ÌâÄ¿ÄÚÈÝ

ij»¯Ñ§Ì½¾¿Ñ§Ï°Ð¡×éÉè¼ÆÈçÏÂͼװÖÃÖÆÈ¡ÏõËᣨ¼Ð³ÖºÍ¼ÓÈÈÒÇÆ÷¾ùÒÑÂÔÈ¥£©¡£ÊµÑéÖпɹ©Ê¹ÓõÄÒ©Æ·ÓУºNa2CO3¡¢NaHCO3¡¢NH4HCO3¡¢Na2O2¡¢NaOHÈÜÒººÍË®¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©×°ÖÃAÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽÊÇ                               ¡£
£¨2£©³·È¥×°ÖÃDÖеļÓÈÈ×°Öú󣬲¬Ë¿ÈÔÈ»±£³ÖºìÈÈ£¬ÕâÊÇÒòΪDÖз¢ÉúµÄ»¯Ñ§·´Ó¦ÊÇÒ»¸ö         £¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©·´Ó¦¡£
£¨3£©×°ÖÃFÖÐÊ¢·ÅµÄÊÇ                  ÈÜÒº£¬Æä×÷ÓÃÊÇ                                          ¡£
£¨4£©ÊµÑé¹ý³ÌÖУ¬ÒªÊ¹NH4HCO3³ä·Öת»¯ÎªHNO3£¬»¹ÒªÔÚ×°ÖÃDÖÐͨÈë¹ýÁ¿µÄÑõÆø¡£¼×ͬѧÌáÒéÔÚC´¦Á¬½ÓÒ»¸öÖÆÈ¡ÑõÆøµÄ×°Öã¬ÒÒͬѧÈÏΪ¿ÉÖ±½ÓÔÚAÖÐÔÙ¼ÓÈëÉÏÊöÌṩҩƷÖеÄÒ»ÖÖÎïÖÊ£¬ÕâÖÖÒ©Æ·µÄ»¯Ñ§Ê½ÊÇ                 
£¨1£©NH4HCO3NH3£«H2O£«CO2¡ü
£¨2£©·ÅÈÈ
£¨3£©NaOH   ÎüÊÕ¿ÉÄÜδ·´Ó¦µÄµªÑõ»¯Î·ÀÖ¹ÎÛȾ¿ÕÆø
£¨4£©NaHCO3
ÖÆÈ¡ÏõËáÉæ¼°·´Ó¦Îª£º
4NH3£«5O24NO£«6H2O   2NO£«O2=2NO2    3NO2£«H2O=2HNO3£«NO
Ëù¸ø×°Öü°Ò©Æ·¿É¿´³ö£ºNH4HCO3NH3£«H2O£«CO2¡ü  2Na2O2£«2H2O=4NaOH£«O2¡ü   2Na2O2£«2CO2=2Na2CO3£«O2
£¨1£©NH4HCO3NH3£«H2O£«CO2¡ü
£¨2£©°±ÆøµÄ´ß»¯Ñõ»¯£º4NH3£«5O24NO£«6H2OΪ·ÅÈÈ·´Ó¦
£¨3£©·´Ó¦¹ý³ÌÖÐÉú³É¿ÉÎÛȾ¿ÕÆøµÄNOºÍNO2£¬Ò»°ã²ÉÓüîÒº½«ÆäÎüÊÕ
£¨4£©¿ÉÒÔ¼ÓÈëNaHCO3£¬Í¨¹ýÆä·Ö½âÉú³ÉµÄH2O¼°CO2£¬ÔÙÓëNa2O2·´Ó¦µÃµ½ÑõÆø
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijÐËȤС×é̽¾¿SO2ÆøÌ廹ԭFe3+¡¢I2£¬ËûÃÇʹÓõÄÒ©Æ·ºÍ×°ÖÃÈçÏÂͼËùʾ£º

£¨1£©SO2ÆøÌ廹ԭFe3+µÄ²úÎïÊÇ        £¨ÌîÀë×Ó·ûºÅ£©£¬²Î¼Ó·´Ó¦µÄSO2ºÍFe3+µÄÎïÖʵÄÁ¿Ö®±ÈÊÇ           ¡£
£¨2£©ÏÂÁÐʵÑé·½°¸ÊÊÓÃÓÚÔÚʵÑéÊÒÖÆÈ¡ËùÐèSO2µÄÊÇ       £¨ÌîÐòºÅ£©¡£
A£®Na2SO3ÈÜÒºÓëHNO3                B£®Na2SO3¹ÌÌåÓëŨÁòËá
C£®¹ÌÌåÁòÔÚ´¿ÑõÖÐȼÉÕ               D£®ÁòÌú¿óÔÚ¸ßÎÂÏÂÓëO2·´Ó¦
£¨3£©×°ÖÃCµÄ×÷ÓÃÊÇ                                               ¡£
£¨4£©ÈôÒª´ÓAÖÐËùµÃÈÜÒºÌáÈ¡¾§Ì壬±ØÐë½øÐеÄʵÑé²Ù×÷²½Ö裺Õô·¢¡¢ÀäÈ´½á¾§¡¢¹ýÂË¡¢×ÔÈ»¸ÉÔÔÚÕâһϵÁвÙ×÷ÖÐûÓÐÓõ½µÄÒÇÆ÷ÓР        £¨ÌîÐòºÅ£©¡£
A£®Õô·¢Ãó   B£®Ê¯ÃÞÍø   C£®Â©¶·   D£®ÉÕ±­    E£®²£Á§°ô   F£®ÛáÛö
£¨5£©ÔÚÉÏÊö×°ÖÃÖÐͨÈë¹ýÁ¿µÄSO2£¬ÎªÁËÑéÖ¤AÖÐSO2ÓëFe3+·¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦£¬ËûÃÇÈ¡AÖеÄÈÜÒº£¬·Ö³ÉÈý·Ý£¬²¢Éè¼ÆÁËÈçÏÂʵÑ飺
·½°¸¢Ù£ºÍùµÚÒ»·ÝÊÔÒºÖмÓÈëKMnO4ÈÜÒº£¬×ϺìÉ«ÍÊÈ¥¡£
·½°¸¢Ú£ºÍùµÚ¶þ·ÝÊÔÒº¼ÓÈëKSCNÈÜÒº£¬²»±äºì£¬ÔÙ¼ÓÈëÐÂÖƵÄÂÈË®£¬ÈÜÒº±äºì¡£
·½°¸¢Û£ºÍùµÚÈý·ÝÊÔÒº¼ÓÈëÓÃÏ¡ÑÎËáËữµÄBaCl2£¬²úÉú°×É«³Áµí¡£
ÉÏÊö·½°¸²»ºÏÀíµÄÊÇ      £¬Ô­ÒòÊÇ                                  ¡£
£¨6£©ÄܱíÃ÷I£­µÄ»¹Ô­ÐÔÈõÓÚSO2µÄÏÖÏóÊÇ                           ¡£
£¨16·Ö£©Ä³»¯Ñ§ÊµÑéС×éµÄͬѧΪ̽¾¿ºÍ±È½ÏSO2ºÍÂÈË®µÄƯ°×ÐÔ£¬Éè¼ÆÁËÈçϵÄʵÑé×°Öã¨Í¼3£©¡£

£¨1£©¢Ù·´Ó¦¿ªÊ¼Ò»¶Îʱ¼äºó£¬¹Û²ìµ½B¡¢DÁ½¸öÊÔ¹ÜÖеÄÆ·ºìÈÜÒº³öÏÖµÄÏÖÏóÊÇ£º
B£º________________________________£¬D£º____________________________¡£
¢ÚֹͣͨÆøºó,ÔÙ¸øB¡¢DÁ½¸öÊԹֱܷð¼ÓÈÈ£¬Á½¸öÊÔ¹ÜÖеÄÏÖÏó·Ö±ðΪ
B£º________________________________£¬D£º____________________________¡£
£¨2£©ÁíÒ»¸öʵÑéС×éµÄͬѧÈÏΪSO2ºÍÂÈË®¶¼ÓÐƯ°×ÐÔ£¬¶þÕß»ìºÏºóµÄƯ°×ÐԿ϶¨»á¸üÇ¿£¬ËûÃǽ«ÖƵõÄSO2ºÍCl2°´1:1ͬʱͨÈ뵽ƷºìÈÜÒºÖУ¬½á¹û·¢ÏÖÍÊɫЧ¹û²¢²»ÏñÏëÏóÄÇÑù¡£ÇëÄã·ÖÎö¸ÃÏÖÏóµÄÔ­Òò£¨Óû¯Ñ§·½³Ìʽ±íʾ£©______________________________      ¡£
£¨3£©×°ÖÃEÖÐÓÃMnO2ºÍŨÑÎËá·´Ó¦ÖƵÃCl2£¬Èô·´Ó¦Éú³ÉµÄCl2Ìå»ýΪ2.24L£¨±ê×¼×´¿ö£©£¬Ôò±»Ñõ»¯µÄHClΪ           mol¡£
£¨4£©ÊµÑé½áÊøºó£¬ÓÐͬѧÈÏΪװÖÃCÖпÉÄܺ¬ÓÐSO32£­¡¢SO42£­¡¢Cl£­¡¢OH£­µÈÒõÀë×Ó£¬ÇëÌîд¼ìÑéÆäÖÐSO42£­ºÍSO32£­µÄʵÑ鱨¸æ¡£
ÏÞÑ¡ÊÔ¼Á£º2 mol¡¤L£­1 HCl£»1 mol¡¤L£­1 H2SO4£»l mol¡¤L£­1 BaCl2£»l mol¡¤L£­1MgCl2
1 mol¡¤L£­1 HNO3£»0.1 mol¡¤L£­1 AgNO3£»ÐÂÖƱ¥ºÍÂÈË®¡£
񅧏
ʵÑé²Ù×÷
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
²½Öè¢Ù
È¡ÉÙÁ¿´ý²âÒºÓÚÊÔ¹ÜÖУ¬µÎÈë
                   ÖÁ¹ýÁ¿
                       £¬Ö¤Ã÷´ý²âÒºÖк¬SO32£­¡£
²½Öè¢Ú
ÔÚ²½Öè¢ÙµÄÈÜÒºÖеÎÈëÉÙÁ¿
                        
 
                                 £¬
Ö¤Ã÷´ý²âÒºÖк¬SO42£­¡£
£¨12·Ö£©
ijѧϰÑо¿Ð¡×é¸ù¾Ý½ðÊôп¡¢ÂÁ¡¢Ìú·Ö±ðÓëÏ¡ÑÎËá·´Ó¦µÄÏà¶ÔËÙÂÊ£¬Ì½¾¿½ðÊôп¡¢ÂÁ¡¢ÌúµÄ½ðÊô»î¶¯ÐÔ¡£
[ʵÑéÉè¼Æ]
ΪÁ˴ﵽʵÑéÄ¿µÄ£®±ØÐë¿ØÖÆʵÑéÌõ¼þ¡£ÄãÈÏΪ¿ØÖƵÄʵÑéÌõ¼þÊÇ£¨¿ÉÌîÂúÒ²¿É²»ÌîÂú
£¨1£©                          £»£¨2£©                          £»
£¨3£©                          £»£¨4£©                          ¡£
[ʵÑé̽¾¿]
¸ÃѧϰÑо¿Ð¡×éÔÚÉÏÊöʵÑéÌõ¼þÏàͬµÄÇé¿öÏ£¬½øÐÐÈý×éʵÑ飺¸÷Á¿È¡50mLÏ¡ÑÎËáÓÚСÉÕ±­ÖУ¬·Ö±ðͬʱ¼ÓÈë×ãÁ¿µÄÈýÖÖ±íÃæÎÞÑõ»¯Ä¤µÄ½ðÊôƬ£¬ÓÃpH¼Æ²â¶¨ÈÜÒºµÄpH²¢½«ÐÅÏ¢´«Êäµ½µçÄÔ£¬µçÄÔ¶¯Ì¬µØ»æÖƳöÈÜÒºÖÐc£¨H+£©Ëæʱ¼ätµÄ±ä»¯ÇúÏߣ¨·´Ó¦¹ý³ÌÖÐÈÜÒºµÄÌå»ýûÓб仯£©¡£ÏÂͼÊÇ·´Ó¦µ½t1sʱµçÄԼǼµÄÇúÏߣº

[½»Á÷Óë±í´ï]
¸ù¾ÝÉÏÊöc(H+)Ëæʱ¼ätµÄ±ä»¯ÇúÏ߻شðÏÂÁÐÎÊÌ⣺   
£¨1£©±íʾ½ðÊô»î¶¯ÐÔ×îÈõµÄ½ðÊôÓëH+·´Ó¦µÄ±ä»¯ÇúÏßÊÇ     £¨Ìî¡°a¡±¡¢¡°b¡±»ò¡°c¡±±àºÅ£©¡£   
£¨2£©ÔÚͬһʱ¼äÄÚ£¬±íʾ²»Í¬·´Ó¦µÄ·´Ó¦ËÙÂÊ¿ÉÒÔÓУºÍ¬ÎÂͬѹÏ·ųöÇâÆøµÄÌå»ýV£¨H2£©£»ÈÜÒºÖÐH+Ũ¶ÈµÄ¼õÉÙ¡÷c(H+)£»¹ÌÌå¼õÉÙµÄÁ¿¡÷m»ò¡÷n¡£ÔÚ0Ò»t1sÖ®¼ä£¬ÉèA1ºÍZn·Ö±ðÓëÑÎËá·´Ó¦µÄËÙÂÊΪv£¨Al£©ºÍv£¨Zn£©£º
AÈôÒÔÈÜÒºÖÐc£¨H+£©¼õÉÙÀ´±íʾ²»Í¬·´Ó¦µÄËÙÂÊ£¬¼´v£¨Al£©ºÍv£¨Zn£©µÄ±ÈֵΪ     £»
BÈôÒÔ¹ÌÌåÖÊÁ¿µÄ¼õÇáÀ´±íʾ²»Í¬·´Ó¦µÄËÙÂÊ£¬¼´v£¨Al£©ºÍv£¨Zn£©µÄ±ÈֵΪ      £»
CÈôÒÔ¹ÌÌåÎïÖʵÄÁ¿¼õÉÙÀ´±íʾ²»Í¬·´Ó¦µÄËÙÂÊ£¬¼´v£¨Al£©ºÍv£¨Zn£©µÄ±ÈֵΪ     ¡£
£¨12·Ö£©Ä³Ñо¿ÐÔѧϰС×éÀûÓÃÊÖ³Ö¼¼Êõ̽¾¿Ç¿¼îºÍ²»Í¬µÄËáÖкͷ´Ó¦µÄ¹ý³ÌÈçÏ£º
(1)ʵÑé²½Ö裺
¢Ù·Ö±ðÅäÖÆŨ¶È¾ùΪ0.1mol¡¤L-1µÄNaOH¡¢HC1¡¢CH3 C00H¡¢H3PO4ÈÜÒº±¸Óá£ÅäÖƹý³ÌÖÐÓõ½µÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢ÈÝÁ¿Æ¿¡¢²£Á§°ô¡¢Ï¸¿ÚÆ¿¡¢____¡¢____¡£
¢ÚÔÚ׶ÐÎÆ¿ÖмÓÈë10mL0.1 mol£®L-1µÄHC1£¬ÔÚ25.00mL____£¨Ìî¡°Ëáʽ¡±¡¢¡°¼îʽ¡±£©µÎ¶¨¹ÜÖмÓÈë0.1 mol£®L-1µÄNaOH£¬Á¬½ÓÊý¾Ý²É¼¯Æ÷ºÍpH´«¸ÐÆ÷¡£
¢ÛÏò׶ÐÎÆ¿ÖеÎÈëNaOH£¬½Ó½ü¹ÀËãµÄNaOHÓÃÁ¿¸½½üʱ£¬¼õÂýµÎ¼ÓËٶȣ¬µÈ¶ÁÊýÎȶ¨ºó£¬ÔÙµÎÏÂÒ»µÎNaOH¡£
¢Ü´æ´¢¼ÆËã»ú»æÖƵÄpH±ä»¯Í¼¡£ÓÃ0.1 mol¡¤L-1µÄCH3 C00H¡¢H3 P04ÈÜÒº´úÌæHC1Öظ´ÉÏÊö¢Ú¡«¢ÜµÄ²Ù×÷¡£
(2)½á¹û·ÖÎö£º20¡æʱNaOH·Ö±ðµÎ¶¨HC1¡¢CH3 C00H¡¢H3 P04µÄpH±ä»¯ÇúÏßÈçÏ¡£

¸ù¾ÝÉÏÊöÇúÏ߻شðÏÂÁÐÎÊÌ⣺
¢Ù20¡æʱ£¬ÈýÖÖËá´ÓÇ¿µ½ÈõµÄ˳ÐòÊÇ            £»
¢ÚµÎ¶¨¿ªÊ¼ºó´×ËáÇúÏ߱仯±ÈÑÎËá¿ìµÄÔ­ÒòÊÇ                     £»
¢Û´×ËáÇ¡ºÃÖкÍʱpH¡Ö8µÄÔ­ÒòÊÇ                £»
¢ÜÄãÈÏΪǰÊöµÄ²½ÖèÖУ¬NaOHµÎ¶¨Á×ËáÊÇ·ñ¿ÉÐУ¿   £¨Ìî¡°¿ÉÐС±¡¢¡°²»¿ÉÐС±£©¡£
£¨13·Ö£©
[̽¾¿Ò»]Ϊ²â¶¨îÑÌú¿ó£¨º¬FeTiO3£¬îÑËáÑÇÌú£©ÖпÉÌáÈ¡ÑõµÄÖÊÁ¿·ÖÊý£¬Ä³Ñо¿ÐÔѧϰС×éÉè¼ÆÁËÒÔÏÂÁ½Ì×ʵÑé×°Ö㺣¨×¢£º¿ÉÌáÈ¡ÑõÖ¸µÄÊǿɱ»H2¡¢Cת»¯ÎªH2O¡¢CO2µÄÑõÔ­×Ó£©

£¨1£©ÒÑÖªFeTiO3±»H2»¹Ô­ºó£¬Éú³ÉFe¡¢TiO2ºÍË®£¬Ñ§Éú¼×ÉèÏëÓÃͼ£­1×°Öã¨AΪµç¼ÓÈÈ×°Öã©´ÓîÑÌú¿óÖÐÌáÈ¡Ñõ¡£Çëд³öFeTiO3ÔÚ¼ÓÈÈʱ±»H2»¹Ô­µÄ»¯Ñ§·½³Ìʽ£º
___________________________________________________________________
£¨2£©ÒÑÖªFeTiO3±»C»¹Ô­ºóÉú³ÉFe¡¢TiO2ºÍCO2£¬Ñ§ÉúÒÒÉèÏëÓÃͼ£­2×°Öã¨AΪµç¼ÓÈÈ×°Öã©´ÓîÑÌú¿óÖÐÌáÈ¡Ñõ£¬²¢ÇÒÈÏΪ¸Ã·½°¸µÄÒâÒåÔÚÓÚ£º·´Ó¦²úÉúµÄCO2¿Éͨ¹ýÂÌÉ«Ö²ÎïµÄ¹âºÏ×÷ÓÃת»¯ÎªÓªÑøÎïÖÊ£¬Í¬Ê±²úÉúÑõÆø£º
6CO2+6H2O C6H12O6£¨ÆÏÌÑÌÇ£©+6O2¡£
ʵÑé¹ý³ÌÖУ¬³ÆµÃ·´Ó¦Ç°×°ÖÃAÖÐîÑÌú¿óµÄÖÊÁ¿Îªag£¬Ì¿·ÛµÄÖÊÁ¿Îªbg£¬·´Ó¦ºó×°ÖÃBÖвúÉúCaCO3µÄÖÊÁ¿Îªcg£¬ÔòîÑÌú¿óÖпÉÌáÈ¡ÑõµÄÖÊÁ¿·ÖÊýµÄ±í´ïʽΪ             ¡£
£¨3£©¾­×¨¼ÒÆÀÉóºó£¬È϶¨Á½¸ö·½°¸¶¼ºÜÓÐÒâÒ壬µ«ÊµÑéÉè¼Æ·½Ã滹´æÔÚ²»×ãÖ®´¦£¬ÀýÈçͼ£­2×°ÖÃÔÚ¼ÓÈÈʱ£¬Éú³ÉµÄCO2²»¿ÉÄÜÈ«²¿±»³ÎÇåʯ»ÒË®ÎüÊÕ£¬Í¬Ê±îÑÌú¿óºÍÌ¿·ÛÔÚ¼ÓÈÈʱ»¹»á²úÉúCO£¬¸øʵÑé´øÀ´Îó²î¡£
¢ÙÇë¶ÔѧÉúÒÒµÄÉèÏëÌá³ö¸Ä½ø´ëÊ©£º¸ÄÓÃͼ£­1×°ÖüÓÈÈîÑÌú¿óºÍÌ¿·Û£¬²¢ÔÚUÐ͹ÜB֮ǰÔö¼ÓÊ¢ÓР      __________µÄȼÉչܣ¬ÔÚBÖ®ºóÔö¼Ó                             ¡£
¢Ú¸Ä½øºó£¬ÊµÑé×ÔʼÖÁÖÕÐè³ÖÐøÏò×°ÖÃAÖÐͨÈëN2µÄÄ¿µÄÊÇ                        ¡£
[̽¾¿¶þ]ÄÉÃ×TiO2×÷ΪһÖÖ¹â´ß»¯¼ÁÔ½À´Ô½Êܵ½ÈËÃǵĹØ×¢£¬ÏÖÕý¹ã·º¿ª·¢ÔËÓá£
£¨4£©Ä³¿ÆÑÐС×éÔÚ²»Í¬µÄÔØÌ壨îÑƬ¡¢ÂÁƬ¡¢ÌÕ´É£©±íÃæÖƱ¸¶þÑõ»¯îѱ¡Ä¤£¬À´¿¼²ì²»Í¬ÔØÌåTiO2±¡Ä¤¹â´ß»¯Ê¹¼×»ù³ÈÍÑÉ«£¬Ã¿´Î¹âÕÕ20minÈ¡Ò»´ÎÑù£¬ÊµÑé½á¹ûÈçͼËùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ          ¡££¨Ë«Ñ¡£©

£¨a£©²»Í¬ÔØÌ壬ÎÞÂÛºÎÖÖζÈÒ»¶¨ÊÇîÑƬ×îºÃ
£¨b£©Ô¼ÔÚ520¡æʱ£¬îÑƬÔØÌåµÄ¹â´ß»¯»îÐÔ×îºÃ
£¨c£©ÎÞÂÛºÎÖÖÔØÌ壬´ß»¯»îÐÔ×ÜÊÇËæζȵÄÉý¸ß¶øÉý¸ß
£¨d£©²»Í¬¸ºÔØTiO2±¡Ä¤µÄ¹â´ß»¯»îÐÔ²»Í¬

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

¾«Ó¢¼Ò½ÌÍø