ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿Ä³»¯Ñ§Ì½¾¿Ð¡×éµÄͬѧ׼±¸ÓÃ̼ËáÄÆÈÜÒºÓëʯ»ÒË®·´Ó¦À´ÖÆÈ¡ÇâÑõ»¯ÄÆ£¬²¢¶Ô±ÈNaOHÓëCa(OH)2µÄÈܽâ¶È¡£
£¨1£©ÊµÑé¹ý³Ì£º
¢Ù°Ñʯ»ÒË®¼ÓÈëÊ¢ÓÐ̼ËáÄÆÈÜÒºµÄÉÕ±Öгä·Ö½Á°è£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__
¢ÚÔÙͨ¹ý__²Ù×÷£¬µÃµ½ÎÞÉ«ÂËÒº£®
¢Û½«ÎÞÉ«ÂËҺͨ¹ý__²Ù×÷£¬µÃµ½°×É«¹ÌÌ壮
£¨2£©²ÂÏ룺µÃµ½µÄ°×É«¹ÌÌåÊÇ´¿ÇâÑõ»¯ÄÆÂð£¿Èýλͬѧ·Ö±ð×÷³öÒÔϲÂÏ룺
СÁ᣺ÊÇ´¿¾»µÄÇâÑõ»¯ÄÆ£¡
¼Ñ¼Ñ£º¿ÉÄÜ»¹º¬ÓÐ̼ËáÄÆ£¡
Сǿ£ºÒ²¿ÉÄܺ¬ÓÐ__£®
£¨3£©ÑéÖ¤£ºÎÒÀ´ÑéÖ¤¼Ñ¼ÑµÄ²ÂÏ룮
ʵÑé²Ù×÷ | ʵÑéÏÖÏó | ʵÑé½áÂÛ |
È¡ÉÙÁ¿°×É«¹ÌÌåÅä³ÉÈÜÒº£¬¼ÓÈë×ãÁ¿__ | __ | __ |
ϱíÊÇCa(OH)2ºÍNaOHµÄÈܽâ¶ÈÊý¾Ý¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
ζÈ/¡æ | 0 | 20 | 40 | 60 | 80 | 100 | |
Èܽâ¶È/g | Ca(OH) | 0.19 | 0.17 | 0.14 | 0.12 | 0.09 | 0.08 |
NaOH | 31 | 91 | 111 | 129 | 313 | 336 |
£¨4£©ÒÀ¾ÝÉϱíÊý¾Ý£¬»æÖÆCa(OH)2ºÍNaOHµÄÈܽâ¶ÈÇúÏߣ¬ÏÂͼÖÐÄܱíʾNaOHÈܽâ¶ÈÇúÏßµÄÊÇ_________£¨ÌîA»òB£©¡£
£¨5£©ÒªÏë°Ñһƿ½Ó½ü±¥ºÍµÄCa(OH)2ÈÜÒº±ä³É±¥ºÍÈÜÒº£¬¾ßÌå´ëÊ©ÓУº
¢Ù¼ÓÈëÇâÑõ»¯¸Æ£¬¢ÚÉý¸ßζȣ¬¢Û½µµÍζȣ¬¢Ü¼ÓÈëË®£¬¢ÝÕô·¢Ë®ºóÔÙ»Ö¸´µ½Ôζȣ¬¢Þ¼ÓÈëÉúʯ»Ò¡£ ÆäÖдëÊ©ÕýÈ·µÄÊÇ_____¡£
A£®¢Ú¢Ü¢Þ B£®¢Û¢Ü C£®¢Ù¢Û¢Ý¢Þ D£®¢Ù¢Ú¢Ý¢Þ
£¨6£©ÏÖÓÐ20¡æʱCa(OH)2µÄ±¥ºÍÈÜÒº£¨¼×ÈÜÒº£©£¬ÏòÆäÖмÓÈëÒ»¶¨Á¿CaOºóµÃµ½µÄÈÜÒº£¨ÒÒÈÜÒº£©£¬´ËʱÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý ÒÒ________¼×£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£
¡¾´ð°¸¡¿Ca(OH)2+Na2CO3 = 2NaOH+CaCO3¡ý ¹ýÂË Õô·¢ ÇâÑõ»¯¸Æ Ï¡ÑÎËá ÆøÅÝ º¬ÓÐ̼ËáÄÆ A D =
¡¾½âÎö¡¿
£¨1£©¢Ùʯ»ÒË®ÖеÄÇâÑõ»¯¸ÆÓë̼ËáÄÆ·´Ó¦Éú³ÉÇâÑõ»¯ÄƺÍ̼Ëá¸Æ³Áµí£¬¹Ê»¯Ñ§·½³ÌʽдΪCa(OH)2+Na2CO3 = 2NaOH+CaCO3¡ý¡£
¢Ú̼Ëá¸ÆÊDz»ÈÜÐÔµÄÎïÖÊ£¬¿ÉÑ¡ÓùýÂ˽«Ì¼Ëá¸ÆÓëÒºÌå·ÖÀ룬¹ÊÌî¹ýÂË¡£
¢Û½«ÎÞÉ«ÂËҺͨ¹ýÕô·¢²Ù×÷£¬µÃµ½°×É«¹ÌÌ壬¹ÊÌîÕô·¢¡£
£¨2£©ÇâÑõ»¯¸ÆÓë̼ËáÄÆ·´Ó¦ÓÐÈýÖÖÇé¿ö£º
¢ÙÇâÑõ»¯¸ÆÓë̼ËáÄÆÇ¡ºÃÍêÈ«·´Ó¦£¬ÈÜÒºÖеÄÈÜÖÊÖ»ÓÐÇâÑõ»¯ÄÆ£»
¢ÚÇâÑõ»¯¸Æ²»×㣬·´Ó¦ºó̼ËáÄÆÓÐÊ£Ó࣬ÈÜÒºÖеÄÈÜÖÊÓÐÇâÑõ»¯ÄƺÍ̼ËáÄÆ£»
¢Û̼ËáÄƲ»×㣬·´Ó¦ºóÇâÑõ»¯¸ÆÓÐÊ£Ó࣬ÈÜÒºÖеÄÈÜÖÊÓÐÇâÑõ»¯ÄƺÍÇâÑõ»¯¸Æ£¬¸ù¾ÝÌâÒ⣬¹ÊÌîÇâÑõ»¯¸Æ¡£
£¨3£©Ì¼ËáÄÆÓëËá·´Ó¦Äܷųö¶þÑõ»¯Ì¼ÆøÌ壬ÓÖÒòΪÈÜÒºÖÐÓÐÉú³ÉµÄÇâÑõ»¯ÄÆ£¬ËùÒÔ¼ÓÈëµÄËáÓ¦¹ýÁ¿£¬¹ÊÌîÏ¡ÑÎË᣻
Èç¹ûÈÜÒºÖÐÓÐ̼ËáÄÆ£¬Ì¼ËáÄÆÓë¼ÓÈëµÄÏ¡ÑÎËáÉú³É¶þÑõ»¯Ì¼²úÉúÆøÅÝ£¬¹ÊÌîÓÐÆøÅÝ£»
ÓÐÆøÅݲúÉú£¬ËµÃ÷ÈÜÒºÖÐÓÐ̼ËáÄÆËùÒÔ£¬¹ÊÌÓÐ̼ËáÄÆ¡£
£¨4£©ÓÉͼÖпÉÖªÇâÑõ»¯ÄƵÄÈܽâ¶ÈËæζÈÉý¸ß¶ø±ä´ó£¬±íÇâÑõ»¯ÄƵÄÈܽâ¶ÈÇúÏßµÄÊÇA£¬¹ÊÑ¡A¡£
£¨5£©¢Ù¼ÓÈëÇâÑõ»¯¸Æ¿ÉÒÔʹ½Ó½ü±¥ºÍµÄCa(OH)2ÈÜÒº±ä³É±¥ºÍÈÜÒº£¬Ñ¡ÏîÕýÈ·£»
¢ÚÉý¸ßζȣ¬ÇâÑõ»¯¸ÆµÄÈܽâ¶È±äС£¬¿ÉÒÔʹ½Ó½ü±¥ºÍµÄCa(OH)2ÈÜÒº±ä³É±¥ºÍÈÜÒº£¬Ñ¡ÏîÕýÈ·£»
¢Û½µµÍζȣ¬ÇâÑõ»¯¸ÆµÄÈܽâ¶È´ó£¬ÈÜÒº»á±äµÃ¸ü²»±¥ºÍ£¬Ñ¡Ïî´íÎó£»
¢Ü¼ÓÈëË®£¬ÈܼÁÔö¶à£¬ÈÜÒº»á±äµÃ¸ü²»±¥ºÍ£¬Ñ¡Ïî´íÎó£»
¢Ý¼õÉÙÈܼÁ£¬¿ÉÒÔʹ½Ó½ü±¥ºÍµÄCa(OH)2ÈÜÒº±ä³É±¥ºÍÈÜÒº£¬Ñ¡ÏîÕýÈ·£»
ÄÜʵÏÖ¢Þ¼ÓÈëÉúʯ»Ò£¬Éúʯ»Ò¿ÉÒÔÓëË®·´Ó¦Éú³ÉÇâÑõ»¯¸ÆÔö¼ÓÈÜÖÊ£¬Í¬Ê±·´Ó¦ÏûºÄË®£¬¼õÉÙÁËÈÜÒºÖеÄÈܼÁ£¬¿ÉÒÔʹ½Ó½ü±¥ºÍµÄCa(OH)2ÈÜÒº±ä³É±¥ºÍÈÜÒº£¬Ñ¡ÏîÕýÈ·£¬¿ÉÒÔʹ½Ó½ü±¥ºÍµÄCa(OH)2ÈÜÒº±ä³É±¥ºÍÈÜÒºµÄ·½·¨ÊǢ٢ڢݢޣ¬¹ÊÑ¡D¡£
£¨6£©Ïò20¡æʱCa(OH)2µÄ±¥ºÍÈÜÒº£¬¼ÓÈëÒ»¶¨Á¿CaOºó£¬CaOÓëË®·´Ó¦Éú³ÉCa(OH)2 £¬Í¬Ê±ÏûºÄÈÜÒºÖеÄÒ»²¿·ÖË®£¬Ï൱ÓÚ¸ø±¥ºÍÈÜÒºÕô·¢ÈܼÁ£¬Îö³ö¾§Ì壬ʣÓàÈÜÒºÈÔΪ¸Ãζȵı¥ºÍÈÜÒº£¬ÈÜÖÊÖÊÁ¿·ÖÊý±£³Ö²»±ä£¬¹ÊÑ¡£½¡£