ij¿ÎÍâ»î¶¯Ð¡×é¶ÔÒ»Ñõ»¯Ì¼»¹Ô­Ñõ»¯ÌúµÄʵÑéÖÐ×îºóµÄ²úÎï²úÉúŨºñµÄÐËȤ£¬ÊÔͨ¹ýʵÑéÀ´Ì½¾¿Æä³É·Ö¡£
¢ñ£®ÊµÑé×°Öãº

ÓÃÒ»Ñõ»¯Ì¼»¹Ô­Ñõ»¯ÌúµÄʵÑé×°ÖÃ
¸Ã×°ÖÃBÖз¢ÉúµÄÀë×Ó·½³ÌʽÊÇ                                       
×°ÖÃBµÄ×÷ÓÃÊÇ                                    
¢ò£®ÊµÑéÏÖÏ󣺲£Á§¹ÜAÖеķÛÄ©ÓɺìÉ«Öð½¥±äΪºÚɫʱ£¬Í£Ö¹¼ÓÈÈ£¬¼ÌÐøͨһÑõ»¯Ì¼£¬ÀäÈ´µ½ÊÒΣ¬Í£Ö¹Í¨Æø£¬Í¬Ê±¹Û²ìµ½³ÎÇåµÄʯ»ÒË®±ä»ë×Ç¡£
¢ó£®ÊµÑé½áÂÛ£º
¼×ÈÏΪ£ºÒÀ¾ÝÉÏÊöʵÑéÏÖÏó¿ÉÒÔÅжϳöÉú³ÉµÄºÚÉ«¹ÌÌåΪ½ðÊôÌú¡£
ÒÒÈÏΪ£º½ö´ÓÉÏÊöʵÑéÏÖÏ󣬲»×ãÒÔÖ¤Ã÷Éú³ÉµÄºÚÉ«¹ÌÌåΪ½ðÊôÌú£¬ËýÔö¼ÓÁËÒ»¸öʵÑ飺ÓôÅÌú¿¿½üÉú³ÉµÄºÚÉ«¹ÌÌ壬¿´µ½ÓкÚÉ«¹ÌÌå±»´ÅÌúÎüÒý¡£ÓÚÊǵóöÉú³ÉµÄºÚÉ«¹ÌÌåΪ½ðÊôÌúµÄ½áÂÛ¡£
ÇëÄãͨ¹ý¸Ã·´Ó¦µÄÏà¹Ø×ÊÁ϶ÔËûÃǽáÂÛ×÷³öÅжϲ¢Í¨¹ýʵÑé¼ìÑéÆäºÏÀíÐÔ£º
£¨1£©ÔÚÒ»¶¨Ìõ¼þÏ£ºÒ»Ñõ»¯Ì¼ÓëÑõ»¯ÌúÔÚ¼ÓÈÈÌõ¼þÏ£¬¿É·¢ÉúÈçÏ·´Ó¦
3Fe2O3+CO2Fe3O4+CO2 
Fe3O4+4CO4Fe+4CO2 
£¨2£©ËÄÑõ»¯ÈýÌú£¨Fe3O4£©ÎªºÚÉ«¹ÌÌ壬ÓÐÇ¿´ÅÐÔ£¬Äܹ»±»´ÅÌúÎüÒý¡£
¼×¡¢ÒÒͬѧµÄ½áÂÛ£º                             Äã¶Ô´ËÆÀ¼ÛµÄÀíÓÉÊÇ£º      
                                                            
¢ô£®ÊµÑé̽¾¿
¶Ô·´Ó¦ºó¹ÌÌå³É·ÖÌá³ö¼ÙÉ裺
¼ÙÉè1£º·´Ó¦ºó¹ÌÌåÖÐÖ»ÓÐFe£»
¼ÙÉè2£º·´Ó¦ºó¹ÌÌåÖÐÖ»ÓÐFe3O4£»
¼ÙÉè3£º·´Ó¦ºó¹ÌÌåÖÐ_______________________
Ϊȷ¶¨ÊµÑéÖÐ×îºóµÄ²úÎïµÄ³É·Ö£¬±ûͬѧÉè¼ÆÈçÏÂʵÑ飬ÇëÄúÀûÓÃÏÞÑ¡ÊÔ¼ÁºÍÒÇÆ÷°ïÖúËûÍê³É¸Ã̽¾¿¹ý³Ì£¬²¢½«´ð°¸Ð´ÔÚ´ðÌ⿨ÏàӦλÖá£
ÏÞÑ¡ÊÔ¼ÁºÍÒÇÆ÷£º 1mol/LCuSO4 ¡¢0.01mol/L KSCNÈÜÒº¡¢1mol/LÑÎËá¡¢0.01mol/LÂÈË®¡¢ÊԹܡ¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü¡£

ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
²½ÖèÒ»£ºÈ¡Ó²Öʲ£Á§¹ÜÖйÌÌå²úÎïÉÙÁ¿·Ö±ðÓÚA¡¢BÊÔ¹ÜÖУ¬¼ÓÈë×ãÁ¿1mol/LCuSO4ÈÜÒº¡¢½Á°èÈܽ⡣
£¨1£©ÈôAÊÔ¹ÜÖкÚÉ«¹ÌÌå²»Èܽ⣬²¢ÇÒûÓй۲쵽ÆäËûÏÖÏó£¬ÔòºÚÉ«¹ÌÌåΪ       
£¨2£©ÈôBÊÔ¹ÜÖÐÓкìÉ«¹ÌÌåÎö³ö£¬Ôò˵Ã÷ºÚÉ«¹ÌÌåÖк¬ÓÐFe¡£
²½Öè¶þ£º¶ÔÊÔ¹ÜBÖÐÈÜÒº¹ýÂË£¬½«ËùµÃ¹ÌÌåÏ´µÓ¸É¾»ºó£¬¼Ó×ãÁ¿1mol/LÑÎËáºó£¬ÔÙÒÀ´Î·Ö±ð¼ÓÈëÊÊÁ¿0.01mol/LÂÈË®¡¢ÉÙÁ¿0.01mol/L KSCNÈÜÒº
£¨1£©ÈôÈÜÒº²»±äºìÉ«£¬Ôò         
£¨2£©ÈôÈÜÒº±äºìÉ«£¬Ôò           
 
¢õ£®ÑÓÉì̽¾¿£º¶¡Í¬Ñ§ÊÔͼͨ¹ý·´Ó¦Ç°ºó¹ÌÌåÖÊÁ¿µÄ±ä»¯À´È·¶¨ºÚÉ«¹ÌÌåµÄ³É·Ö£¬ÄãÈÏΪ¿ÉÐÐÂ𣿣¨¼ÙÉèÑõ»¯ÌúÔÚ·´Ó¦ÖÐÍêÈ«·´Ó¦£©              £¨Ìî¡°ÐС±»ò¡°²»ÐС±£©ÀíÓÉÊÇ                                                              ¡£

£¨16·Ö£©
¢ñ.  Ca2++2OH-+ CO2= CaCO3¡ý+H2O £¨2·Ö£©
ÎüÊÕCOÖеÄCO2£¬±ãÓÚCOȼÉÕ±»³ýÈ¥¡££¨2·Ö£©
¢ó.¼×¡¢ÒÒͬѧµÄ½áÂÛ¾ù²»ÕýÈ·£»ÌúºÍËÄÑõ»¯ÈýÌú¶¼ÊǺÚÉ«ÇÒ¾ùÄܱ»´ÅÌúÎüÒý¡££¨1+2·Ö ¹²3·Ö£©
¢ô.¼ÙÉè3£º·´Ó¦ºó¹ÌÌåÖмȺ¬ÓÐFeÓÖº¬ÓÐFe3O4£¨1·Ö£©

ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
 
£¨1£©ÈôAÊÔ¹ÜÖкÚÉ«¹ÌÌå²»Èܽ⣬²¢ÇÒûÓй۲쵽ÆäËûÏÖÏó£¬ÔòºÚÉ«¹ÌÌåΪFe3O4£¨1·Ö£©
 
£¨1£©ÈôÈÜÒº²»±äºìÉ«£¬Ôò¼ÙÉè1ÕýÈ· £¨2·Ö£©
£¨2£©ÈôÈÜÒº±äºìÉ«£¬Ôò¼ÙÉè3ÕýÈ·£¨2·Ö£©
 
¢õ.ÐУ¨1·Ö£©£»
µÈÖÊÁ¿Ñõ»¯ÌúÍêÈ«·´Ó¦ºóÉú³ÉÌú»òÕßËÄÑõ»¯ÈýÌú»òÌúºÍËÄÑõ»¯ÈýÌú£¬¹ÌÌåÖÊÁ¿µÄ±ä»¯²»Ïàͬ¡££¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º
¢ñ.½áºÏÌâÄ¿ÐÅÏ¢£¬·´Ó¦ºóÆøÌåÖÐÓÐCOºÍCO2£¬¿É֪ʯ»ÒË®µÄ×÷ÓÃÊÇÎüÊÕCO2£¬±ãÓÚCOȼÉÕ±»³ýÈ¥¡£
¢ó.(2) ÓÉÌâÄ¿ÐÅÏ¢£¬ËÄÑõ»¯ÈýÌú£¨Fe3O4£©ÎªºÚÉ«¹ÌÌ壬ÓÐÇ¿´ÅÐÔ£¬Äܹ»±»´ÅÌúÎüÒý£¬¿ÉÒԵõ½´ð°¸¡£
¢ô.²½ÖèÒ»ÖУ¬Èô³öÏÖºìɫ˵Ã÷ÓÐÌúÖû»³öÍ­µ¥ÖÊ£¬ÈôûÓÐÔòºÚÉ«¹ÌÌåΪFe3O4¡£
²½Öè¶þÖУ¬FeÓëÑÎËáÖ»ÄÜÉú³ÉFe2+£¬Fe3O4ÓëÑÎËáÄÜÉú³ÉFe2+ºÍ Fe3+¡£
¢õ.Ñõ»¯ÌúÍêÈ«·´Ó¦ºóÉú³ÉÌú»òÕßËÄÑõ»¯ÈýÌú¹ÌÌåÖÊÁ¿µÄ±ä»¯²»Ïàͬ¡£
¿¼µã£º±¾ÌâÒÔ̽¾¿ÊµÑéΪ»ù´¡£¬¿¼²éÁËÔªËؼ°»¯ºÏÎïÐÔÖÊ¡¢Ì½¾¿»ù±¾·½·¨¡¢·ÖÎöÎÊÌâÄÜÁ¦¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

I£®ÏÂͼ±íʾ´Ó¹ÌÌå»ìºÏÎïÖзÖÀëQµÄ2ÖÖ·½°¸£¬Çë»Ø´ðÓйØÎÊÌâ¡£

£¨1£©Ñ¡Ó÷½°¸(i)ʱ£¬QÓ¦¸Ã¾ßÓеÄÐÔÖÊÊÇ_____________£¬²ÐÁôÎïÓ¦¸Ã¾ß
ÓеÄÐÔÖÊÊÇ__________________________________¡£
£¨2£©Ñ¡Ó÷½°¸(ii)´Óij½ðÊô·ÛÄ©£¨º¬ÓÐAu¡¢AgºÍCu£©ÖзÖÀëAu£¬¼ÓÈëµÄÊÔ¼ÁΪ____________¡£
£¨3£©ÎªÌᴿijFe2O3ÑùÆ·£¨Ö÷ÒªÔÓÖÊÓÐSiO2.Al2O3£©£¬²ÎÕÕ·½°¸(i)ºÍ(ii)£¬ÇëÉè¼ÆÒ»ÖÖÒÔ¿òͼÐÎʽ±íʾµÄʵÑé·½°¸£¨×¢Ã÷ÎïÖʺͲÙ×÷£©£º
______________________________________________________________________________¡£
¢ò£®Ä³ÖÖº¬ÓÐÉÙÁ¿Ñõ»¯ÄƵĹýÑõ»¯ÄÆÑùÆ·£¨¼ºÖªÑùÆ·ÖÊÁ¿Îª1.560g¡¢×¶ÐÎÆ¿ºÍË®µÄÖÊÁ¿Îª
190.720g)£¬ÀûÓÃÓÒÏÂͼËùʾװÖòⶨ»ìºÏÎïÖÐNa2O2µÄÖÊÁ¿·ÖÊý£¬Ã¿¸ôÏàͬʱ¼ä¶ÁµÃµç
×ÓÌìƽµÄÊý¾ÝÈçÏÂ±í£º

£¨4£©Ð´³öNa2O2ÓëH2O·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________________£®
£¨5£©¼ÆËãNa2O2ÖÊÁ¿·ÖÊýʱ£¬±ØÐèµÄÊý¾ÝÊÇ_________________________________________.
²»±Ø×÷µÚ6´Î¶ÁÊýµÄÔ­ÒòÊÇ_____________________________________________________£®
£¨6£©²â¶¨ÉÏÊöÑùÆ·(1.560g)ÖÐNa2O2ÖÊÁ¿·ÖÊýµÄÁíÒ»ÖÖ·½°¸£¬Æä²Ù×÷Á÷³ÌÈçÏ£º

²Ù×÷¢ÚµÄÃû³ÆÊÇ____________£¬¸Ã·½°¸ÐèÖ±½Ó²â¶¨µÄÎïÀíÁ¿ÊÇ_____________ £¬²â¶¨¹ý³ÌÖÐ
ÐèÒªµÄÒÇÆ÷Óеç×ÓÌìƽ¡¢Õô·¢Ã󡢾ƾ«µÆ£¬»¹ÐèÒª___________¡¢__________£¨¹Ì¶¨¡¢¼Ð
³ÖÒÇÆ÷³ýÍ⣩,ÔÚתÒÆÈÜҺʱ£¬ÈçÈÜҺתÒƲ»ÍêÈ«£¬ÔòNa2O2ÖÊÁ¿·ÖÊýµÄ²â¶¨½á¹û_______£¨Ìî
¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ÏÖ´ÓпÖÆÆ·¼Ó¹¤ÆóÒµ»ØÊյķÏÔü£¨º¬ÓÐZnO¡¢FeO¡¢Fe2O3¡¢CuO¡¢Al2O3µÈÔÓÖÊ£©À´ÖÆÈ¡Ñõ»¯Ð¿£¬Á÷³ÌÈçÏ£º

ÓйØÇâÑõ»¯ÎïÍêÈ«³ÁµíµÄpH¼ûÏÂ±í£º

³ÁµíÎï
Al(OH)3
Fe(OH)3
Fe(OH)2
Cu(OH)2
Zn(OH)2
pH
5.2
3.2
9.7
6.7
8.0
 
£¨l£©ÔÚËá½þ¹ý³ÌÖУ¬ÒªÌá¸ßпԪËصĽþ³öÂÊ£¬¿ÉÒÔ²ÉÈ¡         ´ëÊ©¡£
£¨2£©ÉÏÊöÁ÷³ÌÖжദÉæ¼°¡°¹ýÂË¡±£¬ÊµÑéÊÒÖйýÂ˲Ù×÷ÐèҪʹÓõIJ£Á§ÒÇÆ÷ÓР       ¡£
£¨3£©ÔÚ¡°³ýÔÓI¡±²½ÖèÖУ¬½«ÈÜÒºµ÷ÖÁpH=4µÄÄ¿µÄÊÇ              ¡£ÔÚ¡°³ýÔÓII¡±ºó£¬ÈÜÒºµÄpHԼΪ6£¬Ôò´Ë²½¹ýÂËʱÂËÔüÖк¬ÓР              ¡£
£¨4£©ÔÚ¡°Ì¼»¯ºÏ³É¡±ÖУ¬Éú³ÉµÄ²úÎï֮һΪ¼îʽ̼Ëáп[Zn2£¨OH£©2CO3]£¬Í¬Ê±·Å³öCO2£¬Çëд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ                       ¡£
£¨5£©´ÓÂËÒºÖÐÌáÈ¡NaNO3¾§ÌåµÄ²Ù×÷²½ÖèΪ                  ¡£
£¨6£©ÔÚʵÑéÊÒÈçºÎÏ´µÓ¹ýÂ˳öµÄ¼îʽ̼Ëáп£¿                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

FeSO4¡¤7H2OË׳ÆÂÌ·¯£¬¹ã·ºÓÃÓÚÒ½Ò©ºÍ¹¤¼àÁìÓò¡£
£¨1£©ÒÑÖªFeSO4¡¤7H2O¾§ÌåÔÚ¼ÓÈÈÌõ¼þÏ·¢ÉúÈçÏ·´Ó¦£º
¡£ÀûÓÃÏÂͼװÖüìÑé¸Ã·´Ó¦µÄÆøÌå²úÎï¡£

ÇëÌîдÏÂÁпհףº
¢ÙÒÇÆ÷µÄÁ¬½Ó˳ÐòΪ        £¨ÓÃa¡«iµÄ×Öĸ±íʾ£©¡£
¢Ú×°ÖÃCÖеÄÊÔ¼ÁX»¯Ñ§Ê½Îª         £»¸Ã×°ÖÃÖÐÀäË®µÄ×÷ÓÃÊÇ            ¡£
£¨2£©ÒÔÏÂÊÇFeSO4¡¤7H2OµÄʵÑéÖÏÖƱ¸Á÷Àíͼ¡£

¸ù¾ÝÌâÒâÍê³ÉÏÂÁÐÌî¿Õ£º
¢Ù·´Ó¦¢òÐèÒªl00mL l£®5mol¡¤L-1lµÄÏ¡ÁòËáÈܽâ½à¾»µÄÌúм£¬ÓÃÖÊÁ¿·ÖÊýΪ98%¡¢ÃܶÈΪ1.84g¡¤cm-3µÄŨÁòËáÅäÖÆ¡£ËùÓõÄÒÇÆ÷ÓÐÁ¿Í²¡¢ÉÕ±­¡¢²£Á§Åõ¡¢½ºÍ·µÎ¹Ü¼°     £¬Á¿È¡Å¨ÁòËáµÄÌå»ýΪ             ¡£
¢Ú·´Ó¦1ÐèÒª¼ÓÈÈÊý·ÖÖÓ£¬ÆäÔ­ÒòÊÇ                £»²Ù×÷AΪ       ¡£
¢Û²â¶¨FeSO4¡¤7H2O²úÆ·ÖÐFe2+º¬Á¿µÄ³£Ó÷½·¨ÊÇKMnO4ÈÜÒºµÎ¶¨·¨¡£ÒÑÖª³ÆÈ¡3.0g FeSO4¡¤7H2O²úÆ·£¬ÅäÖƳÉÈÜÒº£¬ÓÃÁòËáËữµÄ0.01000moL¡¤L-1 KMnO4ÈÜÒºµÎ¶¨£¬ÏûºÄKMnO4ÈÜÒºµÄÌå»ýΪ200.00mL¡£·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ               £¬¼ÆËãÉÏÊöÑùÆ·ÖÐFeSO4¡¤7H2OµÄÖÊÁ¿·ÖÊýΪ               £¨±£ÁôÁ½Î»ÓÐЧÊý×Ö£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

¶þÑõ»¯ÂÈ£¨ClO2£©ÊÇÄ¿Ç°¹ú¼ÊÉϹ«ÈϵĵÚËÄ´ú¸ßЧ¡¢ÎÞ¶¾µÄ¹ãÆ×Ïû¶¾¼Á£¬ÊÇÒ»ÖÖ»ÆÂÌÉ«µÄÆøÌ壬Ò×ÈÜÓÚË®¡£ÊµÑéÊÒ¿ÉÓÃNH4Cl¡¢ÑÎËá¡¢NaClO2£¨ÑÇÂÈËáÄÆ£©ÎªÔ­ÁÏÀ´ÖƱ¸ClO2£¬ÆäÁ÷³ÌÈçÏ£º

£¨1£©Ð´³öµç½âʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                           ¡£
£¨2£©³ýÈ¥ClO2ÖеÄNH3¿ÉÑ¡ÓõÄÊÔ¼ÁÊÇ       ¡££¨Ìî×Öĸ£©

A£®±¥ºÍʳÑÎË®B£®¼îʯ»ÒC£®Å¨ÁòËáD£®Ë®
£¨3£©²â¶¨ClO2£¨Èçͼ£©µÄ¹ý³ÌÈçÏ£ºÔÚ׶ÐÎÆ¿ÖмÓÈë×ãÁ¿µÄµâ»¯¼Ø£¬ÓÃ100mLË®Èܽâºó£¬ÔÙ¼Ó3mLÁòËáÈÜÒº£»ÔÚ²£Á§Òº·â¹ÜÖмÓÈëË®£»½«Éú³ÉµÄClO2ÆøÌåͨ¹ýµ¼¹ÜÔÚ׶ÐÎÆ¿Öб»ÎüÊÕ£»½«²£Á§·â¹ÜÖеÄË®·âÒºµ¹Èë׶ÐÎÆ¿ÖУ¬¼ÓÈ뼸µÎµí·ÛÈÜÒº£¬ÓÃcmol/LÁò´úÁòËáÄƱê×¼ÈÜÒºµÎ¶¨(I2+2S2O32£­=2I£­+S4O62£­)£¬¹²ÓÃÈ¥VmLÁò´úÁòËáÄÆÈÜÒº¡£

¢Ù×°ÖÃÖв£Á§Òº·â¹ÜµÄ×÷ÓÃÊÇ                       £»                   ¡£
¢ÚÇëд³öÉÏÊö¶þÑõ»¯ÂÈÆøÌåÓëµâ»¯¼ØÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ   ¡¡             ¡£
¢ÛµÎ¶¨ÖÕµãµÄÏÖÏóÊÇ£º                                                   ¡£
¢Ü²âµÃͨÈëClO2µÄÖÊÁ¿m(ClO2)=                  ¡££¨Óú¬c¡¢VµÄ´úÊýʽ±íʾ£©£¨ÒÑÖª£ºClO2µÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª67.5£©
£¨4£©Éè¼ÆʵÑéÀ´È·¶¨ÈÜÒºXµÄ³É·Ö£¬Çë²¹³äÍê³ÉʵÑé²½ÖèºÍÏÖÏó¡£
ʵÑé²½Öè
ʵÑéÏÖÏó
ʵÑé½áÂÛ
¢Ù
 
ÈÜÒºXÖк¬ÓÐNa+
¢Ú
 
ÈÜÒºXÖк¬ÓÐCl£­
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐʵÑé×°ÖÃͼ£¨ÓÐЩͼÖв¿·Ö¼Ð³ÖÒÇÆ÷δ»­³ö£©²»ÄÜ´ïµ½ÆäʵÑéÄ¿µÄµÄÊÇ 



A£®Ö¤Ã÷ËáÐÔ£ºÑÎË᣾̼Ë᣾±½·Ó
B£®ÊµÑéÊÒÖÆÈ¡ÒÒËáÒÒõ¥


C£®Ê¯ÓÍ·ÖÁó
D£®ÊµÑéÊÒÖÆÈ¡Ïõ»ù±½

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐʵÑéºÏÀíµÄÊÇ





A£®Ö¤Ã÷·Ç½ðÐÔ
Cl£¾C£¾Si
B£®ÎüÊÕ°±Æø£¬
²¢·ÀÖ¹µ¹Îü
C£®ÖƱ¸²¢ÊÕ¼¯ÉÙÁ¿NO2ÆøÌå
D£®ÖƱ¸CO2
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

½öÓÃϱíÌṩµÄÒÇÆ÷ºÍÒ©Æ·£¬¾ÍÄÜ´ïµ½ÏàӦʵÑéÄ¿µÄµÄÊÇ

񅧏
ÒÇÆ÷
Ò©Æ·
ʵÑéÄ¿µÄ
A
ÍÐÅÌÌìƽ£¨´øíÀÂ룩¡¢250mLÈÝÁ¿Æ¿¡¢Á¿Í²¡¢ÉÕ±­¡¢Ò©³×¡¢²£²£°ô
NaOH¹ÌÌå¡¢ÕôÁóË®
ÅäÖÆ250mLÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº
B
·ÖҺ©¶·¡¢×¶ÐÎÆ¿¡¢µ¼¹Ü¼°ÏðƤÈû
Ï¡ÁòËᡢ̼ËáÄÆ¡¢¹èËáÄÆÈÜÒº
Ö¤Ã÷·Ç½ðÊôÐÔ£»S£¾C£¾Si
C
¼îʽµÎ¶¨¹Ü¡¢ËáʽµÎ¶¨¹Ü¡¢½ºÍ·µÎ¹Ü¡¢Ìú¼Ų̈£¨´øÌú¼Ð£©¡¢×¶ÐÎÆ¿
¼ºÖªÅ¨¶ÈµÄNaOHÈÜÒº¡¢´ý²âÑÎËá¡¢ÕôÁóË®¡¢°×Ö½
²â¶¨Ï¡ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶È
D
Ìú¼Ų̈£¨´øÌú¼Ð£©¡¢¾Æ¾«µÆ¡¢´óÊԹܡ¢¼¯ÆøÆ¿¡¢µ¼¹Ü¼°ÏðƤÈû
ÂÈ»¯ï§
ÖÆÈ¡°±Æø

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÓйØʵÑéÏÖÏóµÄÔ¤²â»ò×°ÖõÄÑ¡ÓÃÕýÈ·µÄÊÇ

A£®ÓÃ×°ÖÃ(¢ñ)¼ÓÈȲÝËᾧÌå»ñȡijЩÆøÌå(²ÝËᾧÌåµÄÈÛµã101.5 ¡æ£¬·Ö½âζÈԼΪ150 ¡æ)
B£®ÓÃ×°ÖÃ(¢ò)½øÐÐʵÑéʱ£¬ËáÐÔKMnO4ÈÜÒºÖгöÏÖÆøÅÝ£¬ÇÒÑÕÉ«Öð½¥ÍÊÈ¥
C£®ÓÃ×°ÖÃ(¢ó)½øÐÐʵÑéʱ£¬¹ã¿ÚÆ¿ÄÚÏÈÓÐdzºì×ØÉ«ÆøÌå³öÏÖºóÓÖ±äΪÎÞÉ«£¬ÇÒ²»»á²úÉú¿ÕÆøÎÛȾ
D£®ÓÃ×°ÖÃ(¢ô)·ÖÀëäå±½ºÍ±½µÄ»ìºÏÎï

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸