ijѧУÑо¿ÐÔѧϰС×éΪÁ˲ⶨµ±µØ¿óɽʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬È¡À´ÁËһЩ¿óʯÑùÆ·£¬²¢È¡Ï¡ÑÎËá200g£¬Æ½¾ù·Ö³É4·Ý£¬½øÐÐʵÑ飬½á¹ûÈçÏ£º

ʵÑé

1

2

3

4

¼ÓÈëÑùÆ·µÄÖÊÁ¿£¯g

5

10

15

20

Éú³ÉCO2µÄÖÊÁ¿£¯g

1.76

3.52

4.4

m

£¨1£©Äļ¸×é·´Ó¦ÖпóʯÓÐÊ£Óࣿ________¡£

£¨2£©ÉϱíÖÐmµÄÊý¾ÝÊÇ________¡£

£¨3£©ÊÔ¼ÆËãÕâÖÖʯ»Òʯ¿óÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý¡£

 

´ð°¸£º
½âÎö£º

£¨1£©µÚ3¡¢4×é

£¨2£©4.4 g

£¨3£©80£¥

 


Ìáʾ£º


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ʯ»ÒʯÊÇÎÒÊеÄÖ÷Òª¿ó²úÖ®Ò»£®Ä³Ñ§Ð£Ñо¿ÐÔѧϰС×éΪÁ˲ⶨµ±µØ¿óɽÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬È¡À´ÁËһЩ¿óʯÑùÆ·£¬²¢È¡Ï¡ÑÎËá200g£¬Æ½¾ù·Ö³É4·Ý£¬½øÐÐʵÑ飬½á¹ûÈçÏ£º£¨ÔÓÖʼȲ»ÈÜÓÚË®£¬Ò²²»ºÍËá·¢Éú·´Ó¦£©
ʵÑé 1 2 3 4
¼ÓÈëÑùÆ·µÄÖÊÁ¿/g 15 22.5 30 37.5
Éú³ÉµÄCO2ÖÊÁ¿/g 4.4 6.6 8.0 m
£¨1£©ÉϱíÖÐmµÄÊýÖµÊÇ
 
£®
£¨2£©ÊÔ¼ÆËãÕâÖÖʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£®£¨½á¹û¾«È·µ½0.1%£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijѧУÑо¿ÐÔѧϰС×飬ÒÔÒ»¹¤³§Éú²ú¹ý³ÌÖвúÉúµÄ·ÏÆúÎï¡°ºìÉ«ÌúÄࡱΪÑо¿¶ÔÏó£¬Ì½¾¿ºìÉ«ÎïÖʵijɷ֣¬ÇëÄã²ÎÓëËûÃǵĻ£®
[С×éÌÖÂÛ]
¼×¹Ûµã£ººìÉ«ÎïÖÊÊÇÍ­£®    
Òҹ۵㣺ºìÉ«ÎïÖÊÊÇÑõ»¯Ìú£®
±û¹Ûµã£ººìÉ«ÎïÖÊÊÇÇâÑõ»¯Ìú£®
¶¡¹Ûµã£ººìÉ«ÎïÖÊÊÇÑõ»¯ÌúºÍÇâÑõ»¯ÌúµÄ»ìºÏÎ
[²éÔÄ×ÊÁÏ]?
¸Ã³§¡°ºìÉ«ÌúÄࡱÖ÷ÒªÊÇÉú²ú¹ý³ÌÖÐʹÓÃÌú´¥Ã½£¨Ò»ÖÖ´ß»¯¼Á£©ºóµÄ·ÏÆúÎ³ýºìÉ«ÎïÖÊÍ⣬»¹º¬ÓÐͨ³£²»ÓëËá¡¢¼îºÍÑõÆø·´Ó¦µÄÎïÖÊ£®ÇâÑõ»¯ÌúÊÜÈÈÒ×·Ö½â
[Ìá³ö¼ÙÉè]?
ÒÒ¡¢±û¡¢¶¡ÈýÖÖ¹Ûµã´æÔÚºÏÀíÐÔ£®?
[ʵÑéÑéÖ¤]
¢ñÈ¡ÊÊÁ¿¡°ºìÉ«ÌúÄࡱɹ¸É£¬³ÆÁ¿ÆäÖÊÁ¿Îªm1g£»
¢ò½«¡°ºìÉ«ÌúÄࡱÖÃÓÚÛáÛöÖгä·Ö¼ÓÈȲ¢ÔÚ¸ÉÔïÆ÷ÖÐÀäÈ´ºó£¬³ÆÁ¿ÆäÖÊÁ¿Îªm2g£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µ±m1=m2ʱ£¬ÔòºìÉ«ÎïÖÊΪ
 
£®
£¨2£©µ±m1£¾m2£¾160m1/214ʱ£¬Ôò¡°ºìÉ«ÌúÄࡱÖÐÒ»¶¨º¬ÓÐÑõ»¯ÌúÂð£¿
 
£¨Ìîд¡°Ò»¶¨¡±»ò¡°²»Ò»¶¨¡±£©£¬ÀíÓÉÊÇ
 
£®
£¨3£©¼Ù¶¨¡°ºìÉ«ÌúÄࡱÖС°ºìÉ«ÎïÖÊÊÇÇâÑõ»¯Ìú¡±£¬Çëд³ö»ØÊÕ¡°ºìÉ«ÌúÄࡱÖеÄÌúÔªËØÖƵá°Ìúºì¡±£¨»¯Ñ§Ê½ÎªFe2O3£¬ÊǺìÉ«ÓÍÆáµÄÖØÒªÔ­ÁÏ£©¹ý³ÌÖеĻ¯Ñ§·½³Ìʽ£º¢Ù
 
£»    ¢Ú
 
£»?¢Û
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§ÀÏʦ½»¸øijѧУÑо¿ÐÔѧϰС×éÒ»¸öÈÎÎñ£º²â¶¨ÊµÑéÊÒÀïһƿ¾ÃÖõÄNaOH¹ÌÌåÊÇ·ñ±äÖÊ£®¸ÃС×éµÄͬѧ½øÐÐÁËÈçÏÂʵÑé̽¾¿£º
Ìá³öÎÊÌ⣺²ÂÏëI£ºÃ»±äÖÊ£¬È«²¿ÊÇNaOH£»
²ÂÏëII£º²¿·Ö±äÖÊ£¬
 
£»
²ÂÏë¢ó£ºÍêÈ«±äÖÊ£¬È«²¿ÊÇNa2CO3£®
²éÔÄ×ÊÁÏ£º²¿·ÖÈÜÒºÔÚ³£ÎÂϵÄpHÈçÏ£º
 ÑÎÈÜÒº NaCl Na2CO3  BaCl2
pH µÈÓÚ7  ´óÓÚ7 µÈÓÚ7
Éè¼Æ·½°¸²¢½øÐÐʵÑ飺ÇëÄãÓëËûÃÇÒ»Æð¹²Í¬Íê³É£¬²¢»Ø´ðËù¸øÎÊÌ⣮
ʵÑé²½Öè  ÊµÑéÏÖÏó ʵÑé½áÂÛ
¢Ù³ÆÈ¡ÉÏÊöÇâÑõ»¯ÄƹÌÌåÑùÆ·10.0gÈÜÓÚ100ml£¬Ë®Åä³ÉÈÜÒº£¬ÏòÈÜÒºÖеμÓÂÈ»¯±µÈÜÒºÖÁ¹ýÁ¿£¬³ä·Ö·´Ó¦ºó£¬¾²Öã® ²úÉú°×É«³Áµí ˵Ã÷¾ÃÖùÌÌåÖУ¬Ò»¶¨º¬ÓÐ
 
£¨Ìѧʽ£©£®
¢ÚÓò£Á§°ôպȡÉÙÁ¿¢ÙÖгä·Ö·´Ó¦ºóµÄÉϲãÇåÒºµÎÔÚһС¿épHÊÔÖ½ÉÏ£¬Óë±ê×¼±ÈÉ«¿¨¶Ô±È£¬²â³öpH£® pH=11 ˵Ã÷¾ÃÖùÌÌåÖУ¬»¹Ò»¶¨º¬ÓÐ
 
 £¨Ìѧʽ£©£®
ÉÏÊö²½Öè¢ÙÖУ¬³ÆÁ¿ÇâÑõ»¯ÄƹÌÌåÑùÆ·Óõ½µÄ²£Á§ÒÇÆ÷Ãû³ÆÊÇ
 
£¬µÎ¼Ó¹ýÁ¿BaCl2ÈÜÒºµÄÄ¿µÄÊÇ
 

ʵÑé½áÂÛ£ºÍ¨¹ýʵÑ飬˵Ã÷ÉÏÊö²ÂÏëÖÐ
 
ÊÇÕýÈ·µÄ£®
ÍØÕ¹£º¸ÃС×éͬѧΪ²â¶¨³ö¸ÃNaOH¹ÌÌåµÄ±äÖʳ̶ȣ¬¼ÌÐø½«ÉÏÊö°×É«³Áµí¹ýÂË¡¢Ï´¾»¡¢¸ÉÔ³ÆµÃÆäÖÊÁ¿Îª4.0g£¬ÔòÔ­ÑùÆ·ÖÐNaOHµÄÖÊÁ¿·ÖÊýΪ
 
 £¨±£ÁôһλСÊý£©£®
·´Ë¼£º¾ÃÖõÄÇâÑõ»¯ÄƱäÖʵÄÔ­ÒòÊÇ
 
£¨Óû¯Ñ§·½³Ìʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍøʯ»ÒʯÊÇÎÒÊеÄÖ÷Òª¿ó²úÖ®Ò»£®ÎªÁ˲ⶨµ±µØʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬Ä³Ñ§Ð£Ñо¿ÐÔѧϰС×é½øÐÐÁËÈçÏÂʵÑ飺ȡʯ»ÒʯÑùÆ·25g£¬ÏòÆäÖеμÓÏ¡ÑÎËᣬ²â¶¨Éú³ÉÆøÌåCO2µÄÖÊÁ¿£®¼ÓÈëÏ¡ÑÎËáµÄÖÊÁ¿Óë²úÉúCO2ÆøÌåµÄÖÊÁ¿¹ØϵÈçͼËùʾ£®ÊÔ¼ÆË㣨ÔÓÖʲ»ÓëËá·´Ó¦£¬¸ßÎÂÒ²²»·Ö½â£©£º
£¨1£©Ê¯»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ¶àÉÙ£¿
£¨2£©ìÑÉÕ50tÕâÖÖʯ»ÒʯÀíÂÛÉÏ¿ÉÖƵô¿¾»µÄÉúʯ»Ò¶àÉÙ¶Ö£¿£¨CaCO3 
 ¸ßΠ
.
 
CaO+CO2¡ü£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ʯ»ÒʯÊÇijʡµÄÖ÷Òª¿ó²úÖ®Ò»£®Ä³Ñ§Ð£Ñо¿ÐÔѧϰС×éΪÁ˲ⶨµ±µØ¿óɽÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬È¡À´ÁË20g¿óʯÑùÆ·£¬Æ½¾ù·Ö³É4·Ý£¬ÒÀ´Î¼ÓÈëµ½×ãÁ¿µÄÏ¡ÑÎËáÖУ¬½øÐÐʵÑ飬½á¹ûÈçÏ£º
ʵÑé 1 2 3 4
¼ÓÈëÑùÆ·ÖÊÁ¿/g 5 10 15 20
ÉúÆøCO2ÖÊÁ¿/g 1.76 3.52 4.4 m
£¨1£©ÉϱíÖÐmµÄÊýÖµÊÇ
 
£®
£¨2£©ÊÔ¼ÆËãÕâÖÖʯ»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸