¡¾ÌâÄ¿¡¿£¨1£©ÒÔÏÂÁгöµÄÊÇһЩԭ×ÓµÄ2pÄܼ¶ºÍ3dÄܼ¶Öеç×ÓÅŲ¼µÄÇé¿ö£¬ÊÔÅжÏÄÄЩΥ·´ÁËÅÝÀûÔ­Àí________£¬ÄÄЩΥ·´Á˺éÌعæÔò________¡£

£¨2£©¿ÉÓÃÓÚÖÆÔì»ð²ñ£¬Æä·Ö×ӽṹÈçͼËùʾ¡£

·Ö×ÓÖÐÁòÔ­×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ________¡£

ÿ¸ö·Ö×ÓÖк¬Óеŵç×Ó¶ÔµÄÊýĿΪ________¡£

£¨3£©¿Æѧ¼ÒºÏ³ÉÁËÒ»ÖÖÑôÀë×Ó¡°¡±£¬Æä½á¹¹ÊǶԳƵģ¬5¸öNÅųɡ°V¡±ÐΣ¬Ã¿¸öN¶¼´ïµ½8µç×ÓÎȶ¨½á¹¹£¬ÇÒº¬ÓÐ2¸öµªµªÈý¼ü£»´ËºóÓֺϳÉÁËÒ»ÖÖº¬ÓС°¡±µÄ»¯Ñ§Ê½Îª¡°¡±µÄÀë×Ó¾§Ì壬Æäµç×ÓʽΪ________¡£·Ö×ÓÖмüÓë¼üÖ®¼äµÄ¼Ð½ÇΪ£¬²¢ÓжԳÆÐÔ£¬·Ö×ÓÖÐÿ¸öÔ­×ÓµÄ×îÍâ²ã¾ùÂú×ã8µç×ÓÎȶ¨½á¹¹£¬Æä½á¹¹Ê½Îª________________¡£

£¨4£©Ö±Á´¶àÁ×Ëá¸ùÒõÀë×ÓÊÇÓÉÁ½¸ö»òÁ½¸öÒÔÉÏÁ×ÑõËÄÃæÌåͨ¹ý¹²Óö¥½ÇÑõÔ­×ÓÁ¬½ÓÆðÀ´µÄ£¬ÈçͼËùʾ¡£ÔòÓÉn¸öÁ×ÑõËÄÃæÌåÐγɵÄÕâÀàÁ×Ëá¸ùÀë×ÓµÄͨʽΪ________¡£

£¨5£©Ì¼ËáÑÎÖеÄÑôÀë×Ó²»Í¬£¬ÈÈ·Ö½âζȾͲ»Í¬¡£Ï±íΪËÄÖÖ̼ËáÑεÄÈÈ·Ö½âζȺͽðÊôÑôÀë×Ӱ뾶

̼ËáÑÎ

ÈÈ·Ö½âζÈ

402

900

1172

1360

½ðÊôÑôÀë×Ӱ뾶

66

99

112

135

Ëæ׎ðÊôÑôÀë×Ӱ뾶µÄÔö´ó£¬Ì¼ËáÑεÄÈÈ·Ö½âζÈÖð²½Éý¸ß£¬Ô­ÒòÊÇ_____________¡£

£¨6£©Ê¯Ä«µÄ¾§Ìå½á¹¹ºÍ¾§°û½á¹¹ÈçͼËùʾ¡£ÒÑ֪ʯīµÄÃܶÈΪ£¬¼üµÄ¼ü³¤Îª£¬°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪ£¬Ôòʯī¾§ÌåµÄ²ã¼ä¾àΪ________cm¡£

¡£

¡¾´ð°¸¡¿ ̼ËáÑηֽâʵ¼Ê¹ý³ÌÊǾ§ÌåÖÐÑôÀë×Ó½áºÏ̼Ëá¸ùÀë×ÓÖÐÑõÀë×Ó£¬Ê¹Ì¼Ëá¸ùÀë×Ó·Ö½âΪ¶þÑõ»¯Ì¼µÄ¹ý³Ì£¬ÑôÀë×ÓËù´øµçºÉÏàͬʱ£¬ÑôÀë×Ӱ뾶ԽС£¬Æä½áºÏÑõÀë×ÓÄÜÁ¦Ô½Ç¿£¬¶ÔÓ¦µÄ̼ËáÑξÍÔ½ÈÝÒ×·Ö½â

¡¾½âÎö¡¿

£¨1£©¸ù¾ÝÅÝÀû²»ÏàÈÝÔ­Àí¡¢ºéÌعæÔòµÄÄÚÈÝ·ÖÎö»Ø´ð£»

£¨2£©·Ö×ÓÖÐÁòÔ­×ӵŵç×Ó¶ÔÊýÊÇ £¬¼üÊýÊÇ2£»

·Ö×ÓÖÐÁòÔ­×ӵŵç×Ó¶ÔÊýÊÇ¡¢Á×Ô­×ӵŵç×Ó¶ÔÊýÊÇ£»

£¨3£©ÓÉÓÚÑôÀë×Ó¡°¡±£¬Æä½á¹¹ÊǶԳƵģ¬5¸öNÅųɡ°V¡±ÐΣ¬Ã¿¸öN¶¼´ïµ½8µç×ÓÎȶ¨½á¹¹£¬ÇÒº¬ÓÐ2¸öµªµªÈý¼ü£¬ËùÒÔ¸Ã΢Á£´øÓÐÒ»¸öµ¥Î»µÄÕýµçºÉ£¬µç×ÓʽÊÇ£¬º¬ÓС°¡±µÄ»¯Ñ§Ê½Îª¡°¡±µÄÀë×Ó¾§Ì壬Ôò¸ÃÒõÀë×ÓÊÇ¡£

£¨4£©ÀûÓá°Êýѧ¹éÄÉ·¨¡±·ÖÎöͨʽ£»

£¨5£©Ì¼ËáÑηֽâʵ¼Ê¹ý³ÌÊǾ§ÌåÖÐÑôÀë×Ó½áºÏ̼Ëá¸ùÀë×ÓÖÐÑõÀë×Ó£¬Ê¹Ì¼Ëá¸ùÀë×Ó·Ö½âΪ¶þÑõ»¯Ì¼µÄ¹ý³Ì£»

£¨6£©¾§°ûµÄ¸ßµÈÓÚ²ã¼ä¾àµÄ2±¶¡£

ÅÝÀûÔ­ÀíÖ¸µÄÊÇͬһ¸öÔ­×Ó¹ìµÀÄÚ²»¿ÉÄÜÓÐÁ½¸ö×ÔÐý·½ÏòÏàͬµÄµç×Ó£¬Òò´ËÎ¥·´ÅÝÀûÔ­ÀíµÄÊÇ£»ºéÌعæÔòÖ¸µÄÊǵç×Ó¾¡¿ÉÄÜ·ÖÕ¼²»Í¬¹ìµÀ£¬ÇÒ×ÔÐý·½ÏòÏàͬ£¬Òò´ËÎ¥·´ºéÌعæÔòµÄÊÇ£»

£¨2£©·Ö×ÓÖÐÁòÔ­×ӵŵç×Ó¶ÔÊýÊÇ £¬¼üÊýÊÇ2£¬ÔÓ»¯¹ìµÀÊýÊÇ4£¬ËùÒÔÁòÔ­×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ£»

ÿ¸öÁ×Ô­×ӵŵç×Ó¶ÔÊýΪ1£¬Ã¿¸öÁòÔ­×ӵŵç×Ó¶ÔΪ2£¬ËùÒÔÿ¸ö·Ö×ÓÖк¬Óеŵç×Ó¶ÔµÄÊýĿΪ10£»

ÓÉÓÚÑôÀë×Ó¡°¡±£¬Æä½á¹¹ÊǶԳƵģ¬5¸öNÅųɡ°V¡±ÐΣ¬Ã¿¸öN¶¼´ïµ½8µç×ÓÎȶ¨½á¹¹£¬ÇÒº¬ÓÐ2¸öµªµªÈý¼ü£¬ËùÒÔ¸Ã΢Á£´øÓÐÒ»¸öµ¥Î»µÄÕýµçºÉ£¬¸ÃÀë×ӵĵç×ÓʽÊÇ£¬º¬ÓС°¡±µÄ»¯Ñ§Ê½Îª¡°¡±µÄÀë×Ó¾§Ì壬Ôò¸ÃÒõÀë×ÓÊÇ¡£»¯Ñ§Ê½Îª¡°¡±µÄÀë×Ó¾§ÌåµÄµç×ÓʽΪ£»·Ö×ÓÖмüÓë¼üÖ®¼äµÄ¼Ð½ÇΪ£¬²¢ÓжԳÆÐÔ£¬·Ö×ÓÖÐÿ¸öÔ­×ÓµÄ×îÍâ²ã¾ùÂú×ã8µç×ÓÎȶ¨½á¹¹£¬Æä½á¹¹Ê½Îª£»

¸ù¾Ý¸ø¶¨µÄ½á¹¹£¬ÓÉÊýѧ¹éÄÉ·¨¿ÉÖª£¬ÓÉn¸öÁ×ÑõËÄÃæÌåÐγɵÄÕâÀàÁ×Ëá¸ùÀë×ÓµÄͨʽΪ£»

Ëæ׎ðÊôÑôÀë×Ӱ뾶µÄÔö´ó£¬Ì¼ËáÑεÄÈÈ·Ö½âζÈÖð²½Éý¸ß£¬Ô­ÒòÊÇ̼ËáÑηֽâʵ¼Ê¹ý³ÌÊǾ§ÌåÖÐÑôÀë×Ó½áºÏ̼Ëá¸ùÀë×ÓÖÐÑõÀë×Ó£¬Ê¹Ì¼Ëá¸ùÀë×Ó·Ö½âΪ¶þÑõ»¯Ì¼µÄ¹ý³Ì£¬ÑôÀë×ÓËù´øµçºÉÏàͬʱ£¬ÑôÀë×Ӱ뾶ԽС£¬Æä½áºÏÑõÀë×ÓÄÜÁ¦Ô½Ç¿£¬¶ÔÓ¦µÄ̼ËáÑξÍÔ½ÈÝÒ׷ֽ⣻

¸ù¾ÝʯīµÄ½á¹¹·ÖÎö¿É֪ʯī¾§ÌåÖк¬ÓеÄ̼ԭ×ÓÊýΪ£¬ÓÖ¼üµÄ¼ü³¤Îªrcm£¬ËùÒÔ¾§°ûµÄµ×Ãæ»ýΪ£¬¸ù¾Ý¾§°ûµÄÃܶȣ¬½âµÃ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÊÒÎÂʱ£¬ÏòÈÜÒºÖеμÓÈÜÒº£¬µÃµ½ÈÜÒºpHÓëNaOHÈÜÒºÌå»ýµÄ¹ØϵÇúÏßÈçͼËùʾÎÞÆøÌåÒç³ö£¬ÒÔÏÂ˵·¨²»ÕýÈ·µÄÊÇ

A.aµãʱˮµÄµçÀë³Ì¶È×î´ó

B.bµãʱÈÜÒºÖеÄÀë×ÓŨ¶È¹ØϵÓÐ

C.cµãʱÈÜÒºÖеÄÁ£×ÓŨ¶È´óС¹ØϵΪ

D.dµãʱÈÜÒºÖÐ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿2.0molPCl3ºÍ1.0molCl2³äÈëÌå»ý²»±äµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Ìõ¼þÏ·¢ÉúÏÂÊö·´Ó¦£ºPCl3(g)+Cl2(g)PCl5(g)´ïƽºâʱ£¬PCl5Ϊ0.4mol£¬Èç¹û´ËʱÒÆ×ß1.0molPCl3ºÍ0.50molCl2£¬ÔÚÏàͬζÈÏÂÔÙ´ïƽºâʱPCl5µÄÎïÖʵÄÁ¿ÊÇ£¨ £©

A. 0.4mol

B. 0.2mol

C. СÓÚ0.2mol

D. ´óÓÚ0.2mol£¬Ð¡ÓÚ0.4mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿BºÍMgÔÚ²ÄÁÏ¿ÆѧÁìÓòÓй㷺µÄÓ¦Óúͷ¢Õ¹Ç°¾°¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÏÂÁÐBÔ­×Ó¹ìµÀ±í´ïʽ±íʾµÄ״̬ÖУ¬ÄÜÁ¿¸ü¸ßµÄÊÇ________Ìî¡°A¡±»ò¡°B¡±¡£

A. B.

£¨2£©¾§ÌåÅðÖеĻù±¾µ¥ÔªÈçͼ1Ëùʾ£¬ÆäÖк¬ÓÐ12¸öBÔ­×Ó¡£¸Ãµ¥ÔªÖк¬ÓмüµÄÊýĿΪ________¡£

£¨3£©ÊÇÖØÒªµÄ»¹Ô­¼Á¡£ÆäÖÐÒõÀë×ÓµÄÁ¢Ìå¹¹ÐÍΪ________£¬ÖÐÐÄÔ­×ÓµÄÔÓ»¯ÐÎʽΪ________¡£

£¨4£©»ù̬MgÔ­×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª________£»µÚÈýÖÜÆÚÖ÷×åÔªËØÖУ¬µç¸ºÐÔСÓÚMgÔªËصÄÓÐ________ÌîÔªËØ·ûºÅ£¬ÏÂͬ£¬Ô­×ÓµÚÒ»µçÀëÄÜСÓÚMgÔ­×ÓµÄÓÐ________¡£

£¨5£©ºÍ¾ù¿É×÷ΪÄÍ»ð²ÄÁÏ£¬ÆäÔ­ÒòÊÇ________¡£

£¨6£©³£ÓÃÓÚ¹âѧÒÇÆ÷£¬Æ䳤·½ÌåÐ;§°û½á¹¹Èçͼ2Ëùʾ£º

µÄÅäλÊýΪ________¡£

Èô°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪ£¬Ôò¾§ÌåµÄÃܶȿɱíʾΪ________Óú¬a¡¢b¡¢µÄ´úÊýʽ±íʾ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ïû³ýµªÑõ»¯Îï¡¢¶þÑõ»¯ÁòµÈÎïÖÊÔì³ÉµÄÎÛȾÊÇÄ¿Ç°Ñо¿µÄÖØÒª¿ÎÌâ¡£

(1)¹¤ÒµÉϳ£ÓûîÐÔÌ¿»¹Ô­Ò»Ñõ»¯µª£¬Æ䷴ӦΪ£º2NO(g)+C(s) N2(g)+CO2(g)¡£ÏòÈÝ»ý¾ùΪl LµÄ¼×¡¢ÒÒ¡¢±ûÈý¸öºãÈݺãÎÂÈÝÆ÷Öзֱð¼ÓÈë×ãÁ¿µÄ»îÐÔÌ¿ºÍÒ»¶¨Á¿µÄNO£¬²âµÃ¸÷ÈÝÆ÷ÖÐn(NO)Ë淴Ӧʱ¼ätµÄ±ä»¯Çé¿öÈçϱíËùʾ£º

0min

40min

80min

120min

160min

¼×

T¡æ

2mol

1.45 mol

1 mol

1 mol

1 mol

ÒÒ

400¡æ

2 mol

1.5 mol

1.1 mol

0.8 mol

0.8 mol

±û

400¡æ

1 mol

0.8 mol

0.65 mol

0.53 mol

0.45 mol

¼×ÈÝÆ÷·´Ó¦Î¶ÈT¡æ______400¡æ(Ìî¡°>¡±¡°<¡±»ò¡°=¡±)£»ÒÒÈÝÆ÷ÖУ¬0~40minÄÚƽ¾ù·´Ó¦ËÙÂÊv(CO2)=_____________________£»±ûÈÝÆ÷ÖдïƽºâºóNOµÄÎïÖʵÄÁ¿Îª_________mol¡£

(2)»îÐÔÌ¿»¹Ô­NO2µÄ·´Ó¦Îª£º2NO2(g)+2C(s) N2(g)+2CO2(g)£¬ÔÚºãÎÂÌõ¼þÏ£¬l mol NO2ºÍ×ãÁ¿»îÐÔÌ¿·¢Éú¸Ã·´Ó¦£¬²âµÃƽºâʱNO2ºÍCO2µÄÎïÖʵÄÁ¿Å¨¶ÈÓëƽºâ×ÜѹµÄ¹ØϵÈçͼËùʾ£º

¢ÙA¡¢B¡¢CÈýµãÖÐNO2µÄת»¯ÂÊ×î¸ßµÄÊÇ_________µã(Ìî¡°A¡±»ò¡°B¡±»ò¡°C¡±)¡£

¢Ú¼ÆËãCµãʱ¸Ã·´Ó¦µÄѹǿƽºâ³£ÊýKP=_______MPa(KpÊÇÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È¼ÆË㣬·Öѹ=×Üѹ¡ÁÎïÖʵÄÁ¿·ÖÊý)¡£

(3)ȼúÑÌÆøÍÑÁò³£ÓÃÈçÏ·½·¨¡£

·½·¨¢Ù£ºÓÃÉúÎïÖÊÈȽâÆø(Ö÷Òª³É·ÖCO¡¢CH4¡¢H2)½«SO2ÔÚ¸ßÎÂÏ»¹Ô­³Éµ¥ÖÊÁò¡£Éæ¼°µÄ²¿·Ö·´Ó¦ÈçÏ£º

2CO(g)+SO2(g)=S(g)+2CO2(g) ¡÷H1=8.0 kJ¡¤mol-1

2CO(g)+O2(g)=2CO2(g) ¡÷H2=£­566.0kJ¡¤mol-1

2H2(g)+O2(g)=2H2O(g) ¡÷H3=£­483.6 1kJ¡¤mol-1

ÔòH2(g)»¹Ô­SO2(g)Éú³ÉS(g)ºÍH2O(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ£º_____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊÒÎÂÏ£¬ÏòÈÜÒºÖÐÖðµÎ¼ÓÈëÑÎËᣬÈÜÒºº¬R΢Á£µÄÎïÖʵÄÁ¿·ÖÊýÓëpH¹ØϵÈçͼËùʾ²»Îȶ¨£¬Ò×ת»¯ÎªÆøÌåÒݳöÈÜÒº£¬ÆøÌåÒݳöδ»­³ö£¬ºöÂÔÒòÆøÌåÒݳöÒýÆðµÄÈÜÒºÌå»ý±ä»¯ÏÂÁÐ˵·¨´íÎóµÄÊÇ

A.ÈÜÒºÖУº

B.µ±ÈÜҺʱ£¬ÈÜÒº×ÜÌå»ý´óÓÚ40mL

C.ÔÚBµã¶ÔÓ¦µÄÈÜÒºÖУ¬Àë×ÓŨ¶È×î´óµÄÊÇ

D.Aµã¶ÔÓ¦pHԼΪ£¬µÄË®½â³£ÊýÊýÁ¿¼¶Îª

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÖÐèÉè¼ÆÒ»Ì×ʵÑé×°ÖÃÀ´µç½â±¥ºÍʳÑÎË®£¬²¢²âÁ¿µç½â²úÉúµÄÇâÆøµÄÌå»ý(Ô¼6 mL)ºÍ¼ìÑéÂÈÆøµÄÑõ»¯ÐÔ(²»Ó¦½«¶àÓàµÄÂÈÆøÅÅÈë¿ÕÆøÖÐ)¡£

(1)ÊÔ´ÓÏÂͼÖÐÑ¡Óü¸ÖÖ±ØÒªµÄÒÇÆ÷£¬Á¬³ÉÒ»ÕûÌ××°Ö㬸÷ÖÖÒÇÆ÷½Ó¿ÚµÄÁ¬½Ó˳Ðò(Ìî±àºÅ)ÊÇA½ÓG-F-I£¬B½Ó__________¡£

(2)Ìú°ô½ÓÖ±Á÷µçÔ´µÄ________¼«£»Ì¼°ôÉÏ·¢ÉúµÄµç¼«·´Ó¦Îª_____________________________________¡£

(3)ÄÜ˵Ã÷ÂÈÆø¾ßÓÐÑõ»¯ÐÔµÄʵÑéÏÖÏóÊÇ_______________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ò»¶¨Î¶ÈÏÂÔÚÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬ÏÂÁÐÐðÊö²»ÄÜ×÷Ϊ¿ÉÄæ·´Ó¦A(g)£«3B(g) 2C(g)´ïµ½Æ½ºâ״̬±êÖ¾µÄÊÇ ( )

¢ÙCµÄÉú³ÉËÙÂÊÓëCµÄÏûºÄËÙÂÊÏàµÈ

¢Úµ¥Î»Ê±¼äÄÚÉú³Éa mol A£¬Í¬Ê±Éú³É3a mol B

¢ÛA¡¢B¡¢CµÄŨ¶È²»Ôٱ仯

¢ÜCµÄÎïÖʵÄÁ¿²»Ôٱ仯

¢Ý»ìºÏÆøÌåµÄ×Üѹǿ²»Ôٱ仯

¢Þ»ìºÏÆøÌåµÄ×ÜÎïÖʵÄÁ¿²»Ôٱ仯

¢ßµ¥Î»Ê±¼äÏûºÄa mol A£¬Í¬Ê±Éú³É3a mol B

¢àA¡¢B¡¢CµÄ·Ö×ÓÊýÖ®±ÈΪ1¡Ã3¡Ã2

A. ¢Ú¢àB. ¢Ü¢ßC. ¢Ù¢ÛD. ¢Ý¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª£º2RCH2CHORCH2CH=CRCHO£«H2O£»Ë®ÑîËáõ¥EΪ×ÏÍâÎüÊÕ¼Á£¬¿ÉÓÃÓÚÅäÖÆ·Àɹ˪¡£EµÄÒ»ÖֺϳÉ·ÏßÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ò»Ôª´¼AÖÐÑõµÄÖÊÁ¿·ÖÊýԼΪ21.6%¡£ÔòAµÄ·Ö×ÓʽΪ_________£»½á¹¹·ÖÎöÏÔʾAÖ»ÓÐÒ»¸ö¼×»ù£¬AµÄÃû³ÆΪ___________¡£

£¨2£©BÄÜÓëÐÂÖƵÄCu(OH)2·¢Éú·´Ó¦£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________¡£

£¨3£©CÓÐ____Öֽṹ£»ÈôÒ»´ÎÈ¡Ñù£¬¼ìÑéCÖÐËùº¬¹ÙÄÜÍÅ£¬°´Ê¹ÓõÄÏȺó˳Ðòд³öËùÓÃÊÔ¼Á£º______________________________________________________£»

£¨4£©µÚ¢ÛµÄ·´Ó¦ÀàÐÍΪ_________________£»DËùº¬¹ÙÄÜÍŵÄÃû³ÆΪ________________¡£

£¨5£©Ð´³öͬʱ·ûºÏÏÂÁÐÌõ¼þµÄË®ÑîËáËùÓÐͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º_________________________¡£

a£®·Ö×ÓÖк¬ÓÐ6¸ö̼ԭ×ÓÔÚÒ»ÌõÏßÉÏ b£®·Ö×ÓÖÐËùº¬¹ÙÄÜÍÅ°üÀ¨Ë®ÑîËá¾ßÓеĹÙÄÜÍÅ

£¨6£©µÚ¢Ü²½µÄ·´Ó¦Ìõ¼þΪ___________£»Ð´³öEµÄ½á¹¹¼òʽ_________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸