ÏÂÁÐÎåÖÖ¶ÌÖÜÆÚÔªËصÄijЩÐÔÖÊÈç±íËùʾ(ÆäÖÐÖ»ÓÐW£¬Y¡¢ZΪͬÖÜÆÚÔªËØ)¡£

ÔªËØ´úºÅ

X

W

Y

Z

Q

Ô­×Ӱ뾶(¡Á10£­12 m)

37

64

66

70

154

Ö÷Òª»¯ºÏ¼Û

£«1

£­1

£­2

£«5¡¢£­3

£«1

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ÓÉQÓëYÐγɵĻ¯ºÏÎïÖÐÖ»´æÔÚÀë×Ó¼ü

B£®ZÓëXÖ®¼äÐγɵĻ¯ºÏÎï¾ßÓл¹Ô­ÐÔ

C£®ÓÉX¡¢Y¡¢ZÈýÖÖÔªËØÐγɵĻ¯ºÏÎÆ侧ÌåÒ»¶¨ÊÇ·Ö×Ó¾§Ìå

D£®YÓëWÐγɵĻ¯ºÏÎYÏÔ¸º¼Û


½âÎö£ºÓɱíÖÐÊý¾Ý¿ÉÖª£¬W¡¢Y¡¢Z·Ö±ðΪF¡¢O¡¢NÈýÔªËØ£»XΪH£»QΪNa£»¹ÊAÏî¿ÉÄÜ´æÔÚ¹²¼Û¼ü£¬ÈçNa2O2£¬´í£»CÏҲ¿ÉÊÇÀë×Ó¾§Ì壬ÈçNH4NO3£¬C´í£»DÏÓÉÓÚ·úÔªËرÈÑõÔªËطǽðÊôÐÔÇ¿£¬ÆäÐγɵĻ¯ºÏÎÑõÏÔÕý¼Û£¬D´í¡£

´ð°¸£ºB


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


       ½«10.6 g Na2CO3ÈÜÓÚË®Åä³É1 LÈÜÒº

(1)¸ÃÈÜÒºÖÐNa2CO3µÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________£¬ÈÜÒºÖÐNa£«µÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________¡£

(2)Ïò¸ÃÈÜÒºÖмÓÈëÒ»¶¨Á¿NaCl¹ÌÌ壬ʹÈÜÒºÖÐNa£«µÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.4 mol·L£­1(¼ÙÉèÈÜÒºÌå»ý²»±ä)Ðè¼ÓÈëNaClµÄÖÊÁ¿Îª__________£¬Cl£­µÄÎïÖʵÄÁ¿Å¨¶ÈΪ________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ò»¶¨Î¶ÈÏ£¬Ôڹ̶¨Ìå»ýµÄÃܱÕÈÝÆ÷Öз¢ÉúÏÂÁз´Ó¦£º2HI£¨g£©=H2£¨g£©+I2£¨g£©ÈôC(HI£© ÓÉ0.1mol/L½µµ½0.07mol/Lʱ£¬ÐèÒª15S£¬ÄÇôC(HI)ÓÉ0.07mol/L½µµ½0.05mol/Lʱ£¬ËùÐ跴Ӧʱ¼äΪ£¨   £©

   A.µÈÓÚ5S        B.µÈÓÚ10S       C ´óÓÚ10S       DСÓÚ10S

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔªËØÐÔÖʵÄÖÜÆÚÐԱ仯²»°üÀ¨(¡¡¡¡)

A£®Ô­×Ӱ뾶¡¡¡¡¡¡¡¡¡¡¡¡¡¡                          B£®»¯ºÏ¼Û

C£®Ô­×ÓºËÍâµç×ӽṹ                                   D£®½ðÊôÐԺͷǽðÊôÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


X¡¢Y¡¢Z¡¢R¡¢WÊÇ5ÖÖ¶ÌÖÜÆÚÔªËØ£¬Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£¬ËüÃÇ¿É×é³ÉÀë×Ó»¯ºÏÎïZ2YºÍ¹²¼Û»¯ºÏÎïRY3¡¢XW4£¬ÒÑÖªY¡¢RͬÖ÷×壬Z¡¢R¡¢WͬÖÜÆÚ¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®Ô­×Ӱ뾶£ºZ>R>W

B£®X2W6·Ö×ÓÖи÷Ô­×Ó¾ùÂú×ã8µç×ӽṹ

C£®Æø̬Ç⻯ÎïµÄÎȶ¨ÐÔ£ºHmW>HnR

D£®Y¡¢ZÐγɵĻ¯ºÏÎïÖÐÖ»¿ÉÄÜ´æÔÚÀë×Ó¼ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


X¡¢Y¡¢Z¡¢Q¡¢MΪ³£¼ûµÄ¶ÌÖÜÆÚÔªËØ£¬ÆäÔ­×ÓÐòÊýÒÀ´ÎÔö´ó¡£ÓйØÐÅÏ¢ÈçÏÂ±í£º

X

¶¯Ö²ÎïÉú³¤²»¿ÉȱÉÙµÄÔªËØ£¬Êǵ°°×ÖʵÄÖØÒª³É·Ö

Y

µØ¿ÇÖк¬Á¿¾ÓµÚһλ

Z

¶ÌÖÜÆÚÖÐÆäÔ­×Ӱ뾶×î´ó

Q

Éú»îÖдóÁ¿Ê¹ÓÃÆäºÏ½ðÖÆÆ·£¬¹¤ÒµÉÏ¿ÉÓõç½âÆäÑõ»¯ÎïµÄ·½·¨ÖƱ¸

M

º£Ë®ÖдóÁ¿¸»¼¯µÄÔªËØÖ®Ò»£¬Æä×î¸ßÕý¼ÛÓ븺¼ÛµÄ´úÊýºÍΪ6

(1)XµÄÆø̬Ç⻯ÎïµÄ´óÁ¿Éú²úÔø¾­½â¾öÁ˵ØÇòÉÏÒòÁ¸Ê³²»×ã¶øµ¼Öµļ¢¶öºÍËÀÍöÎÊÌ⣬Çëд³ö¸ÃÆø̬Ç⻯ÎïµÄµç×Óʽ£º____¡£

(2)ÒÑÖª37RbºÍ53I¶¼Î»ÓÚµÚÎåÖÜÆÚ£¬·Ö±ðÓëZºÍMλÓÚͬһÖ÷×å¡¢ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ________________(ÌîÐòºÅ)¡£

A£®Ô­×Ӱ뾶£ºRb>I

B£®RbMÖк¬Óй²¼Û¼ü

C£®Æø̬Ç⻯ÎïÈÈÎȶ¨ÐÔ£ºM >I

D£®Rb¡¢Q¡¢MµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï¿ÉÒÔÁ½Á½·¢Éú·´Ó¦

(3)X¡¢Y×é³ÉµÄÒ»ÖÖÎÞÉ«ÆøÌåÓö¿ÕÆø±äΪºì×ØÉ«¡£½«±ê×¼×´¿öÏÂ40 L¸ÃÎÞÉ«ÆøÌåÓë15 LÑõÆøͨÈëÒ»¶¨Å¨¶ÈµÄNaOHÈÜÒºÖУ¬Ç¡ºÃ±»ÍêÈ«ÎüÊÕ£¬Í¬Ê±Éú³ÉÁ½ÖÖÑΡ£Çëд³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÎïÖÊÖУ¬¼Èº¬ÓÐÀë×Ó¼ü£¬ÓÖº¬Óм«ÐÔ¼üµÄÊÇ                   £¨    £©

 A£®CO2                B£®KOH      C£® CaCl2       D£®H2O2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚH£¬H£¬H£¬ Mg£¬Mg£¬O£¬O£¬OÖй²ÓÐ______ÖÖÔªËØ£¬______ÖÖºËËØ£¬ÖÐ×ÓÊý×î¶àµÄÊÇ      ¡£D218OµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÊÇ______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«¼×ÍéÓëÂÈÆø°´Ìå»ý±È1:1»ìºÏÓÚÒ»ÊÔ¹ÜÖв¢µ¹ÖÃÓÚË®²ÛÀÖÃÓÚ¹âÁÁ´¦£¬ÈçͼËùʾ¡£ÏÂÁÐÐðÊöÖУ¬²»ÕýÈ·µÄÊÇ                         

A.Æ¿ÖÐÆøÌåµÄ»ÆÂÌÉ«Öð½¥±ädz                  B.Æ¿ÄÚ±ÚÓÐÓÍ×´ÒºµÎÐγÉ

C.´Ë·´Ó¦µÄÉú³ÉÎïÖ»ÓÐÒ»Âȼ×Íé              D.ÊÔ¹ÜÖÐÒºÃæÉÏÉý

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸