ÅäÖÆ4%µÄÇâÑõ»¯ÄÆÈÜÒº200g£¬²¢ÓôËÈÜÒº²â¶¨Ä³ÁòËáÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý£®
£¨1£©ÅäÖÆ200gÖÊÁ¿·ÖÊýΪ4%µÄÇâÑõ»¯ÄÆÈÜÒº£¬ÐèÒªÇâÑõ»¯ÄƹÌÌåµÄÖÊÁ¿Îª    g£¬Ë®µÄÌå»ýΪ    mL£¨Ë®µÄÃܶȽüËÆ¿´×÷1g/cm3£©£®
£¨2£©ÅäÖÆÇâÑõ»¯ÄÆÈÜҺʱ£¬ÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÁ¿Í²¡¢ÉÕ±­¡¢²£Á§°ô¼°    £®ÅäÖƺõÄÇâÑõ»¯ÄÆÈÜÒºÒªÃÜ·â±£´æ£¬ÆäÔ­ÒòÊÇ£º    £¨Ð´»¯Ñ§·½³Ìʽ£©£®
£¨3£©ÓÃÅäµÃµÄÇâÑõ»¯ÄÆÈÜÒºÓë´ý²âÈÜÖÊÖÊÁ¿·ÖÊýµÄÁòËáÈÜÒº·´Ó¦£¬ÊµÑé¹ý³ÌÖÐÈÜÒºµÄpH±ä»¯ÇúÏßÈçͼËùʾ£º
¢ÙÒªµÃµ½´Ë±ä»¯ÇúÏߣ¬Ëù½øÐеIJÙ×÷ÊÇ    £¨Ìî×Öĸ£©£®
A£®½«ÁòËáÈÜÒºµÎ¼Óµ½ÇâÑõ»¯ÄÆÈÜÒºÖÐ
B£®½«ÇâÑõ»¯ÄÆÈÜÒºµÎ¼Óµ½ÁòËáÈÜÒºÖÐ
¢Úbµã¶ÔÓ¦µÄÈÜÒºÖеÄÈÜÖÊΪ    £¨Ð´»¯Ñ§Ê½£©£®
£¨4£©Ç¡ºÃÍêÈ«·´Ó¦Ê±£¬ÏûºÄ20gÇâÑõ»¯ÄÆÈÜÒººÍ25gÁòËáÈÜÒº£¬ÊÔ¼ÆËã¸ÃÁòËáÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý    £®
¡¾´ð°¸¡¿·ÖÎö£º£¨1£©ÀûÓÃÒ»¶¨Á¿ÈÜÒºµÄÖÊÁ¿·ÖÊý£¬¿É¼ÆËãÈÜÒºÖÐÈÜÖÊÖÊÁ¿ÓëÅäÖƸÃÈÜÒºËùÐèҪˮµÄÌå»ý£»
£¨2£©¸ù¾ÝÅäÖƸÃÇâÑõ»¯ÄÆÈÜÒº¹ý³Ì¼ÆËã-³ÆÁ¿-ÈܽâÖÐËùÐèҪʹÓõÄÒÇÆ÷£¬ÅжÏδд³öµÄÒÇÆ÷£»¸ù¾ÝÇâÑõ»¯ÄÆÈÜÒºÄÜÎüÊÕ¶þÑõ»¯Ì¼µÄÐÔÖÊ£¬½âÊÍÃÜ·â±£´æÇâÑõ»¯ÄÆÈÜÒºµÄÔ­Òò£»
£¨3£©¢ÙÇúÏßËùʾ£¬·´Ó¦ºóÈÜÒºµÄËá¼î¶ÈÖð½¥Ôö´ó£¬ÀûÓÃÁòËáÈÜÒºËá¼î¶ÈСÓÚ7¶øÇâÑõ»¯ÄÆÈÜÒºËá¼î¶È´óÓÚ7£¬ÅжÏËù½øÐеÄÓëͼʾ¶ÔÓ¦µÄ²Ù×÷£»
¢Ú¸ù¾ÝbµÄÈÜÒºµÄËá¼î¶ÈÅжÏÁ½ÈÜÒºµÄ·´Ó¦Çé¿ö£¬ÅжÏÈÜÒºÖÐÈÜÖÊ£»
£¨4£©¸ù¾Ý·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬ÓÉÇ¡ºÃÍêÈ«·´Ó¦ÏûºÄÇâÑõ»¯ÄƵÄÖÊÁ¿£¬¼ÆËãËùÏûºÄ25gÁòËáÈÜÒºÖÐÁòËáµÄÖÊÁ¿£¬ÇóµÃ¸ÃÁòËáÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®
½â´ð£º½â£º
£¨1£©ÅäÖÆ200gÖÊÁ¿·ÖÊýΪ4%µÄÇâÑõ»¯ÄÆÈÜÒº£¬ÐèÒªÇâÑõ»¯ÄƹÌÌåµÄÖÊÁ¿=200g×4%=8g£¬ÐèҪˮµÄÖÊÁ¿=200g-8g=192gºÏ192mL£»
£¨2£©Á¿È¡Ë®Ê±ÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÁ¿Í²¡¢½ºÍ·µÎ¹Ü£¬ÈܽâʱÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô£¬Òò´ËȱÉÙ½ºÍ·µÎ¹Ü£»ÇâÑõ»¯ÄÆÄÜÎüÊÕ¿ÕÆøÖжþÑõ»¯Ì¼¶øÉú³É̼ËáÄƺÍË®£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCO2+2NaOH¨TNa2CO3+H2O£»
£¨3£©¢ÙÇúÏßËùʾ£¬·´Ó¦ºóÈÜÒºµÄËá¼î¶ÈÖð½¥Ôö´ó£¬ËµÃ÷´ËʵÑéÊÇ°ÑÇâÑõ»¯ÄÆÈÜÒºÖðµÎ¼ÓÈëÁòËáÈÜÒºÖУ¬¶øʹÈÜÒºµÄËá¼î¶ÈÖð½¥Ôö´ó£»¹ÊΪB£»
¢ÚbµãÈÜÒºËá¼î¶È´óÓÚ7£¬ËµÃ÷µÎ¼ÓµÄÇâÑõ»¯ÄƹýÁ¿£¬Òò´ËÈÜҺΪÁòËáÄÆÓëÇâÑõ»¯ÄƵĻìºÏÈÜÒº£¬ÆäÈÜÖÊΪNa2SO4 ¡¢NaOH£»
£¨4£©20gÖÊÁ¿·ÖÊýΪ4%µÄÇâÑõ»¯ÄÆÈÜÒºÖÐËùº¬Á¿ÇâÑõ»¯ÄƵÄÖÊÁ¿=20g×4%=0.8g
Éè·´Ó¦ÏûºÄÁòËáµÄÖÊÁ¿Îªx
H2SO4+2NaOH¨TNa2SO4+2H2O
98     80
x      0.8g
=  x=0.98g
¸ÃÁòËáÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý=×100%=3.92%
¹Ê´ð°¸Îª£º
£¨1£©8£»192£»£¨2£©½ºÍ·µÎ¹Ü£»CO2+2NaOH¨TNa2CO3+H2O£»£¨3£©¢ÙB£»¢ÚNa2SO4 ¡¢NaOH£»£¨4£©3.92%£®
µãÆÀ£ºËá¼îÖкͿɸıäÈÜÒºµÄËá¼î¶È£¬ÏòËáÈÜÒºÖеμӼîÈÜҺʱ£¬ÈÜÒºµÄËá¼î¶ÈÖð½¥±ä´óÖÁµÈÓÚ7£¬µÎ¼Ó¹ýÁ¿Ôò´óÓÚ7£»Ïò¼îÈÜÒºÖеμÓËáʱ£¬ÈÜÒºËá¼î¶ÈÖð½¥¼õСÖÁµÈÓÚ7£¬Ëá¹ýÁ¿Á¿Ð¡ÓÚ7£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

25¡¢ÊµÑéÊÒÓûÅäÖÆ1000gÈÜÖÊÖÊÁ¿·ÖÊýΪ4%µÄÇâÑõ»¯ÄÆÈÜÒº£®ÅäÖƲ½ÖèΪ£º
£¨1£©¼ÆË㣺ÐèÇâÑõ»¯ÄƹÌÌå
40
g£¬Ë®
960
mL£¨Ë®µÄÃܶÈÊÇ1g/cm3£©£®
£¨2£©³ÆÁ¿£ºÓÃÖÊÁ¿Îª23.1gµÄÉÕ±­×÷³ÆÁ¿ÈÝÆ÷£¬ÔÚÍÐÅÌÌìƽÉϳÆÈ¡ÇâÑõ»¯ÄƹÌÌåʱ£¬Ê¢ÓÐÇâÑõ»¯ÄƹÌÌåµÄÉÕ±­Ó¦·ÅÔÚ
×ó
ÅÌ£¨Ì¡°×ó¡±»ò¡°ÓÒ¡±£©£¬ÔÚϱíËùÁеÄíÀÂëÖУ¬Ñ¡³öËùÐèíÀÂ루´ò¡°¡Å¡±±íʾѡÓã©£º
íÀÂë/g 100 50   20   20    10   5
´ò¡°V¡±±íʾѡÓà        

²¢ÔÚÏÂÁбê³ßÖÐÑ¡³öÄÜÕýÈ·±íʾÓÎÂëλÖõÄÑ¡Ïî
B
£¨Ìî×Öĸ£©

£¨3£©Èܽ⣺½«ÇâÑõ»¯ÄƹÌÌåÈÜÓÚË®£¬ÓÃ
²£Á§°ô
½Á°è£¬Ê¹ÇâÑõ»¯ÄÆÈ«²¿Èܽ⣬ÀäÈ´ÖÁÊÒΣ®
£¨4£©°ÑÅäºÃµÄÈÜҺװÈëÊÔ¼ÁÆ¿£¬¸ÇºÃÆ¿¸Ç²¢ÌùÉϱêÇ©£¬·ÅÈëÊÔ¼Á¹ñÖУ®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

22¡¢ÊµÑéÊÒÓûÅäÖÆ1 000 gÈÜÖÊÖÊÁ¿·ÖÊýΪ4%µÄÇâÑõ»¯ÄÆÈÜÒº£®Çë»Ø´ðÏÂÁÐÎÊÌ⣻
£¨1£©ÅäÖƲ½Öè
¢Ù¼ÆË㣺ÐèÇâÑõ»¯ÄƹÌÌå
40
g£¬Ë®
960
mL£¨Ë®µÄÃܶÈÊÇ1 g/cm3£©£®
¢Ú³ÆÁ¿£ºÓÃÖÊÁ¿Îª23.1 gµÄÉÕ±­×÷³ÆÁ¿ÈÝÆ÷£¬ÍÐÅÌÌìƽÉϳÆÈ¡ÇâÑõ»¯ÄƹÌÌåʱ£¬Ê¢ÓÐÇâÑõ»¯ÄƹÌÌåµÄÉÕ±­Ó¦·ÅÔÚ
×óÅÌ
ÅÌ£®
¢ÛÈܽ⣺½«ÇâÑõ»¯ÄƹÌÌåÈÜÓÚË®£¬Óò£Á§°ô½Á°è£¬Ê¹ÇâÑõ»¯ÄÆÈ«²¿Èܽ⣬ÀäÈ´ÖÁÊÒΣ®
¢Ü×°Æ¿£º°ÑÅäºÃµÄÈÜҺװÈëÊÔ¼ÁÆ¿£¬¸ÇºÃÆ¿¸Ç²¢ÌùÉϱêÇ©£¬·ÅÈëÊÔ¼Á¹ñÖУ®
ÇëÔÚ×°ÓÐËùÅäÖÆÈÜÒºµÄÊÔ¼ÁÆ¿£¨¼ûͼ£©±êÇ©ÉϱêÃ÷ÏàÓ¦µÄÐÅÏ¢£®
£¨2£©ÔÚÅäÖƹý³ÌÖУ¬µ¼ÖÂÈÜÒºÖÐÇâÑõ»¯ÄÆÖÊÁ¿·ÖÊýСÓÚ4%µÄ¿ÉÄÜÔ­ÒòÊÇ£¨
D
£©
¢ÙÓÃÁ¿Í²Á¿È¡Ë®Ê±¸©ÊÓ¶ÁÊý£»¢ÚÅäÖÆÈÜÒºµÄÉÕ±­ÓÃÉÙÁ¿ÕôÁóË®ÈóÏ´£»¢ÛÔÚÍÐÅÌÌìƽµÄ×óÅ̳ÆÈ¡ÇâÑõ»¯ÄÆʱ£¬ÓÎÂë²»ÔÚÁãλÖþ͵÷½ÚÌìƽƽºâ£¬ºó½«ÓÎÂëÒƶ¯µÃµ½¶ÁÊý£»¢ÜÊ¢×°ÈÜÒºµÄÊÔ¼ÁÆ¿ÓÃÕôÁóË®ÈóÏ´£»¢ÝÇâÑõ»¯ÄƹÌÌå²»´¿£®
A£®¢Ù¢Ú¢Û¢Ü¢Ý£»B£®Ö»ÓТ٢ڢܢݣ»C£®Ö»ÓТ٢ڢࣻD£®Ö»ÓТڢۢܢÝ
£¨3£©ÈôÀÏʦÌṩµÄÒ©Æ·ÓÐ500 g 8%µÄÇâÑõ»¯ÄÆÈÜÒº£¬500 g1%µÄÇâÑõ»¯ÄÆÈÜÒº£¬×ãÁ¿µÄÇâÑõ»¯ÄƹÌÌåºÍË®£¬³ýÁËÉÏÊöÅäÖÆ·½°¸Í⣬Ä㻹¿ÉÒÔÉè¼Æ³öÄÄЩÅäÖÆ·½°¸£¬ÇëÄãд³öÆäÖеÄÒ»ÖÖ£º
500g8%µÄÇâÑõ»¯ÄÆÈÜÒººÍ500gË®
£®£¨Ö»ÒªËµÃ÷ÅäÖÆʱËùÐèµÄ¸÷ÖÖÒ©Æ·¼°ÓÃÁ¿¼´¿É£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

35¡¢ÊµÑéÊÒÓûÅäÖÆ1000gÈÜÖÊÖÊÁ¿·ÖÊýΪ4%µÄÇâÑõ»¯ÄÆÈÜÒº£®ÅäÖò½ÖèΪ£º
£¨1£©¼ÆË㣺ÐèÇâÑõ»¯ÄƹÌÌå
40
g£¬Ë®
960
mL£¨Ë®µÄÃܶÈÊÇ1g/cm3£©£®
£¨2£©³ÆÁ¿£ºÓÃÖÊÁ¿Îª23.1gµÄÉÕ±­×÷³ÆÁ¿ÈÝÆ÷£¬ÔÚÍÐÅÌÌìƽÉϳÆÈ¡ÇâÑõ»¯ÄƹÌÌåʱ£¬Ê¢ÓÐÇâÑõ»¯ÄƹÌÌåµÄÉÕ±­Ó¦·ÅÔÚ
×ó
ÅÌ£¨Ìî¡°×ó¡±»ò¡°ÓÒ¡±£©£¬ÔÚϱíËùÁеÄíÀÂëÖУ¬Ñ¡³öËùÐèíÀÂ루´ò¡°¡Ì¡±±íʾѡÓã©£¬
íÀÂë/g 100 50 20 20 10 5
´ò¡°¡Ì¡±±íʾѡÓà            
²¢ÔÚÏÂÁбê³ßÖÐÑ¡³öÄÜÕýÈ·±íʾÓÎÂëλÖõÄÑ¡Ïî
B
£¨Ìî×Öĸ£©

£¨3£©Èܽ⣺½«ÇâÑõ»¯ÄƹÌÌåÈÜÓÚË®£¬ÓÃ
²£Á§°ô
½Á°è£¬Ê¹ÇâÑõ»¯ÄÆÈ«²¿Èܽ⣬ÀäÈ´ÖÁÊÒΣ®
£¨4£©°ÑÅäºÃµÄÈÜҺװÈëÊÔ¼ÁÆ¿£¬¸ÇºÃÆ¿¸Ç²¢ÌùÉϱêÇ©£¬·ÅÈëÊÔ¼Á¹ñÖУ®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

15¡¢Ä³»¯Ñ§¿ÎÍâÐËȤС×飬¿ªÕ¹ÁËÒÔÏÂÁ½Ïî»î¶¯£º
£¨1£©µÚÒ»×é̽¾¿¿ÕÆøÖÐÑõÆøµÄº¬Á¿£®ËûÃÇÉè¼ÆÁËÈçÓÒͼµÄ×°Öã¬Ñо¿¿ÕÆøÖÐËùº¬ÑõÆøµÄÌå»ý£¬ÓÃÒ»¸ö50mLÁ¿Í²µ¹¿ÛÔÚË®ÖУ¬Ê¹Á¿Í²ÄÚÍâÒºÃæ¾ùλÓÚ40mL´¦£®Á¿Í²ÄÚ¸¡×ÅÒ»¸öÍ­ÖƵÄС´¬£¬´¬ÖзÅÓÐ×ãÁ¿°×Á×£®ÏÖ°ÑŨÁòËỺ»ºµ¹ÈëË®Öв¢½Á°è£®Ò»»á¶ù£¬°×Á×·¢Éú×Ôȼ£¬Á¿Í²ÄÚÒºÃæÉÏÉý£¬ÏòË®²ÛÄÚ¼ÓË®£¬ÖÁË®²ÛÓëÁ¿Í²ÄÚÒºÃæÏàƽ£¬¶Á³ö¶ÁÊý£®
¢Ù¼ÓÈëŨÁòËáµÄµÄÄ¿µÄÊÇ
ŨÁòËáÈÜÓÚË®·ÅÈÈ£¬Ê¹Î¶ȴﵽ°×Á×µÄ×Å»ðµãÒÔÉÏ
£»
¢Ú·Å°×Á×µÄС´¬ÓÃÍ­×öµÄÔ­ÒòÊÇ
Í­µ¼ÈÈÐÔºÃÇÒ»¯Ñ§ÐÔÖÊÎȶ¨
£»
¢Û×îÖÕ£¬ÒºÃæ´óԼλÓÚÁ¿Í²¿Ì¶È
32mL
´¦£»
¢ÜΪºÎÒªÏòË®²ÛÄÚ¼ÓË®ÖÁÄÚÍâÒºÃæÏàƽ
ʹÆøÌåÄÚÍâѹǿÏàµÈ£¬´Ó¶ø²âµÃµÄÆøÌåÌå»ý¸ü¾«È·
£»
¢ÝÊÔÆÀ¼Û¸ÃÐËȤС×éµÄÉè¼ÆÔÚÑ¡ÓÃÒÇÆ÷ÉÏÓкÎȱÏÝ
Á¿Í²²»ÄÜÓÃ×÷·´Ó¦ÈÝÆ÷
£®
£¨2£©ÅäÖÆ4%µÄÇâÑõ»¯ÄÆÈÜÒº200g£¬²¢ÓôËÈÜÒº²â¶¨Ä³ÁòËáÈÜÒºµÄÈÜÖÊÖÊÁ¿·ÖÊý£®
¢ÙÅäÖÆ200gÖÊÁ¿·ÖÊýΪ4%µÄÇâÑõ»¯ÄÆÈÜÒº£¬ÐèÒªÇâÑõ»¯ÄƹÌÌåµÄÖÊÁ¿Îª
8
g£¬Ë®µÄÌå»ýΪ
192
mL£¨Ë®µÄÃܶȽüËÆ¿´×÷1g/cm3£©£®
¢ÚÅäÖÆÇâÑõ»¯ÄÆÈÜҺʱ£¬ÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÁ¿Í²¡¢ÉÕ±­¡¢²£Á§°ô
¼°
½ºÍ·µÎ¹Ü
£®ÅäÖƺõÄÇâÑõ»¯ÄÆÈÜÒºÒªÃÜ·â±£´æ£¬ÆäÔ­ÒòÊÇ£º
CO2+2NaOH=Na2CO3+H2O
£¨Ð´»¯Ñ§·½³Ìʽ£©£®
¢ÛÓÃÅäµÃµÄÇâÑõ»¯ÄÆÈÜÒºÓë´ý²âÈÜÖÊÖÊÁ¿·ÖÊýµÄÁòËáÈÜÒº·´Ó¦£¬
ʵÑé¹ý³ÌÖÐÈÜÒºµÄpH±ä»¯ÇúÏßÈçÓÒͼËùʾ£º
I¡¢ÒªµÃµ½´Ë±ä»¯ÇúÏߣ¬Ëù½øÐеIJÙ×÷ÊÇ
B
£¨Ìî×Öĸ£©£®
A¡¢½«ÁòËáÈÜÒºµÎ¼Óµ½ÇâÑõ»¯ÄÆÈÜÒºÖР   B¡¢½«ÇâÑõ»¯ÄÆÈÜÒºµÎ¼Óµ½ÁòËáÈÜÒºÖÐ
II¡¢bµã¶ÔÓ¦µÄÈÜÒºÖеÄÈÜÖÊΪ
Na2SO4ºÍNaOH
£¨Ð´»¯Ñ§Ê½£©£®
¢ÜÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ÏûºÄ20gÇâÑõ»¯ÄÆÈÜÒººÍ25gÁòËáÈÜÒº£¬ÊÔ¼ÆËã¸ÃÁòËáÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£¨Çëд³ö¼ÆËã¹ý³Ì£©
3.92%
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ʵÑéÊÒÓûÅäÖÆ1000gÈÜÖÊÖÊÁ¿·ÖÊýΪ4%µÄÇâÑõ»¯ÄÆÈÜÒº£®ÅäÖƲ½ÖèΪ£º
£¨1£©¼ÆË㣺ÐèÇâÑõ»¯ÄƹÌÌå
40
40
g£¬Ë®
960
960
mL£¨Ë®µÄÃܶÈÊÇ1g/cm3£©£®
£¨2£©³ÆÁ¿£ºÓÃÖÊÁ¿Îª23.1gµÄÉÕ±­×÷³ÆÁ¿ÈÝÆ÷£¬ÔÚÍÐÅÌÌìƽÉϳÆÈ¡ÇâÑõ»¯ÄƹÌÌåʱ£¬Ê¢ÓÐÇâÑõ»¯ÄƹÌÌåµÄÉÕ±­Ó¦·ÅÔÚ
×ó
×ó
ÅÌ£¨Ì¡°×ó¡±»ò¡°ÓÒ¡±£©£¬ÔÚϱíËùÁеÄíÀÂëÖУ¬Ñ¡³öËùÐèíÀÂ루´ò¡°¡Ì¡±±íʾѡÓã©£º
íÀÂë/g 100 50 20 20 10 5
´ò¡°¡Ì¡±±íʾѡÓÃ
²¢ÔÚÏÂÁбê³ßÖÐÑ¡³öÄÜÕýÈ·±íʾÓÎÂëλÖõÄÑ¡Ïî
B
B
£¨Ìî×Öĸ£©

£¨3£©Èܽ⣺½«ÇâÑõ»¯ÄƹÌÌåÈÜÓÚË®£¬ÓÃ
²£Á§°ô
²£Á§°ô
½Á°è£¬Ê¹ÇâÑõ»¯ÄÆÈ«²¿Èܽ⣮
£¨4£©´æ·Å£ºµÈÇâÑõ»¯ÄÆÈÜÒºÀäÈ´ºó£¬°ÑÆäתÒƵ½²£Á§ÊÔ¼ÁÆ¿ÖУ¬ÌùÉϱêÇ©£®´æ·ÅʱÈû½ôÆ¿×ÓµÄÈû×ÓҪѡÔñ
½ºÈû
½ºÈû
 £¨Ìî¡°²£Á§Èû¡±»ò¡°½ºÈû¡±£©
£¨5£©¼ÙÈçÔÚ¸ÃʵÑéÖУ¬ÄãÅäÖƵÄÇâÑõ»¯ÄÆÈÜÒºÖÊÁ¿·ÖÊýСÓÚ4%£¬Çë·ÖÎöÒ»¸öÔì³ÉСÓÚ4%µÄ¿ÉÄÜÒòËØ£º
³ÆÁ¿Ê±ÇâÑõ»¯ÄƺÍíÀÂë·Å·´ÁË£¬×óÂëÓÒÎºÏÀí¾ù¿É£©
³ÆÁ¿Ê±ÇâÑõ»¯ÄƺÍíÀÂë·Å·´ÁË£¬×óÂëÓÒÎºÏÀí¾ù¿É£©
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸