ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿A¡¢B¡¢C¡¢D¡¢E¡¢FÊǺ˵çºÉÊýÒÀ´ÎÔö´óµÄÁùÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬AÔªËصÄÔ×ÓºËÄÚÖ»ÓÐ1¸öÖÊ×Ó£»BÔªËصÄÔ×Ӱ뾶ÊÇÆäËùÔÚÖ÷×åÖÐ×îСµÄ£¬BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½ÎªHBO3£»CÔªËØÔ×ÓµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ã¶à4£»CµÄÒõÀë×ÓÓëDµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×ÓÅŲ¼£¬Á½ÔªËØ¿ÉÐγɻ¯ºÏÎïD2C£»C¡¢EͬÖ÷×å¡£
(1)BÔÚÖÜÆÚ±íÖеÄλÖÃ______________________________________________
(2)FÔªËصÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïµÄ»¯Ñ§Ê½Îª___________________________________¡£
(3)ÔªËØC¡¢D¡¢EÐγɵļòµ¥Àë×Ӱ뾶ÓÉСµ½´óµÄ˳Ðò________________________(ÓÃÀë×Ó·ûºÅ±íʾ)¡£
(4)Óõç×Óʽ±íʾ»¯ºÏÎïD2CµÄÐγɹý³Ì£º__________________________________________________¡£
C¡¢D»¹¿ÉÐγɻ¯ºÏÎïD2C2£¬D2C2Öк¬ÓеĻ¯Ñ§¼üÊÇ_________________________________________¡£
(5)C¡¢EµÄÇ⻯Î·ÐµãÓɸߵ½µÍ˳ÐòÊÇ£º_______________________________¡£
(6)д³ö̼µ¥ÖÊÓëEµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïŨÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬²¢Óõ¥ÏßÇűêÃ÷µç×ÓµÄתÒÆ·½Ïò_______________¡£µ±×ªÒƵç×ÓΪ0.2molʱ£¬±ê×¼×´¿öÏ·´Ó¦²úÉúÆøÌå_______________L¡£
(7)ÒÑÖªEµ¥ÖʺÍFµ¥ÖʵÄË®ÈÜÒº·´Ó¦»áÉú³ÉÁ½ÖÖÇ¿ËᣬÆäÀë×Ó·½³ÌʽΪ_________________¡£
¡¾´ð°¸¡¿µÚ¶þÖÜÆÚµÚ¢õA×å HClO4 r(Na+) £¼r(O2£)£¼ r(S2£)»òNa+ £¼O2££¼S2£ Àë×Ó¼ü¡¢¹²¼Û¼ü(»ò·Ç¼«ÐÔ¼ü) H2O£¾H2S
3.36 S£«3Cl2£«4H2O = 8H+£«6Cl££«SO42-
¡¾½âÎö¡¿
AÔªËصÄÔ×ÓºËÄÚÖ»ÓÐÒ»¸öÖÊ×Ó£¬ÓÉ´Ë¿ÉÖªAΪH£»¸ù¾Ý BµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ»¯Ñ§Ê½ÎªHBO3 £¬¿ÉÖªBµÄ×î¸ß¼ÛΪ+5¼Û£¬ÊÇVA£¬ÔÙ¸ù¾ÝBÔªËصÄÔ×Ӱ뾶ÊÇÆäËùÔÚÖ÷×åÖÐ×îСµÄ£¬µÃ³öBΪN£»¸ù¾ÝCÔªËØÔ×ÓµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ãµç×ÓÊý¶à4£¬µÃ³öCΪO£»CµÄÒõÀë×ÓÓëDµÄÑôÀë×Ó¾ßÓÐÏàͬµÄµç×ÓÅŲ¼£¬Á½ÔªËØ¿ÉÐγɻ¯ºÏÎïD2C£¬¿ÉÖªDµÄÑôÀë×ÓΪ+1¼ÛÑôÀë×Ó£¬ËùÒÔDΪNa£»C¡¢EͬÖ÷×壬ÇÒEΪ¶ÌÖÜÆÚÔªËØ£¬µÃ³öEΪS£¬FÊǶÌÖÜÆÚÔªËØÅÅÔÚSÔªËغóÃæµÄÖ÷×åÔªËØ£¬FÖ»ÄÜΪCl¡£×ÛÉÏ¿ÉÖªAΪH¡¢BΪN¡¢CΪO¡¢DΪNa£¬EΪS£¬FΪCl¡£¾Ý´Ë»Ø´ð¡£
£¨1£©BΪµªÔªËØ£¬NλÓÚÖÜÆÚ±íµÚ¶þÖÜÆÚµÚVA×壬¹Ê´ð°¸Îª£ºµÚ¶þÖÜÆÚ£¬VA£»
£¨2£©FΪClÔªËØ£¬ClÔªËØ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÊÇHClO4£¬¹Ê´ð°¸Îª£ºHClO4£»
£¨3£©ÔªËØC¡¢D¡¢EÐγɵļòµ¥Àë×Ó·Ö±ðΪO2£¡¢Na+¡¢S2££¬µç×Ó²ãÊý¶àµÄÀë×Ӱ뾶½Ï´ó£¬ºËÍâµç×ÓÅŲ¼ÏàͬµÄ΢Á££¬ÆäÀë×Ӱ뾶ËæÔ×ÓÐòÊýµÄÔö´ó¶ø¼õС£¬ËùÒ԰뾶ÓÉСµ½´óΪr(Na+) £¼r(O2£)£¼ r(S2£)£¬¹Ê´ð°¸Îª£ºr(Na+) £¼r(O2£)£¼ r(S2£)£»
£¨4£©D2CÊÇNa2O£¬Óõç×Óʽ±íʾÐγɹý³ÌΪ£¬¹Ê´ð°¸Îª£º
£»D2C2ΪNa2O2£¬ÆäÖк¬ÓеĻ¯Ñ§¼üΪÀë×Ó¼ü¡¢¹²¼Û¼ü£¬¹Ê´ð°¸Îª£ºÀë×Ó¼ü¡¢¹²¼Û¼ü£»
£¨5£©C¡¢EµÄÇ⻯Îï·Ö±ðΪH2O¡¢H2S£¬ÓÉÓÚH2O·Ö×ÓÖ®¼ä´æÔÚÇâ¼ü£¬µ¼Ö·еãÒª¸ü¸ß£¬ËùÒԷеãH2O£¾H2S£¬¹Ê´ð°¸Îª£ºH2O£¾H2S£»
£¨6£©CÓëŨÁòËá·´Ó¦·½³ÌʽΪ£ºC+2H2SO4(Ũ) 2SO2¡ü+CO2¡ü+2H2O£¬Óá°µ¥ÏßÇÅ¡±±êµç×ÓתÒÆ
£¬µ±×ªÒÆ4mole-ʱÉú³ÉÆøÌåSO2 2molºÍCO2 1mol£¬¹²3mol£¬±ê×¼×´¿öÏÂÌå»ýΪ67.2L£¬µ±×ªÒÆ0.2molµç×Óʱ£¬ÆøÌåÌå»ýΪ
£¬¹Ê´ð°¸Îª£º
£»3.36L£»
£¨7£©SºÍÂÈË®·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ3Cl2+S+ 4H2O=6HCl+H2SO4£¬ËùÒÔÀë×Ó·½³ÌʽΪS£«3Cl2£«4H2O = 8H+£«6Cl££«SO42-£¬¹Ê´ð°¸Îª£ºS£«/span>3Cl2£«4H2O = 8H+£«6Cl££«SO42-¡£
![](http://thumb2018.1010pic.com/images/loading.gif)