¿ÉÄæ·´Ó¦2NO22NO+O2ÔÚÌå»ý²»±äµÄÃܱÕÈÝÆ÷Öз´Ó¦£¬´ïµ½Æ½ºâ״̬µÄ±êÖ¾ÊÇ£¨     £©

¢Ùµ¥Î»Ê±¼äÄÚÉú³Én mol O2 µÄͬʱÉú³É2n molNO   ¢Úµ¥Î»Ê±¼äÄÚÉú³Én mol O2 µÄͬʱÉú³É2n molNO2   ¢ÛÓÃNO2¡¢NO¡¢O2µÄÎïÖʵÄÁ¿Å¨¶È±ä»¯±íʾµÄ·´Ó¦ËÙÂʵıÈΪ2:2:1µÄ״̬   ¢Ü»ìºÏÆøÌåµÄÑÕÉ«²»ÔٸıäµÄ״̬      ¢Ý»ìºÏÆøÌåµÄÃܶȲ»ÔٸıäµÄ״̬

 A.¢Ù¢Ü     ¡¡¡¡   B.¢Ú¢Ü  ¡¡¡¡¡¡      C.¢Ù¢Û¢Ü  ¡¡¡¡¡¡   D.¢Ù¢Ú¢Û¢Ü¢Ý

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ñó¼»ËØÊÇÒ»ÖÖнṹÀàÐ͵Ŀ¹ÒÒÐ͸ÎÑײ¡¶¾ºÍ¿¹°¬×̲¡²¡¶¾µÄ»¯ºÏÎÆä½á¹¹ÈçÏÂͼËùʾ£¬ÓйØÑó¼»ËصÄ˵·¨ÕýÈ·µÄÊÇ£¨    £©

A£®·Ö×ÓÖк¬ÓÐ4¸öÊÖÐÔ̼ԭ×Ó 

B£®²»ÄÜÓëÂÈ»¯ÌúÈÜÒº·¢ÉúÏÔÉ«·´Ó¦

C£®ÄÜ·¢Éúõ¥»¯·´Ó¦µ«²»ÄÜ·¢ÉúË®½â·´Ó¦

D£®1molÑó¼»ËØ×î¶à¿ÉÓë11molNaOH·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¢Ù48 g RO42£­ÖУ¬ºËÍâµç×Ó×ÜÊý±ÈÖÊ×Ó×ÜÊý¶à6.02¡Á1023¸ö£¬ÔòRÔ­×ÓµÄĦ¶ûÖÊÁ¿Îª________¡£

¢ÚÓÐÒ»Õæ¿ÕÆ¿µÄÖÊÁ¿ÎªM1 g£¬¸ÃÆ¿³äÈë¿ÕÆøºó×ÜÖÊÁ¿ÎªM2 g£»ÔÚÏàͬ״¿öÏ£¬Èô¸Ä³äÄ³ÆøÌåAºó£¬×ÜÖÊÁ¿ÎªM3 g£¬ÔòAµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª________¡£

(2)(2014½ì½­Î÷ʦ´ó¸½ÖÐÆÚÄ©)Ò»¶¨Á¿µÄÇâÆøÔÚÂÈÆøÖÐȼÉÕ£¬ËùµÃ»ìºÏÎïÓÃ100 mL 3.00 mol¡¤L£­1µÄNaOHÈÜÒº(ÃܶÈΪ1.12 g¡¤mL£­1)Ç¡ºÃÍêÈ«ÎüÊÕ£¬²âµÃÈÜÒºÖк¬ÓÐNaClOµÄÎïÖʵÄÁ¿Îª0.050 0 mol¡£

¢ÙÔ­NaOHÈÜÒºµÄÖÊÁ¿·ÖÊýΪ________

¢ÚËùµÃÈÜÒºÖÐCl£­µÄÎïÖʵÄÁ¿Îª________mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ëæ×ÅÊÀ½ç¹¤Òµ¾­¼ÃµÄ·¢Õ¹¡¢È˿ڵľçÔö£¬È«ÇòÄÜÔ´½ôÕż°ÊÀ½çÆøºòÃæÁÙÔ½À´Ô½ÑÏÖØµÄÎÊÌ⣬ÈçºÎ½µµÍ´óÆøÖÐCO2µÄº¬Á¿¼°ÓÐЧµØ¿ª·¢ÀûÓÃCO2ÒýÆðÁËÈ«ÊÀ½çµÄÆÕ±éÖØÊÓ¡£

(1)ÓÒͼΪC¼°ÆäÑõ»¯ÎïµÄ±ä»¯¹ØÏµÍ¼£¬Èô¢Ù±ä»¯ÊÇÖû»·´Ó¦£¬ÔòÆä»¯Ñ§·½³Ìʽ¿ÉΪ______________________________________£»

ͼÖб仯¹ý³ÌÄÄЩÊÇÎüÈÈ·´Ó¦________(ÌîÐòºÅ)¡£

(2)¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¾ßÓпª·¢ºÍÓ¦ÓõĹãÀ«Ç°¾°£¬¹¤ÒµÉÏ¿ÉÓÃÈçÏ·½·¨ºÏ³É¼×´¼£º

·½·¨Ò»¡¡CO(g)£«2H2(g)CH3OH(g)

·½·¨¶þ¡¡CO2(g)£«3H2(g)CH3OH(g)£«H2O(g)

ÔÚ25¡æ¡¢101 kPaÏ£¬1¿Ë¼×´¼ÍêȫȼÉÕ·ÅÈÈ22.68 kJ£¬Ð´³ö¼×´¼È¼ÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º_____________________________________________£»

ij»ðÁ¦·¢µç³§CO2µÄÄê¶ÈÅÅ·ÅÁ¿ÊÇ2 200Íò¶Ö£¬Èô½«´ËCO2Íêȫת»¯Îª¼×´¼£¬ÔòÀíÂÛÉÏÓÉ´Ë»ñµÃµÄ¼×´¼ÍêȫȼÉÕ·ÅÈÈÔ¼ÊÇ________kJ(±£ÁôÈýλÓÐЧÊý×Ö)¡£

(3)½ðÊôîÑÒ±Á¶¹ý³ÌÖÐÆäÖÐÒ»²½·´Ó¦Êǽ«Ô­ÁϽðºìʯת»¯£ºTiO2(½ðºìʯ)£«2C£«2Cl2¸ßÎÂ,TiCl4£«2CO¡¡ÒÑÖª£ºC(s)£«O2(g)===CO2(g)¡¡¦¤H£½£­393.5 kJ·mol£­1

2CO(g)£«O2(g)===2CO2(g)¡¡¦¤H£½£­566 kJ·mol£­1

TiO2(s)£«2Cl2(g)===TiCl4(s)£«O2(g)¡¡¦¤H£½£«141 kJ·mol£­1

ÔòTiO2(s)£«2Cl2(g)£«2C(s)===TiCl4(s)£«2CO(g)µÄ¦¤H£½________¡£

(4)³ôÑõ¿ÉÓÃÓÚ¾»»¯¿ÕÆø£¬ÒûÓÃË®Ïû¶¾£¬´¦Àí¹¤Òµ·ÏÎïºÍ×÷ΪƯ°×¼Á¡£³ôÑõ¼¸ºõ¿ÉÓë³ý²¬¡¢½ð¡¢Ò¿¡¢·úÒÔÍâµÄËùÓе¥ÖÊ·´Ó¦¡£È磺

6Ag(s)£«O3(g)===3Ag2O(s)¡¡¦¤H£½£­235.8 kJ·mol£­1£¬

ÒÑÖª£º2Ag2O(s)===4Ag(s)£«O2(g)

¦¤H£½£«62.2 kJ·mol£­1£¬

ÔòO3ת»¯ÎªO2µÄÈÈ»¯Ñ§·½³ÌʽΪ_________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¾Ý±¨µÀ£¬È«ÇòÿÄê·¢Éú½ðÊô¸¯Ê´¶øÔì³ÉµÄÖ±½Ó¾­¼ÃËðʧ´ïÊýǧÒÚÃÀÔª¡£ÏÂÁи÷µç¼«·´Ó¦

ʽÖУ¬ÄܱíʾÌúµÄµç»¯Ñ§¸¯Ê´µÄÊÇ                                        £¨      £©

   ¢Ù Fe£­2e-=Fe2+          ¢Ú 2H++2e-=H2¡ü          ¢Û Fe£­3e-=Fe3+ 

   ¢Ü 2H2O+O2+4e-=4OH-       ¢Ý 4OH-£­4e-=2H2O+O2¡ü
¡¡A.¢Ù¢Ú¢Ý         B.¢Ú¢Û¢Ü             C.¢Ù¢Ú¢Ü           D.¢Ù¢Û¢Ý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÃCl2Éú²úijЩº¬ÂÈÓлúÎïʱ»á²úÉú¸±²úÎïHCl¡£ÀûÓ÷´Ó¦A£º4HCl£«O2 2Cl2£«2H2O£¬¿ÉʵÏÖÂȵÄÑ­»·ÀûÓá£

        ÒÑÖª£º¢ñ£®·´Ó¦AÖУ¬4mol HCl±»Ñõ»¯£¬·Å³ö115.6kJµÄÈÈÁ¿£®
           

¢ò£® Ôò¶Ï¿ª1 mol H—O¼üÓë¶Ï¿ª1 mol H—Cl¼üËùÐèÄÜÁ¿Ïà²îԼΪ              (      )

  A£®16kJ           B£®24kJ           C£®32kJ           D£®48kJ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÒͼËùʾµÄ×°ÖÃÖе缫a¡¢b¾ùΪ̼°ô£¬Á½ÉÕ±­ÖÐËùÊ¢ÈÜÒº¾ùΪ500mL1.0mol/L¡£

¢ÅAΪ       (Ìî¡°Ô­µç³Ø¡±»ò¡°µç½â³Ø¡±)£¬ÆäÖÐAgµç¼«µÄµç¼«·´Ó¦Ê½Îª£º        £»

  ·¢Éú       ·´Ó¦(Ìî¡°Ñõ»¯¡±»ò¡°»¹Ô­¡±)¡£

  ¢ÆB×°ÖÃÖеĵ缫b¼«Îª         ¼«£¬µç¼«·´Ó¦Ê½Îª                              £¬

×Ü·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                                  ¡£

  ¢Ç¹¤×÷Ò»¶Îʱ¼äºó£¬µ±ZnƬÖÊÁ¿¼õÉÙ6.5gʱ£¬a¼«ÒݳöµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ý    L¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Èçͼ£¬a¡¢bÊÇʯīµç¼«£¬Í¨µçÒ»¶Îʱ¼äºó£¬b¼«¸½½üÈÜÒºÏÔºìÉ«¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®X¼«ÊǵçÔ´¸º¼«£¬Y¼«ÊǵçÔ´Õý¼«

B£®a¼«µÄµç¼«·´Ó¦ÊÇ2Cl£­£­2e£­===Cl2¡ü

C£®µç½â¹ý³ÌÖÐCuSO4ÈÜÒºµÄpHÖð½¥Ôö´ó

D£®Pt¼«ÉÏÓÐ6.4 g CuÎö³öʱ£¬b¼«²úÉú2.24 L(±ê×¼×´¿ö)ÆøÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


̼Ëá¸Æ³£ÓÃ×÷ÑÀ¸àµÄĦ²Á¼Á¡£Ä³Í¬Ñ§Éè¼ÆÁËÒ»ÖÖÖÆ±¸Ì¼Ëá¸ÆµÄ·½°¸£¬ÆäÁ÷³ÌͼÈçÏ£º(ËùÓÃʯ»Òʯº¬ÓÐÔÓÖÊSiO2)

 »Ø´ðÏÂÁÐÎÊÌ⣺

(1)³ä·ÖìÑÉÕ110¶Öʯ»ÒʯµÃµ½¹ÌÌå66¶Ö¡£±ê×¼×´¿öÏÂÉú³É¶þÑõ»¯Ì¼µÄÌå»ýΪ______________L£¬Ê¯»ÒʯÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýΪ______________%¡£

(2)¼ÙÉèµÚ¢Ù²½·´Ó¦ÍêÈ«½øÐУ¬ÔòµÚ¢Ú²½·´Ó¦¹ýÂ˺óµÃµ½µÄ²»ÈÜÐÔÂËÔüµÄ³É·ÖΪ________________________¡£

(3)µÚ¢Û²½·´Ó¦Ò»°ã²»²ÉÓÃͨÈëCO2µÄÖ÷ÒªÔ­ÒòÊÇ______________________£¬

ÏàÓ¦µÄÀë×Ó·½³ÌʽΪ_____________________________________¡£

(4)CaCO3ÊÇÒ»ÖÖÄÑÈÜÎïÖÊ£¬25¡æÊ±ÆäKsp£½2.8¡Á10£­9¡£ÏÖ½«µÈÌå»ýµÄCaCl2ÈÜÒºÓëNa2CO3ÈÜÒº»ìºÏ£¬ÈôNa2CO3ÈÜÒºµÄŨ¶ÈΪ2.0¡Á10£­4 mol/L£¬ÔòÉú³É³ÁµíËùÐèCaCl2ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È×îСÊÇ______________¡£

(5)ijѧÉúÓÃʯ»ÒʯΪԭÁÏ£¬Éè¼ÆÁËÁíÒ»ÖÖÖÆ±¸Ì¼Ëá¸ÆµÄʵÑé·½°¸£¬ÆäÁ÷³ÌͼÈçÏ£º

Óëǰһ·½°¸Ïà±È½Ï£¬¸Ã·½°¸µÄÓŵãÊÇ_________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸