ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿£¨6·Ö£©Ï±íÊÇCa(OH)2ºÍNaOHµÄÈܽâ¶ÈÊý¾Ý¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

ζÈ/¡æ

0

20

40

60

80

100

Èܽâ¶È/g

Ca(OH)2

0.19

0.17

0.14[

0.12[

0.09

0.08

NaOH

31

91

111

129

313

336

£¨1£©ÒÀ¾ÝÉϱíÊý¾Ý£¬»æÖÆCa(OH)2ºÍNaOHµÄÈܽâ¶ÈÇúÏߣ¬ÏÂͼÖÐÄܱíʾNaOHÈܽâ¶ÈÇúÏßµÄÊÇ £¨ÌîA»òB£©¡£

£¨2£©ÒªÏë°Ñһƿ½Ó½ü±¥ºÍµÄCa(OH)2ÈÜÒº±ä³É±¥ºÍÈÜÒº£¬¾ßÌå´ëÊ©ÓУº

¢Ù¼ÓÈëÇâÑõ»¯¸Æ£¬¢ÚÉý¸ßζȣ¬¢Û½µµÍζȣ¬¢Ü¼ÓÈëË®£¬¢ÝÕô·¢Ë®ºóÔÙ»Ö¸´µ½Ô­Î¶ȣ¬¢Þ¼ÓÈëÉúʯ»Ò¡£

ÆäÖдëÊ©ÕýÈ·µÄÊÇ ¡£

A£®¢Ú¢Ü¢Þ B£®¢Û¢Ü C£®¢Ù¢Û¢Ý¢Þ D£®¢Ù¢Ú¢Ý¢Þ

£¨3£©20¡æʱ£¬191g ±¥ºÍNaOHÈÜÒº£¬Õô·¢10gË®ºó£¬ÔÙ½µÎµ½20¡æ£¬¿ÉÎö³öNaOH¾§ÌåµÄÖÊÁ¿Îª ¡£

£¨4£©ÏÖÓÐ20¡æʱCa(OH)2µÄ±¥ºÍÈÜÒº£¨¼×ÈÜÒº£©£¬ÏòÆäÖмÓÈëÒ»¶¨Á¿CaOºóµÃµ½µÄÈÜÒº£¨ÒÒÈÜÒº£©£¬´ËʱÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÒÒ ¼×£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£

£¨5£©ÏÖÓÐ60¡æʱº¬Ca(OH)2ºÍNaOHÁ½ÖÖÈÜÖʵı¥ºÍÈÜÒº£¬ÈôÒªµÃµ½½Ï´¿¾»µÄNaOH¾§Ì壬Ӧ²ÉÈ¡µÄÎïÀí·½·¨ÊÇ ¡£

£¨6£©20¡æʱ£¬Óû²â¶¨NaOHÈÜÒºµÄpH£¬ÈôÏȽ«pHÊÔÖ½ÓÃÕôÁóË®Èóʪ£¬ÔÙ½øÐвⶨ£¬ÔòËù²âÈÜÒºµÄpH £¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»ÊÜÓ°Ï족£©¡£

¡¾´ð°¸¡¿£¨1£©A £» £¨2£©D £» £¨3£©9.1 g £¨Â©Ð´µ¥Î»²»µÃ·Ö£© £» £¨4£©£¼ £»

Óйأ¨5£©½µÎ½ᾧ£¬¹ýÂË £» £¨6£©Æ«Ð¡¡£

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©ÇâÑõ»¯ÄƵÄÈܽâ¶ÈËæζȵÄÉý¸ß¶øÔö¼Ó£¬ËùÒÔÄܱíʾÇâÑõ»¯ÄÆÈܽâ¶ÈÇúÏßµÄÊÇͼA£»

£¨2£©ÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæζȵÄÉý¸ß¶ø½µµÍ£¬ËùÒÔҪʹÇâÑõ»¯¸ÆµÄ²»±¥ºÍÈÜÒº±äΪ±¥ºÍÈÜÒº£¬¿É²ÉÈ¡µÄ´ëʩΪ£º¼ÓÈëÈÜÖÊÇâÑõ»¯¸Æ»òÕô·¢Ë®·Ö»òÉý¸ßζȣ»¼ÓÈë̼»¯¸Æʱ£¬Ñõ»¯¸ÆºÍË®·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬ÈܼÁµÄÖÊÁ¿¼õÉÙ£¬²¢ÇÒζÈÉý¸ß£¬ËùÒÔÒ²¿ÉÒÔʹ²»±¥ºÍµÄÇâÑõ»¯¸ÆÈÜÒº±äΪ±¥ºÍÈÜÒº£¬¹Ê´ð°¸ÎªD£»

£¨3£©20¡æʱ£¬ÇâÑõ»¯ÄƵÄÈܽâ¶ÈÊÇ91g ,¼´ 100g×î¶àÈܽâ91gÇâÑõ»¯ÄÆ£¬ÔòÕô·¢10gË®£¬Îö³öÇâÑõ»¯ÄƵÄÖÊÁ¿Îª9.1g£»

£¨4£©ÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæζȵÄÉý¸ß¶ø½µµÍ£¬Ñõ»¯¸ÆÈÜÓÚË®Éú³ÉÇâÑõ»¯¸ÆµÄ·´Ó¦·Å³ö´óÁ¿µÄÈÈ£¬ËùÒÔÏò20¡æʱCa(OH)2µÄ±¥ºÍÈÜÒº£¨¼×ÈÜÒº£©ÖмÓÈëÒ»¶¨Á¿CaOºóµÃµ½µÄÈÜÒº£¨ÒÒÈÜÒº£©£¬´ËʱÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýÊÇ£ºÒÒ<¼×£»

£¨5£©ÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæζȵÄÉý¸ß¶ø½µµÍ£¬ÇâÑõ»¯ÄƵÄÈܽâ¶ÈËæζȵÄÉý¸ß¶øÔö´ó£¬²¢ÇÒÇâÑõ»¯¸ÆµÄÈܽâ¶ÈºÜС£¬ËùÒÔ¿ÉÓýµÎ£¬ÀäÈ´Èȱ¥ºÍÈÜÒºµÄ·½·¨Ï¡ÇâÑõ»¯ÄƽᾧÎö³ö£¬È»ºó¹ýÂ˼´¿ÉµÃµ½½ÏΪ´¿¾»µÄÇâÑõ»¯ÄÆ£»

£¨6£©20¡æʱ£¬Óû²â¶¨NaOHÈÜÒºµÄpH£¬ÈôÏȽ«pHÊÔÖ½ÓÃÕôÁóË®Èóʪ£¬ÔÙ½øÐвⶨ£¬ÔòËù²âÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý¼õС£¬¼îÐÔ¼õÈõ£¬pHֵƫС¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿¸ÖÌúÊÇÖØÒªµÄ½ðÊô²ÄÁÏ£¬ÔÚ¹¤Å©ÒµÉú²úºÍÉú»îÖÐÓ¦Ó÷dz£¹ã·º¡£

¢ñ£®ÌúµÄÓ¦ÓÃ

£¨1£©ÏÂÁÐÌúÖÆÆ·µÄÀûÓÃÓë½ðÊôµ¼ÈÈÐÔÓйصÄÊÇ________________£¨Ìî×Öĸ£¬ÏÂͬ£©¡£

A£®Ìúǯ B£®Ìú¹ø C£®µ¶¾ß

£¨2£©¡°ÄÉÃצÁ-Fe·Û¡±¿ÉÒÔÓÃÓÚʳƷ±£ÏÊ£¬³Æ֮Ϊ¡°Ë«Îü¼Á¡±£¬ÒòΪËüÄÜÎüÊÕ¿ÕÆøÖеÄ________________¡£

¢ò£®ÌúµÄÒ±Á¶

¹¤ÒµÁ¶ÌúµÄÔ­ÀíÊǸßÎÂÏÂÓÃCO×÷»¹Ô­¼Á£¬½«Ìú´ÓÑõ»¯ÎïÖл¹Ô­³öÀ´¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öÒÔ³àÌú¿óΪԭÁÏ£¬ÔÚ¸ßÎÂÏÂÁ¶ÌúµÄ»¯Ñ§·½³Ìʽ£º___________________________£»

£¨2£©¸ß¯Á¶ÌúÖУ¬½¹Ì¿µÄ×÷ÓóýÁË¿ÉÒÔÉú³ÉÒ»Ñõ»¯Ì¼Í⣬»¹ÄÜ_____________________£»

¢ó£®ÊµÑé̽¾¿Á¶ÌúÔ­Àí

ij»¯Ñ§ÐËȤС×éÔÚʵÑéÊÒÖÐÄ£Ä⹤ҵÁ¶ÌúµÄÔ­Àí£¬²¢Ì½¾¿COÓëFe2O3·´Ó¦ºóµÄ²úÎï¡£

ͨ¹ý²éÔÄ×ÊÁÏÖªµÀ£º¢Ù²ÝËᣨH2C2O4£©¹ÌÌåÓëŨÁòËá»ìºÏ¼ÓÈÈ»á²úÉúÒ»Ñõ»¯Ì¼£¬·´Ó¦·½³ÌʽΪ£ºH2C2O4 CO¡ü+ CO2¡ü+ H2O¡£

¢ÚNaOHÈÜÒº¿ÉÒÔÎüÊÕ¶þÑõ»¯Ì¼£¬·´Ó¦·½³ÌʽΪ£º2NaOH+CO2=Na2CO3+H2O

¢Û³£ÎÂÏ£¬Ca(OH)2΢ÈÜÓÚË®£»ÓÚÊÇËûÉè¼ÆÁËÏÂͼµÄʵÑé×°Ö㬽áºÏ×°ÖûشðÏÂÁÐÎÊÌ⣺

£¨1£©Í¼AÊÇÓùÌÌå²ÝËáºÍŨÁòËáÖÆÈ¡COµÄ·´Ó¦×°Öã¬ÄãÈÏΪӦѡÔñÏÂͼÖеÄ_________£¨Ìî×°ÖñàºÅ£©£»

£¨2£©Í¼ÖÐ×°ÖÃC¡¢DµÄ×÷Ó÷ֱðÊÇ ____________¡¢_______________£»

£¨3£©ÎªÁËÖ¤Ã÷²úÎïÖÐÓжþÑõ»¯Ì¼£¬×°ÖÃFÖеÄÊÔ¼ÁÓ¦¸ÃÊÇÉÙÁ¿µÄ_____________£¨ÌîÊÔ¼ÁÃû³Æ£©£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ______________________£»

£¨4£©¸Ã×°ÖÃÉè¼ÆÓÐÒ»¸öÃ÷ÏÔȱÏÝ£¬ÄãÈÏΪÊÇ______________________________¡£

£¨5£©ÊµÑéʱ׼ȷ³ÆÈ¡Ò»¶¨ÖÊÁ¿´¿¾»µÄFe2O3¹ÌÌå°´ÉÏͼ½øÐÐʵÑ飬µ±EÖйÌÌåÈ«²¿±äºÚºó£¬¼ÌÐøͨÈëCOÖ±µ½²£Á§¹ÜÀäÈ´¡£½«µÃµ½ºÚÉ«µÄ¹ÌÌå¼ÓÈëµ½×ãÁ¿µÄÏ¡ÑÎËᣬ·¢ÏÖ¹ÌÌåÈ«²¿Èܽ⣬µ«ÎÞÆøÅݲúÉú¡£

²éÔÄ×ÊÁÏ£ºa. ÌúµÄÑõ»¯ÎïÔÚ×ãÁ¿µÄÏ¡ÑÎËáÖоùÄÜÈ«²¿Èܽ⡣

b. Fe2O3ÓëCO·´Ó¦µÄ¹ÌÌåÉú³ÉÎï¿ÉÄÜÇé¿öÈçÏ£º

¸ù¾Ý¡°ÎÞÆøÅÝ¡±²ÂÏ룬ºÚÉ«·ÛÄ©¿ÉÄÜÊÇ£º¢Ù Fe3O4 £»¢Ú_________ ¢Û ____________¡£

£¨6£©¶¨Á¿·ÖÎö Óõç×ÓÌìƽ³ÆÁ¿µÃ²¿·ÖÊý¾ÝÈçÏ£º

²£Á§¹ÜÖÊÁ¿

²£Á§¹Ü¼°ÆäÖйÌÌåµÄÖÊÁ¿

×°ÖÃF¼°ÆäÖÐÎïÖʵÄ×ÜÖÊÁ¿

·´Ó¦Ç°

28.20 g

33.00 g

300.0 g

·´Ó¦ºó

32.84 g

300.4 g

¸ù¾ÝÉÏÊöÊý¾Ý£¬´¿¾»µÄFe2O3¹ÌÌåÖÊÁ¿Îª_____________g£¬ÇëÑ¡ÔñÓÐЧµÄÊý¾Ý£¬Íƶϳö·´Ó¦ºóºÚÉ«¹ÌÌåµÄ³É·Ö£¨Ð´³ö¼ÆËã¹ý³Ì£©__________________________¡£

¸ù¾ÝÍƶϽá¹û£¬Ð´³ö¸ÃʵÑéÖв£Á§¹ÜÄÚ·¢ÉúµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º__________________________¡£

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

¾«Ó¢¼Ò½ÌÍø