£¨13·Ö£©
ÀûÓÃijº¬¸õ·ÏÒº£Ûº¬½ÏµÍŨ¶ÈµÄNa2Cr2O7¡¢Fe2(SO4)3£ÝÖƱ¸K2Cr2O7¡£
Á÷³ÌÈçÏ£º
¢ñ£®ÓÃNaOHÈÜÒºµ÷pHÖÁ3.6£¬²úÉúºìºÖÉ«³Áµí£¬¹ýÂË£»
¢ò£®ÏòÂËÒºÖмÓÈëNa2SO3£¬Ò»¶¨²Ù×÷ºó·ÖÀë³öNa2SO4£»
¢ó£®½«·ÖÀë³öNa2SO4ºóµÄÈÜÒºµ÷pHԼΪ5£¬µÃµ½Cr(OH)3³Áµí£»
¢ô£®ÔÚKOH´æÔÚÌõ¼þÏ£¬ÏòCr(OH)3ÖмÓÈë×ãÁ¿H2O2ÈÜÒº£¬µÃµ½»ÆÉ«ÈÜÒº£»
¢õ£®Ïò»ÆÉ«ÈÜÒºÖмÓÈëÎïÖÊAºó£¬ÈÜÒº±äΪ³ÈºìÉ«£¬Ò»¶¨²Ù×÷ºóµÃµ½K2Cr2O7¹ÌÌ壻
¢ö£®²â¶¨K2Cr2O7¹ÌÌåµÄ´¿¶È¡£
ÒÑÖª£ºCr2O72£­£¨³ÈºìÉ«£©£«H2O2CrO42£­£¨»ÆÉ«£©£«2H+
£¨1£©²½Öè¢ñÖкìºÖÉ«³ÁµíµÄ»¯Ñ§Ê½ÊÇ     ¡£
£¨2£©²½Öè¢òÖмÓÈëNa2SO3µÄÄ¿µÄÊÇ     ¡£
£¨3£©²½Öè¢ôÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ     ¡£
£¨4£©²½Öè¢õÖмÓÈëµÄÎïÖÊA¿ÉÒÔÊÇ     ¡££¨ÌîÐòºÅ£©
a£®KOH        b£®K2CO3       c£®H2SO4        d£®SO2
£¨5£©²½Öè¢öµÄ²Ù×÷ÊÇ£ºÈ¡0.45 g K2Cr2O7²úÆ·Åä³ÉÈÜÒº£¬ËữºóµÎÈë18.00 mL
0.50 mol/LµÄFeSO4ÈÜÒº£¬Ç¡ºÃʹCr2O72£­Íêȫת»¯ÎªCr3+¡£²úÆ·ÖÐK2Cr2O7µÄ´¿¶ÈÊÇ     ¡££¨×¢£ºK2Cr2O7µÄĦ¶ûÖÊÁ¿Îª294 g/mol£©
£¨6£©Ïò³ÈºìÉ«µÄK2Cr2O7ÈÜÒºÖУ¬µÎ¼ÓBa(NO3)2ÈÜÒº£¬²úÉú»ÆÉ«³Áµí£¬ÈÜÒºpH¼õС¡£ÊÔÍƲâ»ÆÉ«³ÁµíÊÇ     £¬ÈÜÒºpH±äСµÄÔ­ÒòÊÇ     ¡£
£¨13·Ö£©
£¨1£©Fe(OH)3  £¨1·Ö£©         
£¨2£©½«+6¼ÛµÄCr»¹Ô­Îª+3¼Û    £¨2·Ö£©
£¨3£©2Cr(OH)3£«3H2O2£«4OH-£½2CrO42-£«8H2O   £¨2·Ö£©
£¨4£©c       £¨2·Ö£©£»
£¨5£©98%     £¨2·Ö£©
£¨6£©BaCrO4   £¨2·Ö£©£»
K2Cr2O7ÈÜÒºÖдæÔÚƽºâ£ºCr2O72£­£¨³ÈºìÉ«£©+H2O2CrO42£­£¨»ÆÉ«£©+2H+£¬¼ÓÈëBa(NO3)2ÈÜÒººó£¬²úÉúBaCrO4³Áµí£¬c (CrO42£­)½µµÍ£¬Æ½ºâÕýÏòÒƶ¯£¬c (H+)Ôö´ó¡££¨2·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©²½Öè¢ñÖкìºÖÉ«³ÁµíÊÇÇâÑõ»¯Ìú£¬»¯Ñ§Ê½ÊÇFe(OH)3£»
£¨2£©²½Öè¢óÖÐÓÐCr(OH)3³ÁµíÉú³É£¬ËµÃ÷֮ǰµÄÈÜÒºÖдæÔÚCr3+£¬ËùÒÔ¼ÓÈëNa2SO3µÄÄ¿µÄÊǽ«+6¼ÛµÄCr»¹Ô­Îª+3¼Û£»
£¨3£©¸ù¾ÝÒÑÖªµÃ»ÆÉ«ÊÇ2CrO42£­µÄÑÕÉ«£¬ËùÒÔ²½Öè¢ôÖз¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ2Cr(OH)3£«3H2O2£«4OH-£½2CrO42-£«8H2O£»
£¨4£©¸ù¾ÝCr2O72£­£¨³ÈºìÉ«£©£«H2O2CrO42£­£¨»ÆÉ«£©£«2H+¿ÉÖª£¬¼ÓÈëµÄAÎïÖÊ¿ÉʹƽºâÄæÏòÒƶ¯£¬ËùÒÔÓ¦ÊÇËáÐÔÎïÖÊ£¬Åųýab£»¶þÑõ»¯Áò¾ßÓл¹Ô­ÐÔ£¬¿ÉÓëCrO42£­·´Ó¦ÓÖÉú³ÉCr3+£¬¶øÁòËá²»ÓëCrO42£­·´Ó¦£¬ÇÒÄÜʹƽºâÄæÏòÒƶ¯£¬´ð°¸Ñ¡c£»
£¨5£©FeSO4µÄÎïÖʵÄÁ¿ÊÇ0.018L¡Á0.5mol/L=0.009mol£¬±»Ñõ»¯³ÉFe3+ʧµç×ÓµÄÎïÖʵÄÁ¿ÊÇ0.009mol£¬Éè
K2Cr2O7µÄÎïÖʵÄÁ¿ÊÇx£¬»¹Ô­³ÉCr3+£¬µÃµç×ÓµÄÎïÖʵÄÁ¿¶Ô2x¡Á3£¬¸ù¾ÝµÃʧµç×ÓÊغ㣬ÓÐ
2x¡Á3=0.009mol£¬ËùÒÔx=0.0015mol£¬Ôò²úÆ·ÖÐK2Cr2O7µÄ´¿¶ÈÊÇ0.0015mol¡Á294g/mol¡Â0.45g¡Á100%=98%;
£¨6£©ÓÉÀë×ÓµÄÑÕÉ«ÅжϸûÆÉ«³ÁµíÊÇBaCrO4 £»ÈÜÒºÖдæÔÚƽºâ£ºCr2O72£­£¨³ÈºìÉ«£©+H2O2CrO42£­£¨»ÆÉ«£©+2H+£¬¼ÓÈëBa(NO3)2ÈÜÒººó£¬²úÉúBaCrO4³Áµí£¬Ê¹c (CrO42£­)½µµÍ£¬Æ½ºâÕýÏòÒƶ¯£¬c (H+)Ôö´ó£¬pH½µµÍ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ʵÑéÊÒ³£ÀûÓü×È©·¨²â¶¨£¨NH4£©2SO4ÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý£¬Æä·´Ó¦Ô­ÀíΪ£º4NH4++6HCHO¨T3H++6H2O+£¨CH2£©6N4H+[µÎ¶¨Ê±£¬1mol£¨CH2£©6N4H+Óë1molH+Ï൱]£¬È»ºóÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨·´Ó¦Éú³ÉµÄËᣮijÐËȤС×éÓü×È©·¨½øÐÐÁËÈçÏÂʵÑ飺
²½ÖèI£º³ÆÈ¡ÑùÆ·1.500g£®
²½Öè¢ò£º½«ÑùÆ·Èܽâºó£¬ÍêȫתÒƵ½250mLÈÝÁ¿Æ¿ÖУ¬¶¨ÈÝ£¬³ä·ÖÒ¡ÔÈ£®
²½Öè¢ó£ºÒÆÈ¡25.00mLÑùÆ·ÈÜÒºÓÚ250mL׶ÐÎÆ¿ÖУ¬¼ÓÈë10mL 20%µÄÖÐÐÔ¼×È©ÈÜÒº£¬Ò¡ÔÈ¡¢¾²ÖÃ5minºó£¬¼ÓÈë1¡«2µÎ·Ó̪ÊÔÒº£¬ÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣮°´ÉÏÊö²Ù×÷·½·¨ÔÙÖظ´2´Î£®
£¨1£©¸ù¾Ý²½Öè¢óÌî¿Õ£º
¢Ù¼îʽµÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó£¬Ö±½Ó¼ÓÈëNaOH±ê×¼ÈÜÒº½øÐе樣¬Ôò²âµÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊý______£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£®
¢Ú׶ÐÎÆ¿ÓÃÕôÁóˮϴµÓºó£¬Ë®Î´µ¹¾¡£¬ÔòµÎ¶¨Ê±ÓÃÈ¥NaOH±ê×¼ÈÜÒºµÄÌå»ý______£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©£®
¢ÛµÎ¶¨Ê±±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦Ó¦¹Û²ì______£®
A¡¢µÎ¶¨¹ÜÄÚÒºÃæµÄ±ä»¯¡¡¡¡¡¡¡¡B¡¢×¶ÐÎÆ¿ÄÚÈÜÒºÑÕÉ«µÄ±ä»¯
¢ÜµÎ¶¨´ïµ½ÖÕµãʱ£¬·Óָ̪ʾ¼ÁÓÉ______É«±ä³É______É«£®
£¨2£©µÎ¶¨½á¹ûÈçϱíËùʾ£º
µÎ¶¨´ÎÊý´ý²âÈÜÒºµÄÌå»ý/mL±ê×¼ÈÜÒºµÄÌå»ý
µÎ¶¨Ç°¿Ì¶È/mLµÎ¶¨ºó¿Ì¶È/mL
125.001.0221.03
225.002.0021.99
325.000.2020.20
ÈôNaOH±ê×¼ÈÜÒºµÄŨ¶ÈΪ0.1010mol?L-1£¬Ôò¸ÃÑùÆ·ÖеªµÄÖÊÁ¿·ÖÊýΪ______£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

£¨12·Ö£©¡¾»¯Ñ§¡ª»¯Ñ§Óë¼¼Êõ¡¿
¹¤ÒµÉÏÓÃÖظõËáÄÆ£¨Na2Cr2O7£©½á¾§ºóµÄĸҺ£¨º¬ÉÙÁ¿ÔÓÖÊFe3+£©Éú²úÖظõËá¼Ø£¨K2Cr2O7£©£¬¹¤ÒÕÁ÷³Ì¼°Ïà¹ØÎïÖÊÈܽâ¶ÈÇúÏßÈçͼ£º

£¨1£©ÓÉNa2Cr2O7Éú²úK2Cr2O7µÄ»¯Ñ§·½³ÌʽΪ¡¡_________¡¡£¬Í¨¹ýÀäÈ´½á¾§Îö³ö´óÁ¿K2Cr2O7µÄÔ­ÒòÊÇ¡¡_________¡¡£®
£¨2£©ÏòNa2Cr2O7ĸҺÖмӼîÒºµ÷pHµÄÄ¿µÄÊÇ¡¡_________¡¡£®
£¨3£©¹ÌÌåAÖ÷ҪΪ¡¡_________¡¡£¨Ìѧʽ£©£¬¹ÌÌåBÖ÷ҪΪ¡¡_________¡¡£¨Ìѧʽ£©£®
£¨4£©ÓÃÈÈˮϴµÓ¹ÌÌåA£¬»ØÊÕµÄÏ´µÓҺתÒƵ½Ä¸Òº¡¡_________¡¡£¨Ìî¡°¢ñ¡±¡°¢ò¡±»ò¡°¢ó¡±£©ÖУ¬¼ÈÄÜÌá¸ß²úÂÊÓÖ¿ÉʹÄܺÄ×îµÍ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

£¨16·Ö£©Ä³»¯Ñ§ÐËȤС×éÉè¼ÆʵÑ飬ÓÃŨÁòËáÓëÍ­·´Ó¦ÖƵÃSO2²¢½øÐÐÏà¹ØʵÑé̽¾¿¡£

£¨1£©×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
ÊÇ_______£¬×°ÖÃBµÄ×÷ÓÃÊÇ_______¡£
Éè¼Æ×°ÖÃCµÄÄ¿µÄÊÇÑéÖ¤SO2µÄ
_______________ÐÔ£¬DÖÐNaOHÈ«²¿×ª»¯Îª
NaHSO3µÄ±êÖ¾ÊÇ____________¡£
£¨3£©·´Ó¦½áÊøʱÉÕÆ¿ÖÐCuÓÐÊ£Ó࣬ij
ͬѧÈÏΪH2SO4Ò²ÓÐÊ£Ó࣬ËûÉè¼ÆÁËÏÂÁÐʵ
Ñé·½°¸À´²â¶¨Ê£ÓàH2SO4µÄÁ¿¡£¾­ÀäÈ´£¬¶¨
Á¿Ï¡Êͺó½øÐÐÏÂÁÐʵÑ飬ÄܴﵽĿµÄÊÇ________£¨ÌîÐòºÅ£©£»
a£® ÓÃËá¼îÖк͵ζ¨·¨²â¶¨    b£® Óë×ãÁ¿Zn·´Ó¦£¬²âÁ¿Éú³ÉH2µÄÌå»ý
c£® ÓÃPH¼Æ²âÈÜÒºPHÖµ     d£® Óë×ãÁ¿BaCl2ÈÜÒº·´Ó¦£¬³ÆÁ¿Éú³É³ÁµíµÄÖÊÁ¿
£¨4£©ÏòDÆ¿ËùµÃNaHSO3ÈÜÒºÖмÓÈëƯ°×·ÛÈÜÒº£¬·´Ó¦ÓÐÈýÖÖ¿ÉÄÜÇé¿ö£º
I£® HSO3-ÓëClO-¸ÕºÃ·´Ó¦Íꣻ  II£® Ư°×·Û²»×㣻  III£® Ư°×·Û¹ýÁ¿
ͬѧÃÇ·Ö±ðÈ¡ÉÏÊö»ìºÏÈÜÒºÓÚÊÔ¹ÜÖУ¬Í¨¹ýÏÂÁÐʵÑéÈ·¶¨¸Ã·´Ó¦ÊôÓÚÄÄÒ»ÖÖÇé¿ö£¬ÇëÄãÍê³ÉÏÂ±í£º
ʵÑéÐòºÅ
ʵÑé²Ù×÷
ÏÖÏó
·´Ó¦µÄ¿ÉÄÜÇé¿ö
¢Ù
µÎ¼ÓÉÙÁ¿µí·Ûµâ»¯¼ØÈÜÒº£¬Õñµ´
 
III
¢Ú
µÎ¼ÓÉÙÁ¿×غìÉ«µÄKI3ÈÜÒº£¬Õñµ´
 
II
¢Û
µÎÈë¼ÓÉÙÁ¿ËáÐÔKMNO4ÈÜÒº£¬Õñµ´
ÈÜÒº³Ê×ÏÉ«
 
¢Ü
¼ÓÈ뼸С¿éCaCO3¹ÌÌå
ÓÐÆøÅݲúÉú
 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÎÊ´ðÌâ

ijС×éÉè¼ÆÈçͼװÖ㨲¿·Ö¼Ð³Ö×°ÖÃÒÑÂÔÈ¥£©£¬ÒÔ̽¾¿³±ÊªµÄCl2ÓëNa2CO3·´Ó¦µÃµ½¹ÌÌåÎïÖʵijɷ֡£
 
£¨1£©AÖÐÖÆÈ¡Cl2µÄÀë×Ó·½³Ìʽ                                      .
£¨2£©Í¨ÈëÒ»¶¨Á¿³±ÊªµÄCl2·´Ó¦ºó£¬¾­¼ì²â£¬DÖÐÖ»ÓÐCl2OÒ»ÖÖÆøÌ壬CÖÐÖ»º¬Ò»ÖÖÂÈÑÎÍ⣬ͬʱº¬ÓÐNaHCO3µÈ£¬Ä³Í¬Ñ§¶ÔCÖÐËùµÃ¹ÌÌå²ÐÔüµÄ³É·Ö½øÐÐ̽¾¿¡£
¢ÙÌá³öºÏÀí¼ÙÉè¡£
¼ÙÉè1£º´æÔÚÁ½Öֳɷ֣ºNaHCO3ºÍ                            £»
¼ÙÉè2£º´æÔÚÈýÖֳɷ֣ºNaHCO3ºÍ                            ¡£
¢ÚÉè¼Æ·½°¸£¬½øÐÐʵÑ顣д³öʵÑé²½ÖèÒÔ¼°Ô¤ÆÚÏÖÏóºÍ½áÂÛ¡£
ÏÞѡʵÑéÊÔ¼ÁºÍÒÇÆ÷£ºÕôÁóË®¡¢Ï¡HNO3¡¢BaCl2ÈÜÒº¡¢³ÎÇåʯ»ÒË®¡¢AgNO3ÈÜÒº¡¢ÊԹܡ¢Ð¡ÉÕ±­¡£
 £¨3£©ÒÑÖªCÖÐÓÐ0£®1 mol Cl2²Î¼Ó·´Ó¦¡£Èô¼ÙÉèÒ»³ÉÁ¢£¬¿ÉÍÆÖªCÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÎÊ´ðÌâ

ÉòÑôÀí¹¤´óѧÔڷϾɵç³Ø»ØÊÕ´¦ÀíÏÖ×´Ñо¿±¨¸æÖб¨µÀÁËÒ»ÖÖʪ·¨´¦Àí·½·¨£¬ÒÔÏÂÊǶԷϾɵç³Ø²ÄÁÏ(º¬ÓÐNi(OH)2¡¢Ì¼·Û¡¢Ñõ»¯ÌúºÍÂÁ²­µÈ)½øÐÐ×ÊÔ´»ØÊÕËùÉè¼ÆµÄʵÑéÁ÷³Ì

ÒÑÖª£º¢ÙNiCl2Ò×ÈÜÓÚË®£¬Fe3£«²»ÄÜÑõ»¯Ni2£«¡£
¢ÚÒÑ֪ʵÑéζÈʱµÄÈܽâ¶È£ºNiC2O4£¾NiC2O4¡¤H2O£¾NiC2O4¡¤2H2O
¢Û½ðÊôÀë×ÓÔÚ¸ÃʵÑéÁ÷³ÌÏÂÍêÈ«³ÁµíµÄpH£º
Àë×Ó
Al3£«
Fe3£«
Ni2£«
pH
5.2
4.1
9.7
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÎïÖÊA µÄ×÷ÓÃÊǵ÷½ÚÈÜÒºµÄpHÒÔ³ýÈ¥ÔÓÖÊFe3£«ºÍAl3£«£¬ÒÔÏÂ×îÊʺÏ×÷ΪAÎïÖʵÄÊÇ________¡£
A£®NiO  B£®ÑÎËá  C£®NaOH  D£®°±Ë®
(2)ÂËÔü¢ÚµÄÖ÷Òª³É·ÖÊÇ_________________________________¡£
(3)д³ö¼ÓÈëNa2C2O4ÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º _____________¡£
(4)д³öµç½âÂËÒº¢ÛµÄÒõ¼«·´Ó¦Ê½___________________________¡£
¼ìÑéµç½âÂËÒº¢ÛʱÑô¼«²úÉúµÄÆøÌåµÄ·½·¨ÊÇÓÃʪÈóµÄ________ÊÔÖ½¼ìÑé¡£
(5)ÓɳÁµíAÉú³ÉNi(OH)3µÄ¹ý³ÌÊÇÏÈÔÚÂËÔü¢ÛÖмÓÈë¹ýÁ¿NaOHÈÜÒº½øÐгÁµíת»¯£¬´ý·´Ó¦ÍêÈ«ºóÔÙͨÈëµç½âÂËÒº¢Û²úÉúµÄÆøÌå½øÐÐÑõ»¯£¬Çëд³ö¸ÃÑõ»¯¹ý³ÌµÄÀë×Ó·½³Ìʽ£º__________________¡£
(6)ÈçºÎ¼ìÑéNi(OH)3ÊÇ·ñÏ´µÓ¸É¾»£¿
________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

µþµª»¯ÄÆ£¨NaN3£©ÊÇÆû³µ°²È«ÆøÄÒÖеÄÖ÷Òª³É·Ö£¬ÄÜÔÚ·¢ÉúÅöײµÄ˲¼ä·Ö½â²úÉú´óÁ¿ÆøÌ彫ÆøÄÒ¹ÄÆð¡£ÊµÑéÊҲⶨµþµª»¯ÄÆÑùÆ·ÖÐNaN3µÄÖÊÁ¿·ÖÊý¡£ÊµÑé²½ÖèÈçÏ£º
¢Ù³ÆÈ¡Ô¼2.5000gµþµª»¯ÄÆÊÔÑù£¬Åä³É250mLÈÜÒº¡£
¢Ú׼ȷÁ¿È¡25.00mLÈÜÒºÖÃÓÚ׶ÐÎÆ¿ÖУ¬Óõζ¨¹Ü¼ÓÈë50.00mL 0.1000mol¡¤L£­1
(NH4)2Ce(NO3)6£¨ÁùÏõËáîæ泥©£¬
[·¢Éú·´Ó¦Îª£º2(NH4)2Ce(NO3)6 +2NaN3=4NH4NO3+2Ce(NO3)3+2NaNO3+3N2¡ü]£¨ÔÓÖʲ»²ÎÓë·´Ó¦£©¡£
¢Û·´Ó¦ºó½«ÈÜÒºÉÔÏ¡ÊÍ£¬È»ºóÏòÈÜÒºÖмÓÈë5mLŨÁòËᣬµÎÈë2µÎÁÚ·ÆßÜßøָʾҺ£¬ÓÃ0.0500mol¡¤L£­1(NH4)2Fe(SO4)2£¨ÁòËáÑÇÌú泥©±ê×¼µÎ¶¨ÈÜÒºµÎ¶¨¹ýÁ¿µÄCe4+ÖÁÈÜÒºÓɵ­ÂÌÉ«±äΪ»ÆºìÉ«£¨·¢ÉúµÄ·´Ó¦Îª£ºCe4++Fe2+= Ce3++Fe3+£©£¬ÏûºÄÁòËáÑÇÌú隣ê×¼ÈÜÒº24.00mL¡£
£¨1£©²½Öè¢ÙÅäÖƵþµª»¯ÄÆÈÜҺʱ£¬³ýÐèÓõ½ÉÕ±­¡¢²£Á§°ô¡¢Á¿Í²Í⣬»¹Óõ½µÄ²£Á§ÒÇÆ÷ÓР              ¡¢               ¡£
£¨2£©µþµª»¯ÄÆÊÜײ»÷ʱÉú³ÉÁ½ÖÖµ¥ÖÊ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                               ¡£
£¨3£©ÈôÆäËü¶ÁÊýÕýÈ·£¬µÎ¶¨µ½ÖÕµãºó£¬¶ÁÈ¡µÎ¶¨¹ÜÖÐ (NH4)2Fe(SO4)2±ê×¼ÈÜÒºÌå»ý°´Í¼Í¼Ê¾¶ÁÈ¡£¬½«µ¼ÖÂËù²â¶¨ÑùÆ·Öеþµª»¯ÄÆÖÊÁ¿·ÖÊý               £¨Ñ¡Ì¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°²»±ä¡±£©¡£

£¨4£©Í¨¹ý¼ÆËãÈ·¶¨µþµª»¯ÄÆÊÔÑùÖк¬NaN3µÄÖÊÁ¿·ÖÊýΪ¶àÉÙ£¨Ð´³ö¼ÆËã¹ý³Ì£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ÈçͼËùʾ¡°ºÏ³É°±¡±µÄÑÝʾʵÑé(¼Ð³ÖÒÇÆ÷¾ùÒÑÊ¡ÂÔ)¡£ÔÚYÐιܵÄÒ»²àÓÃZnÁ£ºÍÏ¡H2SO4·´Ó¦ÖÆÈ¡H2£¬ÁíÒ»²àÓÃNaNO2¹ÌÌåºÍNH4Cl±¥ºÍÈÜÒº·´Ó¦ÖÆÈ¡N2£¬N2ºÍH2»ìºÏºóͨ¹ý»¹Ô­Ìú·ÛÀ´ºÏ³ÉNH3£¬ÔÙ½«²úÉúµÄÆøÌåͨÈë·Ó̪ÊÔÒºÖУ¬Èô·Ó̪ÊÔÒº±äºì£¬Ôò˵Ã÷²úÉúÁË°±Æø¡£

ij¿ÎÍâ»î¶¯Ð¡×éͨ¹ý²éÔÄ×ÊÁϺͶà´ÎʵÑ飬µÃµ½ÁËÈçÏÂÐÅÏ¢£º
ÐÅÏ¢Ò»£ºNaNO2¹ÌÌåºÍ±¥ºÍNH4ClÈÜÒº»ìºÏ¼ÓÈȵĹý³ÌÖз¢ÉúÈçÏ·´Ó¦£º
¢ÙNaNO2£«NH4Cl NH4NO2£«NaCl
¢ÚNH4NO2NH3¡ü£«HNO2
¢Û2HNO2N2O3¡ü£«H2O
¢Ü2NH3£«N2O32N2£«3H2O
ÐÅÏ¢¶þ£º²éÔÄ×ÊÁÏ£¬²»Í¬Ìå»ý±ÈµÄN2¡¢H2»ìºÏÆøÌåÔÚÏàͬʵÑéÌõ¼þϺϳɰ±£¬Ê¹·Ó̪ÊÔÒº±äºìËùÐèÒªµÄʱ¼äÈçÏ£º
N2ºÍH2µÄÌå»ý±È
5¡Ã1
3¡Ã1
1¡Ã1
1¡Ã3
1¡Ã5
·Ó̪±äºìÉ«ËùÐèʱ¼ä/min
8¡«9
7¡«8
6¡«7
3¡«4
9¡«10
 
¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)YÐιÜ×ó²à¹ÜÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ________________________¡£
(2)Ìú·ÛÈöÔÚʯÃÞÈÞÉϵÄÄ¿µÄÊÇ_________________________________
(3)¿ÎÍâ»î¶¯Ð¡×éµÄͬѧÃÇÈÏΪ£¬¸ÃʵÑéÖм´Ê¹·Ó̪±äºìÒ²²»ÄÜ˵Ã÷N2ºÍH2·´Ó¦ºÏ³ÉÁËNH3£¬µÃ³ö´Ë½áÂÛµÄÀíÓÉÊÇ________________________¡£
ÇëÄãÁíÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑéÑéÖ¤ÄãµÄÀíÓÉ____________________¡£Óû½â¾öÕâÒ»ÎÊÌ⣬¿ÉÒÔÑ¡ÓÃÏÂͼÖеÄ________×°ÖÃÁ¬½ÓÔÚÔ­×°ÖÃÖеÄ________ºÍ________Ö®¼ä¡£

(4)ÔÚÉÏÊöʵÑé¹ý³ÌÖУ¬Îª¾¡¿ì¹Û²ìµ½·Ó̪ÊÔÒº±äºìµÄʵÑéÏÖÏó£¬Ó¦¸Ã¿ØÖÆN2ºÍH2µÄÌå»ý±ÈΪ________±È½ÏÊÊÒË£»µ«¸Ã×°Öû¹ÄÑÒÔʵÏÖ´ËÄ¿µÄ£¬Ô­ÒòÊÇ______________________________________¡£
(5)ʵÑé¹ý³ÌÖÐͨÈëÊÔ¹ÜCÖеÄÆøÌå³É·ÖÓÐ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

¼îʽ̼ËáÑÎA¿ÉÓÃ×÷θҩ£¬Æä×é³É¿É±íʾΪAl2Mg6(OH)x(CO3)y¡¤zH2O¡£Ä³Ð£»¯Ñ§ÐËȤС×éÓû²â¶¨Æ仯ѧʽ£¬ÊµÑéÉè¼ÆÈçÏ£º
ʵÑéI£º³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄA£¬¼ÓÈÈ·Ö½âÖÁºãÖØ¡£
ʵÑé¢ò£º³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄAÓë×ãÁ¿µÄËá·´Ó¦£¬²âÁ¿Éú³ÉCO2ÆøÌåµÄÖÊÁ¿¡£
¿É¹©Ñ¡ÔñµÄÒÇÆ÷ºÍÒ©Æ·ÈçͼËùʾ£º£¨ËáÈÜÒºÏÞÑ¡6mol/LHCl»ò6mol/LH2SO4£¬ÆäËüÊÔ¼ÁÈÎÑ¡¡££©

»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÒÇÆ÷ÒÒµÄÃû³ÆΪ________¡£
(2)ÇëÑ¡Ôñ±ØÒªµÄ×°ÖÃÍê³ÉʵÑéII,ÕýÈ·µÄÁ¬½Ó˳ÐòΪ________ (°´ÆøÁ÷·½Ïò£¬ÓýӿÚ×Öĸ±íʾ£©£»Ñ¡ÓõÄËáÈÜÒºÊÇ________¡£
(3)ÓÐÈËÌá³ö²»²ÉÓÃʵÑéI£¬¿ÉÔÚʵÑéII½áÊøºó£¬ÔÚAÍêÈ«·´Ó¦ºóËùµÃÈÜÒºÖеμÓ×ãÁ¿µÄ°±Ë®£¬ÓÃÎÞ»ÒÂËÖ½¹ýÂË£¬ÓÃÕôÁóˮϴµÓ·´Ó¦ÈÝÆ÷2?3´Î£¬½«Ï´µÓÒº¹ýÂË£¬Ï´µÓ³Áµí2?3´Î£¬½«¸½×ųÁµíµÄÂËÖ½·Åµ½ÛáÛöÖмÓÈÈ·Ö½âÖÁºãÖØ¡£ÅжϳÁµíÒÑÏ´µÓ¸É¾»µÄ·½·¨ÊÇ_________________,ʵ¼ÊÉÏδ²ÉÓø÷½°¸µÄÔ­ÒòÊDz»·ûºÏʵÑéÉè¼ÆµÄ________Ô­Ôò£¨Ìî×Öĸ±àºÅ£©¡£
A£®¿ÆѧÐÔB£®°²È«ÐÔC£®¿ÉÐÐÐÔD£®¼òÔ¼ÐÔ
(4)xÓëyµÄ¹ØϵʽÊÇ________ (ÓõÈʽ±íʾ£©¡£
(5)ͨ¹ýʵÑéIµÄÏà¹ØÊý¾Ý£¬¾­¼ÆËã¿ÉµÃAµÄĦ¶ûÖÊÁ¿Îª602.0g. mol-1¡£ÊµÑéIIÖгƵÃÑùÆ·AµÄÖÊÁ¿Îª9.030g£¬Óë×ãÁ¿ËáÍêÈ«·´Ó¦ºó£¬CO2ÎüÊÕ×°ÖÃÔöÖØ0.660g,ÔòA µÄ»¯Ñ§Ê½Îª________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸