ÌâÄ¿ÄÚÈÝ

5£®ÈçͼÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö

£¨1£©Ð´³öÔªËØ¢âµÄ»ù̬ԭ×ӵĵç×ÓÅŲ¼¼ò»¯Ê½[Ar]3d64s2£¬ÍâΧµç×ÓÅŲ¼Í¼3d64s2£®Ö¸³öËüÔÚÖÜÆÚ±íÖеÄλÖõÚËÄÖÜÆÚµÚ¢õ¢ó×壮º¸½Ó¸Ö¹ìʱ£¬³£ÀûÓââµÄijЩÑõ»¯ÎïÓë¢ßµÄµ¥ÖÊÔÚ¸ßÎÂÏ·¢Éú·´Ó¦£¬ÊÔд³öÆäÖÐÒ»Ìõ·´Ó¦µÄ»¯Ñ§·½³Ìʽ2Al+Fe2O3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$Al2O3+2Fe£®
£¨2£©¢Ù¢Û¢ÝÈýÖÖÔªËØ¿ÉÒÔÐγɶàÖÖÓлú»¯ºÏÎï·Ö×Ó£¬ÆäÖÐ×î¼òµ¥Ô­×ÓÊý×îÉÙµÄÒ»ÖÖÊÇÊÒÄÚ×°äêʱÐγɵÄÖ÷ÒªÆøÌåÎÛȾÎÊÔд³öËüµÄµç×Óʽ£¬ÍƲâ¸Ã·Ö×ӵĿռ乹ÐÍΪƽÃæÈý½ÇÐΣ®
£¨3£©¢Û¢Ü¢Ý¢Þ¢àÎåÖÖÔªËض¼¿ÉÒÔÓëÔªËØ¢ÙÐγɻ¯ºÏÎÆäÖÐÈÛµã×î¸ßµÄÊÇNaH£¨Ð´»¯ºÏÎïµÄ»¯Ñ§Ê½£©£®Èç¹ûÔÚζȽӽü373Kʱ£¬¸ù¾ÝM=m/n²â¶¨¢ÝµÄÆø̬Ç⻯ÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿£¬½á¹û·¢Ïֲⶨ½á¹û×ÜÊDZÈÀíÂÛÖµ¸ß£¬ÆäÔ­ÒòÊÇË®·Ö×Ó¼ä´æÔÚÇâ¼üË®·Ö×ÓÓë·Ö×Ó¼äÐγɵ޺ÏÎ
£¨4£©Ä³Ð©²»Í¬×åÔªËصÄÐÔÖÊÒ²ÓÐÒ»¶¨µÄÏàËÆÐÔ£¬ÈçͼÖÐÔªËØ¢ßÓëÔªËØ¢ÚµÄÇâÑõ»¯ÎïÓÐÏàËƵÄÐÔÖÊ£®Çëд³öÔªËØ¢ÚµÄÇâÑõ»¯ÎïÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽBe£¨OH£©2+2NaOH=Na2BeO2+2H2O£®

·ÖÎö ÓÉÔªËØÔÚÖÜÆÚ±íÖеÄλÖÿÉÖª£¬¢ÙΪH£¬¢ÚΪBe£¬¢ÛΪC£¬¢ÜΪN£¬¢ÝΪO£¬¢ÞΪNa£¬¢ßΪAl£¬¢àΪCl£¬¢áΪCr£¬¢âΪFe£¬
£¨1£©FeµÄÔ­×ÓÐòÊýΪ26£¬¼Ûµç×ÓΪ3d64s2£¬Î»ÓÚµÚËÄÖÜÆÚµÚ¢õ¢ó×壬¢âµÄijЩÑõ»¯ÎïÓë¢ßµÄµ¥ÖÊÔÚ¸ßÎÂÏ·¢ÉúÂÁÈÈ·´Ó¦£»
£¨2£©¢Ù¢Û¢ÝÈýÖÖÔªËØÐγÉÓлúÎïÖÐ×î¼òµ¥Ô­×ÓÊý×îÉÙÎïÖÊÊÇÊÒÄÚ×°äêʱÐγɵÄÖ÷ÒªÆøÌåÎÛȾÎ¸ÃÆøÌåΪ¼×È©£»
£¨3£©¢Û¢Ü¢Ý¢Þ¢àÎåÖÖÔªËض¼¿ÉÒÔÓëÔªËØ¢ÙÐγɻ¯ºÏÎֻÓÐÄÆÓëÇâÐγÉÇ⻯ÄÆÀë×Ó¾§Ì壬ÆäËü¶¼ÊÇ·Ö×Ó¾§Ì壬ËùÒÔÇ⻯ÄƵÄÈÛµã×î¸ß£¬Ë®·Ö×Ó¼ä´æÔÚÇâ¼üÐγɵ޺ÏÎ
£¨4£©ÔªËØ¢ßÓëÔªËØ¢ÚµÄÇâÑõ»¯ÎïÓÐÏàËƵÄÐÔÖÊ£¬ÔªËØ¢ÚµÄÇâÑõ»¯ÎïÓëNaOHÈÜÒº·´Ó¦Éú³ÉNa2BeO2ºÍË®£®

½â´ð ½â£ºÓÉÔªËØÔÚÖÜÆÚ±íÖеÄλÖÿÉÖª£¬¢ÙΪH£¬¢ÚΪBe£¬¢ÛΪC£¬¢ÜΪN£¬¢ÝΪO£¬¢ÞΪNa£¬¢ßΪAl£¬¢àΪCl£¬¢áΪCr£¬¢âΪFe£¬
£¨1£©FeµÄÔ­×ÓÐòÊýΪ26£¬»ù̬ԭ×ӵĵç×ÓÅŲ¼¼ò»¯Ê½[Ar]3d64s2£»¼Ûµç×ÓΪ3d64s2£¬Î»ÓÚµÚËÄÖÜÆÚµÚ¢õ¢ó×壬¢âµÄijЩÑõ»¯ÎïÓë¢ßµÄµ¥ÖÊÔÚ¸ßÎÂÏ·¢ÉúÂÁÈÈ·´Ó¦£¬Èç2Al+Fe2O3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$Al2O3+2Fe£¬
¹Ê´ð°¸Îª£º[Ar]3d64s2£»3d64s2£»µÚËÄÖÜÆÚµÚ¢õ¢ó×壻2Al+Fe2O3$\frac{\underline{\;¸ßÎÂ\;}}{\;}$Al2O3+2Fe£»
£¨2£©¢Ù¢Û¢ÝÈýÖÖÔªËØÐγÉÓлúÎïÖÐ×î¼òµ¥Ô­×ÓÊý×îÉÙÎïÖÊÊÇÊÒÄÚ×°äêʱÐγɵÄÖ÷ÒªÆøÌåÎÛȾÎ¸ÃÆøÌåΪ¼×È©£¬µç×ÓʽΪ£º£¬·Ö×ÓÖÐCÔ­×Ó³Ê2¸öC-H¼ü¡¢1¸öC=O¼ü£¬CÔ­×Ó¼Û²ãµç×Ó¶ÔÊýΪ2+1=3£¬²»º¬¹Âµç×Ó¶Ô£¬ÎªÆ½ÃæÈý½ÇÐΣ¬
¹Ê´ð°¸Îª£º£»Æ½ÃæÈý½ÇÐΣ»
£¨3£©¢Û¢Ü¢Ý¢Þ¢àÎåÖÖÔªËض¼¿ÉÒÔÓëÔªËØ¢ÙÐγɻ¯ºÏÎֻÓÐÄÆÓëÇâÐγÉÇ⻯ÄÆÀë×Ó¾§Ì壬ÈÛµã×î¸ß£¬Ë®·Ö×Ó¼ä´æÔÚÇâ¼üÐγɵ޺ÏÎËùÒԲⶨ½á¹û×ÜÊDZÈÀíÂÛÖµ¸ß£¬¹Ê´ð°¸Îª£ºNaH£»Ë®·Ö×Ó¼ä´æÔÚÇâ¼üË®·Ö×ÓÓë·Ö×Ó¼äÐγɵ޺ÏÎ
£¨4£©ÔªËØ¢ßÓëÔªËØ¢ÚµÄÇâÑõ»¯ÎïÓÐÏàËƵÄÐÔÖÊ£¬ÔªËØ¢ÚµÄÇâÑõ»¯ÎïÓëNaOHÈÜÒº·´Ó¦Éú³ÉNa2BeO2ºÍË®£¬¸Ã·´Ó¦ÎªBe£¨OH£©2+2NaOH=Na2BeO2+2H2O£¬
¹Ê´ð°¸Îª£ºBe£¨OH£©2+2NaOH=Na2BeO2+2H2O£®

µãÆÀ ±¾Ì⿼²éÔªËØÖÜÆÚ±í¼°Ó¦Ó㬰ÑÎÕÔªËØÔÚÖÜÆÚ±íÖеÄλÖá¢Ô­×ӽṹÓëÐÔÖÊ¡¢ÔªËØ»¯ºÏÎï֪ʶΪ½â´ðµÄ¹Ø¼ü£¬²àÖØÔªËØÍƶϼ°»¯Ñ§ÓÃÓïµÄ¿¼²é£¬×¢Ò⣨4£©ÖÐÐÔÖʵÄÏàËÆ£¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
14£®Îª±£»¤»·¾³£¬½ÚÔ¼×ÊÔ´£¬Ä³Ñо¿ÐÔѧϰС×é̽¾¿ÓÃÒ×ק¹ÞÖÆÈ¡Ã÷·¯[KAl£¨SO4£©2•12H2O]£®
²éÔÄ×ÊÁϵÃÖª£ºÀÊ·¯ÈÜÓÚË®£¬²»ÈÜÓÚÒÒ´¼£ºÒ×À­¹ÞÖ÷Òª³É·ÖΪÂÁ£¬Áíº¬ÓÐþ¡¢ÌúµÈ£®
ʵÑé²½ÖèÈçÏ£º
²½Öè1¡¢½«Ò×À­¹Þ¼ô¿ª£¬²Ã³ÉÂÁƬ£®ÓÃɰֽĥȥ±íÃæµÄÓÍÆᣬÑÕÁϼ°Í¸Ã÷ËܽºÄڳģº
²½Öè2¡¢ÖƱ¸NaAlO2
³ÆÁ¿1gÉÏÊö´¦Àí¹ýµÄÂÁƬ£¬ÇÐË飬·ÖÊý´Î·ÅÈëÊ¢ÓÐ40mL5% NaOHÈÜÒºµÄÉÕ±­ÖУº½«ÉÕ±­ÖÃÓÚÈÈˮԡÖмÓÈÈ£®·´Ó¦Íê±Ïºó£¬ÈÜÒº³Ê»ÒºÚÉ«»ë×Ç£¬³ÃÈȹýÂË£®
²½Öè3¡¢ÇâÑõ»¯ÂÁµÄÉú³ÉºÍÏ´µÓ
ÔÚËùµÃÂËÒºÖеμÓ3mol•L-1H2SO4ÈÜÒº£¬ÓÃpHÊÔÖ½¼ìÑ飬µ÷½ÚpHÖÁ8¡«9Ϊֹ£º´ËʱÈÜÒºÖÐÉú³É´óÁ¿µÄ°×É«ÇâÑõ»¯ÂÁ³Áµí£¬¹ýÂË£¬²¢ÓÃÈÈÕôÁóË®¶à´ÎÏ´µÓ³Áµí
²½Öè4¡¢Ã÷·¯µÄÖƱ¸
½«¹ýÂ˺óËùµÃÇâÑõ»¯ÂÁ³ÁµíתÈëÕô·¢ÃóÖУ¬¼Ó10mL 9mol/L H2SO4£¬ÔÙ¼Ó15mLË®£¬Ð¡»ð¼ÓÈÈʹÆäÈܽ⣬¼ÓÈë4gÁòËá¼Ø¼ÌÐø¼ÓÈÈÖÁÈܽ⣬½«ËùµÃÈÜÒºÔÚ¿ÕÆøÖÐ×ÔÈ»ÀäÈ´£¬´ý½á¾§ÍêÈ«ºó£¬¹ýÂË£¬ÓÃÎÞË®¾Æ¾«Ï´µÓ¾§ÌåÁ½´Î£»½«¾§ÌåÓÃÂËÖ½Îü¸É£¬ÖƵÃÃ÷·¯£®
»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©²½Öè2ÖÐÂÁƬÈÜÓÚÉÕ¼îµÄ·´Ó¦½¨ÒéÔÚͨ·ç³÷ÖнøÐУ¬ÓÃÈÈˮԡ¼ÓÈÈ£¬±ÜÃâÃ÷»ð£¬Ô­ÒòÊÇ·´Ó¦ÖÐÉú³ÉµÄÇâÆøÓöµ½Ã÷»ðÒ×±¬Õ¨£»¹ýÂ˺ó£¬ÂËÔüµÄÖ÷Òª³É·ÖÊÇþ¡¢ÌúµÈ£®
£¨2£©²½Öè3ÖУ¬Èôµ÷½ÚpH¹ýµÍ£¬µ¼ÖµĽá¹ûÊDz¿·ÖÇâÑõ»¯ÂÁÈܽ⣻ϴµÓÇâÑõ»¯ÂÁ³ÁµíµÄ·½·¨ÊÇÓò¼ÊÏ©¶·³éÂË£¬²¢ÓÃÈÈˮϴµÓ³Áµí£®
£¨3£©²½Öè4ÖÐÓõ½µÄÖ÷ÒªÒÇÆ÷ÊÇ£ºÌú¼Ų̈£¨´øÌúȦ£©£¬Õô·¢Ãó¡¢Á¿Í²¡¢Â©¶·¡¢ÉÕ±­¡¢¾Æ¾«µÆºÍ²£Á§°ô£»¼ÓÈëÁòËá¼ØºóÖÆÈ¡Ã÷·¯µÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇAl2£¨SO4£©3+K2SO4+24H2O=2KAl£¨SO4£©2•12H2O£®
£¨4£©²»ÓÃË®¶øÓÃÎÞË®¾Æ¾«Ï´µÓÃ÷·¯¾§ÌåµÄÔ­ÒòÊÇÃ÷·¯Ò×ÈÜÓÚË®¡¢²»ÈÜÓÚÎÞË®¾Æ¾«£®
£¨5£©ÊµÑéÑéÖ¤Ã÷·¯¾§ÌåÖк¬ÓÐSO2-4Àë×ӵķ½·¨ÊÇÈ¡ÉÙÁ¿Ã÷·¯ÈÜÓÚË®£¬ÏȼÓÈëÑÎËᣬÎÞÏÖÏó£¬ÔÙ¼ÓÈëÂÈ»¯±µÈÜÒºÉú³É°×É«³Áµí£®
£¨6£©¸ÃʵÑéÖƵÃÃ÷·¯15.8g£¬Ôò³ÆÈ¡µÄ1gÂÁƬÖÐÂÁÔªËغ¬Á¿²»µÍÓÚ90%£®

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

¾«Ó¢¼Ò½ÌÍø