ÌâÄ¿ÄÚÈÝ
[»¯Ñ§¡ªÑ¡ÐÞ2£º»¯Ñ§Óë¼¼Êõ]
ÐÅϢʱ´ú¸øÈËÃǵÄÉú»î´øÀ´Á˼«´óµÄ±ãÀû£¬µ«Í¬Ê±Ò²²úÉúÁË´óÁ¿µÄµç×ÓÀ¬»ø¡£Ä³»¯Ñ§ÐËȤС×齫һÅú·ÏÆúµÄÏß·°å¼òµ¥´¦Àíºó£¬µÃµ½ÁËÖ÷Òªº¬Cu¡¢Al¼°ÉÙÁ¿Fe¡¢AuµÈ½ðÊôµÄ»ìºÏÎ²¢Éè¼ÆÁËÈçÏÂÖÆ±¸ÁòËá;§ÌåºÍÁòËáÂÁ¾§ÌåµÄ·Ïߣº
![]()
²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º
![]()
£¨1£©¹ýÂ˲Ù×÷ÖÐÓõ½µÄ²£Á§ÒÇÆ÷ÓÐ___________¡£
£¨2£©Cu¿ÉÈÜÓÚÏ¡ÁòËáÓëH2O2µÄ»ìºÏÈÜÒº£¬ÆäÀë×Ó·½³ÌʽÊÇ____________¡£
£¨3£©ÂËÔüaµÄÖ÷Òª³É·ÖÊÇ_________________¡£
£¨4£©²½Öè¢ÛÖÐXµÄȡֵ·¶Î§ÊÇ____________________¡£
£¨5£©ÎªÁ˲ⶨÁòËá;§ÌåµÄ´¿¶È£¬¸Ã×é¼×ͬѧ׼ȷ³ÆÈ¡4.0gÑùÆ·ÈÜÓÚË®Åä³É100mLÈÜÒº£¬È¡10mLÈÜÒºÓÚ´øÈû×¶ÐÎÆ¿ÖУ¬¼ÓÊÊÁ¿Ë®Ï¡ÊÍ£¬µ÷½ÚÈÜÒºpH£½3¡«4£¬¼ÓÈë¹ýÁ¿µÄKIºÍµí·Ûָʾ¼Á£¬ÓÃ0.1000 mol¡¤L£1Na2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㡣¹²ÏûºÄ14.00mLNa2S2O3±ê×¼ÈÜÒº¡£ÉÏÊö¹ý³ÌÖз´Ó¦µÄÀë×Ó·½³ÌʽÈçÏ£º
2Cu2+£«4I££½2CuI£¨°×É«£©¡ý£«I2 2S2O32££«I2£½2I££«S4O62£
¢ÙÑùÆ·ÖÐÁòËá;§ÌåµÄÖÊÁ¿·ÖÊýΪ____________¡£
¢Ú¸Ã×éÒÒͬѧÌá³öͨ¹ýÖ±½Ó²â¶¨ÑùÆ·ÖÐSO42£µÄÁ¿Ò²¿ÉÇóµÃÁòËá;§ÌåµÄ´¿¶È£¬ÀÏʦÉóºËºóÓèÒÔ·ñ¾ö£¬ÆäÔÒòÊÇ______________¡£
£¨6£©ÇëÄãÉè¼ÆÒ»¸öÓÉÂËÔücµÃµ½Al2£¨SO4£©3¡¤18H2OµÄʵÑé·½°¸___________¡£
Cl2¡¢Æ¯°×Òº(ÓÐЧ³É·ÖΪNaClO)ÔÚÉú²ú¡¢Éú»îÖй㷺ÓÃÓÚɱ¾ú¡¢Ïû¶¾¡£
£¨1£©µç½âNaClÈÜÒºÉú³ÉÂÈÆøµÄ»¯Ñ§·½³ÌʽΪ ¡£
£¨2£©Cl2ÈÜÓÚH2O¡¢NaOHÈÜÒº¼´»ñµÃÂÈË®¡¢Æ¯°×Òº¡£
¢Ù¸ÉÔïµÄÂÈÆø²»ÄÜÆ¯°×ÎïÖÊ£¬µ«ÂÈˮȴÓÐÆ¯°××÷Óã¬ËµÃ÷ÆðƯ°××÷ÓõÄÎïÖÊÊÇ ¡£
¢Ú25¡æ£¬Cl2ÓëH2O¡¢NaOHµÄ·´Ó¦ÈçÏ£º
·´Ó¦¢ñ | Cl2+H2O |
·´Ó¦¢ò | Cl2+2OH- |
²»Ö±½ÓʹÓÃÂÈË®¶øÊ¹ÓÃÆ¯°×Òº×öÏû¶¾¼ÁµÄÔÒòÊÇ ¡£
£¨3£©¼ÒͥʹÓÃÆ¯°×Һʱ£¬²»ÒËÖ±½Ó½Ó´¥ÌúÖÆÆ·£¬Æ¯°×Òº¸¯Ê´ÌúµÄµç¼«·´Ó¦Îª£ºFe-2e-=Fe2+;ClO·¢ÉúµÄµç¼«·´Ó¦Ê½ÊÇ ¡£
£¨4£©Ñо¿Æ¯°×ÒºµÄÎȶ¨ÐÔ¶ÔÆäÉú²úºÍ±£´æÓÐʵ¼ÊÒâÒå.30¡æÊ±£¬pH=11µÄƯ°×ÒºÖÐNaClOµÄÖÊÁ¿°Ù·Öº¬Á¿ËæÊ±¼ä±ä»¯ÈçͼËùʾ£º
![]()
¢Ù·Ö½âËÙÂÊv(¢ñ) v(¢ò)(Ìî¡°>¡±¡°<¡±»ò¡°=¡±)£¬ÔÒòÊÇ ¡£
¢ÚNaClO·Ö½âµÄ»¯Ñ§·½³ÌʽÊÇ ¡£
¢Û4d-8d,¢ñÖÐv(NaClO)= mol/(L¡¤d)(³£ÎÂÏÂÆ¯°×ÒºµÄÃܶÈԼΪ1g/cm3£¬Çұ仯ºöÂÔ²»¼Æ)