ飬ͬѧÃÇ°´¸÷×ÔÉè¼ÆµÄ·½°¸ÊµÑ飮
¢Ùһλͬѧ½«¶þÑõ»¯Ì¼ÆøÌåͨÈ뺬°±µÄ±¥ºÍʳÑÎË®ÖÐÖƱ¸Ì¼ËáÇâÄÆ£¬ÊµÑé×°ÖÃÈçÏÂͼËùʾ£¨Í¼Öмг֡¢¹Ì¶¨ÓõÄÒÇÆ÷δ»­³ö£©£®

ÊԻشðÏÂÁÐÓйØÎÊÌ⣺
£¨¢ñ£©ÒÒ×°ÖÃÖеÄÊÔ¼ÁÊÇ ±¥ºÍµÄ̼ËáÇâÄÆÈÜÒº£¬×÷ÓÃÊÇ______£»
£¨¢ò£©¶¡×°ÖÃÖÐÏ¡ÁòËáµÄ×÷ÓÃÊÇ______£»
£¨¢ó£©ÊµÑé½áÊøºó£¬·ÖÀë³öNaHCO3¾§ÌåµÄ²Ù×÷ÊÇ______£¨Ìî·ÖÀë²Ù×÷µÄÃû³Æ£©£®
¢ÚÁíһλͬѧÓÃͼÖÐÎì×°Öã¨ÆäËü×°ÖÃδ»­³ö£©½øÐÐʵÑ飮
£¨¢ñ£©ÊµÑéʱ£¬ÐëÏÈ´Ó______¹ÜͨÈë______ÆøÌ壻
£¨¢ò£©ÓÐͬѧ½¨ÒéÔÚÎì×°ÖõÄb¹Ü϶ËÁ¬½Ó¼º×°Öã¬ÀíÓÉÊÇ______£»
£¨3£©ÇëÄãÔÙд³öÒ»ÖÖʵÑéÊÒÖÆÈ¡ÉÙÁ¿Ì¼ËáÇâÄƵķ½·¨£º______£®
£¨4£©¡°´¿¼îÖг£³£»á»ìÓÐÉÙÁ¿ÂÈ»¯ÄÆ£®¡±Ä³Ñо¿ÐÔѧϰС×éÒÔÒ»°ü´¿¼î£¨Ö»¿¼ÂǺ¬ÂÈ»¯ÄÆ£©ÎªÑо¿¶ÔÏó£¬Ì½¾¿´¿¼îÑùÆ·ÖÐ̼ËáÄƵĺ¬Á¿£®
¡¾ÊµÑéÉè¼Æ¡¿
¼×·½°¸
¢ñ£®Éè¼Æ˼·£º¸ù¾ÝÑùÆ·ÓëÂÈ»¯¸ÆÈÜÒº·´Ó¦Éú³É³Áµí̼Ëá¸ÆµÄÖÊÁ¿£¬Çó³ö̼ËáÄƵÄÖÊÁ¿£¬ÔÙ¼ÆËãÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊý£®
¢ò£®²Ù×÷²½Ö裺³ÆÈ¡13.25g´¿¼îÑùÆ·£¬¼ÓÈë¹ýÁ¿µÄÂÈ»¯¸ÆÈÜÒº£¬³ä·Ö½Á°è£®¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔµÃµ½µÄ°×É«³Áµí10.00g£®
¢ó£®Êý¾Ý´¦Àí£ºÇë¸ù¾ÝÉÏÊöʵÑéÊý¾Ý£¬¼ÆËã¸ÃÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊý£®
¼ÆËã¹ý³Ì£º
______

ÒÒ·½°¸
I£®Éè¼Æ˼·£º¸ù¾ÝÑùÆ·£¨ÖÊÁ¿Îªa g£©ÓëÏ¡ÁòËáÍêÈ«·´Ó¦Éú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿£¨ÖÊÁ¿Îªb g£©£¬Çó³ö̼ËáÄƵÄÖÊÁ¿£¬ÔÙ¼ÆËãÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊý£®
II£®ÊµÑé½áÂÛ£º
£¨1£©µ±a¡¢bµÄÖÊÁ¿¹ØϵÂú×ã______£¨Ìîдº¬a¡¢b×ÖĸµÄ±í´ïʽ£¬ÏÂͬ£©Ê±£¬´¿¼îÑùÆ·ÖÐÖ»º¬ÓÐ̼ËáÄÆ£¬ÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊýÊÇ100%£®
£¨2£©µ±a¡¢bµÄÖÊÁ¿¹ØϵÂú×ã______ʱ£¬´¿¼îÑùÆ·ÊÒÓÐ̼ËáÄƺÍÉÙÁ¿ÂÈ»¯ÄÆ×é³ÉµÄ»ìºÏÎÑùÆ·ÖÐ̼ËáÄƵÄÖÊÁ¿·ÖÊýÊÇ______£®
¡¾ÊµÑéÆÀ¼Û¡¿
Îó²î·ÖÎö£º
£¨Ò»£©¼×·½°¸ÖУ¬°×É«³ÁµíµÄÖÊÁ¿ÔÚ¹ýÂË¡¢¸ÉÔïµÈ²Ù×÷¹ý³ÌÖлáÓÐËùËðºÄ£¬Ôì³É¼ÆËã½á¹ûÓëʵ¼ÊÖµÏà±ÈƫС£®Èç¹û½«ÂÈ»¯¸ÆÈÜÒº»»³ÉÂÈ»¯±µÈÜÒº£¬Ôò¿ÉÒÔʹÎó²î¼õС£¬ÀíÓÉÊÇ______£®
£¨¶þ£©ÒÒ·½°¸ÖУ¬Óв¿·ÖÉú³ÉµÄ¶þÑõ»¯Ì¼ÆøÌåÒòÈܽâÓÚË®¶øûÓÐÈ«²¿Òݳö£¬Ôì³É¼ÆËã½á¹ûÓëʵ¼ÊÖµÏà±È______£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢»ò¡°²»±ä¡±£©£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø