ÌâÄ¿ÄÚÈÝ

£¨16·Ö£©Ä³»¯Ñ§ÊµÑéС×éµÄͬѧΪ̽¾¿ºÍ±È½ÏSO2ºÍÂÈË®µÄƯ°×ÐÔ£¬Éè¼ÆÁËÈçϵÄʵÑé×°Öã¨Í¼3£©¡£

£¨1£©¢Ù·´Ó¦¿ªÊ¼Ò»¶Îʱ¼äºó£¬¹Û²ìµ½B¡¢DÁ½¸öÊÔ¹ÜÖеÄÆ·ºìÈÜÒº³öÏÖµÄÏÖÏóÊÇ£º

B£º________________________________£¬D£º____________________________¡£

¢ÚֹͣͨÆøºó,ÔÙ¸øB¡¢DÁ½¸öÊԹֱܷð¼ÓÈÈ£¬Á½¸öÊÔ¹ÜÖеÄÏÖÏó·Ö±ðΪ

B£º________________________________£¬D£º____________________________¡£

£¨2£©ÁíÒ»¸öʵÑéС×éµÄͬѧÈÏΪSO2ºÍÂÈË®¶¼ÓÐƯ°×ÐÔ£¬¶þÕß»ìºÏºóµÄƯ°×ÐԿ϶¨»á¸üÇ¿£¬ËûÃǽ«ÖƵõÄSO2ºÍCl2°´1:1ͬʱͨÈ뵽ƷºìÈÜÒºÖУ¬½á¹û·¢ÏÖÍÊɫЧ¹û²¢²»ÏñÏëÏóÄÇÑù¡£ÇëÄã·ÖÎö¸ÃÏÖÏóµÄÔ­Òò£¨Óû¯Ñ§·½³Ìʽ±íʾ£©______________________________       ¡£

£¨3£©×°ÖÃEÖÐÓÃMnO2ºÍŨÑÎËá·´Ó¦ÖƵÃCl2£¬Èô·´Ó¦Éú³ÉµÄCl2Ìå»ýΪ2.24L£¨±ê×¼×´¿ö£©£¬Ôò±»Ñõ»¯µÄHClΪ            mol¡£

£¨4£©ÊµÑé½áÊøºó£¬ÓÐͬѧÈÏΪװÖÃCÖпÉÄܺ¬ÓÐSO32£­¡¢SO42£­¡¢Cl£­¡¢OH£­µÈÒõÀë×Ó£¬ÇëÌîд¼ìÑéÆäÖÐSO42£­ºÍSO32£­µÄʵÑ鱨¸æ¡£

ÏÞÑ¡ÊÔ¼Á£º2 mol¡¤L£­1 HCl£»1 mol¡¤L£­1 H2SO4£»l mol¡¤L£­1 BaCl2£»l mol¡¤L£­1MgCl2

1 mol¡¤L£­1 HNO3£»0.1 mol¡¤L£­1 AgNO3£»ÐÂÖƱ¥ºÍÂÈË®¡£

񅧏

ʵÑé²Ù×÷

Ô¤ÆÚÏÖÏóºÍ½áÂÛ

²½Öè¢Ù

È¡ÉÙÁ¿´ý²âÒºÓÚÊÔ¹ÜÖУ¬µÎÈë

                    ÖÁ¹ýÁ¿

 

                        £¬Ö¤Ã÷´ý²âÒºÖк¬SO32£­¡£

²½Öè¢Ú

ÔÚ²½Öè¢ÙµÄÈÜÒºÖеÎÈëÉÙÁ¿

                        

 

                                  £¬

Ö¤Ã÷´ý²âÒºÖк¬SO42£­¡£

 

¡¾´ð°¸¡¿

£¨µÚ£¨4£©Ð¡Ìâÿ¿Õ1·Ö£¬ÆäÓàÿ¿Õ2·Ö£©£¨1£©¢ÙB£ºÆ·ºìÈÜÒºÍÊÉ«£¬D£ºÆ·ºìÈÜÒºÍÊÉ«¡£

¢ÚB£ºÍÊÉ«µÄÆ·ºìÓÖ»Ö¸´ÎªºìÉ«£¬D£ºÎÞÃ÷ÏÔÏÖÏó¡£

£¨2£©Cl2£«SO2£«2H2O=2HCl£«H2SO4       £¨3£©0.2

£¨4£©

񅧏

ʵÑé²Ù×÷

Ô¤ÆÚÏÖÏóºÍ½áÂÛ

²½Öè¢Ù

2 mol¡¤L-1HCl

ÓÐÆûÅݲúÉú

²½Öè¢Ú

l mol¡¤L-1BaCl2

Óа×É«³ÁµíÉú³É

¡¾½âÎö¡¿£¨1£©¢ÙSO2¾ßÓÐƯ°×ÐÔ£¬ÄÜʹƷºìÈÜÒºÍÊÉ«£¬ÂÈË®¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Ò²ÄÜʹƷºìÈÜÒºÍÊÉ«¡£

¢ÚÓÉÓÚSO2µÄƯ°×ÐÔÊÇ¿ÉÄæµÄ£¬ËùÒÔ¼ÓÈÈÄÜʹƷºìÈÜÒº»Ö¸´ºìÉ«¡£µ«ÂÈË®µÄƯ°×ÐÔÊÇÑõ»¯µ¼Öµģ¬ÊDz»¿ÉÄæµÄ£¬¼ÓÈȲ»Äָܻ´ºìÉ«¡£

£¨2£©ÓÉÓÚSO2»¹¾ßÓл¹Ô­ÐÔ£¬¶øÂÈÆø¾ßÓÐÑõ»¯ÐÔ£¬¶þÕß·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³ÉÂÈ»¯ÇâºÍÁòËᣬ·½³ÌʽΪCl2£«SO2£«2H2O=2HCl£«H2SO4¡£

£¨3£©2.24LÂÈÆøÔÚ±ê×¼×´¿öϵÄÎïÖʵÄÁ¿ÊÇ0.1mol£¬¸ù¾ÝÂÈÔ­×ÓÊغã¿ÉÖª£¬±»Ñõ»¯µÄÂÈ»¯ÇâÊÇ0.2mol¡£

£¨4£©SO32£­µÄ¼ìÑé¿ÉÓÃÑÎËᣬ·´Ó¦ÖÐÉú³ÉSO2ÆøÌå¡£SO42£­µÄ¼ìÑéÓÃÂÈ»¯±µÈÜÒº£¬Éú³É°×É«³ÁµíÁòËá±µ¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
I£®µªÊǶ¯Ö²ÎïÉú³¤²»¿ÉȱÉÙµÄÔªËØ£¬º¬µª»¯ºÏÎïÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬Ñо¿CO¡¢NOXµÈ´óÆøÎÛȾÆøÌåµÄ´¦Àí¾ßÓÐÖØÒªÒâÒ壮
£¨1£©ÏÂÁйý³ÌûÓÐÆðµ½µªµÄ¹Ì¶¨×÷ÓõÄÊÇ
 
£¨ÌîÑ¡Ï£®
A£®N2ÓëO2·´Ó¦Éú³ÉNO    B£®NH3¾­´ß»¯Ñõ»¯Éú³ÉNO
C£®N2ºÍH2ÔÚÒ»¶¨Ìõ¼þϺϳɰ±D£®¶¹¿ÆÖ²ÎïµÄ¸ùÁö¾ú½«¿ÕÆøÖеªÆøת»¯Îªº¬µª»¯ºÏÎï
£¨2£©ÊµÑéÊÒÀï¿ÉÒÔÑ¡ÔñÏÂÁÐʵÑé×°ÖÃÖеÄ
 
£¨ÌîÑ¡ÏÖÆÈ¡°±Æø£®
¾«Ó¢¼Ò½ÌÍø
д³öʵÑéÊÒÖÆÈ¡°±ÆøµÄ»¯Ñ§·½³Ìʽ
 
£¬¼ìÑé°±ÆøÊÇ·ñÊÕ¼¯ÂúµÄ·½·¨³ýÓÃʪÈóµÄºìɫʯÈïÊÔÖ½ÒÔÍ⻹¿ÉÒÔÓÃ
 
£®
II£®Ä³»¯Ñ§ÊµÑéС×éµÄͬѧΪ̽¾¿ºÍ±È½ÏSO2ºÍÂÈË®µÄƯ°×ÐÔ£¬Éè¼ÆÁËÈçϵÄʵÑé×°Öã®
¾«Ó¢¼Ò½ÌÍø
£¨1£©ÊµÑéÖÐÓÃ×°ÖÃEÖƱ¸Cl2£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
 
£»ÈôÓÐ6molµÄHCl²Î¼Ó·´Ó¦£¬ÔòתÒƵĵç×ÓµÄÎïÖʵÄÁ¿Îª
 
£®
£¨2£©¢Ù·´Ó¦¿ªÊ¼Ò»¶Îʱ¼äºó£¬¹Û²ìµ½B¡¢DÁ½¸öÊÔ¹ÜÖеľ§ºìÈÜÒº³öÏÖµÄÏÖÏó·Ö±ðÊÇ£ºB
 
£¬D
 
£®
¢ÚֹͣͨÆøºó£¬ÔÙ½«B¡¢DÁ½¸öÊԹֱܷð¼ÓÈÈ£¬Á½¸öÊÔ¹ÜÖеÄÏÖÏó·Ö±ðÊÇ£º
B
 
£¬D
 
£®
£¨3£©ÁíÒ»¸öʵÑéС×éµÄͬѧÈÏΪSO2ºÍÂÈË®¶¼ÓÐƯ°×ÐÔ£¬¶þÕß»ìºÏºóµÄƯ°×ÐԿ϶¨  »á¸üÇ¿£®ËûÃǽ«ÖƵõÄSO2ºÍCl2½Ól£º1ͬʱͨÈ뵽ƷºìÈÜÒºÖУ¬½á¹û·¢ÏÖÍÊɫЧ¹û²¢²»ÏñÏëÏóµÄÄÇÑù£®Çë½áºÏÀë×Ó·½³Ìʽ˵Ã÷²úÉú¸ÃÏÖÏóµÄÔ­Òò£º
 

 
£®

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

¾«Ó¢¼Ò½ÌÍø