¡¾ÌâÄ¿¡¿NaIÓÃ×÷ÖƱ¸ÎÞ»úºÍÓлúµâ»¯ÎïµÄÔ­ÁÏ£¬Ò²ÓÃÓÚÒ½Ò©ºÍÕÕÏàµÈ£¬¹¤ÒµÀûÓõ⡢ÇâÑõ»¯ÄƺÍÌúмΪԭÁÏ¿ÉÉú²úNaI£¬ÆäÉú²úÁ÷³ÌÈçÏÂͼ¡£

£¨1£©µâÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃΪ______________________¡£

£¨2£©·´Ó¦¢ÙµÄÀë×Ó·½³ÌʽΪ___________________________________________¡£

£¨3£©·´Ó¦¢Ú¼ÓÈë¹ýÁ¿ÌúмµÄÄ¿µÄÊÇ_________________£¬¹ýÂËËùµÃ¹ÌÌå1ÖгýÊ£ÓàÌúмÍ⣬»¹ÓкìºÖÉ«¹ÌÌ壬Ôò¼ÓÈëÌúмʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________________________¡£

£¨4£©ÈÜÒº2Öгýº¬ÓÐH+Í⣬һ¶¨»¹º¬ÓеÄÑôÀë×ÓÊÇ_______________£»ÊÔÉè¼ÆʵÑéÑéÖ¤ÈÜÒº2ÖиýðÊôÑôÀë×ӵĴæÔÚ£º___________________________________________¡£

£¨5£©ÈÜÒº2¾­Ò»ÏµÁÐת»¯¿ÉÒԵõ½²ÝËáÑÇÌú¾§Ì壨FeC2O4¡¤2H2O£¬Ïà¶Ô·Ö×ÓÖÊÁ¿180£©£¬³ÆÈ¡3.60 g²ÝËáÑÇÌú¾§Ì壬ÓÃÈÈÖØ·¨¶ÔÆä½øÐÐÈȷֽ⣨¸ô¾ø¿ÕÆø¼ÓÈÈ)£¬µÃµ½Ê£Óà¹ÌÌåµÄÖÊÁ¿Ëæζȱ仯µÄÇúÏßÈçÓÒͼËùʾ£º

¢Ù·ÖÎöͼÖÐÊý¾Ý£¬¸ù¾ÝÐÅϢд³ö¹ý³ÌI·¢ÉúµÄ»¯Ñ§·½³Ìʽ£º_________________________________¡£

¢Ú300¡æʱʣÓà¹ÌÌåÖ»º¬Ò»ÖֳɷÖÇÒÊÇÌúµÄÑõ»¯Îд³ö¹ý³ÌII·¢ÉúµÄ»¯Ñ§·½³Ìʽ£º ________________¡£

¡¾´ð°¸¡¿ µÚÎåÖÜÆÚµÚVIIA×å 3I2+6OH-5I-+IO3-+3H2O ½«NaIO3Íêȫת»¯ÎªNaI 2Fe+3H2O+ NaIO3=NaI+2Fe(OH)3¡ý Fe2+ È¡ÉÙÁ¿ÈÜÒº2¼ÓÈëËáÐÔ¸ßÃÌËá¼ØÈÜÒº£¬ÈôÈÜÒºÍÊÉ«£¬Ôòº¬Fe2+£»»òÈ¡ÉÙÁ¿ÈÜÒº2¼ÓÈëK3[Fe(CN)6]ÈÜÒº£¬ÈôÓÐÀ¶É«³ÁµíÉú³É£¬Ôòº¬Fe2+ FeC2O4¡¤2H2OFeC2O4+2H2O FeC2O4¡¤2H2OFeO+CO2¡ü+CO¡ü£«2H2O

¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£ºÓÉÁ÷³Ì¿ÉÖª£¬ÔÚ¼ÓÈÈÌõ¼þϵâÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³ÉÁ˵⻯Äƺ͵âËáÄƵĻìºÏÒº£¬¼ÓÈëÌú·Û°ÑµâËáÄÆ»¹Ô­Îªµâ»¯ÄÆ£¬¹ýÂ˺󣬽«ÂËÒºÕô·¢Å¨Ëõ¡¢½µÎ½ᾧ¡¢¹ýÂ˵õ½²úÆ·¡£¹ÌÌå1ÊÇÇâÑõ»¯ÌúºÍÊ£ÓàµÄÌú·Û£¬ÇâÑõ»¯Ìú±»Ï¡ÁòËáÈܽâºóÉú³ÉÁòËáÌúÈÜÒº£¬ÁòËáÌú¿ÉÒÔ±»Ìú·Û»¹Ô­ÎªÁòËáÑÇÌú¡£

£¨1£©µâÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃΪµÚÎåÖÜÆÚµÚVIIA×å¡£

£¨2£©·´Ó¦¢ÙµÄÀë×Ó·½³ÌʽΪ3I2+6OH-5I-+IO3-+3H2O¡£

£¨3£©·´Ó¦¢Ú¼ÓÈë¹ýÁ¿ÌúмµÄÄ¿µÄÊǽ«NaIO3Íêȫת»¯ÎªNaI£¬¹ýÂËËùµÃ¹ÌÌå1ÖгýÊ£ÓàÌúмÍ⣬»¹ÓкìºÖÉ«¹ÌÌ壬¸ÃºìºÖÉ«¹ÌÌåÊÇÇâÑõ»¯Ìú£¬Ôò¼ÓÈëÌúмʱ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Fe+3H2O+ NaIO3=NaI+2Fe(OH)3¡ý¡£

£¨4£©ÈÜÒº2Öгýº¬ÓÐH+Í⣬һ¶¨»¹º¬ÓеÄÑôÀë×ÓÊÇFe2+£¬ÒòΪFe3+µÄÑõ»¯ÐÔ±ÈH+Ç¿ÇÒÌú·ÛÎÞÊ£Ó࣬ËùÒÔ»¹¿ÉÄܺ¬ÓÐFe3+£¬£»Éè¼ÆʵÑéÑéÖ¤ÈÜÒº2ÖÐFe2+µÄ´æÔÚʱ£¬Òª¿¼ÂÇFe3+µÄ¸ÉÈÅ£¬ËùÒÔ¿ÉÒÔÈ¡ÉÙÁ¿ÈÜÒº2¼ÓÈëËáÐÔ¸ßÃÌËá¼ØÈÜÒº£¬ÈôÈÜÒºÍÊÉ«£¬Ôòº¬Fe2+£»»òÈ¡ÉÙÁ¿ÈÜÒº2¼ÓÈëK3[Fe(CN)6]ÈÜÒº£¬ÈôÓÐÀ¶É«³ÁµíÉú³É£¬Ôòº¬Fe2+¡£

£¨5£©ÓÉÌâÖÐÐÅÏ¢¿ÉÖª£¬3.60 g²ÝËáÑÇÌú¾§ÌåµÄÎïÖʵÄÁ¿Îª0.02mol£¬²ÝËáÑÇÌú¾§ÌåÖÐn(Fe)=0.02mol£¬ÌúÔªËØÖÊÁ¿Îª56g/mol¡Á0.02mol=1.12g£¬½á¾§Ë®µÄÎïÖʵÄÁ¿Îª0.04mol£¬½á¾§Ë®µÄÖÊÁ¿Îª0.04molg/mol=0.72g¡£

¢Ù·ÖÎöͼÖÐÊý¾Ý£¬¹ý³ÌI¹ÌÌåÖÊÁ¿±äΪ2.88g£¬3.60g-2.88g=0.72g£¬ËùÒÔÕâ¸ö¹ý³ÌÖÐʧȥÁËÈ«²¿½á¾§Ë®£¬ËùÒÔ¹ý³ÌI·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪFeC2O4¡¤2H2O FeC2O4 +2H2O ¡£

¢Ú300¡æʱʣÓà¹ÌÌåÖ»º¬Ò»ÖֳɷÖÇÒÊÇÌúµÄÑõ»¯ÎÓÉͼ¿ÉÖªÆäÖÊÁ¿Îª1.44g£¬ÆäÖÐÌúÔªËصÄÖÊÁ¿Îª1.12g£¬ËùÒÔÑõÔªËصÄÖÊÁ¿Îª1.44g-1.12g=0.32g£¬Çó³ön(O)=0.02mol£¬n(Fe): n(O)=1:1£¬ËùÒÔ¸ÃÑõ»¯ÎïΪFeO£¬ÒÀ¾ÝÖÊÁ¿Êغ㶨ÂÉ¿ÉÒÔд³ö¹ý³ÌII·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºFeC2O4¡¤2H2O FeO+CO2¡ü+CO¡ü£«2H2O ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚ373kʱ£¬°Ñ11.5gN2O4ÆøÌåͨÈëÌå»ýΪ500mLµÄÕæ¿ÕÃܱÕÈÝÆ÷ÖУ¬Á¢¼´³öÏÖºì×ØÉ«µÄNO2ÆøÌ壬·´Ó¦Ô­ÀíΪN2O42NO2¡£µ±·´Ó¦½øÐе½2sʱ£¬NO2º¬Á¿Îª0.01mol£¬·´Ó¦½øµ½60sʱ´ïµ½Æ½ºâ£¬´ËʱÈÝÆ÷ÄÚ»ìºÏÆøÌåÃܶÈÊÇÇâÆøÃܶȵÄ28.75±¶¡£ÊÔͨ¹ý¼ÆËãÌî¿Õ£º

£¨1£©¿ªÊ¼2sÄÚ£¬ÒÔN2O4±íʾµÄ·´Ó¦ËÙÂÊΪ___mol¡¤L£­1¡¤s£­1¡£

£¨2£©´ïµ½Æ½ºâʱ£¬ÌåϵµÄѹǿÊÇ¿ªÊ¼Ê±µÄ____±¶¡£

£¨3£©Æ½ºâʱ»¹ÓÐ_______mol N2O4¡£

£¨4£©Æ½ºâºóÈôѹËõÈÝÆ÷Ìå»ý£¬ÔòÔٴﵽƽºâºóNO2µÄŨ¶È½«_______(Ìî¡°Ôö´ó¡±¡¢¡°¼õÉÙ¡±»ò¡°²»±ä¡±)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½øÐÐÏÂÁÐʵÑ飬Ïà¹Ø˵·¨ÕýÈ·µÄÊÇ

A. ͼ¼×£ºÕô¸ÉNH4Cl±¥ºÍÈÜÒºÖƱ¸NH4Cl¾§Ìå

B. ͼÒÒ£ºÐγÉÃÀÀöµÄºìÉ«ÅçȪ£¬Ö¤Ã÷HC1¼«Ò×ÈÜÓÚË®

C. ͼ±û£ºÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº£¬¶¨ÈÝʱÈçͼÔòËùÅäNaOHÈÜҺŨ¶ÈÆ«µÍ

D. ͼ¶¡£ºËùʾװÖÃÓÃÓÚ³ýȥ̼ËáÇâÄƹÌÌåÖеÄÉÙÁ¿Ì¼ËáÄÆ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÎïÖÊÖ»º¬Àë×Ó¼üµÄÊÇ

A. H2O B. Na2O C. NH4Cl D. N2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚʯÓ͵Ä˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)

A. ʯÓÍÊôÓÚ¿ÉÔÙÉú¿óÎïÄÜÔ´ B. ʯÓÍÖ÷Òªº¬ÓÐ̼¡¢ÇâÁ½ÖÖÔªËØ

C. ʯÓ͵ÄÁÑ»¯ÊÇÎïÀí±ä»¯ D. ʯÓÍ·ÖÁóµÄ¸÷Áó·Ö¾ùÊÇ´¿¾»Îï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©´¦Àíº¬CO¡¢SO2Ñ̵ÀÆøÎÛȾµÄÒ»ÖÖ·½·¨Êǽ«ÆäÔÚ´ß»¯¼ÁµÄ×÷ÓÃÏÂת»¯Îªµ¥ÖÊS¡£ÒÑÖª£º

¢ÙCO(g)£«0.5O2(g)£½CO2(g) ¦¤H£½£­283.0 kJ¡¤mol£­1

¢ÚS(s)£«O2(g)£½SO2(g)¡¡ ¦¤H£½£­296.0 kJ¡¤mol£­1

Ôò´¦ÀíCO¡¢SO2µÄ·½·¨µÄÈÈ»¯Ñ§·½³ÌʽÊÇ________________________________¡£

£¨2£©µªÑõ»¯ÎïÊÇÔì³É¹â»¯Ñ§ÑÌÎíºÍ³ôÑõ²ãËðºÄµÄÖ÷ÒªÆøÌå¡£ÒÑÖª£º

CO(g)£«NO2(g)£½NO(g)£«CO2(g) ¦¤H£½£­a kJ¡¤mol£­1(a£¾0)

2CO(g)£«2NO(g)£½N2(g)£«2CO2(g) ¦¤H£½£­b kJ¡¤mol£­1(b£¾0)

ÈôÓñê×¼×´¿öÏÂ3.36 L CO½«NO2»¹Ô­ÖÁN2(COÍêÈ«·´Ó¦)£¬ÔòÕû¸ö¹ý³ÌÖÐתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª______mol£¬·Å³öµÄÈÈÁ¿Îª_____(Óú¬ÓÐaºÍbµÄ´úÊýʽ±íʾ)kJ¡£

£¨3£©ÓÃCH4´ß»¯»¹Ô­NOxÒ²¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÀýÈ磺

CH4(g)£«4NO2(g)£½4NO(g)£«CO2(g)£«2H2O(g) ¦¤H1£½£­574 kJ¡¤mol£­1¡¡ ¢Ù

CH4(g)£«4NO(g)£½2N2(g)£«CO2(g)£«2H2O(g) ¦¤H2¡¡ ¢Ú

Èô1 mol CH4»¹Ô­NO2ÖÁN2£¬Õû¸ö¹ý³ÌÖзųöµÄÈÈÁ¿Îª867 kJ£¬Ôò¦¤H2£½_______¡£

£¨4£©ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º

¢Ù ¡÷H£½£­285.8kJ/mol

¢Ú ¡÷H£½£­241.8kJ/mol

ÔòH2µÄȼÉÕÈÈ£¨¡÷H£©Îª________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÈçÏÂͼÁ½¸öµç½â²ÛÖУ¬A¡¢B¡¢C¡¢D¾ùΪʯīµç¼«¡£Èôµç½â¹ý³ÌÖй²ÓÐ0.02molµç×Óͨ¹ý£¬ÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ£¨ £©

A. ¼×ÉÕ±­ÖÐA¼«ÉÏ×î¶à¿ÉÎö³öÍ­0.64g

B. ¼×ÉÕ±­ÖÐB¼«Éϵ缫·´Ó¦Ê½4OH£­£­4e£­£½ 2H2O+O2¡ü

C. ÒÒÉÕ±­ÖеÎÈë·Ó̪ÊÔÒº£¬D¼«¸½½üÏȱäºì

D. ÉÕ±­ÖÐC¼«Éϵ缫·´Ó¦Ê½Îª4H+ + 4e£­£½2H2¡ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÓйØ˵·¨ÖÐÕýÈ·µÄÊÇ

A£®Ä³Î¶ÈʱµÄ»ìºÏÈÜÒºÖÐc(H+)=mol¡¤L-1£¬ËµÃ÷¸ÃÈÜÒº³ÊÖÐÐÔ(KwΪ¸ÃζÈʱˮµÄÀë×Ó»ý³£Êý)

B£®ÓÉË®µçÀë³öµÄc(H+)=10-12mol¡¤L-1µÄÈÜÒºÖÐ:Na+¡¢Ba2+¡¢HCO3-¡¢Cl-¿ÉÒÔ´óÁ¿¹²´æ

C£®ÒÑÖªKsp(AgCl)=1.56¡Á10-10£¬ Ksp(Ag2CrO4)=9.0¡Á10-12¡£Ïòº¬ÓÐCl-¡¢CrO42-ÇÒŨ¶È¾ùΪ0.010 mol¡¤L-1ÈÜÒºÖÐÖðµÎ¼ÓÈë0.010 mol¡¤L-1µÄAgNO3ÈÜҺʱ£¬CrO42-ÏȲúÉú³Áµí

D£®³£ÎÂÏÂpH=7µÄCH3COOHºÍNaOH»ìºÏÈÜÒºÖУ¬c(Na+)>c(CH3COO-)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿±ê×¼×´¿öÏÂÓТÙ6.72L¼×Íé¡¡¢Ú3.01¡Á1023¸öÂÈ»¯Çâ·Ö×Ó¡¡¢Û13.6gÁò»¯Çâ¢Ü0.2mol NH3 £® ÏÂÁжÔÕâËÄÖÖÆøÌåµÄ¹Øϵ´ÓСµ½´ó±íʾ²»ÕýÈ·µÄÊÇ£¨ £©
A.Ìå»ý£º¢Ü£¼¢Ù£¼¢Ú£¼¢Û
B.Ãܶȣº¢Ù£¼¢Ü£¼¢Û£¼¢Ú
C.ÖÊÁ¿£º¢Ü£¼¢Ù£¼¢Û£¼¢Ú
D.ÇâÔ­×ÓÊý£º¢Ú£¼¢Ü£¼¢Û£¼¢Ù

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸