¡¾ÌâÄ¿¡¿Ä³»î¶¯¿Î³ÌС×éÄâÓÃ50 mL NaOHÈÜÒºÎüÊÕCO2ÆøÌ壬ÖƱ¸Na2CO3ÈÜÒº¡£ÎªÁË·ÀֹͨÈë¹ýÁ¿µÄCO2ÆøÌåÉú³ÉNaHCO3£¬Éè¼ÆÁËÈçÏÂʵÑé²½Ö裺

a.È¡25 mL NaOHÈÜÒºÎüÊÕ¹ýÁ¿µÄCO2ÆøÌ壬ÖÁCO2ÆøÌå²»ÔÙÈܽ⣻

b.С»ðÖó·ÐÈÜÒº1¡«2 min£¬¸Ï×ßÈܽâÔÚÈÜÒºÖеÄCO2ÆøÌ壻

c.Ôڵõ½µÄÈÜÒºÖмÓÈëÁíÒ»°ë(25 mL)NaOHÈÜÒº£¬Ê¹Æä³ä·Ö»ìºÏ·´Ó¦¡£

(1)´Ë·½°¸ÄÜÖƵýϴ¿¾»µÄNa2CO3£¬Ð´³öc²½ÖèµÄÀë×Ó·½³Ìʽ_________¡£´Ë·½°¸µÚÒ»²½µÄʵÑé×°ÖÃÈçͼËùʾ£º

(2)¼ÓÈë·´Ó¦ÎïÇ°£¬ÈçºÎ¼ìÑéÕû¸ö×°ÖõÄÆøÃÜÐÔ£º___________¡£

(3)ÈôÓôóÀíʯÓëÑÎËáÖÆCO2£¬Ôò×°ÖÃBÖÐÊ¢·ÅµÄÊÔ¼ÁÊÇ___________£¬×÷ÓÃÊÇ£º_________¡£

(4)ÔÚʵÑéÊÒͨ³£ÖÆ·¨ÖУ¬×°ÖÃA»¹¿É×÷ΪÏÂÁÐ_________(ÌîÐòºÅ)ÆøÌåµÄ·¢Éú×°Öá£

¢ÙHCl¡¡ ¢ÚH2¡¡ ¢ÛCl2¡¡

(5)ÒÑÖªËùÓÃNaOHÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ40%£¬ÊÒÎÂϸÃÈÜÒºÃܶÈΪ1.44 g/mL£¬¼ÙÉ跴ӦǰºóÈÜÒºµÄÌå»ý²»±ä£¬²»¿¼ÂÇʵÑéÎó²î£¬¼ÆËãÓôËÖÖ·½·¨ÖƱ¸ËùµÃNa2CO3ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_____mol/L

¡¾´ð°¸¡¿HCO3£«OH=CO32£«H2O ÓÃֹˮ¼Ð¼ÐסBÓëÉÕ±­Ö®¼äµÄÈ齺¹Ü£¬È»ºó´Ó©¶·ÖÐ×¢ÈëÒ»¶¨Á¿µÄË®£¬Ê¹Â©¶·ÖеÄË®Ãæ¸ßÓÚ׶ÐÎÆ¿ÄÚµÄË®Ã棬¹ýÒ»¶Îʱ¼ä£¬¹Û²ì©¶·ÄÚÓë׶ÐÎÆ¿ÖеÄÒºÃæ²î£¬Èô±£³Ö²»±ä£¬ËµÃ÷×°Öò»Â©Æø ±¥ºÍ̼ËáÇâÄÆÈÜÒº ÎüÊÕHClÆøÌå ¢Ú 7.2

¡¾½âÎö¡¿

(1)c²½Öè·¢ÉúµÄ·´Ó¦ÎªÌ¼ËáÇâÄÆÄܺÍÇâÑõ»¯ÄÆ·´Ó¦Éú³É̼ËáÄƺÍË®£»

(2)¸ù¾ÝÒº·âÆøÌåÐγÉÒºÃæ¸ß¶È²î£¬»òÀûÓÃÆøÌåµÄÈÈÕÍÀäËõÀ´¼ìÑé×°ÖõÄÆøÃÜÐÔ£»

(3)ÑÎËáÒ×»Ó·¢£¬ÖÆÈ¡µÄ¶þÑõ»¯Ì¼º¬ÓÐHCl£¬BÖÐÊ¢·ÅµÄÊÔ¼ÁÓÃÓÚÎüÊÕHCl£¬µ«²»ÄÜÓë¶þÑõ»¯Ì¼·´Ó¦£¬Éú³É¶þÑõ»¯Ì¼×îºÃ£»

(4)¸ÃÖÆȡװÖÃÊʺϲ»¼ÓÈÈÖÆÈ¡ÆøÌ壻

(5)¼ÆËã50mLNaOHÈÜÒºÖк¬ÓеÄÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿£¬¸ù¾ÝÄÆÔªËØÊغã¿ÉÖª£¬ÈÜÒºÖÐn(Na2CO3)£½0.5n(NaOH)£¬ÔÙ¸ù¾Ýc£½nV¼ÆËã¡£

(1)̼ËáÇâÄÆÄܺÍÇâÑõ»¯ÄÆ·´Ó¦Éú³É̼ËáÄƺÍË®£¬·´Ó¦Àë×Ó·½³ÌʽΪHCO3£«OH=CO32£«H2O£»¹Ê´ð°¸Îª£ºHCO3£«OH=CO32£«H2O£»

(2)ÓÃֹˮ¼Ð¼ÐסBÓëÉÕ±­Ö®¼äµÄÈ齺¹Ü£¬È»ºó´Ó©¶·ÖÐ×¢ÈëÒ»¶¨Á¿µÄË®£¬ÔÚ×°ÖÃÄÚÃÜ·âÆøÌ壬ʹ©¶·ÖеÄË®Ãæ¸ßÓÚ׶ÐÎÆ¿ÄÚµÄË®Ã棬¹ýÒ»»á£¬¹Û²ì©¶·ÄÚÓë׶ÐÎÆ¿ÖеÄÒºÃæ²î£¬Èô±£³Ö²»±ä£¬ËµÃ÷×°Öò»Â©Æø£»

¹Ê´ð°¸Îª£ºÓÃֹˮ¼Ð¼ÐסBÓëÉÕ±­Ö®¼äµÄÈ齺¹Ü£¬È»ºó´Ó©¶·ÖÐ×¢ÈëÒ»¶¨Á¿µÄË®£¬Ê¹Â©¶·ÖеÄË®Ãæ¸ßÓÚ׶ÐÎÆ¿ÄÚµÄË®Ã棬¹ýÒ»¶Îʱ¼ä£¬¹Û²ì©¶·ÄÚÓë׶ÐÎÆ¿ÖеÄÒºÃæ²î£¬Èô±£³Ö²»±ä£¬ËµÃ÷×°Öò»Â©Æø£»

(3)ÑÎËáÒ×»Ó·¢£¬ÖÆÈ¡µÄ¶þÑõ»¯Ì¼º¬ÓÐHCl£¬BÖÐÊ¢·Å±¥ºÍ̼ËáÇâÄÆÈÜÒº£¬ÎüÊÕHClÆøÌ壬ͬʱÉú³É¶þÑõ»¯Ì¼£»

¹Ê´ð°¸Îª£º±¥ºÍ̼ËáÇâÄÆÈÜÒº£»ÎüÊÕHClÆøÌ壻

(4)¸ÃÖÆȡװÖÃÊʺϲ»¼ÓÈÈÖÆÈ¡ÆøÌ壬HCl¡¢Cl2µÄÖƱ¸¶¼ÐèÒª¼ÓÈÈ£¬²»ÄÜʹÓøÃ×°ÖÃÖƱ¸£¬ÖÆÈ¡H2²»ÐèÒª¼ÓÈÈ£¬¿ÉÒÔÑ¡ÓøÃ×°Öã»

¹Ê´ð°¸Îª£º¢Ú£»

(5)m(NaOH)£½50mL¡Á1.44g/mL¡Á40%£½28.8g£¬ËùÒÔn(NaOH)£½£½0.72mol£¬¸ù¾ÝÄÆÔªËØÊغã¿ÉÖª£¬ÈÜÒºÖÐn (Na2CO3)£½0.5 n(NaOH)£½0.36mol£¬¹Êc(Na2CO3)£½£½7.2mol/L£»

¹Ê´ð°¸Îª£º7.2¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿10mL1mol¡¤L£­1µÄÑÎËáÓë¹ýÁ¿µÄп·Û·´Ó¦£¬Èô¼ÓÈëÉÙÁ¿µÄÏÂÁÐÎïÖÊ£¬ÄܼõÂý·´Ó¦ËÙÂʵ«ÓÖ²»Ó°ÏìÇâÆøÉú³ÉÁ¿µÄÊÇ

A.CuSO4(s)B.CH3COONa(s)C.KNO3(aq)D.Na2CO3(aq)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Éú̬¹¤ÒµÔ°ÇøµÄ½¨É裬²»½ö½öÊÇÌåÏÖ»·±£ÀíÄÖØÒªÒÀ¾ÝÑ­»·¾­¼ÃÀíÂۺͳä·Ö¿¼ÂǾ­¼ÃµÄ¿É³ÖÐø·¢Õ¹£¬ÈçͼÊÇijÆóÒµÉè¼ÆµÄÁòËá-Á×ï§-Ë®ÄàÁª²ú£¬º£Ë®-µ­Ë®¶àÓã¬ÑÎ-ÈÈ-µçÁª²úÉúÈý´óÉú̬²úÒµÁ´Á÷³Ìͼ¡£¸ù¾ÝÉÏÊö²úÒµÁ÷³Ì»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©´ÓÔ­ÁÏ¡¢ÄÜÔ´¡¢½»Í¨½Ç¶È¿¼ÂǸÃÆóÒµÓ¦½¨ÔÚ______

AÎ÷²¿É½Çø¡¡¡¡¡¡¡¡BÑغ£µØÇø¡¡¡¡¡¡¡¡¡¡C·¢´ï³ÇÊС¡¡¡¡¡¡¡¡¡¡¡D¶«±±ÄÚ½

£¨2£©¸ÃÁ÷³Ì¢Ù¡¢¢Ú¡¢¢Û¡¢¢Ü¡¢¢ÝΪÄÜÁ¿»òÎïÖʵÄÊäËÍ£¬Çë·Ö±ðд³öÊäË͵ÄÖ÷ÒªÎïÖʵĻ¯Ñ§Ê½»òÄÜÁ¿ÐÎʽ£º¢Ù______¡¢¢Ú______¡¢¢Û______¡¢¢Ü______¡¢¢Ý______¡£

£¨3£©·ÐÌÚ¯·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£»Á׷ʳ§µÄÖ÷Òª²úÆ·ÊÇÆոƣ¬ÆäÖ÷Òª³É·ÖÊÇ______£¨Ìѧʽ£©¡£

£¨4£©Èȵ糧µÄÀäÈ´Ë®ÊÇ______£¬¸ÃÁ÷³ÌÖÐŨËõÑÎË®³ýÌáÈ¡ÑÎÒÔÍ⻹¿ÉÌáÈ¡µÄÎïÖÊÓÐ______£¨Ð´³öÒ»ÖÖ¼´¿É£©¡£

£¨5£©¸ù¾ÝÏÖ´ú»¯¹¤³§Ã»¼ÆÀíÄîÇëÌá³ö¸ß¯Á¶Ìú³§·ÏÆø¡¢·ÏÔü¼°¶àÓàÈÈÄܵÄÀûÓÃÉèÏë¡£______£¬______£¨Ð´³öÁ½µã¼´¿É£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨ £©

A.ÒÑÖª£º¢ÙC(s)£«O2(g)=CO2(g) ¦¤H1=-393.5kJ¡¤mol-1£»¢Ú2CO(g)£«O2(g)=2CO2(g) ¦¤H2=-566kJ¡¤mol-1£»¢ÛTiO2(s)£«2Cl2(g)=TiCl4(s)£«O2(g) ¦¤H3=+141kJ¡¤mol-1£»ÔòTiO2(s)£«2Cl2(g)£«2C(s)=TiCl4(s)£«2CO(g)µÄ¦¤H=2¦¤H1-¦¤H2+¦¤H3

B.ÓлúÎïM¾­¹ýÌ«Ñô¹â¹âÕÕ¿Éת»¯³ÉN£¬¦¤H=£«88.6kJ¡¤mol-1¡£ÔòN±ÈM¸üÎȶ¨

C.ÒÑÖª²ð¿ª1molH¡ªH¼ü£¬1molN¡ªH¼ü£¬1molN¡ÔN¼ü·Ö±ðÐèÒªµÄÄÜÁ¿ÊÇ436kJ¡¢391kJ¡¢946kJ£¬ÔòN2(g)ÓëH2(g)·´Ó¦Éú³ÉNH3(g)µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºN2(g)£«3H2(g)2NH3(g) ¦¤H=-92.0kJ¡¤mol-1

D.ÒÑÖªCH3OH(l)µÄȼÉÕÈȦ¤H=-726.5kJ¡¤mol-1£¬CH3OH(l)£«O2(g)=CO2(g)£«2H2(g) ¦¤H=-akJ¡¤mol-1Ôòa<726.5

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¼îʽÑõ»¯(NiOOH)¿ÉÓ÷ÏÄø´ß»¯¼Á(Ö÷Òªº¬Ni¡¢Al£¬ÉÙÁ¿Cr¡¢FeSµÈ)À´ÖƱ¸£¬Æ乤ÒÕÁ÷³ÌÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¡°½þÅݳýÂÁ¡±Ê±£¬·¢Éú·´Ó¦µÄÀë×Ó·´Ó¦·½³ÌʽΪ_____________________¡£

(2)¡°¹ýÂË1¡±Óõ½µÄ²£Á§ÒÇÆ÷________________________________________¡£

(3)ÒÑÖª¸ÃÌõ¼þϽðÊôÀë×Ó¿ªÊ¼³ÁµíºÍÍêÈ«³ÁµíµÄpHÈçϱí:

¿ªÊ¼³ÁµíµÄpH

ÍêÈ«³ÁµíµÄpH

Ni2+

6.2

8.6

Fe2+

7.6

9.1

Fe3+

2.3

3.3

Cr3+

4.5

5.6

¡°µ÷pH 1¡±Ê±£¬ÈÜÒºpH·¶Î§Îª_______________£»

(4)ÔÚ¿ÕÆøÖмÓÈÈNi(OH)2¿ÉµÃNiOOH£¬Çëд³ö´Ë·´Ó¦µÄ»¯Ñ§·½³Ìʽ________________¡£

(5)ÔÚËáÐÔÈÜÒºÖÐCrO¿ÉÒÔת»¯³ÉCr2O£¬ÓÃÀë×Ó·½³Ìʽ±íʾ¸Ãת»¯·´Ó¦__________£¬ÒÑÖªBaCrO4µÄKsp=1.2¡Á10-10£¬ÒªÊ¹ÈÜÒºÖÐCrO³ÁµíÍêÈ«(c(CrO)¨Q1¡Á10-5mol¡¤L1),ÈÜÒºÖбµÀë×ÓŨ¶ÈÖÁÉÙΪ________mol¡¤L1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¼×¡¢ÒÒ¡¢±ûÈýÖÖÎïÖÊÖ®¼äÓÐÈçÏÂת»¯¹Øϵ£º

¼×ÒÒ±û¼×

(1)Èô¼×ÊDz»ÈÜÓÚË®µÄ°×É«¹ÌÌåÎïÖÊ£¬¼ÈÄÜÈÜÓÚÑÎËáÓÖÄÜÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬Ôò¼×ÊÇ_________(д»¯Ñ§Ê½£¬ÏÂͬ)¡£Ð´³ö¡°ÒÒ±û¡±×ª»¯µÄÀë×Ó·½³Ìʽ£º______________________¡£

(2)ÈôÒÒÈÜÒºÖмÓÈëKSCNÈÜÒº£¬ÓкìÉ«³öÏÖ£¬Ôò¼×ÎïÖÊÊÇ__________¡£Ð´³ö¡°¼×ÒÒ¡±×ª»¯µÄÀë×Ó·½³Ìʽ£º_________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿²¿·Ö±»Ñõ»¯µÄFe£­CuºÏ½ðÑùÆ·(Ñõ»¯²úÎïΪFe2O3¡¢CuO)¹²5£®92 g£¬¾­ÈçÏ´¦Àí£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ¢ÙÂËÒºAÖеÄÑôÀë×ÓΪFe2+¡¢Fe3+¡¢H+ £»¢ÚÑùÆ·ÖÐÑõÔªËصÄÎïÖʵÄÁ¿Îª0£®03 mol£»¢ÛÈܽâÑùÆ·µÄ¹ý³ÌÖÐÏûºÄÁòËáµÄ×ÜÎïÖʵÄÁ¿Îª0£®04 mol£»¢ÜV£½224£»¢ÝV£½336¡£

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

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿³£ÎÂÏ£¬ÓÃ1.0mol¡¤L-1µÄNaOHÈÜÒºÖкÍijŨ¶ÈµÄH2SO4ÈÜÒº£¬ËùµÃÈÜÒºµÄpHºÍËùÓÃNaOHÈÜÒºÌå»ýµÄ¹ØϵÈçͼËùʾ£¬ÔòÔ­H2SO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¼°ÍêÈ«·´Ó¦ºóÈÜÒºµÄÌå»ý(ºöÂÔ·´Ó¦Ç°ºóÈÜÒºÌå»ýµÄ±ä»¯)·Ö±ðÊÇ£¨ £©

A.1.0mol¡¤L-1£¬20mLB.0.5mol¡¤L-1£¬40mL

C.0.5mol¡¤L-1£¬80mLD.1.0mol¡¤L-1£¬80mL

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¢ñ.¼×´¼ÊÇÖØÒªµÄ»¯Ñ§¹¤Òµ»ù´¡Ô­ÁϺÍÇå½àÒºÌåȼÁÏ£¬¹¤ÒµÉÏ¿ÉÀûÓÃCO»òCO2À´Éú²úȼÁϼ״¼¡£ÒÑÖª¼×´¼ÖƱ¸µÄÓйػ¯Ñ§·´Ó¦ÒÔ¼°ÔÚ²»Í¬Î¶ÈϵĻ¯Ñ§·´Ó¦Æ½ºâ³£ÊýÈç±íËùʾ£º

»¯Ñ§·´Ó¦

ƽºâ³£Êý

ζȡæ

500

800

¢Ù2H2(g)+CO(g)CH3OH(g)

K1

2.5

0.15

¢ÚH2(g)+CO2(g)H2O(g)+CO(g)

K2

1.0

2.50

¢Û3H2(g)+CO2(g)CH3OH(g)+H2O(g)

K3

£¨1£©¾Ý·´Ó¦¢ÙÓë¢Ú¿ÉÍƵ¼³öK1¡¢K2ÓëK3Ö®¼äµÄ¹Øϵ£¬ÔòK3£½__________£¨ÓÃK1¡¢K2±íʾ£©

£¨2£©·´Ó¦¢ÛµÄ¦¤S__________0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±£©£»·´Ó¦¢ÛµÄ¦¤H__________0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±£©

£¨3£©500¡æʱ²âµÃ·´Ó¦¢ÛÔÚijʱ¿Ì£¬H2(g)¡¢CO2(g)¡¢CH3OH(g)¡¢H2O(g)µÄŨ¶È£¨mol/L£©·Ö±ðΪ0.8¡¢0.1¡¢0.3¡¢0.15£¬Ôò´ËʱVÕý__________VÄ棨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©

¢ò.Ò»¶¨Ìõ¼þÏ£¬ÔÚÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖгäÈëlmolCOÓë2molH2ºÏ³É¼×´¼£¬Æ½ºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼËùʾ£º

£¨1£©p1__________p2£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©¡£

£¨2£©¸Ã·´Ó¦´ïµ½Æ½ºâʱ£¬·´Ó¦Îïת»¯ÂʵĹØϵÊÇCO____________H2£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©¡£

£¨3£©Èô100¡æP1ʱ´ïƽºâËùÓõÄʱ¼äΪ5min£¬Ôò´Ó¿ªÊ¼µ½Æ½ºâÕâ¶Îʱ¼äÓÃH2±íʾµÄËÙÂÊΪ_______________________¡£

£¨4£©¸Ã¼×´¼ºÏ³É·´Ó¦ÔÚAµãµÄƽºâ³£ÊýK£½___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸