£¨2008?ÍþÐÅÏØ£©Çë×ÐϸÔĶÁÏÂÁÐijÖÖ¼Ó¸ÆʳÑΰü×°´üÉϵIJ¿·ÖÎÄ×Ö£¬²¢Ìî¿Õ»Ø´ðÎÊÌ⣮
ÅäÁÏ±í£ºÂÈ»¯ÄÆ¡¢Ê³ÓÃ̼Ëá¸Æ¡¢µâËá¼Ø
¾»º¬Á¿£º500g
³É·Ö±í£ºÂÈ»¯ÄÆ¡Ý88%
                ¸Æ£¨ÒÔCa¼Æ£©£¨0.5¡«1.3£©%
                µâ£¨ÒÔI¼Æ£©£¨20¡«50£©mg/kg
£¨1£©Èô¼ìÑé´ËʳÑÎÖÐÊÇ·ñº¬Ì¼Ëá¸Æ£¬ÔÚ¼ÒÍ¥³ø·¿ÖпÉÑ¡ÓÃ______£¬½øÐмìÑ飮
£¨2£©ÎªÁ˲ⶨ´ËÑÎÖÐ̼Ëá¸ÆµÄº¬Á¿£¬È¡10gÕâÖÖÑÎÈÜÓÚÊÊÁ¿Ë®£¬¼ÓÈë×ãÁ¿µÄÑÎËᣬÉú³É0.132g¶þÑõ»¯Ì¼£¨¼ÙÉèÎÞËðʧ£©£¬Ôò´Ë¼Ó¸ÆʳÑÎÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ¶àÉÙ£¿
¡¾´ð°¸¡¿·ÖÎö£º£¨1£©Ì¼Ëá¸Æº¬ÓÐ̼Ëá¸ù£¬ÓöËáÄܷųö¶þÑõ»¯Ì¼ÆøÌ壻¼ÒÍ¥³ø·¿ÖÐʳ´×º¬ÓÐÒ»¶¨Á¿µÄ´×Ëᣬ´×ËáÄÜÓë̼ËáÑη´Ó¦Éú³É¶þÑõ»¯Ì¼£»
£¨2£©´Ë¼Ó¸ÆʳÑÎÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý=×100%£»¸ù¾Ý¼Ó¸ÆʳÑεÄÅäÁϱí¿ÉÖª£¬Ïò¸ÃÑÎÖмÓÈëÑÎËᣬÑÎËáÄÜÓë¸ÃÑÎÖеÄ̼Ëá¸Æ·´Ó¦·Å³öÆøÌå¶þÑõ»¯Ì¼£»Òò´Ë£¬¸ù¾Ý·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬ÓÉÉú³É¶þÑõ»¯Ì¼µÄÖÊÁ¿¿É¼ÆËã³öËùÈ¡ÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿£®
½â´ð£º½â£º£¨1£©¸ù¾Ý̼ËáÑεļìÑé·½·¨£¬¼ìÑé̼Ëá¸ÆÐèҪѡȡËᣬÔÚ¼ÒÍ¥³ø·¿ÀïʳËẬÓд×Ëᣬ¿ÉÓë̼Ëá¸Æ·´Ó¦²úÉúÆøÌå¶þÑõ»¯Ì¼£¬Òò´Ë¿ÉÓô׼ìÑé¸ÃÑÎÖÐÊÇ·ñº¬ÓÐ̼Ëá¸Æ£»
¹Ê´ð°¸Îª£º´×£¨»òʳ´×£©£»
£¨2£©Éè10g¸ÃÑÎÑùÆ·Öк¬Ì¼Ëá¸ÆµÄÖÊÁ¿Îªx
CaCO3+2HCl=CaCl2+H2O+CO2¡ü
100                  44
 x                  0.132g
  x=0.3g
10g¸ÃÑÎÑùÆ·Öк¬Ì¼Ëá¸ÆµÄÖÊÁ¿·ÖÊý=×100%=3%
´ð£º´Ë¼Ó¸ÆʳÑÎÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ3%£®
µãÆÀ£ºCO32-¼ìÑé·½·¨£ºÈ¡ÑùÆ·£¬µÎ¼ÓÏ¡ÑÎËᣬ²úÉúÆøÌåͨÈë³ÎÇåʯ»ÒË®£¬ÓÐÆøÌå²úÉúÇÒ³ÎÇåʯ»ÒË®±ä»ë×Ç£¬ËµÃ÷ÑùÆ·Öк¬ÓÐCO32-£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2008?ÍþÐÅÏØ£©[ʵÑé̽¾¿]ÇâÑõ»¯ÄÆÈÜÒºÖÐÄÄÒ»ÖÖÁ£×Ó£¨H2O¡¢K+¡¢OH-£©ÄÜʹָʾ¼Á±äÉ«£®
[ʵÑé²½Öè]
£¨1£©ÔÚµÚÒ»Ö§ÊÔ¹ÜÖмÓÈëÔ¼2mLÕôÁóË®£¬µÎÈ뼸µÎÎÞÉ«·Ó̪ÊÔÒº£¬¹Û²ìÏÖÏó£®
£¨2£©ÔÚµÚ¶þÖ§ÊÔ¹ÜÖмÓÈëÔ¼2mLNaCl£¨pH=7£©ÈÜÒº£¬µÎÈ뼸µÎÎÞÉ«·Ó̪ÊÔÒº£¬¹Û²ìÏÖÏó£®
£¨3£©ÔÚµÚÈýÖ§ÊÔ¹ÜÖмÓÈëÔ¼2mLNaOHÈÜÒº£¬µÎÈ뼸µÎÎÞÉ«·Ó̪ÊÔÒº£¬¹Û²ìÏÖÏó£®
£¨ÏÂÊö¡°ÏÖÏó¡±ÊÇÖ¸£ºÎÞÉ«·Ó̪ÊÔÒºÊÇ·ñ±äºì£©£º
ͨ¹ýÉÏÊöʵÑé̽¾¿£¬»Ø´ðÏÂÁÐÎÊÌ⣺
ʵÑ飨1£©¿´µ½µÄÏÖÏóÊÇ
ÎÞÉ«·Ó̪²»±äÉ«
ÎÞÉ«·Ó̪²»±äÉ«
£¬ÄãµÃµ½µÄ½áÂÛÊÇ
Ë®²»ÄÜʹָʾ¼Á±äÉ«
Ë®²»ÄÜʹָʾ¼Á±äÉ«
£»
ʵÑ飨2£©¿´µ½µÄÏÖÏóÊÇ
ÎÞÉ«·Ó̪²»±äÉ«
ÎÞÉ«·Ó̪²»±äÉ«
£¬ÄãµÃµ½µÄ½áÂÛÊÇ
Na+²»ÄÜʹָʾ¼Á±äÉ«
Na+²»ÄÜʹָʾ¼Á±äÉ«
£»
ʵÑ飨3£©¿´µ½µÄÏÖÏóÊÇ
ÎÞÉ«·Ó̪±äºìÉ«
ÎÞÉ«·Ó̪±äºìÉ«
£¬ÄãµÃµ½µÄ½áÂÛÊÇ
OH-ʹָʾ¼Á±äÉ«
OH-ʹָʾ¼Á±äÉ«
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2008?ÍþÐÅÏØ£©¼ø±ðÏÂÁи÷×éÎïÖÊ£¬Ëù¼ÓÊÔ¼Á²»ÄܴﵽĿµÄµÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2008?ÍþÐÅÏØ£©ÎÒ¹úÓ®µÃÁË2008ÄêµÄ°ÂÔË»á¾Ù°ìȨ£®ÎªÁËÏòÊÀ½çÕ¹ÏÖÒ»¸öеÄÐÎÏó£¬ÍõÏèͬѧÌá³öÁËÏÂÁн¨Ò飺£¨1£©¿ª·¢ÐÂÄÜÔ´£¬¼õÉÙ»¯Ê¯È¼ÁϵÄȼÉÕ£»£¨2£©¿ª·¢Éú²úÎÞ¹¯µç³Ø£»£¨3£©·ÖÀà»ØÊÕÀ¬»ø£»£¨4£©ÌᳫʹÓÃÒ»´ÎÐÔ·¢ÅÝËÜÁϲ;ߺÍËÜÁÏ´ü£»£¨5£©ÌᳫʹÓÃÊÖÅÁ£¬¼õÉٲͽíÖ½µÄʹÓ㻣¨6£©Ìᳫ²»Ê¹Óú¬Á×Ï´Ò·ۣ»£¨7£©Å©ÒµÉÏ´óÁ¿Ê¹ÓÃɱ³æ¼Á¡¢É±¾ú¼ÁµÈÅ©Ò©£¬ÒÔ¼õÉÙ²¡³æº¦£»£¨8£©ÌᳫÉú²úСÅÅÁ¿Æû³µ£»£¨9£©²»ÔÊÐí·ÙÉսոѣ¬·ÀÖ¹ÎÛȾ¿ÕÆø£®´Ó¡°ÂÌÉ«°ÂÔË¡±½Ç¶È·ÖÎö£¬ÄãÈÏΪ¿ÉÒÔ²ÉÄɵÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2008?ÍþÐÅÏØ£©Ï±íÁгöÁ½ÖÖʳƷ²¿·ÖÓªÑø³É·Ö£¬Çë×ÐϸÔĶÁºó£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ã¿100¿ËʳƷÖÐËùº¬ÓªÑø³É·ÖÖÊÁ¿×î´óµÄÊÇ
ʳƷ¼×ÖеÄÌÇÀà
ʳƷ¼×ÖеÄÌÇÀà
£¬±íÖÐËùÁÐÓªÑø³É·ÖÊôÓÚÓлúÎïµÄ×ܹ²ÓÐ
Èý
Èý
À࣮
£¨2£©Ä³Í¬Ñ§×¼±¸µ½Ò°Íâ̽ÏÕ£¬ÔÚÉÏÊöʳƷÖнøÐÐÑ¡Ôñ£¬Ä㽨ÒéËûЯ´øʳƷ
¼×
¼×
£¨Ñ¡Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©£¬Ä㽨ÒéµÄÒÀ¾ÝÊÇ
ÿ100¿ËʳƷÄܲ¹³äµÄÄÜÁ¿ºÍÓªÑø³É·Ö¶¼±ÈÒҸߣ®
ÿ100¿ËʳƷÄܲ¹³äµÄÄÜÁ¿ºÍÓªÑø³É·Ö¶¼±ÈÒҸߣ®
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2007Äê½­ËÕÊ¡ÄϾ©ÊÐÁùÇøÁªºÏ¾ÅÄ꼶Öп¼»¯Ñ§Ä£ÄâÊÔ¾í£¨2£©£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ

£¨2008?ÍþÐÅÏØ£©½«ÏÂÁÐÎïÖÊ·ÅÈëË®ÖУ¬³ä·ÖÕñµ´£¬ÄÜÐγÉÈÜÒºµÄÊÇ£¨ £©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸