(20·Ö)
18-I(6·Ö)ÒÑÖª£º £¬Èç¹ûÒªºÏ³É ËùÓõÄÔʼÔÁÏ¿ÉÒÔÊÇ
A. 2-¼×»ù-l£¬3-¶¡¶þÏ©ºÍ2-¶¡È² B£®1£¬3-Îì¶þÏ©ºÍ2-¶¡È²
C£®2£¬3-¶þ¼×»ù-1£¬3-Îì¶þÏ©ºÍÒÒȲ D¡¢2£¬3-¶þ¼×»ù-l£¬3-¶¡¶þÏ©ºÍ±ûȲ
18-II(14·Ö)A¡«G¶¼ÊÇÓлú»¯ºÏÎËüÃǵÄת»¯¹ØϵÈçÏ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÒÑÖª£º6.0g»¯ºÏÎïEÍêȫȼÉÕÉú³É8.8gC02ºÍ3.6gH20£»EµÄÕôÆøÓëÇâÆøµÄÏà¶ÔÃܶÈΪ30£¬ÔòEµÄ·Ö×ÓʽΪ_______________£º
(2)AΪһȡ´ú·¼Ìþ£¬BÖк¬ÓÐÒ»¸ö¼×»ù¡£ÓÉBÉú³ÉCµÄ»¯Ñ§·½³ÌʽΪ_______________£»
(3)ÓÉBÉú³ÉD¡¢ÓÉCÉú³ÉDµÄ·´Ó¦Ìõ¼þ·Ö±ðÊÇ_______________¡¢_______________£»
(4)ÓÉAÉú³ÉB¡¢ÓÉDÉú³ÉGµÄ·´Ó¦ÀàÐÍ·Ö±ðÊÇ_______________¡¢_______________£»
(5)F´æÔÚÓÚèÙ×ÓÏãÓÍÖУ¬Æä½á¹¹¼òʽΪ_______________£»
(6)ÔÚGµÄͬ·ÖÒì¹¹ÌåÖУ¬±½»·ÉÏÒ»Ïõ»¯µÄ²úÎïÖ»ÓÐÒ»ÖֵĹ²ÓÐ___________¸ö£¬ÆäÖк˴Ź²
ÕñÇâÆ×ÓÐÁ½×é·å£¬ÇÒ·åÃæ»ý±ÈΪl£º1µÄÊÇ_______________(Ìî½á¹¹¼òʽ)¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê¸ß¿¼»¯Ñ§ÊÔÌâ·ÖÏîרÌâÊ®Èý Óлú»¯ºÏÎï ÌâÐÍ£ºÌî¿ÕÌâ
(20·Ö)
18-I(6·Ö)ÒÑÖª£º £¬Èç¹ûÒªºÏ³ÉËùÓõÄÔʼÔÁÏ¿ÉÒÔÊÇ
A. 2-¼×»ù-l£¬3-¶¡¶þÏ©ºÍ2-¶¡È² B£®1£¬3-Îì¶þÏ©ºÍ2-¶¡È²
C£®2£¬3-¶þ¼×»ù-1£¬3-Îì¶þÏ©ºÍÒÒȲ D¡¢2£¬3-¶þ¼×»ù-l£¬3-¶¡¶þÏ©ºÍ±ûȲ
18-II(14·Ö)A¡«G¶¼ÊÇÓлú»¯ºÏÎËüÃǵÄת»¯¹ØϵÈçÏ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÒÑÖª£º6.0g»¯ºÏÎïEÍêȫȼÉÕÉú³É8.8g C02ºÍ3.6g H20£»EµÄÕôÆøÓëÇâÆøµÄÏà¶ÔÃܶÈΪ30£¬ÔòEµÄ·Ö×ÓʽΪ_______________£º[À´Ô´:ѧ¿ÆÍøZXXK]
(2)AΪһȡ´ú·¼Ìþ£¬BÖк¬ÓÐÒ»¸ö¼×»ù¡£ÓÉBÉú³ÉCµÄ»¯Ñ§·½³ÌʽΪ_______________£»
(3)ÓÉBÉú³ÉD¡¢ÓÉCÉú³ÉDµÄ·´Ó¦Ìõ¼þ·Ö±ðÊÇ_______________¡¢_______________£»
(4)ÓÉAÉú³ÉB¡¢ÓÉDÉú³ÉGµÄ·´Ó¦ÀàÐÍ·Ö±ðÊÇ_______________¡¢_______________£»
(5)F´æÔÚÓÚèÙ×ÓÏãÓÍÖУ¬Æä½á¹¹¼òʽΪ_______________£»
(6)ÔÚGµÄͬ·ÖÒì¹¹ÌåÖУ¬±½»·ÉÏÒ»Ïõ»¯µÄ²úÎïÖ»ÓÐÒ»ÖֵĹ²ÓÐ___________¸ö£¬ÆäÖк˴Ź²
ÕñÇâÆ×ÓÐÁ½×é·å£¬ÇÒ·åÃæ»ý±ÈΪl£º1µÄÊÇ_______________ (Ìî½á¹¹¼òʽ)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄê¹ã¶«Ê¡Ã¯ÃûÊеڶþ´Î¸ß¿¼Ä£Ä⿼ÊÔÀí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ
²ÝËáÑÇÌú£¨FeC2O4•2H2O£©ÓÃ×÷·ÖÎöÊÔ¼Á¼°ÏÔÓ°¼ÁºÍÐÂÐ͵ç³Ø²ÄÁÏÁ×ËáÑÇÌú﮵ÄÉú²ú¡£ÒÑÖª£ºCOÄÜÓëÂÈ»¯îÙ£¨PdCl2£©ÈÜÒº·´Ó¦Éú³ÉºÚÉ«µÄîÙ·Û¡£»Ø´ðÏÂÁÐÎÊÌ⣺
I£®ÐËȤС×é¶Ô²ÝËáÑÇÌúµÄ·Ö½â²úÎï½øÐÐʵÑéºÍ̽¾¿¡£
£¨1£©½«ÆøÌå²úÎïÒÀ´Îͨ¹ýA¡¢³ÎÇåʯ»ÒË®£¬B¡¢ÂÈ»¯îÙ£¬¹Û²ìµ½AÖгÎÇåʯ»ÒË®¶¼±ä»ë×Ç£¬BÖгöÏÖºÚÉ«ÎïÖÊÉú³É£¬ÔòÉÏÊöÏÖÏó˵Ã÷ÆøÌå²úÎïÖÐÓÐ ¡£
£¨2£©Ì½¾¿·Ö½âµÃµ½µÄ¹ÌÌå²úÎïÖÐÌúÔªËصĴæÔÚÐÎʽ¡£
¢ÙÌá³ö¼ÙÉè
¼ÙÉè1£º________£» ¼ÙÉè2£ºFeO£» ¼ÙÉè3£ºFeOºÍFe»ìºÏÎï
¢ÚÉè¼ÆʵÑé·½°¸Ö¤Ã÷¼ÙÉè3¡£
ÏÞÑ¡ÊÔ¼Á£º 1.0 mol•L£1ÑÎËá¡¢3% H2O2¡¢0.1 mol•L£1CuSO4¡¢20% KSCN¡¢ÕôÁóË®¡£
ʵÑé²½Öè |
ÏÖÏóÓë½áÂÛ |
²½Öè1 £ºÏòÊÔ¹ÜÖмÓÈëÉÙÁ¿¹ÌÌå²úÎÔÙ¼ÓÈë×ãÁ¿_________________£¬³ä·ÖÕñµ´ |
ÈôÈÜÒºÑÕÉ«Ã÷ÏԸı䣬ÇÒÓÐ_______Éú³É£¬ÔòÖ¤Ã÷ÓÐÌúµ¥ÖÊ´æÔÚ |
²½Öè2£º ½«²½Öè1Öеõ½µÄ×ÇÒº¹ýÂË£¬²¢ÓÃÕôÁóˮϴµÓÖÁÏ´µÓÒºÎÞÉ« |
|
²½Öè3£ºÈ¡²½Öè2µÃµ½µÄÉÙÁ¿¹ÌÌåÓëÊÔ¹ÜÖУ¬ µÎ¼Ó___________________________________ _______________________________________
|
__________________________________ ___________________________________
|
II£®Ä³²ÝËáÑÇÌúÑùÆ·Öк¬ÓÐÉÙÁ¿²ÝËᣨΪ·½±ãÓÚ¼ÆË㣬²ÝËáÑÇÌúÖвÝËá¸ùºÍ²ÝËá·Ö×Ó¾ùÓÃC2O42£´úÌ棩¡£ÏÖÓõζ¨·¨²â¶¨¸ÃÑùÆ·ÖÐFeC2O4µÄº¬Á¿¡£µÎ¶¨·´Ó¦·Ö±ðÊÇ£º5Fe2++MnO4£+8H+=5Fe3+ +Mn2++4H2O¡¢5C2O42£+2MnO4£+16H+=10CO2¡ü+2Mn2++8H2O¡£
£¨3£©ÊµÑé·½°¸Éè¼ÆΪ£º
¢Ù½«×¼È·³ÆÁ¿µÄ0.20g²ÝËáÑÇÌúÑùÆ·ÖÃÓÚ250 mL׶ÐÎÆ¿ÄÚ£¬¼ÓÈëÊÊÁ¿2 mol/LµÄH2SO4ÈÜÒº£¬Ê¹ÑùÆ·Èܽ⣬¼ÓÈÈÖÁ70¡æ×óÓÒ£¬Á¢¼´ÓÃŨ¶ÈΪ0.02000 mol/LµÄ¸ßÃÌËá¼Ø±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬¼Ç¼Êý¾Ý¡£Öظ´µÎ¶¨2´Î¡£Æ½¾ùÏûºÄV1 mL¡£
¢ÚÏòÉÏÊöµÎ¶¨»ìºÏÒºÖмÓÈëÊÊÁ¿µÄZn·ÛºÍ¹ýÁ¿µÄ2 mol/LµÄH2SO4ÈÜÒº£¬Öó·Ð5¡«8min£¬ÓÃKSCNÈÜÒºÔÚµãµÎ°åÉϼìÑéµãµÎÒº£¬Ö±ÖÁÈÜÒº²»Á¢¿Ì±äºì¡£½«ÂËÒº¹ýÂËÖÁÁíÒ»¸ö׶ÐÎÆ¿ÖУ¬¼ÌÐøÓÃ0.02000 mol/LµÄ¸ßÃÌËá¼Ø±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬¼Ç¼Êý¾Ý¡£Öظ´µÎ¶¨2´Î¡£Æ½¾ùÏûºÄV2 mL¡£
¢ÛÈôijС×éµÄÒ»´Î²â¶¨Êý¾Ý¼Ç¼ÈçÏ£º V1= 18.90mL£¬V2=6.20mL¡£¸ù¾ÝÊý¾Ý¼ÆËã0.20g ÑùÆ·ÖУºn£¨Fe2+£©= £» n£¨C2O42££©= £»FeC2O4 µÄÖÊÁ¿·ÖÊýΪ £¨¾«È·µ½0.01%£¬FeC2O4µÄʽÁ¿Îª144£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÆÕͨ¸ßµÈѧУÕÐÉúÈ«¹úͳһ¿¼ÊÔ»¯Ñ§ÊÔÌ⣨º£ÄÏ¾í£© ÌâÐÍ£ºÌî¿ÕÌâ
(20·Ö)
18-I(6·Ö)ÒÑÖª£º £¬Èç¹ûÒªºÏ³É ËùÓõÄÔʼÔÁÏ¿ÉÒÔÊÇ
A. 2-¼×»ù-l£¬3-¶¡¶þÏ©ºÍ2-¶¡È² B£®1£¬3-Îì¶þÏ©ºÍ2-¶¡È²
C£®2£¬3-¶þ¼×»ù-1£¬3-Îì¶þÏ©ºÍÒÒȲ D¡¢2£¬3-¶þ¼×»ù-l£¬3-¶¡¶þÏ©ºÍ±ûȲ
18-II(14·Ö)A¡«G¶¼ÊÇÓлú»¯ºÏÎËüÃǵÄת»¯¹ØϵÈçÏ£º
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÒÑÖª£º6.0g»¯ºÏÎïEÍêȫȼÉÕÉú³É8.8g C02ºÍ3.6g H20£»EµÄÕôÆøÓëÇâÆøµÄÏà¶ÔÃܶÈΪ30£¬ÔòEµÄ·Ö×ÓʽΪ_______________£º[À´Ô´:ZXXK]
(2)AΪһȡ´ú·¼Ìþ£¬BÖк¬ÓÐÒ»¸ö¼×»ù¡£ÓÉBÉú³ÉCµÄ»¯Ñ§·½³ÌʽΪ_______________£»
(3)ÓÉBÉú³ÉD¡¢ÓÉCÉú³ÉDµÄ·´Ó¦Ìõ¼þ·Ö±ðÊÇ_______________¡¢_______________£»
(4)ÓÉAÉú³ÉB¡¢ÓÉDÉú³ÉGµÄ·´Ó¦ÀàÐÍ·Ö±ðÊÇ_______________¡¢_______________£»
(5) F´æÔÚÓÚèÙ×ÓÏãÓÍÖУ¬Æä½á¹¹¼òʽΪ_______________£»
(6)ÔÚGµÄͬ·ÖÒì¹¹ÌåÖУ¬±½»·ÉÏÒ»Ïõ»¯µÄ²úÎïÖ»ÓÐÒ»ÖֵĹ²ÓÐ___________¸ö£¬ÆäÖк˴Ź²
ÕñÇâÆ×ÓÐÁ½×é·å£¬ÇÒ·åÃæ»ý±ÈΪl£º1µÄÊÇ_______________ (Ìî½á¹¹¼òʽ)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÒÑÖª£ºN2£¨g£©£«3H2£¨g£©2NH3£¨g£©¡÷H£½£92.4kJ/mol£¬ÕâÊÇÄ¿Ç°ÆÕ±éʹÓõÄÈ˹¤¹ÌµªµÄ·½·¨¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)450¡æʱ£¬ÍùÒ»¸ö2LµÄÃܱÕÈÝÆ÷ÖгäÈë2. 6mol H2ºÍ1mol N2£¬ ·´Ó¦¹ý³ÌÖжÔNH3µÄŨ¶È½øÐмì²â£¬µÃµ½µÄÊý¾ÝÈçϱíËùʾ£º
ʱ¼ä/min | 5 | 10 | 15 | 20 | 25 | 30 |
c(NH3)/mol·L—1 | 0.08 | 0.14 | 0.18 | 0.20 | 0.20 | 0.20 |
¢Ù´ËÌõ¼þϸ÷´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK=_______________£»·´Ó¦´ïµ½Æ½ºâºó£¬ÈôÍùƽºâÌåϵÖмÓÈëH2¡¢N2ºÍNH3¸÷2mol£¬´Ëʱ¸Ã·´Ó¦µÄv£¨N2£©Õý_______________v£¨N2£©Ä棨Ìîд¡°>¡±¡¢¡°=¡±»ò¡°<¡±=£©¡£
¢ÚÈô¸Ä±äijһÌõ¼þ£¬´ïÐÂƽºâʱn(H2)=1.60mol £¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_____________¡£
A.ƽºâÒ»¶¨ÕýÏòÒƶ¯ B.¿ÉÄÜÊÇÏòÈÝÆ÷ÖмÓÈëÁËÒ»¶¨Á¿µÄH2ÆøÌå
C.¿ÉÄÜÊǽµµÍÁËÈÝÆ÷µÄÎÂ¶È D.¿ÉÄÜÊÇËõСÁËÈÝÆ÷µÄÌå»ý
£¨2£©450¡æʱ£¬ÔÚÁíÒ»ÃܱÕÈÝÆ÷ÖнøÐÐÉÏÊöºÏ³É°±µÄ·´Ó¦£¬¸÷ÎïÖʵÄÆðʼŨ¶ÈºÍƽºâŨ¶ÈÈçϱíËùʾ£º
N2 | H2 | NH3 | |
ÆðʼŨ¶È£¨mol/L£© | 0.2 | 0.3 | 0.2 |
ƽºâŨ¶È£¨mol/L£© | a | b | c |
Çë»Ø´ð£º
¢ÙaµÄÈ¡Öµ·¶Î§ÊÇ£º_______________£®
¢ÚÇëÓÃÊýѧ±í´ïʽ±íʾÏÂÁÐÁ¿Ö®¼äµÄ¹Øϵ£º
(I)aÓëbµÄ¹Øϵ£º_______________¡£(¢ò)a¡¢b¡¢cµÄ¹Øϵ£º_______________¡£
¢Û·´Ó¦´ïµ½Æ½ºâºó£¬¸Ä±äijһÍâ½çÌõ¼þ£¬·´Ó¦ËÙÂÊÓëʱ¼äµÄ¹ØϵÈçÏÂͼËùʾ£¬ÆäÖÐt2﹑t7ʱ¿ÌËù¶ÔÓ¦µÄʵÑéÌõ¼þ¸Ä±ä·Ö±ðÊÇ£ºt2 £»t7 ¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com