ÌâÄ¿ÄÚÈÝ
ÏÂÁÐÎïÖÊ»ìºÏ£¬Äܵõ½Èé×ÇÒºµÄÊÇ( )
A. ÉÙÁ¿¾Æ¾«·ÅÈëË®ÖкóÕñµ´
B. ÉÙÁ¿ÇâÑõ»¯ÄÆ·ÅÈëË®ÖкóÕñµ´
C. ÉÙÁ¿ÆûÓÍ·ÅÈëË®ÖкóÕñµ´
D. ÉÙÁ¿¶þÑõ»¯Ì¼Í¨Èë³ÎÇåʯ»ÒË®ÖÐ
C ¡¾½âÎö¡¿A¡¢¾Æ¾«ÓëË®ÄÜÒÔÈÎÒâ±ÈÀý»¥ÈÜ£¬¹ÊÐγɵÄÊÇÈÜÒº£»B¡¢ÇâÑõ»¯ÄÆÈÜÓÚË®£¬ÐγɵÄÊÇÈÜÒº£»C¡¢ÆûÓÍÊÇÒºÌå²¢ÇÒ²»ÈÜÓÚË®£¬ËùÒԵõ½µÄÊÇÈé×ÇÒº£»D¡¢¶þÑõ»¯Ì¼ºÍÇâÑõ»¯¸Æ·´Ó¦Éú³É̼Ëá¸ÆºÍË®£¬Ì¼Ëá¸Æ¹ÌÌå²»ÈÜÓÚË®ËùÒԵõ½µÄÊÇÐü×ÇÒº¡£¹ÊÑ¡C¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ΪÁ˲ⶨij´¿¼îÑùÆ·(ÔÓÖÊΪÂÈ»¯ÄÆ)ÖÐ̼ËáÄÆµÄÖÊÁ¿·ÖÊýÊýС½Í¬Ñ§¶ÔÑùÆ·½øÐÐÁËÈçÏÂͼ¶¨Á¿ÊµÑé:
ʵʼ²½Öè | ¢Ù³ÆÈ¡ÉÕ±µÄÖÊÁ¿ | ¢Ú½«ÊÊÁ¿ÑÎËá¼ÓÈëÉÕ±Öв¢³ÆÖØ | ¢Û³ÆÈ¡ÉÙÁ¿´¿¼îÑùÆ·¼ÓÈëÉÕ±ÖУ¬Ê¹Ö®Óë¹ýÁ¿Ï¡ÑÎËá·´Ó¦ | ¢Ü´ý·´Ó¦ÍêÈ«ºó£¬³ÆÖØ |
ʵÑéͼʾ |
| |||
ʵÑéÊý¾Ý | ÉÕ±µÄÖÊÁ¿Îª50.0 | ÉÕ±ºÍÑÎËáµÄÖÊÁ¿Îª£º100.0 | ÑùÆ·µÄÖÊÁ¿11.0g | ÉÕ±ºÍÆäÖлìºÏÎïµÄÖÊÁ¿106.6g |
£¨1£©´¿¼îµÄË®ÈÜÒºÏÔ__________(Ìî¡°ËáÐÔ¡±¡¢¡°¼îÐÔ¡±»ò¡°ÖÐÐÔ¡±Ö®Ò»)
£¨2£©¸ÃʵÑéÖÐÉú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿Îª___________;
£¨3£©Í¨¹ý»¯Ñ§·½³Ìʽ¼ÆËãÑùÆ·ÖÐ̼ËáÄÆµÄÖÊÁ¿·ÖÊý(¼ÆËã½á¹û¾«È·µ½0.1g)______;
¼îÐÔ 4.4¿Ë 96.4% ¡¾½âÎö¡¿(1) £¨1£©´¿¼îµÄË®ÈÜÒºÏÔ¼îÐÔ£»£¨2£©¸ù¾ÝÖÊÁ¿Êغ㶨Âɵ㺷´Ó¦·Å³öCO2µÄÖÊÁ¿=(100.0g+11.0g)?106.6g=4.4g£» (2)Éè11.0g̼ËáÄÆÑùÆ·Öк¬Na2CO3ÖÊÁ¿Îªx Na2CO3+2HCl¨TNaCl+CO2¡ü+H2O 106 44 x 4.4g ...