NH3¾­Ò»ÏµÁз´Ó¦¿ÉÒԵõ½HNO3ºÍNH4NO3£¬ÈçÏÂͼËùʾ¡£

(1)¢ñÖУ¬NH3ºÍO2ÔÚ´ß»¯¼Á×÷ÓÃÏ·´Ó¦£¬Æ仯ѧ·½³ÌʽÊÇ_________________________¡£

(2)¢òÖУ¬2NO(g)£«O2(g) 2NO2(g)¡£ÔÚÆäËûÌõ¼þÏà

ͬʱ£¬·Ö±ð²âµÃNOµÄƽºâת»¯ÂÊÔÚ²»Í¬Ñ¹Ç¿(p1¡¢p2)ÏÂËæζȱ仯µÄÇúÏß(Èçͼ)¡£

¢Ù±È½Ïp1¡¢p2µÄ´óС¹Øϵ£º________¡£

¢ÚËæζÈÉý¸ß£¬¸Ã·´Ó¦Æ½ºâ³£Êý±ä»¯µÄÇ÷ÊÆÊÇ________¡£

(3)¢óÖУ¬½µµÍζȣ¬½«NO2(g)ת»¯ÎªN2O4(l)£¬ÔÙÖƱ¸Å¨ÏõËá¡£

¢ÙÒÑÖª£º2NO2(g) N2O4(g)¡¡¦¤H1         2NO2(g)N2O4(l)¡¡¦¤H2

ÏÂÁÐÄÜÁ¿±ä»¯Ê¾ÒâͼÖУ¬ÕýÈ·µÄÊÇ(Ñ¡Ìî×Öĸ)________¡£

¡¡

¡¡¡¡¡¡A¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡B                   C

¢ÚN2O4ÓëO2¡¢H2O»¯ºÏµÄ»¯Ñ§·½³ÌʽÊÇ________________________________________¡£

(4)¢ôÖУ¬µç½âNOÖƱ¸NH4NO3£¬Æ乤×÷Ô­ÀíÈçͼËùʾ¡£ÎªÊ¹µç½â²úÎïÈ«²¿×ª»¯ÎªNH4NO3£¬Ðè²¹³äA¡£AÊÇ________£¬ËµÃ÷ÀíÓÉ£º________________________________________¡£


(1)4NH3£«5O24NO£«6H2O

(2)¢Ùp1£¼p2¡¡¢Ú¼õС

(3)¢ÙA¡¡¢Ú2N2O4£«O2£«2H2O===4HNO3

(4)NH3¡¡¸ù¾Ý·´Ó¦£º8NO£«7H2O3NH4NO3£«2HNO3£¬µç½â²úÉúµÄHNO3¶à

[½âÎö] (1)°±µÄ´ß»¯Ñõ»¯µÄ·´Ó¦·½³ÌʽΪ

4NH3£«5O24NO£«6H2O¡£

(2)¢ÙÓÉ2NO(g)£«O2(g) 2NO2(g)¿ÉÖª¸Ã·´Ó¦ÎªÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬Î¶ÈÏàͬ£¬Ôö´óѹǿ£¬Æ½ºâÕýÏòÒƶ¯£¬NOµÄƽºâת»¯ÂÊÔö´ó£¬¸ù¾ÝͼʾºÍ×ø±êº¬Ò壬ÅжÏp1<p2£»¢ÚÔÙ¿´Í¬Ò»Ñ¹Ç¿Ïߣ¬Î¶ÈÉý¸ß£¬NOµÄƽºâת»¯ÂʽµµÍ£¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬ÔòÕý·´Ó¦Îª·ÅÈÈ·´Ó¦£¬Î¶ÈÉý¸ß£¬Æ½ºâ³£Êý¼õС¡£

(3)¢Ù½µµÍζȣ¬NO2(g)ת±äΪN2O4(l)£¬Ôò¦¤H2<0£¬·´Ó¦ÎïÄÜÁ¿¸ßÓÚÉú³ÉÎïÄÜÁ¿£¬C´íÎó£» N2O4(g)ת±äΪN2O4(l)ÐèÒª·Å³öÈÈÁ¿£¬ËùÒÔNO2(g)ת±äΪN2O4(g)±ÈNO2(g)ת±äΪN2O4(l)·Å³öµÄÈÈÁ¿ÉÙ£¬B´íÎó£¬ËùÒÔÕýÈ·µÄͼʾΪA£»¢ÚN2O4¡¢O2ºÍH2O·´Ó¦Éú³ÉÏõËáµÄ·´Ó¦·½³ÌʽΪ2N2O4£« O2£«2H2O===4HNO3¡£

(4)¸ù¾Ý¹¤×÷Ô­Àí×°ÖÃͼ£¬¿ÉÒÔÈ·¶¨Ñô¼«ÎªNOʧȥµç×Óת±äΪNO£¬Òõ¼«NOת±äΪNH£¬¸ù¾Ýµç¼«·´Ó¦Êéдµç¼«·´Ó¦Ê½Îª£º

Ñô¼«£ºNO£­3e£­£«2H2O===NO£«4H£«

Òõ¼«£ºNO£«5e£­£« 6H£«===NH£« H2O

È»ºó¸ù¾ÝµÃʧµç×ÓÊغ㣬ÏõËá¸ùÀë×ÓÎïÖʵÄÁ¿±È笠ùÀë×ÓÎïÖʵÄÁ¿¶à£¬ËùÒÔÐèÒªÏòÈÜÒºÖмÓÈëµÄÎïÖÊΪNH3(¼´8NO£«7H2O3NH4NO3£«2HNO3)¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


»¯ºÏÎïAX3ºÍµ¥ÖÊX2ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦¿ÉÉú³É»¯ºÏÎïAX5¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÒÑÖªAX3µÄÈÛµãºÍ·Ðµã·Ö±ðΪ£­93.6 ¡æºÍ76 ¡æ£¬AX5µÄÈÛµãΪ167 ¡æ¡£ÊÒÎÂʱAX3ÓëÆøÌåX2·´Ó¦Éú³É1 mol AX5£¬·Å³öÈÈÁ¿123.8 kJ¡£¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ____________________________________________¡£

(2)·´Ó¦AX3(g)£«X2(g) AX5(g)ÔÚÈÝ»ýΪ10 LµÄÃܱÕÈÝÆ÷ÖнøÐС£ÆðʼʱAX3ºÍX2¾ùΪ0.2 mol¡£·´Ó¦ÔÚ²»Í¬Ìõ¼þϽøÐУ¬·´Ó¦Ìåϵ×ÜѹǿËæʱ¼äµÄ±ä»¯ÈçͼËùʾ¡£

¢ÙÁÐʽ¼ÆËãʵÑéa´Ó·´Ó¦¿ªÊ¼ÖÁ´ïµ½Æ½ºâʱµÄ·´Ó¦ËÙÂÊv(AX5)£½______________________¡£

¢ÚͼÖÐ3×éʵÑé´Ó·´Ó¦¿ªÊ¼ÖÁ´ïµ½Æ½ºâʱµÄ·´Ó¦ËÙÂÊv(AX5)ÓÉ´óµ½Ð¡µÄ´ÎÐòΪ____________(ÌîʵÑéÐòºÅ)£»ÓëʵÑéaÏà±È£¬ÆäËûÁ½×é¸Ä±äµÄʵÑéÌõ¼þ¼°ÅжÏÒÀ¾ÝÊÇ£ºb________________________________________________£¬c____________________________________________¡£

¢ÛÓÃp0±íʾ¿ªÊ¼Ê±×Üѹǿ£¬p±íʾƽºâʱ×Üѹǿ£¬¦Á±íʾAX3µÄƽºâת»¯ÂÊ£¬Ôò¦ÁµÄ±í´ïʽΪ______________£»ÊµÑéaºÍcµÄƽºâת»¯ÂÊ£º¦ÁaΪ________£¬¦ÁcΪ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ÈôÔÚº£ÂÖÍâ¿ÇÉϸ½×ÅһЩͭ¿é£¬Ôò¿ÉÒÔ¼õ»ºº£ÂÖÍâ¿ÇµÄ¸¯Ê´

B£®2NO(g)£«2CO(g)===N2(g)£«2CO2(g)ÔÚ³£ÎÂÏÂÄÜ×Ô·¢½øÐУ¬Ôò¸Ã·´Ó¦µÄ¦¤H>0

C£®¼ÓÈÈ0.1 mol¡¤L£­1Na2CO3ÈÜÒº£¬COµÄË®½â³Ì¶ÈºÍÈÜÒºµÄpH¾ùÔö´ó

D£®¶ÔÓÚÒÒËáÓëÒÒ´¼µÄõ¥»¯·´Ó¦(¦¤H<0)£¬¼ÓÈëÉÙÁ¿Å¨ÁòËá²¢¼ÓÈÈ£¬¸Ã·´Ó¦µÄ·´Ó¦ËÙÂʺÍƽºâ³£Êý¾ùÔö´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Áò»¯ÇâµÄת»¯ÊÇ×ÊÔ´ÀûÓúͻ·¾³±£»¤µÄÖØÒªÑо¿¿ÎÌâ¡£ÓÉÁò»¯Çâ»ñµÃÁòµ¥ÖÊÓжàÖÖ·½·¨¡£

(1)½«ÉÕ¼îÎüÊÕH2SºóµÄÈÜÒº¼ÓÈëµ½ÈçͼËùʾµÄµç½â³ØµÄÑô¼«Çø½øÐеç½â¡£µç½â¹ý³ÌÖÐÑô¼«Çø·¢ÉúÈçÏ·´Ó¦£º

S2£­£­2e£­===S¡¡(n£­1)S£«S2£­===S

¢Ùд³öµç½âʱÒõ¼«µÄµç¼«·´Ó¦Ê½£º________________¡£

¢Úµç½âºóÑô¼«ÇøµÄÈÜÒºÓÃÏ¡ÁòËáËữµÃµ½Áòµ¥ÖÊ£¬ÆäÀë×Ó·½³Ìʽ¿Éд³É__________________________¡£

(2)½«H2SºÍ¿ÕÆøµÄ»ìºÏÆøÌåͨÈëFeCl3¡¢FeCl2¡¢CuCl2µÄ»ìºÏÈÜÒºÖз´Ó¦»ØÊÕS£¬ÆäÎïÖÊת»¯ÈçͼËùʾ¡£

¢ÙÔÚͼʾµÄת»¯ÖУ¬»¯ºÏ¼Û²»±äµÄÔªËØÊÇ________¡£

¢Ú·´Ó¦Öе±ÓÐ1 mol H2Sת»¯ÎªÁòµ¥ÖÊʱ£¬±£³ÖÈÜÒºÖÐFe3£«µÄÎïÖʵÄÁ¿²»±ä£¬ÐèÏûºÄO2µÄÎïÖʵÄÁ¿Îª________¡£

¢ÛÔÚζÈÒ»¶¨ºÍ²»²¹¼ÓÈÜÒºµÄÌõ¼þÏ£¬»ºÂýͨÈë»ìºÏÆøÌ壬²¢³ä·Ö½Á°è¡£ÓûʹÉú³ÉµÄÁòµ¥ÖÊÖв»º¬CuS£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ________________¡£

(3)H2SÔÚ¸ßÎÂÏ·ֽâÉú³ÉÁòÕôÆøºÍH2¡£Èô·´Ó¦ÔÚ²»Í¬Î¶ÈÏ´ﵽƽºâʱ£¬»ìºÏÆøÌåÖи÷×é·ÖµÄÌå»ý·ÖÊýÈçͼËùʾ£¬H2SÔÚ¸ßÎÂÏ·ֽⷴӦµÄ»¯Ñ§·½³ÌʽΪ__________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijͬѧ×é×°ÁËÈçͼËùʾµÄµç»¯Ñ§×°Ö㬵缫¢ñΪAl£¬ÆäËûµç¼«¾ùΪCu£¬Ôò(¡¡¡¡)

ͼ4

A£®µçÁ÷·½Ïò£ºµç¼«¢ô¨D¡ú¨D¡úµç¼«¢ñ

B£®µç¼«¢ñ·¢Éú»¹Ô­·´Ó¦

C£®µç¼«¢òÖð½¥Èܽâ

D£®µç¼«¢óµÄµç¼«·´Ó¦£ºCu2£«£«2e£­===Cu

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijÐîµç³ØµÄ·´Ó¦ÎªNiO2£«Fe£«2H2OFe(OH)2£«Ni(OH)2¡£

(1)¸ÃÐîµç³Ø³äµçʱ£¬·¢Éú»¹Ô­·´Ó¦µÄÎïÖÊÊÇ________(ÌîÏÂÁÐ×Öĸ)¡£·ÅµçʱÉú³ÉFe(OH)2µÄÖÊÁ¿Îª18 g£¬ÔòÍâµç·ÖÐתÒƵĵç×ÓÊýÊÇ____________________¡£

A£®NiO2  B£®Fe

C£®Fe(OH)2  D£®Ni(OH)2

(2)Ϊ·ÀÖ¹Ô¶ÑóÂÖ´¬µÄ¸ÖÌú´¬ÌåÔÚº£Ë®Öз¢Éúµç»¯Ñ§¸¯Ê´£¬Í¨³£ÔÚ´¬ÌåÏâǶZn¿é£¬»òÓë¸ÃÐîµç³ØµÄ______¼«(Ìî¡°Õý¡±»ò¡°¸º¡±)ÏàÁ¬¡£

ͼK19­7

(3)ÒÔ¸ÃÐîµç³Ø×÷µçÔ´£¬ÓÃÈçͼK19­7ËùʾµÄ×°ÖÃÔÚʵÑéÊÒÄ£ÄâÂÁÖÆÆ·±íÃæ¡°¶Û»¯¡±´¦ÀíµÄ¹ý³ÌÖУ¬·¢ÏÖÈÜÒºÖð½¥±ä»ë×Ç£¬Ô­ÒòÊÇ(ÓÃÏà¹ØµÄµç¼«·´Ó¦Ê½ºÍÀë×Ó·½³Ìʽ±íʾ)________________________________________________________________________

________________________________________________________________________¡£

(4)¾«Á¶Í­Ê±£¬´ÖÍ­Ó¦ÓëÖ±Á÷µçÔ´µÄ______¼«(Ìî¡°Õý¡±»ò¡°¸º¡±)ÏàÁ¬¡£¾«Á¶¹ý³ÌÖУ¬µç½âÖÊÈÜÒºÖеÄc(Fe2£«)¡¢c(Zn2£«)»áÖð½¥Ôö´ó¶øÓ°Ïì½øÒ»²½µç½â¡£¼×ͬѧÉè¼ÆÈçϳýÔÓ·½°¸£º

 

ÒÑÖª£º

³ÁµíÎï

Fe(OH)3

Fe(OH)2

Cu(OH)2

Zn(OH)2

¿ªÊ¼³ÁµíʱµÄpH

2.3

7.5

5.6

6.2

ÍêÈ«³ÁµíʱµÄpH

3.9

9.7

6.4

8.0

Ôò¼ÓÈëH2O2µÄÄ¿µÄÊÇ________________________________________________________________________¡£

ÒÒͬѧÈÏΪӦ½«·½°¸ÖеÄpHµ÷½Úµ½8£¬ÄãÈÏΪ´Ë¹Ûµã____________(Ìî¡°ÕýÈ·¡±»ò¡°²»ÕýÈ·¡±)£¬ÀíÓÉÊÇ______________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚÎÒÃǵÄÈÕ³£Éú»îÖгöÏÖÁË¡°¼ÓµâʳÑΡ±¡¢¡°ÔöÌú½´ÓÍ¡±¡¢¡°¸ß¸ÆÅ£ÄÌ¡±¡¢¡°¸»Îø²èÒ¶¡±¡¢¡°º¬    ·ú

ÑÀ¸à¡±µÈÉÌÆ·¡£ÕâÀïµÄµâ¡¢Ìú¡¢¸Æ¡¢Îø¡¢·úÓ¦Àí½âΪ                  

   A£®ÔªËØ            B£®µ¥ÖÊ           C£®·Ö×Ó          D£®Ñõ»¯Îï

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÃÓÚºôÎüÓþßÉϺÍDZˮͧÖÐ×÷ΪÑõÆøµÄÀ´Ô´µÄÎïÖÊÊÇ                    ¡¡

    A£®Na2O2     B£®NaHCO3       C£®SO3(¹Ì)      D£®KMnO4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¡¡Y¡¢Z¡¢W¾ùΪ¶ÌÖÜÆÚÔªËØ£¬ËüÃÇÔÚÖÜÆÚ±íÖеÄλÖÃÈçͼËùʾ¡£ÒÑÖªYÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ3±¶£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ(¡¡¡¡)

A£®Ô­×Ӱ뾶£ºW£¾Z£¾Y£¾X

B£®×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄËáÐÔ£ºZ£¾W£¾X

C£®Ç⻯ÎïµÄÎȶ¨ÐÔ£ºX£¾Y£¾Z

D£®ËÄÖÖÔªËصĵ¥ÖÊÖУ¬Zµ¥ÖʵÄÈÛ¡¢·Ðµã×î¸ß

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸