ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÊµÑé¿ÎÉÏ£¬Í¬Ñ§ÃÇÀûÓóÎÇåʯ»ÒË®¼ìÑé¿ÕÆøÖеĶþÑõ»¯Ì¼£¬ÊµÑéÖз¢ÏÖ¸÷×éʯ»ÒË®»ë×dz̶ȴæÔڽϴó²îÒ죬ÓÚÊÇÉè¼ÆʵÑé̽¾¿Ó°Ïìʯ»ÒË®»ë×dz̶ȵÄÒòËØ¡£

£¨½øÐÐʵÑ飩װÖÃͼÈçÏÂËùʾ¡£

ʵÑé1£ºÈ¡±¥ºÍ³ÎÇåʯ»ÒË®½øÐÐʵÑé¡£

ʵÑéÐòºÅ

1-1

1-2

1-3

1-4

±¥ºÍ³ÎÇåʯ»ÒË®Ìå»ý/mL

2.0

1.5

1.0

0.5

»ºÂýͨÈë2 L¿ÕÆøºóµÄ»ë×dz̶È

²»Ã÷ÏÔ

Ã÷ÏÔ

·Ç³£Ã÷ÏÔ

Ã÷ÏÔ

ʵÑé2£ºÀûÓÃÏò±¥ºÍʯ»ÒË®ÖÐÌí¼ÓÕôÁóË®À´µ÷½Úʯ»ÒˮŨ¶È½øÐÐʵÑ飨ÈÜҺŨ¶È±ä»¯ÒýÆðµÄÈÜÒºÃܶȱ仯ºöÂÔ²»¼Æ£©¡£

ʵÑéÐòºÅ

2-1

2-2

2-3

2-4

±¥ºÍ³ÎÇåʯ»ÒË®Ìå»ý/mL

2.0

1.5

1.0

0.5

¼ÓÈëÕôÁóË®Ìå»ý/mL

0

a

b

c

»ºÂýͨÈë3 L¿ÕÆøºóµÄ»ë×dz̶È

·Ç³£Ã÷ÏÔ

Ã÷ÏÔ

²»Ã÷ÏÔ

ÎÞÏÖÏó

£¨½âÊÍÓë½áÂÛ£©

£¨1£©Ð¡ÊÔ¹ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________¡£

£¨2£©ÊµÑé1µÄÄ¿µÄΪ____________¡£

£¨3£©ÊµÑé2ÖÐa¡¢b¡¢cÈý´¦Êý¾Ý£¬Éè¼ÆºÏÀíµÄÊÇ________¡£

A 0.5 0.5 0.5 B 0.5 1.0 1.5

C 1.0 1.5 2.0 D 1.0 2.0 3.0

£¨4£©ÊµÑé2µÄ½áÂÛÊÇ________¡£

£¨·´Ë¼ÓëÆÀ¼Û£©

£¨5£©Í¨¹ý¶Ô±ÈʵÑé1-1ºÍ2-1£¬¿ÉµÃ³öÓ°Ïìʯ»ÒË®»ë×dz̶ȵÄÁíÒ»ÒòËØÊÇ________£¬ÀíÓÉÊÇ________¡£

¡¾´ð°¸¡¿CO2 + Ca(OH)2 === CaCO3¡ý + H2O ̽¾¿Ê¯»ÒË®Ìå»ý¶Ô»ë×dz̶ȵÄÓ°Ïì¡£ B ÔÚʯ»ÒË®Ìå»ý¡¢Í¨Èë¿ÕÆøÌå»ýÏàͬµÄÌõ¼þÏ£¬Ê¯»ÒˮŨ¶ÈÔ½´ó£¬»ë×dz̶ÈÔ½Ã÷ÏÔ¡£ ͨÈë¿ÕÆøµÄÌå»ý ʵÑé1-1ºÍ2-1ÖУ¬Ê¯»ÒË®Ìå»ýºÍŨ¶ÈÏàͬ£¬Í¨Èë¿ÕÆøµÄÌå»ý²»Í¬£¬»ë×dz̶Ȳ»Í¬¡£

¡¾½âÎö¡¿

£¨1£©³ÎÇåʯ»ÒË®Óë¿ÕÆøÖеĶþÑõ»¯Ì¼·´Ó¦Éú³É̼Ëá¸ÆºÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCO2 + Ca(OH)2 === CaCO3¡ý + H2O£»

£¨2£©ÊµÑé1µÄµÄ±äÁ¿ÊdzÎÇåʯ»ÒË®µÄÌå»ý£¬Ä¿µÄΪ̽¾¿Ê¯»ÒË®Ìå»ý¶Ô»ë×dz̶ȵÄÓ°Ïì¡£

£¨3£©ÊµÑé2ÊÇ̽¾¿±¥ºÍʯ»ÒˮŨ¶È¶Ô»ë×dz̶ȵÄÓ°Ï죬ËùÒÔÓ¦¸Ã±£³Ö±¥ºÍ³ÎÇåʯ»ÒË®Ìå»ý¼ÓÉÏÕôÁóË®Ìå»ýÏàͬ£¬¹ÊÑ¡B¡£

£¨4£©ÊµÑé2µÄ½áÂÛÊÇÔÚʯ»ÒË®Ìå»ý¡¢Í¨Èë¿ÕÆøÌå»ýÏàͬµÄÌõ¼þÏ£¬Ê¯»ÒˮŨ¶ÈÔ½´ó£¬»ë×dz̶ÈÔ½Ã÷ÏÔ¡£

£¨5£©ÊµÑé1-1ºÍ2-1ÖУ¬Ê¯»ÒË®Ìå»ýºÍŨ¶ÈÏàͬ£¬µ«ÊÇʵÑé1ͨÈë¿ÕÆøµÄÌå»ýΪ2L£¬ÊµÑé2ͨÈëµÄͨÈë¿ÕÆøµÄÌå»ýΪ3L£¬¿ÉµÃ³öÓ°Ïìʯ»ÒË®»ë×dz̶ȵÄÁíÒ»ÒòËØÊÇͨÈë¿ÕÆøµÄÌå»ý£¬ÀíÓÉÊÇʵÑé1-1ºÍ2-1ÖУ¬Ê¯»ÒË®Ìå»ýºÍŨ¶ÈÏàͬ£¬Í¨Èë¿ÕÆøµÄÌå»ý²»Í¬£¬»ë×dz̶Ȳ»Í¬¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø