¡¾ÌâÄ¿¡¿ÓÉÓÚþºÏ½ð¾ßÓÐÓ²¶È´ó¡¢ÃܶÈС¡¢É¢ÈÈÐԺᢿ¹ÕðÐԺõÈÓÅÒìÐÔÄÜËü±»ÓÃÓÚÖƱʼDZ¾µçÄÔÍâ¿Ç¡¢¾ºÈü×ÔÐгµ³µ¼ÜµÈ¡£ÏÖ³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄþÂÁºÏ½ðÑùÆ··ÅÈë500 mLÏ¡ÁòËáÖУ¬¹ÌÌåÈ«²¿ÈܽⲢ·Å³öÆøÌå¡£´ý·´Ó¦ÍêÈ«ºó£¬ÏòËùµÃÈÜÒºÖмÓÈëNaOHÈÜÒº£¬Éú³É³ÁµíµÄÎïÖʵÄÁ¿Óë¼ÓÈëNaOHÈÜÒºµÄÌå»ý¹ØϵÈçÏÂͼËùʾ¡£

(1)ºÏ½ðÖÐAlµÄÖÊÁ¿Îª__________________¡£

(2)NaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________________¡£

(3)Ï¡ÁòËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________________¡£

¡¾´ð°¸¡¿5.4g 4.0 mol/L 0.8 mol/L

¡¾½âÎö¡¿

£¨1£©¸ù¾ÝͼÏó¿ÉÖª£¬0¡«25 mL·¢ÉúËá¼îÖкͣ¬25¡«200 mL·¢ÉúÀë×ÓÓë¼îÉú³É³ÁµíµÄ·´Ó¦£¬200¡«240mL·¢ÉúAl(OH)3+NaOH=NaAlO2+2H2O£¬200 mLʱÉú³É³Áµí×î¶à£¬ÈÜÒºÖеÄÈÜÖÊΪÁòËáÄÆ£¬ÓÉͼÏó¿ÉÖª£¬ÇâÑõ»¯Ã¾µÄÎïÖʵÄÁ¿Îª0.15 mol£¬ÈܽâµÄÇâÑõ»¯ÂÁµÄÎïÖʵÄÁ¿Îª£º0.35 mol-0.15 mol=0.2 mol£¬¸ù¾ÝÂÁÔ­×ÓÊغã¿ÉµÃAlµÄÖÊÁ¿£»

£¨2£©¸ù¾Ý·´Ó¦Al(OH)3+NaOH=NaAlO2+2H2O¿ÉÇóµÃÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿Å¨¶È£»

£¨3£©ÈÜÒº×îºóµÄÈÜÖÊΪÁòËáÄÆ£¬¸ù¾ÝÔªËØÊغã¿ÉÖª£¬n(H2SO4) = n(SO42-) = n(Na+)/2 = n(NaOH)/2£¬¾Ý´Ë·ÖÎö×÷´ð¡£

£¨1£©Ã¾ÂÁºÏ½ðÑùÆ··ÅÈë500 mLÏ¡ÁòËáÖУ¬¹ÌÌåÈ«²¿ÈܽâÉú³ÉþÀë×Ó¡¢ÂÁÀë×Ó£¬Ïò´ËÈÜÒºÖмÓÈëNaOHÈÜÒº£¬´ÓÉú³É³ÁµíͼÏó·ÖÎö¿ÉÖª£¬ÇâÑõ»¯ÄÆÌå»ý´Ó0-25 mLʱûÓгÁµíÉú³É£¬ËµÃ÷Ô­ÈÜÒºÖдæÔÚÇâÀë×Ó¼´ÁòËá¹ýÁ¿£¬200-250 mL¶Î³Áµí²¿·ÖÏûʧ£¬·¢ÉúµÄ·´Ó¦ÊÇ£ºAl(OH)3+OH-=AlO2-+2H2O£¬ÓÉÂÁÔªËØÊغãµÃ£ºn(Al) = n[Al(OH)3] = (0.35-0.15) mol = 0.2 mol£¬m(Al)= 0.2mol ¡Á27g/mol = 5.4 g£¬

¹Ê´ð°¸£º5.4 g£»

£¨2£©n(Al) = n[Al(OH)3] = n(OH-) = n(NaOH) = (250-200) mL¡Á10-3¡Ác(NaOH) = 0.2mol£¬c(NaOH)= 4.0 mol/L£»

£¨3£©ÔÚ200 mLʱ£¬³ÁµíÊÇÇâÑõ»¯Ã¾ºÍÇâÑõ»¯ÂÁ£¬ÈÜÖÊΪÁòËáÄÆ£¬´ËʱV(NaOH)=200 mL£¬n(NaOH)=0.2 L¡Á4.0 mol/L=0.8 mol£¬Ôòn(H2SO4) = n(SO42-) = n(Na+)/2= n(NaOH)/2 = 0.4 mol£¬c(H2SO4)= 0.4 mol/0.5 L=0.8 mol/L¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖªA¡¢B¡¢C¡¢D¡¢E¡¢F¶¼ÊÇÔªËØÖÜÆÚ±íÇ°ËÄÖÜÆÚµÄÔªËØ£¬ËüÃǵĺ˵çºÉÊý£ºB£¼A£¼C£¼D£¼E£¼F¡£B¡¢CÁ½ÖÖÔªËض¼ÄÜÒýÆðË®Ì帻ӪÑø»¯¡£EÔ­×ӵõ½Ò»¸öµç×Óºó3p¹ìµÀÈ«³äÂú¡£A£«±ÈEÔ­×ÓÐγɵÄÀë×ÓÉÙ1¸öµç×Ӳ㡣D¿ÉÒÔÐγÉÁ½ÖÖÑõ»¯ÎÆäÖÐÒ»ÖÖÑõ»¯ÎïÊÇÐγÉËáÓêµÄÖ÷ÒªÆøÌåÖ®Ò»¡£FµÄÔ­×ÓÐòÊýΪ26¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)C¡¢D¡¢EµÄµÚÒ»µçÀëÄÜÓÉСµ½´óµÄ˳ÐòΪ________(ÓÃÔªËØ·ûºÅ±íʾ)¡£

(2)д³öBµÄÇ⻯ÎïÓëË®·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________£¬BµÄÇ⻯ÎKÒ×ÈÜÓÚË®µÄÔ­ÒòÊÇ____________________________________________________________¡£

(3)»¯ºÏÎïBE3µÄ·Ö×ӿռ乹ÐÍΪ________________¡£

(4)FÔªËØÔ­×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª________¡£FµÄÒ»ÖÖ³£¼û»¯ºÏÎïF(CO)5ÔÚ³£ÎÂϳÊҺ̬£¬ÈÛµãΪ£­20.5¡æ£¬·ÐµãΪ103¡æ£¬Ò×ÈÜÓڷǼ«ÐÔÈܼÁ£¬ÔòF(CO)5µÄ¾§ÌåÀàÐÍΪ__________________¡£

(5)½«FE3µÄ±¥ºÍÈÜÒºµÎÈë·ÐË®ÖУ¬Çëд³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ£º___________________________¡£

(6)±È½ÏB¡¢CÁ½ÖÖÔªËصÄÇ⻯ÎïÎȶ¨ÐÔ²¢ËµÃ÷ÀíÓÉ£º____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿25¡æʱ£¬ÓйØÎïÖʵĵçÀëƽºâ³£ÊýÈçÏ£º

»¯Ñ§Ê½

CH3COOH

H2CO3

H2SO3

µçÀëƽºâ³£Êý

K=1.8¡Á10£­5

K1=4.3¡Á10£­7

K2=5.6¡Á10£­11

K1=1.5¡Á10£­2

K2=1.02¡Á10£­7

(1)Çëд³öH2SO3µÄµçÀëƽºâ³£ÊýK1µÄ±í´ïʽ£º________________¡£

(2) ³£ÎÂÏ£¬½«Ìå»ýΪ10mL pH=2µÄ´×ËáÈÜÒºÓëÑÇÁòËáÈÜÒº·Ö±ð¼ÓÕôÁóˮϡÊÍÖÁ1000mL£¬Ï¡ÊͺóÈÜÒºµÄpH£¬Ç°Õß_____ºóÕߣ¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©¡£

(3)Ò»¶¨Ìõ¼þÏ£¬±ù´×Ëá¼ÓˮϡÊ͹ý³ÌÖÐÈÜÒºµ¼µçÄÜÁ¦IËæ¼ÓË®Ìå»ýV±ä»¯ÇúÏßÈçÓÒͼËùʾ£¬Ôòa¡¢b¡¢cÈýµãÈÜÒº´×ËáµÄµçÀë³Ì¶ÈÓÉ´óµ½Ð¡Îª____________________¡£

(4)ÏÂÁÐÀë×ÓCH3COO£­¡¢CO32£­¡¢HSO3£­¡¢SO32£­ÔÚÈÜÒºÖнáºÏH£«µÄÄÜÁ¦ÓÉ´óµ½Ð¡µÄ¹ØϵΪ___________¡£

(5)Ìå»ýÏàͬ¡¢c(H£«)ÏàͬµÄ¢ÙCH3COOH£»¢ÚHCl£»¢ÛH2SO4 ÈýÖÖËáÈÜÒº·Ö±ðÓëͬŨ¶ÈµÄNaOHÈÜÒºÍêÈ«ÖкÍʱ£¬ÏûºÄNaOHÈÜÒºµÄÌå»ýÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòÊÇ_____(ÌîÐòºÅ)¡£

(6)ÒÑÖª£¬H£«(aq) + OH£­(aq) == H2O(l) ¦¤H =£­57.3 kJ/mol¡£ÊµÑé²âµÃÏ¡´×ËáÓëÏ¡NaOHÈÜÒº·´Ó¦Éú³É1 mol H2Oʱ·Å³ö57 kJµÄÈÈ£¬Ôò´×ËáÈÜÒºÖУ¬´×ËáµçÀëµÄÈÈ»¯Ñ§·½³ÌʽΪ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁнṹͼÖУ¬¡ñ´ú±íÇ°¶þÖÜÆÚÔªËصÄÔ­×Óʵ£¨Ô­×ÓʵÊÇÔ­×Ó³ýÈ¥×îÍâ²ãµç×ÓºóÊ£ÓàµÄ²¿·Ö£©£¬Ð¡ºÚµã´ú±íδÓÃÓÚÐγɹ²¼Û¼üµÄ×îÍâ²ãµç×Ó£¬¶ÌÏß´ú±í¼Û¼ü¡£Ê¾Àý£º£¨Í¼ÖÐF2£©¸ù¾Ý¸÷ͼ±íʾµÄ½á¹¹Ìص㣬ÏÂÁÐÓйØÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©

A. ÉÏÊö½á¹¹Í¼Öй²³öÏÖ6ÖÖÔªËØ B. ¼×¡¢ÒÒ¡¢±ûΪ·Ç¼«ÐÔ·Ö×Ó£¬¶¡Îª¼«ÐÔ·Ö×Ó

C. ¼×Ó붡¿ÉÒÔ·¢Éú»¯ºÏ·´Ó¦Éú³ÉÀë×Ó»¯ºÏÎï D. ÏòCaCl2ÈÜÒºÖÐͨÈë±ûÓа×É«³Áµí²úÉú

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©147 g H2SO4µÄÎïÖʵÄÁ¿ÊÇ________£»0.5 mol H2SO4µÄÖÊÁ¿ÊÇ________g£»

£¨2£©Ñõ»¯Í­ÔÚ¸ßÎÂÏ·¢Éú·Ö½â·´Ó¦£º4CuO2Cu2O+O2¡ü£¬Èô·´Ó¦¹ý³ÌÖÐÉú³É1molO2·Ö×Ó£¬ÔòתÒƵç×ÓÊýΪ____________mol¡£

£¨3£©·´Ó¦£º2FeCl3£«2KI=2FeCl2£«2KCl£«I2£¬ÆäÖÐ_____ÔªËØ»¯ºÏ¼ÛÉý¸ß£¬Ôò¸ÃÔªËر»_______£¬_____ÔªËØ»¯ºÏ¼Û½µµÍ£»__________×ö»¹Ô­¼Á£¬_______·¢Éú»¹Ô­·´Ó¦¡£

£¨4£©ÄÆÔÚ¿ÕÆøÖÐȼÉյĻ¯Ñ§·½³Ìʽ£º________________________________£¬1molÄÆÍêȫȼÉÕתÒƵç×ӵĸöÊýΪ____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ìú¡¢ÂÁµÄ»ìºÏÎï½øÐÐÈçÏÂʵÑ飺

£¨1£©²Ù×÷XµÄÃû³ÆÊÇ________________£»

£¨2£©ÆøÌåAÊÇ___________£¨Ìѧʽ£©£»

£¨3£©ÈÜÒºBÖÐÒõÀë×Ó³ýOH- Í⻹ÓÐ__________£¨ÌîÀë×Ó·ûºÅ£©£¬ÈÜÒºDÖдæÔڵĽðÊôÀë×ÓΪ_________£¨ÌîÀë×Ó·ûºÅ£©£»

£¨4£©¼ÓÈë×ãÁ¿NaOHÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º_________£»¼ÓÈëÏ¡ÑÎËá·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º___________________________________________£»

£¨5£©ÏòÈÜÒºD¼ÓÈëNaOHÈÜÒº£¬¹Û²ìµ½²úÉúµÄ°×É«Ðõ×´³ÁµíѸËÙ±äΪ»ÒÂÌÉ«£¬×îÖÕ±äΪºìºÖÉ«£¬Çëд³ö³Áµíת»¯µÄ»¯Ñ§·½³Ìʽ£º___________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÕàÊÇÒ»ÖÖÖØÒªµÄ°ëµ¼Ìå²ÄÁÏ¡£¹¤ÒµÉÏÓþ«ÁòÕà¿ó£¨Ö÷Òª³É·ÖΪGeS2£¬ÔÓÖʲ»·´Ó¦£©ÖÆÈ¡Ge£¬Æ乤ÒÕÁ÷³ÌÈçͼËùʾ£º

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¿ªÊ¼½«¾«ÁòÕà¿ó·ÛËéµÄÄ¿µÄÊÇ______¡£

£¨2£©¸ßαºÉÕ¾«ÁòÕà¿óµÄ»¯Ñ§·½³ÌʽΪ______¡£

£¨3£©ÈÈ»¹Ô­Õæ¿Õ»Ó·¢Â¯ÄÚ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______¡£

£¨4£©ÔÚ³ÁÕà¹ý³ÌÖУ¬µ±Î¶ÈΪ90¡æ£¬pHΪ14ʱ£¬¼ÓÁÏÁ¿£¨CaCl2/GeÖÊÁ¿±È£©¶Ô³ÁÕàµÄÓ°ÏìÈç±íËùʾ£¬Ñ¡Ôñ×î¼Ñ¼ÓÁÏÁ¿Îª______£¨Ìî¡°10-15¡±¡°15-20¡±»ò¡°20-25¡±£©£¬ÀíÓÉÊÇ______¡£

񅧏

¼ÓÁÏÁ¿£¨CaCl2/Ge£©

ĸҺÌå»ý

£¨mL£©

¹ýÂ˺óÒºº¬Õࣨmg/L£©

¹ýÂ˺óÒº

pH

Õà³ÁµíÂÊ£¨%£©

1

10

500

76

8

93.67

2

15

500

20

8

98.15

3

20

500

2

11

99.78

4

25

500

1.5

12

99.85

£¨5£©Ä³Î¶Èʱ£¬³ÁÕàµÃµ½µÄCaGeO3ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏßÈçͼËùʾ¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ______¡£

a£®nµãÓëpµã¶ÔÓ¦µÄKspÏàµÈ

b£®Í¨¹ýÕô·¢¿ÉÒÔʹÈÜÒºÓÉqµã±äµ½pµã

c£®qµãÎÞCaGeO3³ÁµíÉú³É

d£®¼ÓÈëNa2GeO3¿ÉÒÔʹÈÜÒºÓÉnµã±äµ½mµã

£¨6£©CaGeO3ÓëÇ¿Ëá·´Ó¦¿ÉµÃµ½H2GeO3¡£0.l molL£­1µÄNaHGeO3ÈÜÒºpH_____£¨Ìî¡°£¾¡±¡°=¡±»ò¡°£¼¡±£©7£¬ÅжÏÀíÓÉÊÇ______£¨Í¨¹ý¼ÆËã±È½Ï£©¡££¨25¡æʱ£¬H2GeO3µÄKa1=1.7¡Á10£­9£¬Ka2=1.9¡Á10£­13¡££©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Óá°Ë«ÇÅ·¨¡±±íʾÏÂÁÐÑõ»¯»¹Ô­·´Ó¦Öеç×ÓתÒƵķ½ÏòºÍÊýÄ¿£¬²¢Ö¸³öÑõ»¯¼ÁºÍ»¹Ô­¼Á¡£2KClO3 === 2KCl+3O2¡ü _______Ñõ»¯¼Á_____¡¡»¹Ô­¼Á_____

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¼ÓÈÈHCOONa¹ÌÌ壬·¢ÉúµÄ·´Ó¦ÓУº2HCOONa Na2C2O4+H2¡ü ¢Ù

2HCOONa Na2CO3 +H2¡ü+CO¡ü ¢Ú

Na2C2O4 Na2CO3 + CO ¡ü ¢Û

HCOONa¼ÓÈÈ·Ö½âʱ£¬¹ÌÌåʧÖØÂÊÓëζȵĹØϵÈçÓÒͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ( )

A. T£¼415¡æʱ£¬Ö»Óз´Ó¦¢Ù·¢Éú

B. ·´Ó¦¢Ù¡¢¢Ú²»¿ÉÄÜͬʱ·¢Éú

C. 570¡æ£¼T£¼600¡æʱ£¬²ÐÁô¹ÌÌåµÄÖ÷Òª³É·ÖÊÇNa2CO3

D. ²ÐÁô¹ÌÌåÖÐm(Na2C2O4)=m( Na2CO3)ʱ£¬·´Ó¦¢Ù¡¢¢ÚµÄ·´Ó¦ËÙÂÊÏàµÈ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸