ÌâÄ¿ÄÚÈÝ

15£®Ä³Í¬Ñ§½øÐÐʵÑéÑо¿Ê±£¬ÐèÒªÅäÖÆ1000mL 0.2mol/LNaClÈÜÒº£®
£¨1£©¾­¹ý¼ÆË㣬ʹÓÃÍÐÅÌÌìÆ½Ó¦¸Ã³ÆÁ¿11.7g NaCl¹ÌÌ壻
£¨2£©ÅäÖÆÈÜҺʱ£¬³ýÁËÍÐÅÌÌìÆ½¡¢Ò©³×¡¢Á¿Í²¡¢ÉÕ±­¡¢²£Á§°ôÍ⣬»¹ÐèÒªµÄÒÇÆ÷ÓÐ1000 mLÈÝÁ¿Æ¿£» ½ºÍ·µÎ¹Ü£»
£¨3£©ÅäÖÆÈÜҺʱ£¬Ðè¾­¹ý³ÆÁ¿¡¢ÈÜ½â¡¢×ªÒÆÈÜÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ¡¢×°Æ¿µÈ²Ù×÷£®
ÏÂÁÐͼʾ¶ÔÓ¦µÄ²Ù×÷¹æ·¶µÄÊÇB£»
            
A£®³ÆÁ¿B£®ÈܽâC£®×ªÒÆD£®Ò¡ÔÈ
£¨4£©ÈôÅäÖÆÈÜҺʱ£¬Ï´¸É¾»µÄÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏß֮ϲÐÁôÓÐÉÙÁ¿ÕôÁóË®£¬ÔòËùÅäÈÜÒºµÄŨ¶ÈÎÞÓ°Ï죻ÈôÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ¼ÓÈëÉÙÁ¿ÕôÁóË®Óë¿Ì¶ÈÏßÏàÆ½£¬ÔòËùÅäÈÜÒºµÄŨ¶ÈÆ«µÍ£®£¨ÌîÆ«¸ß¡¢Æ«µÍ¡¢ÎÞÓ°Ï죩

·ÖÎö £¨1£©ÒÀ¾Ým=CVM¼ÆËãÐèÒªÈÜÖʵÄÖÊÁ¿£»
£¨2£©ÒÀ¾ÝÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÒ»°ã²½ÖèÑ¡ÔñÐèÒªµÄÒÇÆ÷£»
£¨3£©ÒÀ¾ÝÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÕýÈ·µÄ²Ù×÷²½Öè½â´ð£»
£¨4£©·ÖÎö²Ù×÷¶ÔÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºÌå»ýµÄÓ°Ï죬ÒÀ¾ÝC=$\frac{n}{V}$½øÐзÖÎö£®

½â´ð ½â£º£¨1£©ÅäÖÆ1000mL 0.2mol/LNaClÈÜÒº£¬ÐèÒªÂÈ»¯ÄƵÄÖÊÁ¿m=0.2mol/L¡Á1L¡Á58.5g/mol=11.7g£»
¹Ê´ð°¸Îª£º11.7£»
£¨2£©ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÒ»°ã²½Ö裺¼ÆËã¡¢³ÆÁ¿¡¢ÈÜ½â¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ£¬Óõ½µÄÒÇÆ÷£º
ÍÐÅÌÌìÆ½¡¢Ò©³×¡¢Á¿Í²¡¢ÉÕ±­¡¢²£Á§°ô¡¢1000 mLÈÝÁ¿Æ¿£» ½ºÍ·µÎ¹Ü£¬ËùÒÔ»¹È±ÉÙµÄÒÇÆ÷£º
1000 mLÈÝÁ¿Æ¿£» ½ºÍ·µÎ¹Ü£»
¹Ê´ð°¸Îª£º1000 mLÈÝÁ¿Æ¿£» ½ºÍ·µÎ¹Ü£»
£¨3£©ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºÒ»°ã²½Ö裺¼ÆËã¡¢³ÆÁ¿¡¢ÈÜ½â¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔÈ£»
A£®³ÆÁ¿Ê±Ó¦×ñÑ­×óÎïÓÒÂ룬¹ÊA´íÎó£»
B£®ÈܽâʱӦÓò£Á§°ô½Á°è£¬¹ÊBÕýÈ·£»
C£®ÒÆÒºÊ±Ó¦Óò£Á§°ôÒýÁ÷£¬¹ÊC´íÎó£»
D£®Ò¡ÔÈʱÓÃʳָ¶¥×¡Æ¿Èû£¬ÉÏÏÂ×óÓҵߵ¹£¬¹ÊDÕýÈ·£»
¹Ê´ð°¸Îª£º¶¨ÈÝ£» BD£»
£¨4£©ÈôÅäÖÆÈÜҺʱ£¬Ï´¸É¾»µÄÈÝÁ¿Æ¿µÄ¿Ì¶ÈÏß֮ϲÐÁôÓÐÉÙÁ¿ÕôÁóË®£¬¶ÔÈÜÖʵÄÎïÖʵÄÁ¿ºÍÈÜÒºÌå»ý¶¼²»²úÉúÓ°Ï죬ÈÜҺŨ¶È²»±ä£»
ÈôÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬ÓÖ¼ÓÈëÉÙÁ¿ÕôÁóË®Óë¿Ì¶ÈÏßÏàÆ½£¬µ¼ÖÂÈÜÒºÌå»ýÆ«´ó£¬ÈÜҺŨ¶ÈÆ«µÍ£»
¹Ê´ð°¸Îª£ºÎÞÓ°Ï죻 Æ«µÍ£»

µãÆÀ ±¾Ì⿼²éÁËÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈÈÜÒºµÄÅäÖÆ£¬Ã÷È·ÅäÖÆ¹ý³Ì¼°²Ù×÷²½ÖèÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶȲ»´ó£¬×¢ÒâÎó²î·ÖÎöµÄ·½·¨ºÍ¼¼ÇÉ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
7£®¶Ô¹¤Òµ·ÏË®ºÍÉú»îÎÛË®½øÐд¦ÀíÊÇ·ÀֹˮÌåÎÛȾ¡¢¸ÄÉÆË®ÖʵÄÖ÷Òª´ëʩ֮һ£®
Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©ÁòËṤ³§µÄËáÐÔ·ÏË®ÖÐÉ飨As£©ÔªËØ£¨Ö÷ÒªÒÔÈõËáH3AsO3ÐÎʽ´æÔÚ£©º¬Á¿¼«¸ß£¬Îª¿ØÖÆÉéµÄÅÅ·Å£¬Ä³¹¤³§²ÉÓû¯Ñ§³Á½µ·¨´¦Àíº¬Éé·ÏË®£®
¢ÙÒÑÖªÉéÊǵªµÄͬ×åÔªËØ£¬±ÈµªÔ­×Ó¶à2¸öµç×Ӳ㣬ÉéÔÚÔªËØÖÜÆÚ±íµÄλÖÃΪµÚËÄÖÜÆÚ£¬µÚ¢õA×壬AsH3µÄÎȶ¨ÐÔ±ÈNH3µÄÎȶ¨ÐÔÈõ£¨Ìî¡°Ç¿¡¯¡¯»ò¡°Èõ¡¯¡¯£©£»
¢Ú¹¤ÒµÉϲÉÓÃÁò»¯·¨£¨Í¨³£ÓÃÁò»¯ÄÆ£©È¥³ý·ÏË®ÖеÄÉ飬Éú³ÉÄÑÈܵÄÈýÁò»¯¶þÉ飬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ2H3AsO3+3S2-+6H+=As2S3+6H2O£»
£¨2£©ÉéµÄ³£¼ûÑõ»¯ÎïÓÐAs2O3ºÍAs2O5£¬ÆäÖÐAs2O5ÈÈÎȶ¨ÐԲ¸ù¾ÝÏÂͼд³öAs2O5·Ö½âΪAs2O3µÄÈÈ»¯Ñ§·½³Ìʽ£ºAs2O5£¨s£©=As2O3£¨s£©+O2£¨g£©¡÷H=+295.4kJ•mol-1£®
£¨3£©ÉéËáÑοɷ¢ÉúÈçÏ·´Ó¦£ºAsO43-+2I-+2H+?AsO33-+I2+H2O£®ÏÂͼװÖÃÖУ¬C1¡¢C2ÊÇʯīµç¼«£®
¢ÙAÖÐÊ¢ÓÐרɫµÄKIºÍI2µÄ»ìºÏÈÜÒº£¬BÖÐÊ¢ÓÐÎÞÉ«µÄNa3AsO4ºÍNa3AsO3µÄ»ìºÏÈÜÒº£¬µ±Á¬½Ó¿ª¹ØK£¬²¢ÏòBÖеμÓŨÑÎËáʱ·¢ÏÖÁéÃôµçÁ÷¼ÆGµÄÖ¸ÕëÏòÓÒÆ«×ª£®´ËʱC2ÉÏ·¢ÉúµÄµç¼«·´Ó¦ÊÇAsO43-+2H++2e-=AsO33-+H2O£®
¢Ú¸Ãµç³Ø¹¤×÷ʱ£¬µ±ÍâµçÂ·×ªÒÆ4NA e-ʱÉú³É2mol I2£®
£¨4£©ÀûÓã¨3£©Öз´Ó¦¿É²â¶¨º¬As2O3ºÍAs2O5µÄÊÔÑùÖеĸ÷×é·Öº¬Á¿£¨Ëùº¬ÔÓÖʶԲⶨÎÞÓ°Ï죩£¬¹ý³ÌÈçÏ£º
¢Ù½«ÊÔÑùÈÜÓÚNaOHÈÜÒº£¬µÃµ½º¬AsO43-ºÍAsO33-µÄ»ìºÏÈÜÒº£®ÒÑÖª£ºAs2O3ÓëNaOHÈÜÒº·´Ó¦Éú³ÉAsO33-£¬ÔòAs2O5ÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇAs2O5+6OH-¨T2AsO43-+3H2O£»
¢ÚÉÏÊö»ìºÏÒºÓÃ0.02500mol•L-1µÄI2ÈÜÒºµÎ¶¨£¬ÏûºÄI2ÈÜÒº20.00mL£®µÎ¶¨Íê±Ïºó£¬Ê¹ÈÜÒº³ÊËáÐÔ£¬¼ÓÈë¹ýÁ¿µÄKI£¬Îö³öµÄI2ÓÖÓÃ0.1000mol•L-1µÄNa2S2O3ÈÜÒºµÎ¶¨£¬ÏûºÄNa2S2O3ÈÜÒº30.00mL£®£¨ ÒÑÖª2Na2S2O3+I2=Na2S4O6+2NaI£¬MAs=75£©ÊÔÑùÖÐAs2O5µÄÖÊÁ¿ÊÇ0.115g£®

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

¾«Ó¢¼Ò½ÌÍø