¡¾ÌâÄ¿¡¿¢ñ.Ϊ̽¾¿ÁòËáÍ­¾§Ìå(CuSO4¡¤xH2O)ÊÜÈÈ·Ö½âºóËùµÃ²úÎÉè¼ÆʵÑé×°ÖÃÈçͼËùʾ£¬²¿·ÖʵÑéÏÖÏóΪ£ºAÖÐÀ¶É«¾§ÌåÖð½¥±ä³É°×É«·ÛÄ©£¬¼ÌÐø¼ÓÈÈ×îÖÕ±ä³ÉºÚÉ«£»BÖвúÉú°×É«³Áµí£»DÖÐÈÜÒº±ä³ÉºìÉ«¡£(Ï´ÆøÆ¿ÖÐÊÔ¼Á¾ù×ãÁ¿)

(1)·ÖÎöÍƲâ¸ÃÁòËáÍ­¾§Ìå×îÖÕ·Ö½â²úÎï¿ÉÄÜÓÐ_______________________________¡£

(2)DÖеķ´Ó¦·ÖÁ½²½½øÐУ¬Ð´³öµÚÒ»²½·´Ó¦µÄÀë×Ó·½³Ìʽ_____________________¡£

II.²â¶¨ÁòËáÍ­¾§Ìå(CuSO4¡¤xH2O)ÖнᾧˮxµÄÖµ£¬ÊµÑé×°Öú͹ý³ÌÈçÏ£ºÈ¡ÁòËáÍ­¾§Ìå7.23 gÖÃÓÚÓ²ÖÊÊÔ¹ÜÖУ¬ÏÈͨN2ÅųýÌåϵÄÚ¿ÕÆø£¬¾Æ¾«ÅçµÆ¸ßμÓÈȳä·Ö£¬´ýAÖÐÀ¶É«¾§Ìå×îÖÕ±ä³ÉºÚÉ«£¬Í£Ö¹¼ÓÈÈ£¬ÔٴιÄÈëN2ÖÁ×°ÖÃÀäÈ´µ½ÊÒΡ£(Ï´ÆøÆ¿ÖÐÊÔ¼Á¾ù×ãÁ¿)

(1)È¡BÖа×É«³Áµí¾­¹ýÂËÏ´µÓ¸ÉÔï³ÆÁ¿µÃ¹ÌÌå6.99 g£¬¾­¼ÆËã¿ÉµÃCuSO4¡¤xH2OÖÐx=__________£¬ÔٴιÄÈëN2µÄÄ¿µÄÊÇ____________________¡£

(2)ijͬѧÌá³ö£¬Òª²â¶¨¾§ÌåÖнᾧˮxµÄÖµ£¬Ò²¿É½«B×°ÖÃÓÃ×°ÓÐŨÁòËáµÄÏ´ÆøÆ¿Ìæ»»£¬×îÖÕ²âŨÁòËáÔöÖؼ´¿É¼ÆËãµÃµ½½á¹û£¬ÆÀ¼Û¸ÃͬѧµÄ·½°¸ÊÇ·ñ¿ÉÐУ¿(Èç¹û²»¿ÉÐУ¬Çë˵Ã÷ÀíÓÉ)_____________________________

¡¾´ð°¸¡¿CuO¡¢SO3¡¢SO2¡¢O2¡¢H2O 4Fe2++O2+4H+ = 4Fe3++2H2O 4.5 ʹ·Ö½â²úÉúµÄÆøÌåÈ«²¿±»Ï´ÆøÆ¿ÖÐÊÔ¼ÁÎüÊÕ ²»¿ÉÐУ¬SO3Ò²ÄÜÈܽâÔÚŨÁòËáÖУ¬Å¨ÁòËáµÄÔöÖز»Ö¹Ö»ÊÇË®µÄÖÊÁ¿

¡¾½âÎö¡¿

I¡¢¸ÃʵÑéµÄÄ¿µÄÊÇ̽¾¿ÁòËáÍ­¾§Ìå(CuSO4¡¤xH2O)ÊÜÈÈ·Ö½âºóËùµÃ²úÎAÖÐÀ¶É«¾§ÌåÖð½¥±ä³É°×É«·ÛÄ©£¬¼ÌÐø¼ÓÈÈ×îÖÕ±ä³ÉºÚÉ«£¬ËµÃ÷¾§ÌåÊÜÈÈ·Ö½â²úÉúÁËCuO£»BÖÐËáÐÔBaCl2ÈÜÒº²úÉú°×É«³Áµí£¬Ôò·Ö½â²úÎïÖÐÓÐSO3£»DÖÐÈÜÒº±äºì£¬ËµÃ÷Fe2+±»Ñõ»¯ÎªFe3+£¬¼´²úÎïÖÐÓÐÑõ»¯ÐÔÎïÖÊ£¬¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ¿ÉÖª£¬¸ÃÑõ»¯ÐÔÎïÖÊΪO2£¬Ôò·Ö½â²úÎïÖÐÓÐSO2Éú³É£¨CuSO4ÖУ¬OΪ-2¼Û£¬ÈôµÃµ½O2£¬ÐèÒªÉý¸ß»¯ºÏ¼Û£¬ËùÒÔSµÄ»¯ºÏ¼ÛÒ»¶¨½µµÍ£¬Éú³ÉSO2£©£¬ËùÒÔBÖеÄÏÖÏóÊÇÆ·ºìÈÜÒºÍÊÉ«£¬EµÄ×÷ÓÃÊÇÎüÊÕ¶àÓàµÄÓк¦ÆøÌ壬·ÀÖ¹ÎÛȾ»·¾³£»

II¡¢¸ù¾ÝʵÑéIÍƳöÁòËáÍ­¾§Ìå²úÎïÓÐSO2¡¢SO3£¬BÖк¬ÓÐCl2ºÍBaCl2£¬Cl2ÔÚË®ÖпÉÒÔ½«SO2Ñõ»¯ÎªSO42-£¬¹ÊCuSO4ÖеÄSÈ«²¿×ª»¯ÎªÁËBaSO4£¬Ôò¿ÉÒÔͨ¹ýBaSO4µÄÖÊÁ¿À´¼ÆËãxµÄÖµ¡£

I¡¢£¨1£©¾­·ÖÎö£¬CuSO4ÊÜÈÈ·Ö½â²úÉúµÄ²úÎïÓÐCuO¡¢SO2¡¢SO3¡¢O2£»¶ÔÓÚ¾§Ìå¶øÑÔ£¬ÊÜÈÈ·Ö½â²úÎﻹÓÐH2O£¬¹Ê¿ÉÍƲâ¸ÃÁòËáÍ­¾§Ìå×îÖÕ·Ö½â²úÎï¿ÉÄÜÓÐCuO¡¢SO3¡¢SO2¡¢O2¡¢H2O£»

£¨2£©DÖУ¬Fe2+Ïȱ»O2Ñõ»¯ÎªFe3+£¬Fe3+ÔÙºÍSCN-½áºÏÉú³ÉѪºìÉ«ÎïÖÊ£¬Éæ¼°µÄÀë×Ó·´Ó¦Îª£º4Fe2++O2+4H+=4Fe3++2H2O¡¢Fe3++3SCN-=Fe(SCN)3£»

II¡¢£¨1£©¸ù¾ÝÌâÒâ¿ÉÖª£¬n(BaSO4)==0.03mol£¬Ôòn(CuSO4¡¤xH2O)=0.03mol£¬ËùÒÔÓÐ0.03mol¡Á(150+18x)g/mol=7.23g£¬½âµÃx=4.5£»ÔÙ´ÎͨÈëN2µÄÄ¿µÄÊÇʹ·Ö½â²úÉúµÄÆøÌåÈ«²¿±»Ï´ÆøÆ¿ÖÐÊÔ¼ÁÎüÊÕ£»

£¨2£©²»¿ÉÐУ¬·Ö½â²úÎïSO3¿ÉÒÔÈܽâÔÚŨÁòËáÖУ¬Å¨ÁòËáµÄÔöÖز»Ö¹Ö»ÊÇË®µÄÖÊÁ¿¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¸ù¾ÝÒªÇó»Ø´ðÏÂÁÐÎÊÌâ¡£

¢ÙCaBr2¡¡¢ÚH2O¡¡¢ÛNH4Cl¡¡¢ÜH2O2¡¡¢ÝNa2O2¡¡¢ÞCa(OH)2¡¡¢ßHClO¡¡¢àCO2

(1)д³öNa2O2µÄµç×Óʽ________£¬´æÔڵĻ¯Ñ§¼üÀàÐÍÓÐ________________¡£

(2)д³öHClOµç×Óʽ________________,д³öCO2µÄ½á¹¹Ê½_______¡£

(3)º¬Óй²¼Û¼üµÄÀë×Ó»¯ºÏÎïÊÇ_______(ÌîÐòºÅ)

(4)Óõç×Óʽ±íʾCaBr2µÄÐγɹý³Ì_________________________________ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿CH3OHÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ,¹¤ÒµÉÏÓÃCOÓëH2ÔÚ´ß»¯¼Á×÷ÓÃϺϳÉCH3OH,Æ䷴ӦΪ:CO£¨g£©+2H2£¨g£©CH3OH£¨g£©¡£°´n(CO):n(H2)=1:2£¬ÏòÃܱÕÈÝÆ÷ÖгäÈë·´Ó¦Îï,²âµÃƽºâʱ»ìºÏÎïÖÐCH3OHµÄÌå»ý·ÖÊýÔÚ²»Í¬Ñ¹Ç¿ÏÂËæζȵı仯ÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÖÐ,ÕýÈ·µÄÊÇ

A. p1<p2

B. ¸Ã·´Ó¦µÄ¦¤H>0

C. ƽºâ³£Êý:K(A)=K(B)

D. ÔÚCµãʱ,COת»¯ÂÊΪ75%

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿·¼Ï㻯ºÏÎïFÊÇÓлú»¯¹¤µÄÖØÒªÔ­ÁÏ£¬Ò²¿ÉÖƶ¾£¬Êܹ«°²²¿¹ÜÖÆ¡£

ÒÑÖª£º±½»·ÉÏÓÐÍéÌþ»ùʱ£¬ÐÂÒýÈëµÄÈ¡´ú»ùÁ¬ÔÚÍéÌþ»ùµÄÁÚ¶ÔλÉÏ£»±½»·ÉÏÓÐôÈ»ùʱ£¬ÐÂÒýÈëµÄÈ¡´ú»ùÁ¬ÔÚôÈ»ùµÄ¼äλÉÏ¡£

£¨1£©DÎïÖʵÄÃû³ÆΪ___£¬GÖйÙÄÜÍŵÄÃû³ÆΪ___¡¢___¡£

£¨2£©A¡¢FµÄ½á¹¹¼òʽ·Ö±ðΪ___¡¢___£¬EÉú³ÉFµÄ·´Ó¦ÀàÐÍÊÇ___¡£

£¨3£©Ð´³öA¡úB²úÉú½Ï¶àµÄ¸±²úÎïµÄ½á¹¹¼òʽ___£¬¸ÃÁ÷³Ìδ²ÉÓüױ½Ö±½ÓÏõ»¯µÄ·½·¨ÖÆÈ¡D£¬¶øÊǾ­Àú¼¸²½·´Ó¦²ÅÖƵÃDµÄÄ¿µÄÊÇ___¡£

£¨4£©FÔÚÒ»¶¨Ìõ¼þÏÂÄÜ·¢Éú¾ÛºÏ·´Ó¦£¬Çëд³öÆ仯ѧ·´Ó¦·½³Ìʽ___¡£

£¨5£©FµÄͬ·ÖÒì¹¹ÌåÖУ¬°±»ùÖ±½ÓÁ¬ÔÚ±½»·ÉÏ¡¢ÇÒÄÜ·¢ÉúÒø¾µ·´Ó¦µÄ·¼Ïã×廯ºÏÎï¹²ÓÐ___ÖÖ£¨²»º¬Á¢Ìå½á¹¹£©¡£

£¨6£©ÇëÒÔ¼×±½ÎªÔ­ÁÏ£¬Ð´³öºÏ³É¼ä°±»ù±½¼×ËáµÄÁ÷³Ìͼ___£¨ÎÞ»úÊÔ¼ÁÈÎÑ¡£©¡£ºÏ³ÉÁ÷³Ìͼ±íʾ·½·¨Ê¾Àý£ºCH2=CH2CH3CH2BrCH3CH2OH

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¼×´¼±»³ÆΪ2lÊÀ¼ÍµÄÐÂÐÍȼÁÏ£¬¹¤ÒµÉÏͨ¹ýÏÂÁз´Ó¦¢ñºÍ¢ò£¬ÓÃCH4ºÍH2OΪԭÁÏÀ´ÖƱ¸¼×´¼¡£

£¨1£©½«1.0molCH4ºÍ2.0molH2O£¨g£©Í¨ÈëÈÝ»ýΪ10LµÄ·´Ó¦ÊÒ£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦¢ñ£ºCH4£¨g£©+H2O£¨g£©CO£¨g£©+3H2£¨g£©£¬CH4µÄת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼ£º

¢ÙÒÑÖª100¡æʱ´ïµ½Æ½ºâËùÐèµÄʱ¼äΪ5min£¬ÔòÓÃH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ______¡£

¢ÚÔÚÆäËüÌõ¼þ²»±äµÄÇé¿öÏÂÉý¸ßζȣ¬»¯Ñ§Æ½ºâ³£Êý½«_______£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©¡£

¢ÛͼÖеÄP1_____P2£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©£¬100¡æʱƽºâ³£ÊýΪ________¡£

¢Ü±£³Ö·´Ó¦ÌåϵΪ100¡æ£¬5minºóÔÙÏòÈÝÆ÷ÖгäÈëH2O¡¢H2¸÷0.5mol£¬»¯Ñ§Æ½ºâ½«Ïò_____Òƶ¯£¨Ìî¡°Ïò×󡱡°ÏòÓÒ¡±»ò¡°²»¡±£©¡£

£¨2£©ÔÚѹǿΪ0.1MPaÌõ¼þÏ£¬½«amolCOÓë3amol H2µÄ»ìºÏÆøÌåÔÚ´ß»¯¼Á×÷ÓÃÏÂÄÜ×Ô·¢·¢Éú·´Ó¦¢ò£º

CO£¨g£©+2H2£¨g£©CH3OH£¨g£©

¢Ù¸Ã·´Ó¦µÄ¡÷H_____0£¬¡÷S_____0£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©¡£

¢ÚÈôÈÝÆ÷ÈÝ»ý²»±ä£¬ÏÂÁдëÊ©¿ÉÒÔÌá¸ßCOת»¯ÂʵÄÊÇ______¡£

A£®Éý¸ßζÈ

B£®½«CH3OH£¨g£©´ÓÌåϵÖзÖÀë³öÀ´

C£®³äÈëHe£¬Ê¹Ìåϵ×ÜѹǿÔö´ó

D£®ÔÙ³äÈë1molCOºÍ3molH2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½«ÎïÖʵÄÁ¿Ö®±ÈΪ1:3µÄµªÆøºÍÇâÆø³äÈëºãÈÝÃܱÕÈÝÆ÷ÖУ¬²â¶¨²»Í¬Î¶ȡ¢Ñ¹Ç¿ÏÂƽºâ»ìºÏÎïÖа±µÄÎïÖʵÄÁ¿·ÖÊý£¬½á¹ûÈçͼËùʾ¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©

A.

B.¸Ã·´Ó¦

C.µã£¬µÄת»¯ÂÊΪ

D.ºÏ³É°±¹¤ÒµÊµÏÖÁËÈ˹¤¹Ìµª

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿NA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÊýÖµ¡£ÏÂÁеÄ˵·¨ÖУ¬ÕýÈ·µÄÊÇ£¨ £©

A. 4.6g½ðÊôÄÆÓÉÔ­×ÓÍêÈ«±äΪNa+ Àë×Óʱ£¬Ê§È¥µÄµç×ÓÊýΪ0.1NA

B. NA ¸öÑõÆø·Ö×ÓÓëNA ¸öÇâÆø·Ö×ÓµÄÖÊÁ¿±ÈΪ8©U1

C. 0.2 NA¸öÁòËá·Ö×ÓÓë19.6gÁ×ËᣨÏà¶Ô·Ö×ÓÖÊÁ¿£º98£©º¬ÓÐÏàͬµÄÑõÔ­×ÓÊý

D. 22.4LµÄµªÆøËùº¬ÓеÄÔ­×ÓÊýΪ2NA

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Î¶ÈÏ£¬¿ÉÄæ·´Ó¦X£¨g£©£«3Y£¨g£©2Z£¨g£©´ïµ½»¯Ñ§Æ½ºâ״̬µÄ±êÖ¾ÊÇ

A.ZµÄÉú³ÉËÙÂʺÍXµÄ·Ö½âËÙÂÊÏàµÈ

B.µ¥Î»Ê±¼äÄÚÉú³ÉnmolX£¬Í¬Ê±Éú³ÉÁË3nmolY

C.X¡¢Y¡¢ZµÄŨ¶È²»Ôٱ仯

D.X¡¢Y¡¢ZµÄ·Ö×Ó¸öÊýÖ®±ÈΪ1¡Ã3¡Ã2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»¯Ñ§·´Ó¦A2£¨g£©£«B2£¨g£©===2AB£¨g£©µÄÄÜÁ¿±ä»¯ÈçͼËùʾ¡£ÏÂÁÐÓйØÐðÊöÕýÈ·µÄÊÇ

A.ÿÉú³É2molAB£¨g£©ÎüÊÕbkJÈÈÁ¿

B.¶ÏÁÑ1molA¡ªA¼üºÍ1molB¡ªB¼ü£¬·Å³öakJÄÜÁ¿

C.¸Ã·´Ó¦Öз´Ó¦ÎïµÄ×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿

D.·´Ó¦ÈȦ¤H£½£«£¨a£­b£©kJ¡¤mol£­1

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸