ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©Å©´å»¹ÓÐЩ¼ÒÍ¥Ö±½ÓÓúÓË®»ò³ØÌÁË®»ò¾®Ë®ÁúÍ·×÷ΪÉú»îÒûÓÃË®£¬Ò»°ãÊÇ°ÑË®µ£»Ø¼Òµ¹ÈëË®¸×£¬¼ÓÃ÷·¯Ê¹Ë®³ÎÇ壬¼ÓƯ°×·Û½øÐÐÏû¶¾£¬Öó·ÐÏû³ýÔÝʱӲ¶È£¬Ã÷·¯µÄ»¯Ñ§Ê½Îª    £¬Ð´³öËüÈÜÓÚË®ºóÉú³É½ºÌåµÄÀë×Ó·½³Ìʽ£º                        ¡£Æ¯°×·ÛµÄ³É·ÖÊÇ                   £¬ËüÔÚ×ÔÈ»Ìõ¼þÏÂÄܲúÉú´ÎÂÈËᣬ´ÎÂÈËáµÄÑõ»¯ÐÔºÜÇ¿£¬ÄÜɱËÀ²¡¾ú¡£Ð´³öƯ°×·ÛÔÚ¿ÕÆøÖÐÉú³É´ÎÂÈËáµÄ»¯Ñ§·½³Ìʽ£º                     ¡£Öó·ÐʱÐγÉË®¹¸µÄ³É·ÝÊÇ                            £¨Ìѧʽ£©
KAl(SO4)2¡¤12H2O       Al3++3H2OAl(OH)3(½ºÌå)+3H+ 
CaCl2¡¢Ca(C1O)2     Ca(C1O)2+CO2+2H2O=CaCO3¡ý+2HClO
CaCO3¡¢Mg(OH)2
Ã÷·¯µÄ»¯Ñ§Ê½ÎªKAl(SO4)2¡¤12H2O ¡£Ã÷·¯ÈÜÓÚºó£¬ÂÁÀë×ÓË®½âË®Éú³ÉÇâÑõ»¯ÂÁ½ºÌ壬·½³ÌʽΪAl3++3H2OAl(OH)3(½ºÌå)+3H+¡£Æ¯°×·ÛÊÇ»ìºÏÎº¬ÓÐCaCl2¡¢Ca(C1O)2¡£´ÎÂÈËá¸ÆÄÜÎüÊÕ¿ÕÆøÖеÄË®ºÍCO2£¬Éú³É´ÎÂÈËᣬ·½³ÌʽΪCa(C1O)2+CO2+2H2O=CaCO3¡ý+2HClO¡£Ë®¹¸µÄ³É·ÝÊÇ̼Ëá¸ÆºÍÇâÑõ»¯Ã¾¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨15·Ö£©ÂÈÆøµÄʵÑéÊÒÖÆ·¨ÊÇMnO2ºÍŨÑÎËáÔÚ¼ÓÈÈÌõ¼þÏ·´Ó¦£¬ÓÃKMnO4¡¢KClO3Ñõ»¯Å¨ÑÎËá¿É¿ìËÙÖÆÈ¡ÂÈÆø£®¸ù¾ÝÉÏÊö·´Ó¦Ô­Àí£¬ÓÐÈËÌá³öÄÜ·ñÀûÓÃNa2O2µÄÇ¿Ñõ»¯ÐÔÑõ»¯Å¨ÑÎËáµÃµ½ÂÈÆøÄØ£®Ä³¿ÎÍâС×éÔÚʵÑéÊÒ½øÐÐÁË̽Ë÷ÐÔʵÑ飬Éè¼ÆÁËÈçͼװÖãº

²Ù×÷²½Öè¼°ÏÖÔÚÈçÏ£º
¢Ù×é×°ºÃ×°Ö㬼ì²é×°ÖõÄÆøÃÜÐÔ£¬¼ÓÈëÒ©Æ·£®
¢Ú»ºÂýͨÈëÒ»¶¨Á¿µÄN2ºó£¬½«×°ÖÃEÁ¬½ÓºÃ£¨µ¼¹ÜδÉìÈ뼯ÆøÆ¿ÖУ©£¬ÔÙÏòÔ²µ×ÉÕÆ¿ÖлºÂýµÎ¼ÓŨÑÎËᣬ·´Ó¦¾çÁÒ£¬²úÉú»ÆÂÌÉ«ÆøÌ壮
¢ÛÒ»¶Îʱ¼äºó£¬½«µ¼¹ÜÄ©¶ËÉìÈ뼯ÆøÆ¿ÖÐÊÕ¼¯ÆøÌ壮װÖÃEÖÐÊÕ¼¯µ½ÄÜʹ´ø»ðÐǵÄľÌõ¸´È¼µÄÎÞÉ«ÆøÌ壮
¢Ü·´Ó¦½áÊøºó£¬¹Ø±Õ·ÖҺ©¶·µÄ»îÈû£¬ÔÙͨÈëÒ»¶¨Á¿µÄN2£¬ÖÁ×°ÖÃÖÐÆøÌåÎÞÉ«£®
»Ø´ðÏÂÁÐÎÊÌ⣺
¢Åд³öKClO3Ñõ»¯Å¨ÑÎËáÖÆÈ¡Cl2µÄÀë×Ó·½³Ìʽ                            
¢Æ×°ÖÃBÖÐʪÈóµÄºìÉ«²¼ÌõÍÊÉ«£¬ËµÃ÷AÖÐÓС¡¡¡¡¡¡¡£¨Ìѧʽ£©Éú³É£¬×°ÖÃCÖÐΪʪÈóµÄKI£­µí·ÛÊÔÖ½£¬ÄÜ·ñ½öͨ¹ýÊÔÖ½±äÀ¶Ö¤Ã÷ÉÏÊö½áÂÛ£¬ÇëÓÃÀë×Ó·½³Ìʽ˵Ã÷Ô­Òò¡¡¡¡¡¡
¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
¢Ç×°ÖÃDµÄ×÷ÓÃÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
¢Èд³öÉú³ÉO2µÄ¿ÉÄܵķ´Ó¦·½³Ìʽ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
¢ÉʵÑéÖ¤Ã÷£¬Na2O2ÉõÖÁÄÜÓë¸ÉÔïµÄHCl·´Ó¦Éú³ÉÂÈÆø£¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ¡¡¡¡¡¡¡¡
×ÛºÏÉÏÊöʵÑ飬Çë·ÖÎöʵÑéÊÒ        £¨ÌÄÜ¡±»ò¡±·ñ¡±)ÀûÓÃNa2O2ÓëŨÑÎËá·´Ó¦ÖƱ¸´¿¾»µÄCl2£¬ÀíÓÉÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡
(7·Ö)ÒÑÖª£ºCH3CH2OH+NaBr+H2SO4(Ũ)  CH3CH2Br+NaHSO4 +H2O¡£
ʵÑéÊÒÖƱ¸äåÒÒÍ飨·ÐµãΪ38.4¡æ£©µÄ×°ÖúͲ½ÖèÈçÏ£º

¢Ù°´ÓÒͼËùʾÁ¬½ÓÒÇÆ÷£¬¼ì²é×°ÖõÄÆøÃÜÐÔ£¬È»ºóÏòUÐιܺʹóÉÕ±­Àï¼ÓÈë±ùË®£»¢ÚÔÚÔ²µ×ÉÕÆ¿ÖмÓÈë10mL95£¥ÒÒ´¼¡¢28mLŨÁòËᣬȻºó¼ÓÈëÑÐϸµÄ13gä廯Äƺͼ¸Á£Ëé´ÉƬ£»¢ÛС»ð¼ÓÈÈ£¬Ê¹Æä³ä·Ö·´Ó¦¡£
ÊԻشðÏÂÁÐÎÊÌ⣺
(1)·´Ó¦Ê±Èôζȹý¸ß¿É¿´µ½Óкì×ØÉ«ÆøÌå²úÉú£¬¸ÃÆøÌåµÄ»¯Ñ§Ê½Îª
            ¡£
(2)ΪÁ˸üºÃµÄ¿ØÖÆ·´Ó¦Î¶ȣ¬³ýÓÃͼʾµÄС»ð¼ÓÈÈ£¬¸üºÃµÄ¼ÓÈÈ·½Ê½ÊÇ__________¡£
(3)·´Ó¦½áÊøºó£¬UÐιÜÖдÖÖƵÄäåÒÒÍé³Ê×Ø»ÆÉ«¡£½«UÐιÜÖеĻìºÏÎïµ¹Èë·ÖҺ©¶·ÖУ¬¾²Ö㬴ýÒºÌå·Ö²ãºó£¬·ÖÒº£¬È¡        (Ìî¡°Éϲ㡱»ò¡°Ï²㡱)ÒºÌ塣ΪÁ˳ýÈ¥ÆäÖеÄÔÓÖÊ£¬×îºÃÑ¡ÔñÏÂÁÐÊÔ¼ÁÖеĠ    (ÌîÐòºÅ)¡£
A£®Na2SO3ÈÜÒºB£®H2OC£®NaOHÈÜÒºD£®CCl4
(4)Òª½øÒ»²½ÖƵô¿¾»µÄC2H5Br£¬¿ÉÔÙÓÃˮϴ£¬È»ºó¼ÓÈëÎÞË®CaCl2¸ÉÔÔÙ½øÐР     (Ìî²Ù×÷Ãû³Æ)¡£
(5)ÏÂÁм¸ÏîʵÑé²½Ö裬¿ÉÓÃÓÚ¼ìÑéäåÒÒÍéÖеÄäåÔªËØ£¬ÆäÕýÈ·µÄ²Ù×÷˳ÐòÊÇ£ºÈ¡ÉÙÁ¿äåÒÒÍ飬Ȼºó                  (ÌîÐòºÅ)¡£
¢Ù¼ÓÈÈ   ¢Ú¼ÓÈëAgNO3ÈÜÒº   ¢Û¼ÓÈëÏ¡HNO3Ëữ   ¢Ü¼ÓÈëNaOHÈÜÒº   ¢ÝÀäÈ´

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

¾«Ó¢¼Ò½ÌÍø