ÌâÄ¿ÄÚÈÝ

º¬Ã¾3%¡«5%µÄÂÁþºÏ½ðÊÇÂÖ´¬ÖÆÔì¡¢»¯¹¤Éú²ú¡¢»úеÖÆÔìµÈÐÐÒµµÄÖØÒª²ÄÁÏ£®ÏÖÓÐÒ»¿éÖÊÁ¿Îªm gµÄÂÁþºÏ½ð£¬Óû²â¶¨ÆäÖÐþµÄÖÊÁ¿·ÖÊý£¬¼¸Î»Í¬Ñ§Éè¼ÆÁ˲»Í¬µÄʵÑé·½°¸£®
·½°¸1£ºÂÁþºÏ½ð²â¶¨Éú³ÉÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý£¨V1L£©
·½°¸2£ºÂÁþºÏ½ð³ä·Ö·´Ó¦ºó²â¶¨Ê£Óà¹ÌÌåµÄÖÊÁ¿£¨W1g£©
·½°¸3£ºÂÁþºÏ½ðÈÜÒº¹ýÂË£¬²â¶¨³ÁµíµÄÖÊÁ¿£¨W2g£©
£¨¢ñ£©Ä³»¯Ñ§ÊµÑéС×éÀûÓÃÓÒͼËùʾʵÑé×°Ö㬰´ÕÕ·½°¸1½øÐÐÁËʵÑ飬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù×°ÖÃÖÐÒÇÆ÷aµÄÃû³ÆÊÇ______£®
¢ÚʵÑé×°ÖÃÖÐÓÐÒ»´¦Ã÷ÏÔ´íÎó£¬ÇëÖ¸³ö______£®
£¨¢ò£©·½°¸2Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®
£¨¢ó£©·½°¸3¡°¹ýÂË¡±²Ù×÷ÖÐÓõ½µÄ²£Á§°ôµÄ×÷ÓÃÊÇ______£®
Èô°´·½°¸3½øÐÐÊÔÑ飬²âµÃþµÄÖÊÁ¿·ÖÊýΪ______£®
£¨¢ô£©Ä³Í¬Ñ§ÓÐÉè¼ÆÁËÓëÉÏÊö·½°¸²»Í¬µÄ·½°¸4£¬Ò²²âµÃÁËþµÄÖÊÁ¿·ÖÊý£®ÇëÄãÔÚÀ¨ºÅÄÚÌîÉϺÏÊʵÄÄÚÈÝ£¬½«¸Ã·½°¸²¹³äÍêÕû£®
·½°¸4£ºÂÁþºÏ½ð²â¶¨Éú³ÉÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý£¨V2L£©£®

¡¾´ð°¸¡¿·ÖÎö£º£¨¢ñ£©¢ÙaÊÇ·ÖҺ©¶·£»
¢Ú¹ã¿ÚÆ¿µÄÅÅË®µ¼¹Ü½øÈëµÄÓ¦¸ÃÔÚÒºÃæÒÔÉÏ£¬³öµÄµ¼¹ÜÓ¦¸ÃÔÚÒºÃæÒÔÏ£»
£¨¢ò£©½ðÊôÂÁÄܹ»ºÍÇ¿¼îÈÜÒº·´Ó¦Éú³ÉÆ«ÂÁËáÑΣ»
£¨¢ó£©¸ù¾Ý²£Á§°ôÔÚ¹ýÂËÖеÄ×÷ÓÃÍê³É£»¸ù¾Ý³ÁµíÊÇÇâÑõ»¯Ã¾£¬Çó³öÇâÑõ»¯Ã¾µÄÎïÖʵÄÁ¿£¬ÀûÓÃþԭ×ÓÊغ㣬¼ÆËã³öºÏ½ðÖÐþµÄÖÊÁ¿·ÖÊý£»
£¨¢ô£©¿ÉÒÔÀûÓÃÂÁºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉµÄÇâÆøÌå»ý¼ÆËãºÏ½ðÖÐþµÄÖÊÁ¿·ÖÊý£®
½â´ð£º½â£º£¨¢ñ£©£©¢Ù¸ù¾Ýͼʾ²»ÄÑ¿´³ö£¬ÒÇÆ÷aÃû³ÆÊÇ·ÖҺ©¶·£¬¹Ê´ð°¸Îª£º·ÖҺ©¶·£»
¢ÚÓÉÓÚÊǸù¾ÝÅųöµÄË®µÄÌå»ýÀ´²â¶¨Éú³ÉµÄÆøÌåµÄÌå»ý£¬Òò´Ë¹ã¿ÚÆ¿ÖеĽøÆø¹ÜÓ¦¸Ã¸Õ¶³öÏðƤÈû£¬¶ø³öÆø¹ÜÓ¦ÉìÈ뵽ƿµ×£¬
¹Ê´ð°¸Îª£º¹ã¿ÚÆ¿ÖнøÆøµ¼¹Ü²åÈëµ½ÁËË®ÖУ¬ÅÅË®µ¼¹Üδ²åÖÁ¹ã¿ÚÆ¿µ×²¿£»
£¨¢ò£©ÂÁºÍÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü£¬¹Ê´ð°¸Îª£º2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü£»
£¨¢ó£©²£Á§°ôÔÚ¹ýÂËÖеÄ×÷Óã¬ÈÃÂËÒºÑØ×Ų£Á§°ôÁ÷Ï£¬Æðµ½ÒýÁ÷×÷Óã»ÓÉÓÚÇâÑõ»¯ÄƹýÁ¿£¬W2g³ÁµíÊÇÇâÑõ»¯Ã¾£¬ÎïÖʵÄÁ¿ÊÇ£ºmol£¬
þµÄÖÊÁ¿·ÖÊýΪ£º×100%=×100%
¹Ê´ð°¸Îª£ºÒýÁ÷×÷Óã»×100%£»
£¨¢ô£©ÀûÓÃÇâÑõ»¯ÄÆÈÜÒººÍ½ðÊôÂÁ·´Ó¦²úÉúÇâÆø£¬¸ù¾Ý±ê¿öÏÂÇâÆøµÄÌå»ý£¬¼ÆËã³öÂÁµÄÎïÖʵÄÁ¿£¬ÔÙÇó³öþµÄÖÊÁ¿·ÖÊý£¬
¹Ê´ð°¸Îª£º¹ýÁ¿ÇâÑõ»¯ÄÆÈÜÒº£®
µãÆÀ£º±¾Ì⿼²éÁ˲ⶨþÂÁºÏ½ðÖÐþµÄÖÊÁ¿·ÖÊý·½·¨£¬Éæ¼°ÁËijЩ½ðÊôµÄ»¯Ñ§ÐÔÖʼ°ÓйصĻ¯Ñ§·½³ÌʽµÄ¼ÆË㣬ÅàÑøѧÉúµÄ·ÖÎöÄÜÁ¦ºÍÀí½âÄÜÁ¦£¬±¾ÌâÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
º¬Ã¾3%¡«5%µÄÂÁþºÏ½ðÊÇÂÖ´¬ÖÆÔì¡¢»¯¹¤Éú²ú¡¢»úеÖÆÔìµÈÐÐÒµµÄÖØÒª²ÄÁÏ£®ÏÖÓÐÒ»¿éÖÊÁ¿Îªm gµÄÂÁþºÏ½ð£¬Óû²â¶¨ÆäÖÐþµÄÖÊÁ¿·ÖÊý£¬¼¸Î»Í¬Ñ§Éè¼ÆÁ˲»Í¬µÄʵÑé·½°¸£®
·½°¸1£ºÂÁþºÏ½ð
¹ýÁ¿ÑÎËá
²â¶¨Éú³ÉÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý£¨V1L£©
·½°¸2£ºÂÁþºÏ½ð
¹ýÁ¿NaOHÈÜÒº
³ä·Ö·´Ó¦ºó²â¶¨Ê£Óà¹ÌÌåµÄÖÊÁ¿£¨W1g£©
·½°¸3£ºÂÁþºÏ½ð
¹ýÁ¿ÑÎËá
ÈÜÒº
¹ýÁ¿NaOHÈÜÒº
¹ýÂË£¬²â¶¨³ÁµíµÄÖÊÁ¿£¨W2g£©
£¨¢ñ£©Ä³»¯Ñ§ÊµÑéС×éÀûÓÃÓÒͼËùʾʵÑé×°Ö㬰´ÕÕ·½°¸1½øÐÐÁËʵÑ飬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù×°ÖÃÖÐÒÇÆ÷aµÄÃû³ÆÊÇ
·ÖҺ©¶·
·ÖҺ©¶·
£®
¢ÚʵÑé×°ÖÃÖÐÓÐÒ»´¦Ã÷ÏÔ´íÎó£¬ÇëÖ¸³ö
¹ã¿ÚÆ¿ÖнøÆøµ¼¹Ü²åÈëµ½ÁËË®ÖУ¬ÅÅË®µ¼¹Üδ²åÖÁ¹ã¿ÚÆ¿µ×²¿
¹ã¿ÚÆ¿ÖнøÆøµ¼¹Ü²åÈëµ½ÁËË®ÖУ¬ÅÅË®µ¼¹Üδ²åÖÁ¹ã¿ÚÆ¿µ×²¿
£®
£¨¢ò£©·½°¸2Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ
2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü
2Al+2OH-+2H2O¨T2AlO2-+3H2¡ü
£®
£¨¢ó£©·½°¸3¡°¹ýÂË¡±²Ù×÷ÖÐÓõ½µÄ²£Á§°ôµÄ×÷ÓÃÊÇ
ÒýÁ÷
ÒýÁ÷
£®
Èô°´·½°¸3½øÐÐÊÔÑ飬²âµÃþµÄÖÊÁ¿·ÖÊýΪ
3W2
5m
¡Á100%
3W2
5m
¡Á100%
£®
£¨¢ô£©Ä³Í¬Ñ§ÓÐÉè¼ÆÁËÓëÉÏÊö·½°¸²»Í¬µÄ·½°¸4£¬Ò²²âµÃÁËþµÄÖÊÁ¿·ÖÊý£®ÇëÄãÔÚÀ¨ºÅÄÚÌîÉϺÏÊʵÄÄÚÈÝ£¬½«¸Ã·½°¸²¹³äÍêÕû£®
·½°¸4£ºÂÁþºÏ½ð
()
²â¶¨Éú³ÉÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý£¨V2L£©£®
ijÑо¿ÐÔѧϰС×éΪ²â¶¨Ä³º¬Ã¾3%Ò»5%µÄÂÁþºÏ½ð£¨²»º¬ÆäËüÔªËØ£©ÖÐþµÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÁËÏÂÁÐÈýÖÖ²»Í¬ÊµÑé·½°¸½øÐÐ̽¾¿£¬Çë¸ù¾ÝËûÃǵÄÉè¼Æ»Ø´ðÓйØÎÊÌ⣮
¡¾Ì½¾¿Ò»¡¿ÊµÑé·½°¸£ºÂÁþºÏ½ð
NaOHÈÜÒº
²â¶¨Ê£Óà¹ÌÌåÖÊÁ¿£®
ÎÊÌâÌÖÂÛ£º£¨1£©ÊµÑéÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
 
£®
£¨2£©ÈôʵÑéÖгÆÈ¡5.4gÂÁþºÏ½ð·ÛÄ©ÑùÆ·£¬Í¶ÈëVmL2.0mol/L NaOHÈÜÒºÖУ¬³ä·Ö·´Ó¦£®ÔòNaOHÈÜÒºµÄÌå»ýV¡Ý
 
mL£®
£¨3£©ÊµÑéÖУ¬µ±ÂÁþºÏ½ð³ä·Ö·´Ó¦ºó£¬ÔÚ³ÆÁ¿Ê£Óà¹ÌÌåÖÊÁ¿Ç°£¬»¹Ðè½øÐеÄʵÑé²Ù×÷°´Ë³ÐòÒÀ´ÎΪ
 
£®
¡¾Ì½¾¿¶þ¡¿ÊµÑé·½°¸£º³ÆÁ¿xgÂÁþºÏ½ð·ÛÄ©£¬·ÅÔÚÈçͼ1ËùʾװÖõĶèÐÔµçÈÈ°åÉÏ£¬Í¨µçʹÆä³ä·Ö×ÆÉÕ£®
¾«Ó¢¼Ò½ÌÍø
ÎÊÌâÌÖÂÛ£º£¨4£©Óû¼ÆËãMgµÄÖÊÁ¿·ÖÊý£¬¸ÃʵÑéÖл¹Ðè²â¶¨µÄÊý¾ÝÊÇ
 
£®
£¨5£©¼ÙÉèʵÑéÖвâ³ö¸ÃÊý¾ÝΪyg£¬ÔòÔ­ÂÁþºÏ½ð·ÛÄ©ÖÐþµÄÖÊÁ¿·ÖÊýΪ
 
£¨Óú¬x¡¢y´úÊýʽ±íʾ£©£®
¡¾Ì½¾¿Èý¡¿ÊµÑé·½°¸£ºÂÁþºÏ½ð
Ï¡ÁòËá
²â¶¨Éú³ÉÆøÌåµÄÌå»ý£®
ÎÊÌâÌÖÂÛ£º£¨6£©Í¬Ñ§ÃÇÄâÑ¡ÓÃÈçͼ2µÄʵÑé×°ÖÃÍê³ÉʵÑ飬ÄãÈÏΪ×î¼òÒ×µÄ×°ÖÃÆäÁ¬½Ó˳ÐòÊÇ£º
a½Ó
 
£®£¨Ìî½Ó¿Ú×Öĸ£¬ÒÇÆ÷²»Ò»¶¨È«Ñ¡£®£©
£¨7£©Í¬Ñ§ÃÇ×Ðϸ·ÖÎö£¨6£©ÖÐÁ¬½ÓµÄʵÑé×°Öúó£¬ÓÖÉè¼ÆÁËͼ3ËùʾµÄʵÑé×°Öã®
¢Ù×°ÖÃÖе¼¹ÜaµÄ×÷ÓÃÊÇ
 
£®
¢ÚʵÑéÇ°ºó¼îʽµÎ¶¨¹ÜÖÐÒºÃæ¶ÁÊý·Ö±ðÈçͼ4£¬Ôò²úÉúÇâÆøµÄÌå»ýΪ
 
mL£®
¢ÛÓëÉÏ×óͼװÖÃÏà±È£¬Óã¨6£©ÖÐÁ¬½ÓµÄ×°ÖýøÐÐʵÑéʱ£¬ÈÝÒ×ÒýÆðÎó²îµÄÔ­ÒòÊÇ
 
£¨ÈÎдһµã£©£®

Çë°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌâ:

£¨1£©.ijºÏ×÷ѧϰС×éÌÖÂÛ±æÎöÒÔÏÂ˵·¨£¬ÆäÖÐ˵·¨ÕýÈ·µÄÊÇ  _____________

     A¸ù¾ÝËá·Ö×ÓÖк¬ÓеÄÇâÔ­×Ó¸öÊý£¬½«Ëá·ÖΪһԪËá¡¢¶þÔªËáµÈ

B ½«Na2O2ͶÈëFeCl2ÈÜÒºÖÐ, ¿É¹Û²ìµ½µÄÏÖÏóÊÇÉú³ÉºìºÖÉ«³Áµí¡¢ÓÐÆøÅݲúÉú                

C ½ðÊôÔªËصÄÔ­×ÓÖ»Óл¹Ô­ÐÔ£¬Àë×ÓÖ»ÓÐÑõ»¯ÐÔ

       D²»Ðâ¸ÖºÍÄ¿Ç°Á÷ͨµÄÓ²±Ò¶¼ÊǺϽ𣻠  

E  NH3¡¢CO2µÄË®ÈÜÒº¾ùÄܵ¼µç£¬ËùÒÔNH3¡¢CO2¾ùÊǵç½âÖÊ

£¨2£©½«5mol/LµÄMg£¨NO3£©2ÈÜÒºa mLÏ¡ÊÍÖÁb mL£¬Ï¡ÊͺóÈÜÒºÖÐNO3£­µÄÎïÖʵÄÁ¿Å¨¶ÈΪ_____________

£¨3£©ÔÚK2Cr2O7+14HCl==2KCl+2CrCl3+3Cl2¡ü+7H2O·´Ó¦ÖУ¬

Ñõ»¯¼ÁÊÇ_____   _£¨Ìѧʽ£© µ±ÓÐ14.6gHCl±»Ñõ»¯Ê±£¬µç×ÓתÒÆΪ    mol¡£

£¨4£©Ñ§Ð£Ñо¿ÐÔѧϰС×éµÄͬѧ£¬Îª²â¶¨Ä³º¬Ã¾3%¡«5%µÄÂÁþºÏ½ð(²»º¬ÆäËûÔªËØ)ÖÐþµÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÏÂÁÐʵÑé·½°¸½øÐÐ̽¾¿£¬ÌîдÏÂÁпհףº

ʵÑé·½°¸£ºÂÁþºÏ½ð ²â¶¨Ê£Óà¹ÌÌåÖÊÁ¿¡£

¢Ù ʵÑéÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ__________________¡£

¢Ú ʵÑé²½Ö裺³ÆÈ¡5.4 gÂÁþºÏ½ð·ÛÄ©ÑùÆ·£¬Í¶ÈëV mL 2.0 mol/L NaOH ÈÜÒºÖгä·Ö·´Ó¦¡£¼ÆËãËùÓÃNaOHÈÜÒºµÄÌå»ýV¡Ý______________¡£

¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡¢³ÆÁ¿¹ÌÌå¡£¸Ã²½ÖèÖÐÈôδϴµÓ¹ÌÌ壬²âµÃþµÄÖÊÁ¿·ÖÊý½«________(  Ìî  ²»±ä  Æ«¸ß »ò Æ«µÍ )¡£

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

¾«Ó¢¼Ò½ÌÍø