¡¾ÌâÄ¿¡¿²¿·ÖÈõËáµÄµçÀëƽºâ³£ÊýÈç±í£º
£¨1£©ÊÒÎÂÏ¢Ù0.1 mol¡¤L£1 HCOONa£¬¢Ú0.1 mol¡¤L£1 NaClO£¬¢Û0.1 mol¡¤L£1 Na2CO3£¬¢Ü0.1 mol¡¤L£1
NaHCO3ÈÜÒºµÄpHÓÉ´óµ½Ð¡µÄ¹ØϵΪ_____________________________________¡£
£¨2£©Å¨¶È¾ùΪ0.1 mol¡¤L£1µÄNa2SO3ºÍNa2CO3µÄ»ìºÏÈÜÒºÖУ¬SO¡¢CO¡¢HSO¡¢HCOŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ___________________________________¡£
£¨3£©ÏÂÁÐÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ________(Ìî×Öĸ)¡£
a.2ClO££«H2O£«CO2===2HClO£«CO b.2HCOOH£«CO===2HCOO££«H2O£«CO2¡ü
c.H2SO3£«2HCOO£===2HCOOH£«SO d.Cl2£«H2O£«2CO===2HCO£«Cl££«ClO£
£¨4£©Ä³Î¶È(T ¡æ)ϵÄÈÜÒºÖУ¬c(H£«)£½10£xmol¡¤L£1£¬c(OH£)£½10£y mol¡¤L£1£¬xÓëyµÄ¹ØϵÈçͼËùʾ.
¢Ù ´ËζÈÏ£¬0.01mol/LµÄNaOHÈÜÒºÖÐË®µçÀë³öµÄOH-Ũ¶ÈΪ_____¡£
¢ÚÔÚ´ËζÈÏ£¬0.1 mol¡¤L£1µÄNaHSO4ÈÜÒºÓë0.1 mol¡¤L£1µÄBa(OH)2ÈÜÒº°´Ï±íÖмס¢ÒÒ¡¢±û¡¢¶¡²»Í¬·½Ê½»ìºÏ£º
¼× | ÒÒ | ±û | ¶¡ | |
0.1 mol¡¤L£1Ba(OH)2ÈÜÒºÌå»ý/mL | 10 | 10 | 10 | 10 |
0.1 mol¡¤L£1NaHSO4ÈÜÒºÌå»ý/mL | 5 | 10 | 15 | 20 |
°´¶¡·½Ê½»ìºÏºó£¬ËùµÃÈÜÒºÏÔ________(Ìî¡°Ëᡱ¡°¼î¡±»ò¡°ÖС±)ÐÔ£®Ð´³ö°´ÒÒ·½Ê½»ìºÏºó£¬·´Ó¦µÄÀë×Ó·½³Ìʽ£º__________________________¡£°´¼×·½Ê½»ìºÏºó£¬ËùµÃÈÜÒºµÄpHΪ________¡£
¡¾´ð°¸¡¿ 3241 SO32->CO32-> HCO3- >HSO3- bd 10-10mol/L ÖÐÐÔ ÖÐÐÔ 11
¡¾½âÎö¡¿£¨1£©ËÄÖÖÑÎÈÜÒº¾ùΪǿ¼îÈõËáÑΣ¬Ë®½â¾ùÏÔ¼îÐÔ£¬ÐγɸÃÑεÄËáÔ½Èõ£¬¸ÃÑεÄË®½âÄÜÁ¦Ô½Ç¿£¬¼îÐÔ¾ÍԽǿ£»´Ó±íÖеóöËáÐÔ´óС¹ØϵΪ£ºHCOOH>H2CO3>HClO>HCO3-,ËùÒÔÑÎÈÜÒºµÄpHÓÉ´óµ½Ð¡µÄ¹ØϵΪ¢Û>¢Ú>¢Ü>¢Ù£»ÕýÈ·´ð°¸£º¢Û>¢Ú>¢Ü>¢Ù¡£
£¨2£©¸ù¾ÝÁ½ÖÖÑÎÈÜÒº¾ùΪǿ¼îÈõËáÑΣ¬Ë®½â¾ùÏÔ¼îÐÔ£¬ÐγɸÃÑεÄËáÔ½Èõ£¬¸ÃÑεÄË®½âÄÜÁ¦Ô½Ç¿£¬¼îÐÔ¾ÍԽǿ£»´Ó±íÖеóöËáÐÔ´óС¹ØϵΪ£ºHSO3->HCO3-,ËùÒÔNa2CO3ÈÜҺˮ½âÄÜÁ¦Ç¿£¬Ê£ÓàµÄc£¨CO£©Å¨¶ÈС£¬Éú³ÉµÄc(HCO)¶à£¬Òò´ËËÄÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºSO32->CO32-> HCO3- >HSO3- £»ÕýÈ·´ð°¸£ºSO32->CO32-> HCO3- >HSO3-¡£
£¨3£©ÓÉÓÚËáÐÔH2CO3>HClO>HCO3-,¸ù¾ÝÇ¿ËáÖƱ¸ÈõËá¹æÂÉ£¬ClO££«H2O£«CO2===HClO£«HCO3-£¬a´íÎó£»ÓÉÓÚËáÐÔHCOOH> H2CO3£¬¸ù¾ÝÇ¿ËáÖƱ¸ÈõËá¹æÂÉ£¬·´Ó¦¿ÉÒÔ·¢Éú£¬bÕýÈ·£»ÓÉÓÚËáÐÔ.H2SO3> HCOOH> HSO3-£¬H2SO3£« HCOO£===HCOOH£«HSO3-£¬c´íÎó£»ÓÉÓÚËáÐÔHCl>H2CO3>HClO>HCO3-,¸Ã·´Ó¦Äܹ»·¢Éú£¬dÕýÈ·£»ÕýÈ·Ñ¡Ïîbd¡£
£¨4£©¢Ù¸ù¾ÝͼÏñ¿ÉÖª£ºKW=10-12£¬¸ù¾Ý¹«Ê½£º£¨c(OH-)(¼î)+ c(OH-)(Ë®)£©¡Ác(H£«)£¨Ë®£©= KW=10-12£¬£¨0.01+ c(OH-)(Ë®)£©¡Ác(H£«)£¨Ë®£©=10-12£¬ÓÉÓÚc(OH-)(Ë®)= c(H£«)£¨Ë®£©£¬ËùÒÔ½üËƼÆËãµÃµ½c(OH-)(Ë®)= c(H£«)£¨Ë®£©=10-10 mol¡¤L£1£»ÕýÈ·´ð°¸10-10 mol¡¤L£1¡£
¢Ú0.1 mol¡¤L£1µÄNaHSO4ÈÜÒº20 mLºÍ0.1 mol¡¤L£1µÄBa(OH)2ÈÜÒº10 mL»ìºÏºó£¬ÇâÀë×ÓºÍÇâÑõ¸ùÀë×ÓÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉÁòËáÄÆ¡¢ÁòËá±µºÍË®£¬ÈÜÒº³ÊÖÐÐÔ£»°´ÒÒ·½Ê½»ìºÏºó
ÇâÑõ¸ùÀë×ÓÓÐÊ£Ó࣬ÈÜÒºÏÔ¼îÐÔ£¬Àë×Ó·½³ÌʽΪBa2++SO42-+H++OH-=BaSO4¡ý+H2O£»¼×·½Ê½»ìºÏºó£¬ÈÜÒºÏÔ¼îÐÔ£¬Ê£Óàc(OH-)=£¨0.1¡Á2¡Á10¡Á10-3-0.1¡Á5¡Á10-3£©/(10+5)¡Á10-3=0.1 mol¡¤L£1,ÓÉÓÚKW=10-12£¬ËùÒÔc(H£«)=10-11 mol¡¤L£1, ËùµÃÈÜÒºµÄpHΪ11£»ÕýÈ·´ð°¸£ºÖÐÐÔ£»11¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿·Ç½ðÊôµ¥ÖʾÈçͼËùʾµÄ¹ý³Ìת»¯Îªº¬ÑõËá,ÒÑ֪ΪǿËá,Çë»Ø´ðÏÂÁÐÎÊÌâ:
(1)ÈôÔÚ³£ÎÂÏÂΪ¹ÌÌå, ÊÇÄÜʹƷºìÈÜÒºÍÊÉ«µÄÓд̼¤ÐÔÆøζµÄÎÞÉ«ÆøÌå¡£
¢ÙµÄ»¯Ñ§Ê½ÊÇ__________¡£
¢ÚÔÚ¹¤ÒµÉú²úÖÐ, ÆøÌåµÄ´óÁ¿Åŷű»ÓêË®ÎüÊÕºóÐγÉÁË__________¶øÎÛȾÁË»·¾³¡£
(2)ÈôÔÚ³£ÎÂÏÂΪÆøÌå, ÊǺì×ØÉ«µÄÆøÌå¡£
¢ÙµÄ»¯Ñ§Ê½ÊÇ__________¡£
¢ÚµÄŨÈÜÒºÔÚ³£ÎÂÏ¿ÉÓëÍ·´Ó¦²¢Éú³ÉÆøÌå,Çëд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿½«11.2gÌúͶÈë200mLijŨ¶ÈµÄÑÎËáÖУ¬ÌúºÍÑÎËáÇ¡ºÃÍêÈ«·´Ó¦¡£Çó£º
£¨1£©11.2gÌúµÄÎïÖʵÄÁ¿
£¨2£©ËùÓÃÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶È
£¨3£©·´Ó¦ÖÐÉú³ÉµÄH2ÔÚ±ê×¼×´¿öϵÄÌå»ý
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¼×¡¢ÒÒÁ½·ÝµÈÖÊÁ¿µÄ̼ËáÇâÄƾ§Ì壬½«¼×ÓÃÛáÛö³ä·Ö¼ÓÈȺóÀäÈ´ºó£¬½«¹ÌÌåÍêȫתÒƵ½Ò»ÊÔ¹ÜÖУ¬ ÔÙ¼ÓÈë×ãÁ¿µÄÑÎË᣻ÒÒ²»¾¼ÓÈÈÖÃÓÚÁíÒ»ÊÔ¹ÜÖУ¬Ò²¼ÓÈë×ãÁ¿µÄÑÎËá¡£·´Ó¦ÍêÈ«ºó£¬Á½ÊÔ¹ÜÖÐʵ¼Ê²Î ¼Ó·´Ó¦µÄHC1ÖÊÁ¿Ö®±ÈΪ( )
A. 2:3 B. 2:1 C. 1£º2 D. 1:1
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÖÓк¬ÓÐÉÙÁ¿KC1¡¢K2SO4¡¢K2CO3ÔÓÖʵÄKNO3ÈÜÒº£¬Ñ¡ÔñÊʵ±µÄÊÔ¼Á³ýÈ¥ÔÓÖÊ£¬µÃµ½´¿¾»µÄKNO3¹ÌÌ壬ʵÑéÁ÷³ÌÈçÏÂͼËùʾ¡£
×¢£ºKNO3¹ÌÌåÈÝÒ×ÊÜÈÈ·Ö½â
(1)³ÁµíAµÄÖ÷Òª³É·ÖÊÇ___________¡¢___________(Ìѧʽ)£»
(2)ΪÁ˳ýÈ¥ÈÜÒº3ÖеÄÔÓÖÊ£¬¿ÉÒÔÏòÆäÖмÓÈëÊÊÁ¿µÄ___________£»³ýÔÓºó´ÓÈÜÒº3»ñµÃKNO3¾§ÌåµÄ²Ù×÷ÊÇ___________¡¢___________¡¢¹ýÂË£»
(3)²½Öè¢Û¼ÓÈë¹ýÁ¿K2CO3ÈÜÒºµÄÄ¿µÄÊÇ___________£»
(4)ʵÑéÊÒÓÃÉÏÊöʵÑé»ñµÃµÄKNO3ÌåÅäÖÆ450mL0.40 mol/L KNO3ÈÜÒº£¬Ðè³ÆÈ¡KNO3¹ÌÌåµÄÖÊÁ¿ÊÇ_____g£»
(5)ÏÂÁвÙ×÷»áµ¼ÖÂËùÅäÈÜҺŨ¶ÈÆ«´óµÄÊÇ£¨___________£©
A.ʹÓÃÁËÉúÐâµÄíÀÂë
B.¶¨ÈÝʱÑöÊӿ̶ÈÏß
C.ÈÝÁ¿Æ¿Ê¹ÓÃÇ°ÓÃÕôÁóˮϴ¾»µ«Ã»ÓиÉÔï
D.¹ÌÌåÈܽâºóδÀäÈ´µ½ÊÒξÍתÒƵ½ÈÝÁ¿Æ¿ÖÐ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÓÃÏÂÁÐʵÑé×°ÖýøÐÐÏàӦʵÑ飬Éè¼ÆÕýÈ·ÇÒÄܴﵽʵÑéÄ¿µÄµÄÊÇ
A. ¼×ÓÃÓÚʵÑéÊÒÖÆÈ¡ÉÙÁ¿CO2 B. ÒÒÓÃÓÚÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÁòËá
C. ±ûÓÃÓÚÄ£ÄâÉúÌúµÄµç»¯Ñ§¸¯Ê´ D. ¶¡ÓÃÓÚÕô¸ÉA1Cl3ÈÜÒºÖƱ¸ÎÞË®AlC13
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿È«·°ÒºÁ÷µç³Ø×°ÖÃÈçͼ£¬µç½âÒºÔÚµç½âÖÊ´¢¹ÞºÍµç³Ø¼ä²»¶ÏÑ»·¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ
A. ³äµçʱ£¬ÇâÀë×Óͨ¹ý½»»»Ä¤ÒÆÏòÓÒ²à
B. ³äµçʱ£¬µçÔ´¸º¼«Á¬½Óaµç¼«
C. ·ÅµçʱװÖ÷¢ÉúµÄ×Ü·´Ó¦Îª£ºVO2+£«V2+£«2H+£½VO2+£«V3+£«H2O
D. ÖÊ×Ó½»»»Ä¤¿É×èÖ¹VO2+ÓëV2+Ö±½Ó·¢Éú·´Ó¦
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Í¼Ó÷ÖÀà·¨±íʾÁËһЩÎïÖÊ»ò¸ÅÄîÖ®¼äµÄ´ÓÊô»ò°üº¬¹Øϵ£¬²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
X | Y | Z | |
A | NaAlO2 | ÑÎ | ´¿¾»Îï |
B | ½ºÌå | ·Öɢϵ | »ìºÏÎï |
C | Al2O3 | Á½ÐÔÑõ»¯Îï | Ñõ»¯Îï |
D | µ¥ÖʲÎÓë·´Ó¦ | Öû»·´Ó¦ | Ñõ»¯»¹Ô·´Ó¦ |
A. A B. B C. C D. D
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿°±Æø¼°º¬µª»¯ºÏÎïÔÚ»¯¹¤Éú²úºÍ¹ú·À¹¤ÒµÖоßÓй㷺ӦÓá£Çë»Ø´ð£º
£¨1£©ÒÑÖª£º(i)ÇâÆøµÄȼÉÕÈÈΪ286.0 kJ¡¤mol-1
(ii)4NH3(g)+3O2(g)2N2(g)+6H2O (l) ¦¤H=- 1530.6 kJ¡¤mol-1¡£
ºÏ³É°±·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_____________________________¡£
£¨2£©ºãκãÈÝÌõ¼þÏ£¬Æðʼ°´ÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã1ÏòÃܱÕÈÝÆ÷ÖгäÈëN2(g)ºÍH2(g)£¬·¢ÉúºÏ³É°±µÄ·´Ó¦¡£´ïƽºâºó£¬N2(g)µÄÌå»ý·ÖÊýΪ_______________£»È»ºóÖ»½µµÍζȣ¬N2(g)µÄÌå»ý·ÖÊý»á_________£¨ÌîÑ¡Ïî×Öĸ£©¡£
A.Ôö´ó B.¼õС C.²»±ä D.²»ÄÜÅжÏ
£¨3£©T¡æʱ£¬CO2(g)ºÍNH3(g)ºÏ³ÉÄòËصÄÔÀíΪ2NH3(g)+CO2(g)CO(NH2)2(s)+H2O(1)¡£ÔÚ2 LºãÈÝÃܱÕÈÝÆ÷ÖУ¬Í¨Èë1.2 mol NH3(g)ºÍ0.6 mol CO2(g)£¬2 minʱ·´Ó¦Ç¡ºÃ´ïµ½Æ½ºâ£¬²âµÃc(NH3)=0.2 mol¡¤L-1¡£
¢Ù0~2minÄÚ£¬ÓÃNH3±íʾµÄ·´Ó¦ËÙÂʦÔ(NH3)=___________£»·´Ó¦µÄƽºâ³£ÊýK=____________¡£
¢ÚÈôÆäËûÌõ¼þ²»±ä£¬2 minʱ½«ÈÝÆ÷Ìå»ýѸËÙѹËõµ½1L£¬ÔÚ3 minʱÖØдﵽƽºâ£¬ÇëÔÚͼ1Öл³ö2~3 minÄÚc(NH3)Ëæʱ¼ä(t)±ä»¯µÄÇúÏß¹Øϵͼ¡£
£¨4£©¼îÐÔ°±ÆøȼÁϵç³ØµÄ×°ÖÃÈçͼ2 Ëùʾ£¬Ð´³ö¸º¼«µÄµç¼«·´Ó¦Ê½____________________¡£µ±µç·ÖÐÿͨ¹ý3.6 mol e-£¬ÔòÐèÒª±ê¿öÏ¿ÕÆøµÄÌå»ý________________L¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com