¡¾ÌâÄ¿¡¿ÏÂÁÐÓйصç½âÖÊÈÜÒºÖÐ΢Á£µÄÎïÖʵÄÁ¿Å¨¶È¹ØϵÕýÈ·µÄÊÇ(¡¡¡¡)

A. pH=3µÄÒ»ÔªËáºÍpH=11µÄһԪǿ¼îµÈÌå»ý»ìºÏºóµÄÈÜÒºÖУºc(OH£­)=c(H+)

B. ÔÚ0£®1 mol¡¤L£­1Na2CO3ÈÜÒºÖУºc(OH£­)£­c(H£«)£½c(HCO3£­)£«2c(H2CO3)

C. ÒÑÖªµþµªËá(HN3)Óë´×ËáËáÐÔÏà½ü£¬ÔòÔÚNaN3Ë®ÈÜÒºÖÐÀë×ÓŨ¶È´óС˳ÐòΪ£ºc(Na+)£¾c(OH)£¾c(N3)£¾c(H£«)

D. 0£®2 mol¡¤L£­1NaHCO3ÈÜÒºÖмÓÈëµÈÌå»ý0£®1 mol¡¤L£­1NaOHÈÜÒº£ºc(CO32£­)£¾c(HCO3£­)£¾c(OH£­)£¾c(H£«)

¡¾´ð°¸¡¿B

¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£ºA£®ÈôpH=3µÄÒ»ÔªËáÊÇÇ¿ËᣬÔòËüºÍpH=11µÄһԪǿ¼îµÈÌå»ý»ìºÏºóµÄÈÜÒºÖÐc(OH£­)=c(H+)£¬ÈôpH=3µÄÒ»ÔªËáÊÇÈõËᣬÓÉÓÚÈõËᲿ·ÖµçÀ룬ËáµÄŨ¶È´óÓÚH+µÄŨ¶È£¬ÔòËüºÍpH=11µÄһԪǿ¼îµÈÌå»ý»ìºÏºó£¬¹ýÁ¿µÄËá»á¼ÌÐøµçÀë²úÉúH+,ʹµÄÈÜÒºÖÐc(OH£­)<c(H+)£¬´íÎó£»B£®¸ù¾ÝÖÊ×ÓÊغã¿ÉÖª£ºÔÚ0£®1 mol¡¤L£­1Na2CO3ÈÜÒºÖУºc(OH£­)£­c(H£«)£½c(HCO3£­)£«2c(H2CO3)£¬ÕýÈ·£»C£®µþµªËá(HN3)Óë´×ËáËáÐÔÏà½ü£¬ÔòµþµªËáÊÇÈõËᣬNaN3ÊÇÇ¿¼îÈõËáÑΣ¬ÔÚNaN3Ë®ÈÜÒºÖÐN3-·¢ÉúË®½â·´Ó¦ÏûºÄË®µçÀë²úÉúµÄH+£¬Ê¹ÈÜÒºÏÔ¼îÐÔ£¬c(OH)£¾c(H£«)£¬ÓÉÓÚN3-²»¶ÏÏûºÄ£¬ËùÒÔÀë×ÓŨ¶Èc(Na+)£¾c(N3)¡£µ«ÊÇÑÎË®½âµÄ³Ì¶ÈÊÇ΢ÈõµÄ£¬Òò´Ëc(N3)£¾c(OH)¡£ËùÒÔÀë×ÓŨ¶È´óС˳ÐòΪ£ºc(Na+)£¾c(N3)£¾c(OH)£¾c(H£«)£¬´íÎó£»D£®0£®2 mol¡¤L£­1NaHCO3ÈÜÒºÖмÓÈëµÈÌå»ý0£®1 mol¡¤L£­1NaOHÈÜÒº£¬·´Ó¦ºóµÃµ½NaHCO3¡¢Na2CO3µÈŨ¶ÈµÈÌå»ýµÄ»ìºÏÈÜÒº¡£ÓÉÓÚNa2CO3¡¢NaHCO3¶¼·¢ÉúË®½â·´Ó¦£¬Ê¹ÈÜÒºÏÔ¼îÐÔ£¬ËùÒÔc(OH£­)£¾c(H£«)£»ÑεÄË®½â³Ì¶ÈCO32£­> HCO3£­;c(CO32£­)<c(HCO3£­)£»ÑεÄË®½â³Ì¶ÈÊÇ΢ÈõµÄ£¬ËùÒÔc(CO32£­)£¾c(OH£­)£¬ËùÒÔÀë×ÓŨ¶È´óС¹ØϵÊÇ£ºc(HCO3£­)£¾c(CO32£­)£¾c(OH£­)£¾c(H£«)£¬´íÎó¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊµÑéÊÒͨ³£ÓÃMnO2×÷´ß»¯¼Á·Ö½âH2O2£¬ÒÑÖªCuSO4ÈÜÒº¶Ô¹ýÑõ»¯ÇâµÄ·Ö½âÒ²¾ßÓд߻¯×÷Óã¬Ä³ÊµÑéÐËȤС×éͬѧ²ÂÏëÆäËûÑÎÈÜÒºÒ²¿ÉÄÜÔÚÕâ¸ö·´Ó¦ÖÐÆðͬÑùµÄ×÷Óã¬ÓÚÊÇËûÃÇ×öÁËÒÔÏÂ̽¾¿¡£ÇëÄã°ïÖúËûÃÇÍê³ÉʵÑ鱨¸æ£º

(1)ʵÑé¹ý³Ì£ºÔÚÒ»Ö§ÊÔ¹ÜÖмÓÈë5 mL 5%µÄH2O2ÈÜÒº£¬È»ºóµÎÈëÊÊÁ¿µÄFeCl3ÈÜÒº£¬°Ñ´ø»ðÐǵÄľÌõÉìÈëÊԹܡ£

ʵÑéÏÖÏó£º____________________________¡£

ʵÑé½áÂÛ£ºFeCl3ÈÜÒº¿ÉÒÔ´ß»¯·Ö½âH2O2¡£

(2)ÒÑÖªFeCl3ÔÚË®ÖпɵçÀë³öFe3£«ºÍCl£­£¬Í¬Ñ§ÃÇÌá³öÒÔϲÂÏ룺

´Ù½øH2O2·Ö½âµÄ΢Á£

¼×ͬѧ

H2O

ÒÒͬѧ

Fe3£«

±ûͬѧ

Cl£­

ÄãÈÏΪ×î²»¿ÉÄܵÄÊÇ__________ͬѧµÄ²ÂÏ룬ÀíÓÉÊÇ________________________¡£

(3)ͬѧÃǶÔÓàϵÄÁ½¸ö²ÂÏëÓÃʵÑé½øÐÐÁË̽¾¿£¬ÇëÄã×Ðϸ·ÖÎöºóÌî±í£º

ʵÑé¹ý³Ì

ʵÑéÏÖÏó

½áÂÛ

ÏòÊ¢ÓÐ5 mL 5%µÄH2O2ÈÜÒºµÄÊÔ¹ÜÖмÓÈëÉÙÁ¿µÄÑÎËᣬ°Ñ´ø»ðÐǵÄľÌõÉìÈëÊÔ¹Ü

____________

____________

ÏòÊ¢ÓÐ5 mL 5%µÄH2O2ÈÜÒºµÄÊÔ¹ÜÖмÓÈëÉÙÁ¿µÄFe2(SO4)3ÈÜÒº£¬°Ñ´ø»ðÐǵÄľÌõÉìÈëÊÔ¹Ü

_____________

____________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ëæ×ÅÉú»îˮƽµÄÌá¸ß£¬ÈËÃÇÔ½À´Ô½¹Ø×¢×ÔÉí½¡¿µ¡£ÏÖÓÐÏÂÁÐÎåÖÖÎïÖÊ£º¢ÙÅ£ÄÌ¢ÚÆ»¹û¢ÛÆÏÌÑÌÇ¢Üζ¾«¢ÝÑÇÏõËáÄÆ¡£Çë°´ÏÂÁÐÒªÇóÌî¿Õ£¨ÌîÐòºÅ£©¡£

£¨1£©¸»º¬µ°°×ÖʵÄÊÇ________¡£

£¨2£©ÊôÓÚ·À¸¯¼ÁµÄÊÇ________¡£

£¨3£©ÊôÓÚµ÷ζ¼ÁµÄÊÇ________¡£

£¨4£©¸»º¬Î¬ÉúËØCµÄÊÇ________¡£

£¨5£©¿ÉÖ±½Ó½øÈëѪҺ£¬²¹³äÄÜÁ¿µÄÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚ 101kPaʱ£¬ H 2ÔÚ 1molO 2ÖÐÍêȫȼÉÕÉú³É 2molҺ̬ H 2O£¬·Å³ö 571.6kJµÄÈÈÁ¿¡£

£¨1£©H2µÄȼÉÕÈÈΪ ______£¬±íʾ H 2ȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ ___________ ¡£

£¨2£©1L 1mol/L H 2SO 4ÈÜÒºÓë 2L 1mol/L NaOHÈÜÒºÍêÈ«·´Ó¦£¬·Å³ö 114.6kJµÄÈÈÁ¿£¬¸Ã·´Ó¦µÄÖкÍÈÈΪ ___________£¬±íʾÆäÖкÍÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ ______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÃºÈ¼Éյķ´Ó¦ÈÈ¿Éͨ¹ýÒÔÏÂÁ½¸ö;¾¶À´ÀûÓãºa.ÀûÓÃúÔÚ³ä×ãµÄ¿ÕÆøÖÐÖ±½ÓȼÉÕ²úÉúµÄ·´Ó¦ÈÈ£»b.ÏÈʹúÓëË®ÕôÆø·´Ó¦µÃµ½ÇâÆøºÍÒ»Ñõ»¯Ì¼£¬È»ºóʹµÃµ½µÄÇâÆøºÍÒ»Ñõ»¯Ì¼ÔÚ³ä×ãµÄ¿ÕÆøÖÐȼÉÕ¡£ÕâÁ½¸ö¹ý³ÌµÄÈÈ»¯Ñ§·½³ÌʽΪ£º

a£®C(s)£«O2(g)===CO2(g) ¦¤H£½E1¢Ù

b£®C(s)£«H2O(g)===CO(g)£«H2(g) ¦¤H£½E2¢Ú

H2(g)£«1/2O2(g)===H2O(g) ¦¤H£½E3¢Û

CO(g)£«1/2O2(g)===CO2(g) ¦¤H£½E4¢Ü

ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©Óë;¾¶aÏà±È£¬Í¾¾¶bÓн϶àµÄÓŵ㣬¼´___________________________________¡£

£¨2£©ÉÏÊöËĸöÈÈ»¯Ñ§·½³ÌʽÖЦ¤H>0µÄ·´Ó¦ÓÐ_____________________________________ ___________________________________¡£

£¨3£©µÈÖÊÁ¿µÄú·Ö±ðͨ¹ýÒÔÉÏÁ½Ìõ²»Í¬µÄ;¾¶²úÉúµÄ¿ÉÀûÓõÄ×ÜÄÜÁ¿¹ØϵÕýÈ·µÄÊÇ________¡£

A£®a±Èb¶à B£®a±ÈbÉÙ

C£®aÓëbÔÚÀíÂÛÉÏÏàͬ D£®Á½ÕßÎÞ·¨±È½Ï

£¨4£©¸ù¾ÝÄÜÁ¿Êغ㶨ÂÉ£¬E1¡¢E2¡¢E3¡¢E4Ö®¼äµÄ¹ØϵΪ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚ200mL0.5molL-1NaOHÈÜÒºÖУ¬¼ÓÈëÒ»¶¨Á¿SO3£¬ÔÚÒ»¶¨Ìõ¼þÏÂÕô·¢ËùµÃÈÜÒº£¬Îö³öµÄ¹ÌÌåÎïÖÊÖ»ÓÐ5.0g£¬¸Ã¹ÌÌåÎïÖʳɷֿÉÄÜÊÇ£¨¹ÌÌå²»º¬½á¾§Ë®£©£¨ £©

A. Na2SO4 B. NaHSO4 C. NaOHºÍNa2SO4 D. Na2SO4ºÍNaHSO4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÎªÁ˲ⶨijͭÒøºÏ½ðµÄ³É·Ö£¬½«30.0gºÏ½ðÈÜÓÚ80mL13.5mol/LµÄŨÏõËáÖУ¬´ýºÏ½ðÍêÈ«Èܽâºó£¬ÊÕ¼¯µ½ÆøÌå6.72L£¨±ê×¼×´¿ö£©£¬²¢²âµÃÈÜÒºÖеÄc£¨H+£©=1mol/L¡£¼ÙÉè·´Ó¦ºóÈÜÒºµÄÌå»ýÈÔΪ80mL£¬ÊÔ¼ÆË㣺

£¨1£©±»»¹Ô­µÄÏõËáµÄÎïÖʵÄÁ¿£»

£¨2£©ºÏ½ðÖÐÒøµÄÖÊÁ¿·ÖÊý£»

£¨3£©ÈôÐèÒª½«È«²¿Î²ÆøÓÃË®ÎüÊÕ£¬ÐèÒªµÄÑõÆøÌå»ý£¨±ê×¼×´¿ö£©£»

£¨4£©Èç¹û½«ËùµÃÈÜÒºÖÐÈ«²¿½ðÊôÑôÀë×Ó³Áµí£¬ÖÁÉÙÐè¼ÓÈë1mol/LNaOHÈÜÒºµÄÌå»ý¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¹éÄÉÓëÍÆÀíÊÇ»¯Ñ§Ñ§Ï°³£Óõķ½·¨¡£ÏÂÁÐÍÆÀíÕýÈ·µÄÊÇ

A. ÒòΪϡÁòËáÄÜÓëÌú·´Ó¦·Å³öH2£¬ËùÒÔÏ¡ÏõËáÓëÌú·´Ó¦Ò²Ò»¶¨ÄܷųöH2

B. ÒòΪNO2ÈÜÓÚË®ÐγÉHNO3£¬ËùÒÔNO2ÊÇËáÐÔÑõ»¯Îï

C. ÕáÌǼÓÈëŨÁòËáºó±äºÚ£¬ËµÃ÷ŨÁòËá¾ßÓÐÍÑË®ÐÔ

D. ÒòΪSO2¿ÉÒÔʹËáÐÔKMnO4ÈÜÒºÍÊÉ«£¬ËùÒÔSO2¾ßÓÐƯ°×ÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿°´ÒªÇóÌî¿Õ¡£

£¨1£©Å¨¶È¾ùΪ0.1 mol/LµÄ¢ÙÁòËá ¢Ú´×Ëá ¢ÛÇâÑõ»¯ÄÆ ¢ÜÂÈ»¯ï§ËÄÖÖÈÜÒºÖÐÓÉË®µçÀë³öµÄH£«Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ_________________________£¨ÌîÐòºÅ£©¡£

£¨2£©ÏÖ½«×ãÁ¿µÄ¹ÌÌåAgCl·Ö±ð·ÅÈëÏÂÁÐÎïÖÊÖУº ¢Ù30 mL 0.02 mol/L CaCl2ÈÜÒº ¢Ú20 mL 0.01 mol/L KClÈÜÒº¢Û40 mL0.03mol/L HClÈÜÒº ¢Ü10 mLÕôÁóË® ¢Ý50 mL 0.05 mol/L AgNO3ÈÜÒº¡£AgClµÄÈܽâ¶ÈÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòÊÇ________________________£¨ÌîÐòºÅ£©

£¨3£©Ïò±¥ºÍFeCl3ÈÜÒºÖмÓÈëCaCO3·ÛÄ©£¬·¢ÏÖ̼Ëá¸ÆÖð½¥Èܽ⣬ͬʱ»¹²úÉúµÄÏÖÏóÓÐ____________________________________________________¡£

£¨4£©³£ÎÂÏ£¬½«0.2 mol/L CH3COOHºÍ0.1 mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpH£¼7£¬¸Ã»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ______________________________¡£

£¨5£©ÏàͬŨ¶ÈµÄÏÂÁÐÈÜÒºÖУ¬c(CO32£­)µÄ´óС˳ÐòÊÇ_____________________£¨ÌîÐòºÅ£©

¢ÙNa2CO3 ¢ÚNaHCO3 ¢ÛH2CO3 ¢Ü(NH4)2CO3 ¢ÝNH4HCO3

£¨6£©³£ÎÂÏ£¬ÓÃ0.01 mol/L HClÈÜÒºÍêÈ«ÖкÍpH£½11µÄÏÂÁÐÈÜÒº¸÷100mL£¬ÐèHClÈÜÒºÌå»ýµÄ´óС¹ØϵÊÇ_________________________________________£¨ÌîÐòºÅ£©

¢ÙNaOH ¢ÚBa(OH)2 ¢ÛNH3¡¤H2O

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸