SO2ÊÇÁòËṤҵβÆøµÄÖ÷Òª³É·Ö£®ÊµÑéÊÒÖУ¬ÄâÓÃÏÂͼËùʾÁ÷³Ì£¬²â¶¨±ê×¼×´¿öÏ£¬Ìå»ýΪV LµÄÁòËṤҵβÆøÖÐSO2µÄº¬Á¿£º
¾«Ó¢¼Ò½ÌÍø

¾«Ó¢¼Ò½ÌÍø

£¨1£©²½Öè¢ÙÖмÓÈëH2O2ÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£¬1mol H2O2²Î¼Ó·´Ó¦£¬×ªÒƵĵç×ÓÊýΪ______£®
£¨2£©²½Öè¢ÛµÄ²Ù×÷ÒÀ´ÎÊÇ£º¹ýÂË______¡¢______¡¢______¡¢³ÆÖØ£®
£¨3£©Ò»¶¨Î¶ÈÏ£¬BaSO4µÄ³ÁµíÈܽâƽºâÇúÏßÈçͼËùʾ£®²½Öè¢ÚÖðµÎ¼ÓÈëBa£¨OH£©2ÈÜÒºµÄ¹ý³ÌÖУ¬BaSO4µÄÈܶȻý³£Êý______£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢¡°²»±ä¡±Ö®Ò»£©£¬ÈÜÒºÖÐS
O2-4
Ũ¶ÈµÄ±ä»¯Çé¿öΪ______£¨ÌîÐòºÅ£©
¢Ùd¡úc¡úe   ¢Úb¡úc¡úd   ¢Ûa¡úc¡úe   ¢Üd¡úc¡úa
£¨4£©¸ÃV LβÆøÖÐSO2µÄÌå»ý·ÖÊýΪ______£¨Óú¬ÓÐV¡¢mµÄ´úÊýʽ±íʾ£©£®
£¨1£©²½Öè¢ÙÖмÓÈëH2O2ÈÜÒº£¬¹ýÑõ»¯Çâ¾ßÓÐÑõ»¯ÐÔ£¬¶þÑõ»¯Áò¾ßÓл¹Ô­ÐÔ±»Ñõ»¯ÎªÁòËᣬ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪH2O2+SO2=2H+++SO42-£¬1mol¹ýÑõ»¯ÇⷴӦתÒƵç×ÓÎïÖʵÄÁ¿Îª2mol£¬×ªÒƵĵç×ÓÊýΪ2¡Á6.02¡Á1023=1.204¡Á1024£»
¹Ê´ð°¸Îª£ºH2O2+SO2=2H+++SO42-£¬1.204¡Á1024£»
£¨2£©²½Öè¢ÛµÄ²Ù×÷ÊÇ´ÓÈÜÒºÖзÖÀë³ö³ÁµíÁòËá±µ£¬²Ù×÷²½ÖèÊǹýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢³ÆÖØ£»
¹Ê´ð°¸Îª£ºÏ´µÓ£¬¸ÉÔ
£¨3£©³ÁµíÈܽâƽºâÖÐÁòËá¸ùÀë×ÓŨ¶ÈºÍ±µÀë×ÓŨ¶È³Ë»ýΪ³£Êý£¬Ëæ׿ÓÈëµÄ±µÀë×ÓŨ¶ÈÔö´ó£¬ÁòËá¸ùÀë×ÓŨ¶È¼õС£»Ê¼ÖÕÊDZ¥ºÍÈÜÒºÖеijÁµíÈܽâƽºâ£¬Ó¦ÔÚÇúÏßÉϱ仯£»bdµã²»ÊǸÃζÈϵı¥ºÍÈÜÒº£»
¹Ê´ð°¸Îª£º²»±ä£¬¢Û£»
£¨4£©mgÊÇÁòËá±µµÄÖÊÁ¿£¬ÁòËá±µµÄÎïÖʵÄÁ¿Îª
mg
233g/mol
=
m
233
mol£¬¸ù¾ÝÁòÔªËØÊغã¿ÉÖª¶þÑõ»¯ÁòµÄÌå»ýΪ
m
233
mol¡Á22.4L/mol=
22.4m
233
L£¬¹ÊβÆøÖжþÑõ»¯ÁòµÄÌå»ý·ÖÊý=
22.4m
233
L
VL
=
m¡Á22.4
233V
¡Á100%£»
¹Ê´ð°¸£º
m¡Á22.4
233V
¡Á100%£»
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2012?¼ÃÄ϶þÄ££©SO2ÊÇÁòËṤҵβÆøµÄÖ÷Òª³É·Ö£®ÊµÑéÊÒÖУ¬ÄâÓÃÏÂͼËùʾÁ÷³Ì£¬²â¶¨±ê×¼×´¿öÏ£¬Ìå»ýΪV LµÄÁòËṤҵβÆøÖÐSO2µÄº¬Á¿£º
£¨1£©²½Öè¢ÙÖмÓÈëH2O2ÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
H2O2+SO2=2H++SO42-
H2O2+SO2=2H++SO42-
£¬1mol H2O2²Î¼Ó·´Ó¦£¬×ªÒƵĵç×ÓÊýΪ
1.204¡Á1024
1.204¡Á1024
£®
£¨2£©²½Öè¢ÛµÄ²Ù×÷ÒÀ´ÎÊÇ£º¹ýÂË
¹ýÂË
¹ýÂË
¡¢
Ï´µÓ
Ï´µÓ
¡¢
¸ÉÔï
¸ÉÔï
¡¢³ÆÖØ£®
£¨3£©Ò»¶¨Î¶ÈÏ£¬BaSO4µÄ³ÁµíÈܽâƽºâÇúÏßÈçͼËùʾ£®²½Öè¢ÚÖðµÎ¼ÓÈëBa£¨OH£©2ÈÜÒºµÄ¹ý³ÌÖУ¬BaSO4µÄÈܶȻý³£Êý
²»±ä
²»±ä
£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢¡°²»±ä¡±Ö®Ò»£©£¬ÈÜÒºÖÐS
O
2-
4
Ũ¶ÈµÄ±ä»¯Çé¿öΪ
¢Û
¢Û
£¨ÌîÐòºÅ£©
¢Ùd¡úc¡úe   ¢Úb¡úc¡úd   ¢Ûa¡úc¡úe   ¢Üd¡úc¡úa
£¨4£©¸ÃV LβÆøÖÐSO2µÄÌå»ý·ÖÊýΪ
m¡Á22.4
233V
¡Á100%
m¡Á22.4
233V
¡Á100%
£¨Óú¬ÓÐV¡¢mµÄ´úÊýʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§¶þÂÖרÌâ³å´ÌµÚ10½² ·Ç½ðÊô¼°Æ仯ºÏÎïÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

SO2ÊÇÁòËṤҵβÆøµÄÖ÷Òª³É·Ö¡£ÊµÑéÊÒÖУ¬ÄâÓÃÈçͼËùʾÁ÷³Ì£¬²â¶¨±ê×¼×´¿öÏÂÌå»ýΪV LµÄÁòËṤҵβÆøÖÐSO2µÄº¬Á¿¡£

(1)²½Öè¢ÙÖмÓÈëH2O2ÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________£¬1 mol H2O2²Î¼Ó·´Ó¦£¬×ªÒƵĵç×ÓÊýΪ________¡£

(2)²½Öè¢ÛµÄ²Ù×÷ÒÀ´ÎÊǹýÂË¡¢________¡¢________¡¢³ÆÖØ¡£

(3)Ò»¶¨Î¶ÈÏ£¬BaSO4µÄ³ÁµíÈܽâƽºâÇúÏßÈçͼËùʾ¡£²½Öè¢ÚÖðµÎ¼ÓÈëBa(OH)2ÈÜÒºµÄ¹ý³ÌÖУ¬BaSO4µÄÈܶȻý³£Êý________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)£¬ÈÜÒºÖÐSO42-Ũ¶ÈµÄ±ä»¯Çé¿öΪ________(ÌîÐòºÅ)¡£

¢Ùd¡úc¡úe??????? ¢Úb¡úc¡úd????? ¢Ûa¡úc¡úe???? ¢Üd¡úc¡úa

(4)¸ÃV LβÆøÖÐSO2µÄÌå»ý·ÖÊýΪ________(Óú¬ÓÐV¡¢mµÄ´úÊýʽ±íʾ)¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

SO2ÊÇÁòËṤҵβÆøµÄÖ÷Òª³É·Ö£®ÊµÑéÊÒÖУ¬ÄâÓÃÏÂͼËùʾÁ÷³Ì£¬²â¶¨±ê×¼×´¿öÏ£¬Ìå»ýΪV LµÄÁòËṤҵβÆøÖÐSO2µÄº¬Á¿£º
£¨1£©²½Öè¢ÙÖмÓÈëH2O2ÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£¬1mol H2O2²Î¼Ó·´Ó¦£¬×ªÒƵĵç×ÓÊýΪ______£®
£¨2£©²½Öè¢ÛµÄ²Ù×÷ÒÀ´ÎÊÇ£º¹ýÂË______¡¢______¡¢______¡¢³ÆÖØ£®
£¨3£©Ò»¶¨Î¶ÈÏ£¬BaSO4µÄ³ÁµíÈܽâƽºâÇúÏßÈçͼËùʾ£®²½Öè¢ÚÖðµÎ¼ÓÈëBa£¨OH£©2ÈÜÒºµÄ¹ý³ÌÖУ¬BaSO4µÄÈܶȻý³£Êý______£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢¡°²»±ä¡±Ö®Ò»£©£¬ÈÜÒºÖÐÊýѧ¹«Ê½Å¨¶ÈµÄ±ä»¯Çé¿öΪ______£¨ÌîÐòºÅ£©
¢Ùd¡úc¡úe¡¡ ¢Úb¡úc¡úd¡¡ ¢Ûa¡úc¡úe¡¡ ¢Üd¡úc¡úa
£¨4£©¸ÃV LβÆøÖÐSO2µÄÌå»ý·ÖÊýΪ______£¨Óú¬ÓÐV¡¢mµÄ´úÊýʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012Äêɽ¶«Ê¡¼ÃÄÏÊи߿¼»¯Ñ§¶þÄ£ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º½â´ðÌâ

SO2ÊÇÁòËṤҵβÆøµÄÖ÷Òª³É·Ö£®ÊµÑéÊÒÖУ¬ÄâÓÃÏÂͼËùʾÁ÷³Ì£¬²â¶¨±ê×¼×´¿öÏ£¬Ìå»ýΪV LµÄÁòËṤҵβÆøÖÐSO2µÄº¬Á¿£º
£¨1£©²½Öè¢ÙÖмÓÈëH2O2ÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£¬1mol H2O2²Î¼Ó·´Ó¦£¬×ªÒƵĵç×ÓÊýΪ______£®
£¨2£©²½Öè¢ÛµÄ²Ù×÷ÒÀ´ÎÊÇ£º¹ýÂË______¡¢______¡¢______¡¢³ÆÖØ£®
£¨3£©Ò»¶¨Î¶ÈÏ£¬BaSO4µÄ³ÁµíÈܽâƽºâÇúÏßÈçͼËùʾ£®²½Öè¢ÚÖðµÎ¼ÓÈëBa£¨OH£©2ÈÜÒºµÄ¹ý³ÌÖУ¬BaSO4µÄÈܶȻý³£Êý______£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢¡°²»±ä¡±Ö®Ò»£©£¬ÈÜÒºÖÐŨ¶ÈµÄ±ä»¯Çé¿öΪ______£¨ÌîÐòºÅ£©
¢Ùd¡úc¡úe   ¢Úb¡úc¡úd   ¢Ûa¡úc¡úe   ¢Üd¡úc¡úa
£¨4£©¸ÃV LβÆøÖÐSO2µÄÌå»ý·ÖÊýΪ______£¨Óú¬ÓÐV¡¢mµÄ´úÊýʽ±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸