ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿¶à²ÊµÄ¡°Ì¼¡±£¬¶à×˵ÄÉú»î£¬ÈÃÎÒÃÇÒ»Æð×ß½ø¡°Ì¼¡±µÄÊÀ½ç¡£
(1)ÌîдÓйغ¬Ì¼ÎïÖʵĶÔÓ¦ÌØÐÔ¡£
ÎïÖÊÓÃ; | ½ð¸ÕʯÇиÁ§ | ʯī×÷µç¼« |
¶ÔÓ¦ÌØÕ÷ | ¢Ù___________________ | ¢Ú______________ |
(2)ÔÚ440¡ãCºÍ¸ßѹÌõ¼þÏ£¬½ðÊôÄÆÓë¶þÑõ»¯Ì¼·´Ó¦ÄÜÉú³É½ð¸Õʯ(C) ºÍ̼ËáÄÆ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________¡£
(3)¡° ̼º£ÃࡱÊÇÒÑÖª×îÇáµÄ¹ÌÌå²ÄÁÏ£¬ÓÉ̼ԪËØ×é³É£¬¾ßÓжà¿×½á¹¹£¬µ¯ÐԺá£Ëü¶ÔʯÓÍÓкÜÇ¿µÄÎü¸½ÄÜÁ¦(²»ÎüË®)£¬½«ÎüÈëµÄʯÓͼ·³öºóÈԿɻָ´Ô×´¡£ÏÂÁйØÓÚ̼º£ÃàµÄ˵·¨ÕýÈ·µÄÊÇ____________(Ìî×ÖĸÐòºÅ)¡£
A ¾ßÓÐÎü¸½ÐÔ
B ¿ÉÖظ´Ê¹ÓÃ
C ¿É´¦Àíº£ÉÏʯÓÍй©.
(4)Һ̬¶þÑõ»¯Ì¼Ãð»ðÆ÷¿ÉÓÃÓÚÆ˾ȵµ°¸×ÊÁÏ·¢ÉúµÄ»ðÔÖ£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÓÐ____________¡£
AҺ̬¶þÑõ»¯Ì¼Æø»¯ºó²»»áÎÛȾµµ°¸×ÊÁÏ
B ¶þÑõ»¯Ì¼¿É¸²¸ÇȼÉÕÎï±íÃ棬¸ô¾ø¿ÕÆø
C Һ̬¶þÑõ»¯Ì¼Æø»¯Ê±ÎüÈÈ£¬½µµÍÁË¿ÉȼÎïµÄ×Å»ðµã
¡¾´ð°¸¡¿Ó²¶È´ó µ¼µçÐÔÁ¼ºÃ 3CO2+4Na2Na2CO3+C ABC AB
¡¾½âÎö¡¿
£¨1£©½ð¸ÕʯµÄÓ²¶È´ó£¬ËùÒÔ¿ÉÓÃÇиÁ§£»Ê¯Ä«¾ßÓÐÁ¼ºÃµÄµ¼µçÐÔ£¬¿ÉÓÃÓÚ×÷µç¼«£»¹ÊÌӲ¶È´ó£»µ¼µçÐÔÁ¼ºÃ£»
£¨2£©¶þÑõ»¯Ì¼ÓëÄÆÔÚ440¡æºÍ¸ßѹÌõ¼þÏ£¬·´Ó¦Éú³É½ð¸ÕʯºÍ̼ËáÄÆ£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º3CO2+4Na2Na2CO3+C£®¹ÊÌ3CO2+4Na2Na2CO3+C£»
£¨3£©Óɶԡ°Ì¼º£ÃࡱÃèÊö¿ÉÖª£¬Æä¾ßÓжà¿×½á¹¹£¬µ¯ÐԺã®Ëü¶ÔʯÓÍ£¨¿É´¦Àíº£ÉÏʯÓÍй©£©ÓкÜÇ¿µÄÎü¸½ÄÜÁ¦£¨¾ßÓÐÎü¸½ÐÔ£©£¬½«ÎüÈëµÄʯÓͼ·³öºóÈԿɻָ´Ô×´£¨¿ÉÖظ´Ê¹Óã©£¬¹Ê´ð°¸Îª£ºABC£»
£¨4£©A¡¢ÒºÌ¬¶þÑõ»¯Ì¼Æø»¯ºó²»»áÎÛȾµµ°¸×ÊÁÏ£¬¹ÊAÕýÈ·£»
B¡¢¶þÑõ»¯Ì¼ÃܶȱȿÕÆø´ó£¬¿É¸²¸ÇÔÚȼÉÕÎï±íÃ棬¸ô¾ø¿ÕÆø£¬´Ó¶øÆðµ½Ãð»ðµÄ×÷Ó㬹ÊBÕýÈ·£»
C¡¢ÒºÌ¬¶þÑõ»¯Ì¼Æø»¯Ê±ÎüÈÈ£¬µ«ÊÇ¿ÉȼÎïµÄ×Å»ðµãÊDz»±äµÄ£¬²»ÄܽµµÍ£¬¹ÊC´íÎó£»
¹ÊÌAB¡£
¡¾ÌâÄ¿¡¿ÒÑÖª20¡ãCʱCa (OH) 2µÄÈܽâ¶ÈΪ0.165g,ÏÖÓÐ20¡ãCʱº¬ÓÐ7.4gÈÜÖʵı¥ºÍ³ÎÇåʯ»ÒË®¡£Çë¼ÆËã:
(1)ÉÏÊö³ÎÇåʯ»ÒË®µÄÖÊÁ¿Îª____________g (¾«È·µ½1g)¡£
(2)ÏòÉÏÊö³ÎÇåʯ»ÒË®ÖÐͨÈëCO2,³ÁµíµÄÖÊÁ¿´ïµ½×î´óֵʱ£¬Í¨ÈëCO2µÄÖÊÁ¿Îª¶àÉÙ?____ (д ³ö¼ÆËã¹ý³Ì)
(3)ÒÑÖª: £¬ Ca (HCO3) 2 Ò×ÈÜÓÚË®¡£ÇëÔÚͼÖл³öͨÈëCO2¹ý³ÌÖгÁµíÖÊÁ¿µÄ±ä»¯ÇúÏß_____¡£
(4)ÁíÈ¡Ò»¶¨Á¿µÄ±¥ºÍ³ÎÇåʯ»ÒË®£¬Í¨ÈëÒ»¶Îʱ¼äµÄCO2,·´Ó¦ÎïÓëÉú³ÉÎïµÄÖÊÁ¿Èçϱí
ÎïÖÊ | Ca (OH) 2 | CO2 | CaCO3 | X | H2O |
ÖÊÁ¿/g | 14.8 | 13.2 | 10 | a | 1.8 |
Ôòa=______________£»·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________________¡£