ÌâÄ¿ÄÚÈÝ

³£ÎÂÏ£¬Ïò100 mL 0.01 mol¡¤L£­1 HAÈÜÒºÖÐÖðµÎ¼ÓÈë0.02 mol¡¤L£­1 MOHÈÜÒº£¬Í¼ÖÐËùʾÇúÏß±íʾ»ìºÏÈÜÒºµÄpH±ä»¯Çé¿ö(Ìå»ý±ä»¯ºöÂÔ²»¼Æ)¡£

 

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÓÉͼÖÐÐÅÏ¢¿ÉÖªHAΪ________Ëá(Ìî¡°Ç¿¡±»ò¡°Èõ¡±)£¬ÀíÓÉÊÇ________________________________________________¡£

(2)³£ÎÂÏÂÒ»¶¨Å¨¶ÈµÄMAÏ¡ÈÜÒºµÄpH£½a£¬Ôòa________________________________________________________7

(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆäÔ­ÒòΪ_____________________________________________________

´Ëʱ£¬ÈÜÒºÖÐÓÉË®µçÀë³öµÄc(OH£­)£½________¡£

(3)Çëд³öKµãËù¶ÔÓ¦µÄÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС¹Øϵ£º_________________________________________¡£

(4)Kµã¶ÔÓ¦µÄÈÜÒºÖУ¬c(M£«)£«c(MOH)________2c(A£­)(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)£»Èô´ËʱÈÜÒºÖУ¬pH£½10£¬Ôòc(M£«)£­c(OH£­)£½________mol¡¤L£­1¡£

 

(1)Ç¿¡¡0.01 mol¡¤L£­1 HAµÄpHΪ2£¬ËµÃ÷HAÍêÈ«µçÀë

(2)<¡¡M£«£«H2O??MOH£«H£«¡¡1¡Á10£­a mol¡¤L£­1

(3)c(M£«)>c(A£­)>c(OH£­)>c(H£«)

(4)£½¡¡0.005

¡¾½âÎö¡¿(1)ÓÉͼÏñ¿´³ö0.01 mol/L HAÈÜÒº£¬pH£½2£¬ËµÃ÷ÆäÍêÈ«µçÀ룬¹ÊΪǿµç½âÖÊ¡£(2)ÓÉÌâĿͼÏñ¿ÉÖª100 mL 0.01 mol/L HAÈÜÒºÖеμÓ51 mL 0.02 mol/L MOHÈÜÒº£¬pH£½7£¬ËµÃ÷MOHÊÇÈõ¼î£¬¹ÊÆä¶ÔÓ¦µÄMAÊÇÈõ¼îÇ¿ËáÑΣ¬Ë®½âÏÔËáÐÔ£¬ÈÜÒºÖÐH£«È«²¿ÊÇË®µçÀë³öÀ´µÄ£¬¹ÊË®µçÀë³öµÄc(OH£­)£½1¡Á10£­a mol¡¤L£­1¡£(3)¡«(4)ÔÚKµãʱÏò100 mL 0.01 mol/L HAÈÜÒºÖеμÓ100 mL 0.02 mol/L MOHÈÜÒº·´Ó¦£¬·´Ó¦ºóµÄÈÜҺΪµÈŨ¶ÈµÄMAºÍMOHÈÜÒº£¬¹Êc(M£«)>c(A£­)>c(OH£­)>c(H£«)¡£ÓÉÎïÁÏÊغã¿ÉµÃc(M£«)£«c(MOH)£½2c(A£­)£¬½áºÏµçºÉÊغ㣺c(M£«)£«c(H£«)£½c(A£­)£«c(OH£­)£¬c(M£«)£­c(OH£­)£½c(A£­)£­c(H£«)¡Ö0.005 mol¡¤L£­1

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

»¯Ñ§ÄÜÓëµçÄÜÖ®¼äµÄÏ໥ת»¯ÓëÈËÀàµÄÉú»îʵ¼ÊÃÜÇÐÏà¹Ø£¬ÔÚÉú²ú¡¢Éú»îÖÐÓÐÖØÒªµÄÓ¦Óã¬Í¬Ê±Ò²ÊÇѧÉúÐγɻ¯Ñ§Ñ§¿ÆËØÑøµÄÖØÒª×é³É²¿·Ö¡£

£¨1£©ÈÛÈÚ״̬Ï£¬ÄƵĵ¥ÖʺÍÂÈ»¯ÑÇÌúÄÜ×é³É¿É³äµçµç³Ø£¬Èçͼ9£­8¹¤×÷Ô­ÀíʾÒâͼ£¬·´Ó¦Ô­ÀíΪ2Na£«FeCl2Fe£«2NaCl£¬¸Ãµç³Ø·Åµçʱ£¬Õý¼«·´Ó¦Ê½Îª________________________________________________________________________£»

³äµçʱ£¬____________£¨Ð´ÎïÖÊÃû³Æ£©µç¼«½ÓµçÔ´µÄ¸º¼«£»¸Ãµç³ØµÄµç½âÖÊΪ________¡£

 

£¨2£©Ä³Í¬Ñ§ÓÃͭƬ¡¢Ê¯Ä«×÷µç¼«µç½âÒ»¶¨Å¨¶ÈµÄÁòËáÍ­ÈÜÒº£¬¹¤×÷Ô­ÀíʾÒâͼÈçͼËùʾ£¬Ò»¶Îʱ¼äֹͣͨµçÈ¡³öµç¼«¡£ÈôÔÚµç½âºóµÄÈÜÒºÖмÓÈë0.98 gÇâÑõ»¯Í­·ÛÄ©Ç¡ºÃÍêÈ«Èܽ⣬¾­²â¶¨ËùµÃÈÜÒºÓëµç½âÇ°ÍêÈ«Ïàͬ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙYµç¼«²ÄÁÏÊÇ________£¬·¢Éú________£¨Ìî¡°Ñõ»¯¡±»ò¡°»¹Ô­¡±£©·´Ó¦¡£

¢Úµç½â¹ý³ÌÖÐXµç¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½ÊÇ_______________________________________________________________________¡£

¢ÛÈçÔÚµç½âºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄСËÕ´ò£¬³ä·Ö·´Ó¦ºó²úÉúÆøÌåÔÚ±ê×¼×´¿öÏÂËùÕ¼µÄÌå»ýÊÇ__________¡£

 

(1)ÒÑÖª£º

¢ÙFe(s)£«O2(g)=FeO(s)¡¡¦¤H£½£­272.0 kJ¡¤mol£­1

¢Ú2Al(s)£«O2(g)=Al2O3(s)¡¡¦¤H£½£­1675.7 kJ¡¤mol£­1

AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ____________________________________

 

(2)ij¿ÉÄæ·´Ó¦ÔÚ²»Í¬Ìõ¼þϵķ´Ó¦Àú³Ì·Ö±ðΪA¡¢B(ÈçÉÏͼËùʾ)¡£

¢Ù¸ù¾ÝͼÅжϸ÷´Ó¦´ïµ½Æ½ºâºó£¬ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȣ¬·´Ó¦ÎïµÄת»¯ÂÊ________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)£»

¢ÚÆäÖÐBÀú³Ì±íÃ÷´Ë·´Ó¦²ÉÓõÄÌõ¼þΪ________(Ñ¡ÌîÐòºÅ)¡£

A£®Éý¸ßζȡ¡¡¡¡¡¡¡ B£®Ôö´ó·´Ó¦ÎïµÄŨ¶È C£®½µµÍÎÂ¶È D£®Ê¹Óô߻¯¼Á

(3)1000 ¡æʱ£¬ÁòËáÄÆÓëÇâÆø·¢ÉúÏÂÁз´Ó¦£ºNa2SO4(s)£«4H2(g)Na2S(s)£«4H2O(g)

¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ________________________________£»

ÒÑÖªK1000 ¡æ<K1200 ¡æ£¬Èô½µµÍÌåϵζȣ¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿½«»á________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)¡£

(4)³£ÎÂÏ£¬Èç¹ûÈ¡0.1 mol¡¤L£­1 HAÈÜÒºÓë0.1 mol¡¤L£­1 NaOHÈÜÒºµÈÌå»ý»ìºÏ(»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ)£¬²âµÃ»ìºÏÒºµÄpH£½8¡£

¢Ù»ìºÏÒºÖÐÓÉË®µçÀë³öµÄOH£­Å¨¶ÈÓë0.1 mol¡¤L£­1 NaOHÈÜÒºÖÐÓÉË®µçÀë³öµÄOH£­Å¨¶ÈÖ®±ÈΪ________£»

¢ÚÒÑÖªNH4AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na2CO3ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶÏ(NH4)2CO3ÈÜÒºµÄpH________7(Ìî¡°<¡±¡°>¡±»ò¡°£½¡±)£»ÏàͬζÈÏ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐËÄÖÖÑÎÈÜÒº°´pHÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòΪ(ÌîÐòºÅ)________¡£

a£®NH4HCO3 b£®NH4A c£®(NH4)2CO3 d£®NH4Cl

 

(1)ijÎÞÉ«Ï¡ÈÜÒºXÖУ¬¿ÉÄܺ¬ÓÐϱíËùÁÐÀë×ÓÖеÄij¼¸ÖÖ¡£

ÒõÀë×Ó

CO32¡ª¡¢SiO32¡ª¡¢AlO2¡ª¡¢Cl£­

ÑôÀë×Ó

Al3£«¡¢Cu2£«¡¢Mg2£«¡¢NH4+¡¢Na£«

 

ÏÖÈ¡¸ÃÈÜÒºÊÊÁ¿£¬ÏòÆäÖмÓÈëijÊÔ¼ÁY£¬²úÉú³ÁµíµÄÎïÖʵÄÁ¿(n)Óë¼ÓÈëÊÔ¼ÁYµÄÌå»ý(V)µÄ¹ØϵÈçͼËùʾ¡£

¢ÙÈôYÊÇÑÎËᣬÔòÈÜÒºÖк¬ÓеĽðÊôÑôÀë×ÓÊÇ_________________________________£¬

ab¶Î·¢Éú·´Ó¦µÄ×ÜÀë×Ó·½³ÌʽΪ___________________________________

±íÖÐOa¶ÎÓëYÈÜÒº·´Ó¦µÄÀë×ÓµÄÎïÖʵÄÁ¿Ö®±ÈΪ__________[Òª±êÃ÷Àë×Ó·ûºÅ£¬Èçn(Na£«)]¡£

¢ÚÈôYÊÇNaOHÈÜÒº£¬Ôòbc¶Î·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________________________

Èô²»¿¼ÂÇÀë×ÓµÄË®½âÒòËØ£¬ºöÂÔH£«ºÍOH£­µÄÓ°Ï죬ÇÒÈÜÒºÖÐÖ»´æÔÚ4ÖÖÀë×Ó£¬ÔòËüÃǵÄÀë×Ó¸öÊý±ÈΪ___________________________________[°´ÑôÀë×ÓÔÚÇ°£¬ÒõÀë×ÓÔں󣬸߼ÛÔÚÇ°£¬µÍ¼ÛÔÚºóµÄ˳ÐòÅÅÁÐ]¡£

(2)ÎýΪµÚ¢ôA×åÔªËØ£¬ÎýµÄµ¥Öʺͻ¯ºÏÎïÓëijЩÎïÖʵĻ¯Ñ§ÐÔÖÊÉÏÓÐÐí¶àÏàËÆÖ®´¦¡£ÒÑÖªÎýÔªËؾßÓÐÈçÏÂÐÔÖÊ£º

Sn4£«£«Sn=2Sn2£«£»

2Sn2£«£«O2£«4H£«=2Sn4£«£«2H2O£»

2H£«£«SnO22¡ªSn(OH)2Sn2£«£«2OH£­¡£

ÊԻشð£º

¢ÙÎýÈÜÓÚÑÎËᣬÔÙÏò·´Ó¦ºóµÄÈÜÒºÖÐͨÈëÂÈÆø£¬Óйط´Ó¦ÀàËÆÓÚÌúµÄÏàÓ¦±ä»¯£¬ÊÔд³öÓйط´Ó¦µÄÀë×Ó·½³Ìʽ£º______________________________________£¬

¢Ú½«¢ÙÖÐÈÜÒºÕô¸Éºó¼ÌÐø¼ÓÈÈËùµÃ¹ÌÌ壬±ä»¯¹ý³ÌÀàËÆÓÚFeCl3ÈÜÒºÏàÓ¦µÄ±ä»¯£¬Ôò×îºóµÃµ½µÄ¹ÌÌåÎïÖÊÊÇ(·Ö×Óʽ)__________¡£

¢ÛÈô¿ÉÓÃSnCl2ÈÜÒºÓë¹ýÁ¿µÄ¼îÈÜÒº·´Ó¦µÄ·½·¨ÖÆSn(OH)2, ¸Ã¼îÊÇ__________¡£

 

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø