ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿FeSe ¡¢MgB2µÈ³¬µ¼²ÄÁϾßÓйãÀ«µÄÓ¦ÓÃÇ°¾°¡£

£¨1£©»ù̬FeÔ­×Ó¼Û²ãµç×ӵĵç×ÓÅŲ¼Í¼£¨¹ìµÀ±í´ïʽ£©Îª________£¬»ù̬SeÔ­×ӵĵç×ÓÕ¼¾Ý×î¸ßÄܼ¶µÄµç×ÓÔÆÂÖÀªÍ¼Îª________ÐΡ£

£¨2£©ÏòFeSeÖÐǶÈëßÁण¨£©Äܵõ½¾ßÓÐÓÅÒìÐÔÄܵij¬µ¼²ÄÁÏ¡£ßÁà¤ÖеªÔ­×ÓµÄÔÓ»¯ÀàÐÍΪ________£»¸Ã·Ö×ÓÄÚ´æÔÚ________£¨Ìî±êºÅ£©¡£

A£®¦Ò¼ü B£®¦Ð¼ü C£®Åäλ¼ü D£®Çâ¼ü

£¨3£©½«½ðÊôï®Ö±½ÓÈÜÓÚÒº°±£¬µÃµ½¾ßÓкܸ߷´Ó¦»îÐԵĽðÊôµç×ÓÈÜÒº£¬ÔÙͨ¹ýϵÁз´Ó¦¿ÉÖƵÃFeSe»ù³¬µ¼²ÄÁÏLi0.6(NH2)0.2(NH3)0.8Fe2Se2¡£

¢ÙNH2£­µÄ¿Õ¼ä¹¹ÐÍΪ________¡£

¢ÚÒº°±ÊÇ°±ÆøÒº»¯µÄ²úÎï,°±ÆøÒ×Òº»¯µÄÔ­ÒòÊÇ________¡£

¢Û½ðÊôï®ÈÜÓÚÒº°±Ê±·¢Éú·´Ó¦£ºLi + (m+n)NH3=X+e£­(NH3)n¡£XµÄ»¯Ñ§Ê½Îª________¡£

£¨4£©MgB2¾§Ìå½á¹¹ÈçͼËùʾ¡£BÔ­×Ó¶ÀÁ¢ÎªÒ»²ã£¬¾ßÓÐÀàËÆÓÚʯīµÄ½á¹¹£¬Ã¿¸öBÔ­×ÓÖÜΧ¶¼ÓÐ________¸öÓëÖ®µÈ¾àÀëÇÒ×î½üµÄBÔ­×Ó£»ÁùÀâÖùµ×±ß±ß³¤Îªa cm£¬¸ßΪc cm£¬°¢·ü¼ÓµÂÂÞ³£ÊýµÄֵΪNA £¬¸Ã¾§ÌåµÄÃܶÈΪ________ g¡¤cm£­3£¨Áгö¼ÆËãʽ£©¡£

¡¾´ð°¸¡¿ ÑÆÁ壨·Ä´¸£© sp2 ÔÓ»¯ AB v ÐÍ °±·Ö×Ó¼ä´æÔÚÇâ¼ü£¬·Ö×Ó¼ä×÷ÓÃÁ¦½ÏÇ¿£¬ÈÝÒ×Òº»¯ Li(NH3)m+ 3

¡¾½âÎö¡¿

£¨1£©FeÔ­×Ӻ˵çºÉÊýΪ26£¬»ù̬FeÔ­×Óµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d64s2£»SeÔ­×Ӻ˵çºÉÊýΪ34£¬»ù̬SeÔ­×Óµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d104s24p4£¬ÒÔ´Ë·ÖÎö½â´ð£»

£¨2£©ßÁà¤NºÍÁ½¸ö̼³É¼ü£¬ÓÐÒ»¸öδ³É¶Ôµç×ÓºÍÆäËüÎå¸ö̼ԭ×ÓÐγɴó¦Ð¼ü£¬Òò´Ë·¢Éú²»µÈÐÔµÄSp2ÔÓ»¯£»¸Ã·Ö×ÓÄÚ´æÔÚ¦Ò¼üºÍ¦Ð¼ü£»

£¨3£©¢ÙNH2£­ÖÐNÔ­×Ó¼Û²ãµç×Ó¶ÔÊý==4£¬ÓÐ2¶Ô¹Âµç×Ó¶Ô£¬ËùÒԿռ乹ÐÍΪv ÐÍ£»

¢Ú°±·Ö×Ó¼ä´æÔÚÇâ¼ü£¬·Ö×Ó¼ä×÷ÓÃÁ¦½ÏÇ¿£¬ÈÝÒ×Òº»¯£»

¢ÛLi + (m+n)NH3=X+e£­(NH3)n£¬¸ù¾ÝÔ­×ÓÊغãºÍµçºÉÊغã¿ÉÈ·¶¨XµÄ»¯Ñ§Ê½£»

£¨4£©ÓÉÓÚBÔ­×Ó¶ÀÁ¢ÎªÒ»²ã£¬¾ßÓÐÀàËÆÓÚʯīµÄ½á¹¹£¬¶øʯī¾§ÌåÖÐÿ¸ö̼ԭ×ÓÓëÆäËü3¸ö̼ԭ×ӵȾàÇÒ×î½ü£¬¹Ê¸Ã¾§ÌåÖÐÿ¸öBÔ­×ÓÖÜΧ¶¼ÓÐ3¸öÓëÖ®µÈ¾àÀëÇÒ×î½üµÄBÔ­×Ó£»¸ù¾Ý=¼ÆË㾧ÌåµÄÃܶȡ£

£¨1£©FeÔ­×Ӻ˵çºÉÊýΪ26£¬»ù̬FeÔ­×Óµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d64s2£¬Ôò»ù̬FeÔ­×ӵĺËÍâ¼Ûµç×ÓÅŲ¼Í¼Îª£»SeÔ­×Ӻ˵çºÉÊýΪ34£¬»ù̬SeÔ­×Óµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d104s24p4£¬»ù̬SeÔ­×Óµç×ÓÕ¼¾ÝµÄÄܼ¶ÓÐ1s¡¢2s¡¢2p¡¢3p¡¢3d¡¢4s¡¢4P£¬×î¸ßÄܼ¶Îª4p£¬Æäµç×ÓÔÆÂÖÀªÍ¼ÎªÑÆÁåÐΣ¬

¹Ê´ð°¸Îª£º£»ÑÆÁ壻

£¨2£©ßÁà¤NºÍÁ½¸ö̼³É¼ü£¬ÓÐÒ»¸öδ³É¶Ôµç×ÓºÍÆäËüÎå¸ö̼ԭ×ÓÐγɴó¦Ð¼ü£¬Òò´Ë·¢Éú²»µÈÐÔµÄSp2ÔÓ»¯£»¸Ã·Ö×ÓÄÚ´æÔÚ¦Ò¼üºÍ¦Ð¼ü£¬²»´æÔÚÅäλ¼üºÍÇâ¼ü£¬¹ÊÑ¡AB£»

¹Ê´ð°¸Îª£ºsp2 ÔÓ»¯£»AB£»

£¨3£©¢ÙNH2£­ÖÐNÔ­×Ó¼Û²ãµç×Ó¶ÔÊý==4£¬ÓÐ2¶Ô¹Âµç×Ó¶Ô£¬ËùÒԿռ乹ÐÍΪv ÐÍ£»

¢Ú°±·Ö×Ó¼ä´æÔÚÇâ¼ü£¬·Ö×Ó¼ä×÷ÓÃÁ¦½ÏÇ¿£¬ÈÝÒ×Òº»¯£»

¢ÛLi + (m+n)NH3=X+e£­(NH3)n£¬¸ù¾ÝÔ­×ÓÊغãºÍµçºÉÊغã¿ÉÈ·¶¨XµÄ»¯Ñ§Ê½ÎªLi(NH3)m+£»

¹Ê´ð°¸Îª£ºv ÐÍ£»°±·Ö×Ó¼ä´æÔÚÇâ¼ü£¬·Ö×Ó¼ä×÷ÓÃÁ¦½ÏÇ¿£¬ÈÝÒ×Òº»¯£»Li(NH3)m+£»

£¨4£©¾§°ûÖÐMgµÄ¸öÊýΪ12+2=3£¬BµÄ¸öÊýΪ6£¬¾§°ûÌå»ý=6cm3=6 cm3£¬¾§°ûÃܶÈ== g¡¤cm£­3£¬

¹Ê´ð°¸Îª£º¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿½üÄêÀ´£¬Ñо¿ÈËÔ±Ìá³öÀûÓú¬ÁòÎïÖÊÈÈ»¯Ñ§Ñ­»·ÊµÏÖÌ«ÑôÄܵÄת»¯Óë´æ´¢¡£¹ý³ÌÈçÏ£º

£¨1£©·´Ó¦¢ñ£º2H2SO4(l)=2SO2(g)+2H2O(g)+O2(g) ¦¤H1=+551 kJ¡¤mol£­1

·´Ó¦¢ó£ºS(s)+O2(g)=SO2(g) ¦¤H3=£­297 kJ¡¤mol£­1

·´Ó¦¢òµÄÈÈ»¯Ñ§·½³Ìʽ£º________________________________¡£

£¨2£©I£­¿ÉÒÔ×÷ΪˮÈÜÒºÖÐSO2Æ绯·´Ó¦µÄ´ß»¯¼Á£¬¿ÉÄܵĴ߻¯¹ý³ÌÈçÏ¡£½«ii²¹³äÍêÕû¡£

i£®SO2+4I£­+4H+=S¡ý+2I2+2H2O

ii£®I2+2H2O+_________=_________+_______+2 I£­_____________

£¨3£©Ì½¾¿i¡¢ii·´Ó¦ËÙÂÊÓëSO2Æ绯·´Ó¦ËÙÂʵĹØϵ£¬ÊµÑéÈçÏ£º·Ö±ð½«18 mL SO2±¥ºÍÈÜÒº¼ÓÈëµ½2 mLÏÂÁÐÊÔ¼ÁÖУ¬ÃܱշÅÖù۲ìÏÖÏó¡££¨ÒÑÖª£ºI2Ò×ÈܽâÔÚKIÈÜÒºÖУ©

ÐòºÅ

A

B

C

D

ÊÔ¼Á×é³É

0.4 mol¡¤L£­1 KI

a mol¡¤L£­1 KI

0.2 mol¡¤L£­1 H2SO4

0.2 mol¡¤L£­1 H2SO4

0.2 mol¡¤L£­1 KI

0.0002 mol I2

ʵÑéÏÖÏó

ÈÜÒº±ä»Æ£¬Ò»¶Îʱ¼äºó³öÏÖ»ë×Ç

ÈÜÒº±ä»Æ£¬³öÏÖ»ë×ǽÏA¿ì

ÎÞÃ÷ÏÔÏÖÏó

ÈÜÒºÓÉ×غÖÉ«ºÜ¿ìÍÊÉ«£¬±ä³É»ÆÉ«£¬³öÏÖ»ë×ǽÏA¿ì

¢ÙBÊÇAµÄ¶Ô±ÈʵÑ飬Ôòa=__________¡£

¢Ú±È½ÏA¡¢B¡¢C£¬¿ÉµÃ³öµÄ½áÂÛÊÇ______________________¡£

¢ÛʵÑé±íÃ÷£¬SO2µÄÆ绯·´Ó¦ËÙÂÊD£¾A£¬½áºÏi¡¢ii·´Ó¦ËÙÂʽâÊÍÔ­Òò£º________________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø