ÌâÄ¿ÄÚÈÝ

(1)Ç⻯ÑÇÍ­(CuH)ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬ÓÃCuSO4ÈÜÒººÍ¡°ÁíÒ»ÎïÖÊ¡±ÔÚ40¡ª50 ¡æʱ·´Ó¦¿ÉÉú³ÉËü¡£CuH¾ßÓеÄÐÔÖÊÓУº²»Îȶ¨£¬Ò׷ֽ⣬ÔÚÂÈÆøÖÐÄÜȼÉÕ£»ÓëÏ¡ÑÎËá·´Ó¦ÄÜÉú³ÉÆøÌ壺Cu+ÔÚËáÐÔÌõ¼þÏ·¢ÉúµÄ·´Ó¦ÊÇ2Cu+Cu2++Cu¡£

¸ù¾ÝÒÔÉÏÐÅÏ¢£¬½áºÏ×Ô¼ºËùÕÆÎյĻ¯Ñ§ÖªÊ¶£¬»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÔÚCuSO4ÈÜÒººÍ¡°ÁíÒ»ÎïÖÊ¡±ÖÆCuHµÄ·´Ó¦ÖУ¬ÓÃÑõ»¯»¹Ô­¹Ûµã·ÖÎö£¬Õâ¡°ÁíÒ»ÎïÖÊ¡±ÔÚ·´Ó¦ÖÐËùÆðµÄ×÷ÓÃÊÇ_______________(Ìî¡°Ñõ»¯¼Á¡±»ò¡°»¹Ô­¼Á¡±)¡£

¢ÚÈç¹û°ÑCuHÈܽâÔÚ×ãÁ¿µÄÏ¡ÏõËáÖÐÉú³ÉµÄÆøÌåÖ»ÓÐNO£¬Çëд³öCuHÈܽâÔÚ×ãÁ¿Ï¡ÏõËáÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ(²»±ØÅäƽ)£º_________________________________¡£¸Ã·´Ó¦ÖеÄÑõ»¯²úÎïÊÇ________________¡£

(2)ÏÂͼÖУ¬PΪһ¸ö¿ÉÒÔ×ÔÓÉ»¬¶¯µÄ»îÈû¡£¹Ø±ÕK£¬·Ö±ðÏòA¡¢BÁ½ÈÝÆ÷Öи÷³äÈë2 mol XºÍ2 mol Y£¬ÔÚÏàͬζȺÍÓд߻¯¼Á´æÔÚµÄÌõ¼þÏ£¬Á½ÈÝÆ÷Öи÷×Ô·¢ÉúÈçÏ·´Ó¦£º

2X(g)+Y(g)2Z(g)+2W(g)

ÒÑÖª£ºÆðʼʱ£¬VA=1 L£¬VB=0.8 L(Á¬Í¨¹ÜÄÚµÄÌå»ýºöÂÔ²»¼Æ)£¬´ïµ½Æ½ºâʱ£¬VB=0.9 L¡£»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙA¡¢BÁ½ÈÝÆ÷´ïµ½Æ½ºâµÄʱ¼ätA________________tB(Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)£»

¢Ú¼ÆËã¸ÃζÈÏ·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK=___________£»

¢Ûµ±A¡¢B·Ö±ð´ïµ½Æ½ºâʱ£¬BÈÝÆ÷ÖÐYµÄת»¯ÂÊΪ____________£¬Á½ÈÝÆ÷ÖÐWµÄÌå»ý·ÖÊý¹ØϵΪA____________B(Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)£»

¢Ü´ò¿ªK£¬Ò»¶Îʱ¼äºó·´Ó¦Ôٴδﵽƽºâ£¬Ôò´ËʱBµÄÌå»ýΪ____________¡£

(1)¢Ù»¹Ô­¼Á

¢ÚCuH+3HNO3Cu(NO3)2+NO¡ü+2H2O  Cu(NO3)2ºÍH2O

(2)¢Ù´óÓÚ  ¢Ú0.74 mol¡¤L-1  ¢Û25%  ´óÓÚ  ¢Ü0.8 L

½âÎö£º(1)£¬CuµÄ»¯ºÏ¼Û½µµÍ£¬ÁíÒ»ÖÖÎïÖʵĻ¯ºÏ¼Û¿Ï¶¨Éý¸ß£¬ËùÒÔÁíÒ»ÖÖÎïÖÊÔÚ·´Ó¦ÖÐ×÷»¹Ô­¼Á¡£

Ñõ»¯²úÎïÊÇCu(NO3)2¡¢H2O¡£

(2)Ũ¶ÈÔ½´ó£¬·´Ó¦ËÙÂÊÔ½¿ì£¬×îÖÕV(A)=1 L£¬V(B)=0.9 L£¬ËùÒÔtA£¾tB¡£

                                  2X(g)+Y(g)2Z(g)+2W(g)

³õʼÎïÖʵÄÁ¿/mol           2        2               0          0

ת»¯ÎïÖʵÄÁ¿/mol           2x      x               2x        2x

ƽºâÎïÖʵÄÁ¿/mol           2-2x  2-x             2x        2x

  x=0.5

K= mol¡¤L-1=0.74 mol¡¤L-1

YµÄת»¯ÂÊΪ¡Á100%=25%

AÈÝÆ÷Ï൱ÓÚÔÚBÈÝÆ÷µÄ»ù´¡ÉϼõСѹǿ£¬Ôö´óÌå»ý£¬Æ½ºâÕýÏòÒƶ¯£¬WµÄÌå»ý·ÖÊýA£¾B¡£

´ò¿ª»îÈûK£¬Ï൱ÓÚA¡¢B¶¼±äΪºãѹϵÄÈÝÆ÷£¬´ïµ½Æ½ºâʱ£¬V(A)=V(B)=0.9 L£¬V (A)+V(B)=1.8 L£¬ËùÒÔ´ËʱBÈÝÆ÷µÄÌå»ýΪ1.8 L-1 L=0.8 L¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?ÏÃÃŶþÄ££©Ñõ»¯ÑÇÍ­ÊÇ´óÐÍË®Ãæ½¢´¬·À»¤Í¿²ãµÄÖØÒªÔ­ÁÏ£®Ä³Ð¡×é½øÐÐÈçÏÂÑо¿£¬ÇëÌîдÏÂÁпհף®
ʵÑé1šâ»¯ÑÇÍ­µÄÖÆÈ¡Ñõ»¯ÑÇÍ­¿ÉÓÃÆÏÌÑÌǺÍÐÂÖÆÇâÑõ»¯Í­Ðü×ÇÒº·´Ó¦ÖÆÈ¡£®ÎÄÏ×±íÃ÷£¬Ìõ¼þ¿ØÖƲ»µ±Ê±»áÓÐÉÙÁ¿CuOÉú³É£®
£¨1£©ÊµÑéÊÒÖÆÈ¡ÇâÑõ»¯Í­Ðü×ÇÒºµÄÀë×Ó·½³ÌʽΪ
Cu2++2OH-=Cu£¨OH£©2¡ý
Cu2++2OH-=Cu£¨OH£©2¡ý

£¨2£©ÊµÑéÊÒÓô˷½·¨ÖÆÈ¡²¢»ñµÃÉÙÁ¿Ñõ»¯ÑÇÍ­¹ÌÌ壬ÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÊԹܡ¢¾Æ¾«µÆ¡¢ÉÕ ±­
©¶·
©¶·
ºÍ
²£Á§°ô
²£Á§°ô

£¨3£©ÈôҪ̽¾¿¸Ã·´Ó¦·¢ÉúµÄ×îµÍζȣ¬Ó¦Ñ¡ÓõļÓÈÈ·½Ê½Îª
ˮԡ¼ÓÈÈ
ˮԡ¼ÓÈÈ
£®
ʵÑé2²â¶¨šâ»¯ÑÇÍ­µÄ´¿¶È
·½°¸1£º³ÆȡʵÑé1ËùµÃ¹ÌÌåm g£¬²ÉÓÃÈçÏÂ×°ÖýøÐÐʵÑ飮

£¨4£©×°ÖÃaÖÐËù¼ÓµÄËáÊÇ
H2SO4
H2SO4
 £¨Ìѧʽ£©£®
£¨5£©Í¨¹ý²â³öÏÂÁÐÎïÀíÁ¿£¬ÄܴﵽʵÑéÄ¿µÄÊÇ
BC
BC
£®
A£®·´Ó¦Ç°ºó×°ÖÃaµÄÖÊÁ¿B£®×°ÖÃc³ä·Ö·´Ó¦ºóËùµÃ¹ÌÌåµÄÖÊÁ¿
C£®·´Ó¦Ç°ºó×°ÖÃdµÄÖÊÁ¿D£®·´Ó¦Ç°ºó×°ÖÃeµÄÖÊÁ¿
£¨6£©ÔÚÇâÆøÑé´¿ºó£¬µãȼװÖÃcÖоƾ«µÆ֮ǰÐèÒª¶ÔK1¡¢K2½øÐеIJÙ×÷ÊÇ
´ò¿ªK2£¬¹Ø±ÕK1
´ò¿ªK2£¬¹Ø±ÕK1

·½°¸2£º½«ÊµÑé1ËùµÃ¹ÌÌåmgÈÜÓÚ×ãÁ¿Ï¡ÁòËᣬ¾­¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïºó³Æ³ö²»ÈÜÎïµÄÖÊÁ¿ £¨×ÊÁÏ£ºCu2O+2H+=Cu2++Cu+H2O£©
£¨7£©ÅжϾ­¸ÉÔïÆ÷¸ÉÔïºóµÄ²»ÈÜÎïÊÇ·ñÒÑÍêÈ«¸ÉÔïµÄ²Ù×÷·½·¨ÊÇ
½«¸Ã²»ÈÜÎïÔٴθÉÔïºó³ÆÁ¿£¬Ö±µ½×îºóÁ½´ÎÖÊÁ¿»ù±¾Ïàͬ£»
½«¸Ã²»ÈÜÎïÔٴθÉÔïºó³ÆÁ¿£¬Ö±µ½×îºóÁ½´ÎÖÊÁ¿»ù±¾Ïàͬ£»

£¨8£©ÈôʵÑéËùµÃ²»ÈÜÎïΪ
n
n
g£¬Ôò¸ÃÑùÆ·ÖÐÑõ»¯ÑÇÍ­µÄÖÊÁ¿·ÖÊýΪ
9n
4m
¡Á100%£®
9n
4m
¡Á100%£®
£®

£¨15·Ö£©ÔÚʵÑéÊÒÖÐijʵÑéС×éͬѧ¹ØÓÚÁòËáÍ­µÄÖÆÈ¡ºÍÓ¦Óã¬Éè¼ÆÁËÒÔÏÂʵÑ飺
£¨1£©Í­ÓëŨÁòËá·´Ó¦£¬ÊµÑé×°ÖÃÈçͼËùʾ¡£
            
¢Ù×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ              £»
¢ÚÉÏͼװÖÃÖе¼¹ÜBµÄ×÷ÓÃÄãÈÏΪ¿ÉÄÜÊÇ£¨Ð´³öÒ»ÖÖ£©                  ¡£
£¨2£©Îª·ûºÏÂÌÉ«»¯Ñ§µÄÒªÇó£¬Ä³Í¬Ñ§½øÐÐÈçÏÂÉè¼Æ£º½«Í­·ÛÔÚ                  £¨ÌîÒÇÆ÷Ãû³Æ£©Öз´¸´×ÆÉÕ£¬Ê¹Í­Óë¿ÕÆø³ä·Ö·´Ó¦Éú³ÉÑõ»¯Í­£¬ÔÙ½«Ñõ»¯Í­ÓëÏ¡ÁòËá·´Ó¦£¬·´Ó¦ºóÈÜÒº¾­¹ý            ¡¢            ¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ¼´¿ÉµÃµ½²úÆ·CuSO4¡¤5H2O¾§Ì壬¸ÉÔïÐèÒª¿ØÖÆζÈСÓÚ100¡æ£¬Èôζȹý¸ß£¬Ôò»áµ¼Ö              ¡£
£¨3£©½«¿ÕÆø»òÑõÆøÖ±½ÓͨÈ뵽ͭ·ÛÓëÏ¡ÁòËáµÄ»ìºÏÎïÖУ¬·¢ÏÖÔÚ³£ÎÂϼ¸ºõ²»·´Ó¦¡£Ïò·´Ó¦ÒºÖмÓÉÙÁ¿£ºFeSO4×÷´ß»¯¼Á£¬¼´·¢Éú·´Ó¦£¬Éú³ÉÁòËáÍ­¡£Æä·´Ó¦¹ý³ÌµÄµÚ2²½ÊÇ£º2Fe3++Cu=2Fe2++Cu2+£¬Çëд³öÆäµÚl²½·´Ó¦µÄÀë×Ó·½³Ìʽ                  ¡£
£¨4£©ÖÆÈ¡µÄCuSO4ÈÜÒººÍ¡°ÁíÒ»ÎïÖÊ¡±ÔÚ40-50¡æ»ìºÏʱÉú³ÉÁËÒ»ÖÖÄÑÈÜÎïÖÊÇ⻯ÑÇÍ­£¨CuH£©¡£½«CuHÈܽâÔÚÏ¡ÑÎËáÖÐʱÉú³ÉÁËÒ»ÖÖÆøÌ壬ÕâÖÖÆøÌåÊÇ                  £¬¾­ÊÕ¼¯²â¶¨Éú³ÉµÄ¸ÃÆøÌåΪ±ê¿öÏÂ11£®2 L£¬Ôò±»»¹Ô­µÄÀë×ӵõç×ÓµÄÎïÖʵÄÁ¿ÊÇ                  £¨ÒÑÖªCu+ÔÚËáÐÔÌõ¼þÏ·¢Éú·´Ó¦2Cu+=Cu2++Cu£©¡£

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

¾«Ó¢¼Ò½ÌÍø