¡¾ÌâÄ¿¡¿A¡¢B¡¢CÊǵ¥ÖÊ£¬ÆäÖÐAÊǽðÊô£¬¸÷ÖÖÎïÖʼäµÄת»¯¹ØϵÈçͼ£º
¸ù¾Ýͼʾת»¯¹Øϵ»Ø´ð£º
(1)д³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½¡£A_______£¬B___________£¬ÒÒ__________£¬¶¡_______¡£
(2)д³öÏÂÁб仯µÄ·½³Ìʽ¡£
¢ÙAÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________¡£
¢Ú¼×ÓëNaOHÈÜÒº·´Ó¦Éú³ÉÒҺͱûµÄÀë×Ó·½³Ìʽ_______________________¡£
(3)½«Ò»¶¨Á¿µÄA¼ÓÈëµ½NaOHÈÜÒºÖУ¬²úÉúµÄCÔÚ±ê×¼×´¿öϵÄÌå»ýΪ3.36 L£¬ÔòÏûºÄµÄAµÄÎïÖʵÄÁ¿Îª________£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª________¡£
¡¾´ð°¸¡¿Al O2¡¡ NaAlO2 Al(OH)3 2Al£«2NaOH£«2H2O===2NaAlO2£«3H2¡ü Al2O3£«2OH£===2AlO2££«H2O 0.1 mol 0.3 mol
¡¾½âÎö¡¿
AÊǽðÊôµ¥ÖÊ£¬AÄÜÓëNaOHÈÜÒº·´Ó¦£¬½ðÊôµ¥ÖÊÄÜÓëNaOHÈÜÒº·´Ó¦µÄΪAl£¬ÔòAΪAl£»AlÓëNaOHÈÜÒº·´Ó¦Éú³ÉNaAlO2ºÍH2£¬CÊǵ¥ÖÊ£¬CΪH2£¬ÒÒΪNaAlO2£»BÊǵ¥ÖÊ£¬Al+B¡ú¼×£¬¼×+NaOH¡ú±û+NaAlO2£¬H2+B¡ú±û£¬ÔòBΪO2£¬¼×ΪAl2O3£¬±ûΪH2O¡£ÏòÆ«ÂÁËáÄÆÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼Éú³ÉÇâÑõ»¯ÂÁ³Áµí£¬Ôò¶¡ÊÇAl(OH)3£¬¾Ý´ËÅжϡ£
¸ù¾ÝÒÔÉÏ·ÖÎö¿ÉÖªAΪAl£¬BΪO2£¬CΪH2£¬¼×ΪAl2O3£¬ÒÒΪNaAlO2£¬±ûΪH2O£¬¶¡ÊÇAl(OH)3£¬Ôò
£¨1£©AµÄ»¯Ñ§Ê½ÎªAl£¬BµÄ»¯Ñ§Ê½ÎªO2£¬ÒҵĻ¯Ñ§Ê½ÎªNaAlO2£¬¶¡µÄ»¯Ñ§Ê½ÎªAl(OH)3¡£
£¨2£©¢ÙAlÓëNaOHÈÜÒº·´Ó¦Éú³ÉÆ«ÂÁËáÄƺÍÇâÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Al+2NaOH+2H2O£½2NaAlO2+3H2¡ü¡£
¢ÚAl2O3ÓëNaOHÈÜÒº·´Ó¦Éú³ÉÆ«ÂÁËáÄƺÍË®£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪAl2O3£«2OH££½2AlO2££«H2O¡£
£¨3£©n£¨H2£©=3.36L¡Â22.4L/mol=0.15mol£¬¸ù¾Ý·´Ó¦2Al+2NaOH+2H2O£½2NaAlO2+3H2¡ü¿ÉÖªn£¨Al£©=2/3¡Á0.15mol=0.1mol£»·´Ó¦ÏûºÄ1molAlתÒÆ3molµç×Ó£¬Ôò¸Ã·´Ó¦ÖÐתÒƵç×ÓÎïÖʵÄÁ¿Îª0.3mol¡£
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¿¡¿ÏÖÓк¬ÓÐÉÙÁ¿NaCl¡¢Na2SO4¡¢Na2CO3¡¢Äàɳ£¨SiO2£©µÈÔÓÖʵÄNaNO3¹ÌÌ壬ѡÔñÊʵ±µÄÊÔ¼Á³ýÈ¥ÔÓÖÊ£¬µÃµ½´¿¾»µÄNaNO3¹ÌÌ壬ʵÑéÁ÷³ÌÈçÏÂͼËùʾ¡£
£¨1£©²Ù×÷1¡¢²Ù×÷3µÄÃû³Æ·Ö±ðÊÇ_________¡¢_________¡£
£¨2£©ÊÔ¼Á1ºÍÊÔ¼Á2·Ö±ðÊÇ_________¡¢_________¡£¹ÌÌå2ÖгýÁ˺¬ÓÐAgCl£¬»¹ÓÐ_________£¨´ðÈ«£©¡£
£¨3£©ÊÔ¼Á3¼ÓÈëºó·¢ÉúµÄÀë×Ó·´Ó¦·½³ÌʽÊÇ_________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤ÔÁÏ£¬¿ÉÓÃÓںϳɿɽµ½âµÄ¸ß¾ÛÎïPESÊ÷Ö¬ÒÔ¼°¾ßÓп¹Ö×Áö»îÐԵĻ¯ºÏÎïK¡£
ÒÑÖª:i.R1CO18OR2+R3OHR1COOR3+R218OH
ii.
¢£. (R1¡¢R2¡¢R3´ú±íÌþ»ù)
£¨1£©AµÄÃû³ÆÊÇ_________£»CµÄ¹ÙÄÜÍŵÄÃû³ÆÊÇ_________¡£
£¨2£©B·Ö×ÓΪ»·×´½á¹¹£¬ºË´Å¹²ÕñÇâÆ×Ö»ÓÐÒ»×é·å£¬BµÄ½á¹¹¼òʽΪ_________ .
£¨3£©E·Ö×ÓÖк¬ÓÐÁ½¸öõ¥»ù£¬ÇÒΪ˳ʽ½á¹¹£¬EµÄ½á¹¹¼òʽΪ_________ .
£¨4£©·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽΪ_________ ¡£
£¨5£©ÊÔ¼ÁaµÄ½á¹¹¼òʽΪ_________£»·´Ó¦¢ÚËùÊôµÄ·´Ó¦ÀàÐÍΪ________·´Ó¦¡£
£¨6£©ÒÑÖª: ¡£ÒÔ1£¬3-¶¡¶þϩΪÆðʼÔÁÏ£¬½áºÏÒÑÖªÐÅϢѡÓñØÒªµÄÎÞ»úÊÔ¼ÁºÏ³É¡£½«ÒÔϺϳÉ·Ïß²¹³äÍêÕû:_________________
£¨7£©ÒÑÖª°±»ù(-NH2)ÓëôÇ»ùÀàËÆ£¬Ò²ÄÜ·¢Éú·´Ó¦i¡£ÔÚÓÉJÖƱ¸KµÄ¹ý³ÌÖУ¬³£»á²úÉú¸±²úÎïL¡£L·Ö×ÓʽΪC16H13NO3£¬º¬Èý¸öÁùÔª»·£¬ÔòLµÄ½á¹¹¼òʽΪ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¸ßÌúËáÑÎÔÚË®ÈÜÒºÖÐÓÐËÄÖÖº¬ÌúÐÍÌ壬25¡æʱ£¬ËüÃǵÄÎïÖʵÄÁ¿·ÖÊýËæpHµÄ±ä»¯ÈçͼËùʾ¡£ÏÂÁÐÐðÊö´íÎóµÄÊÇ
A. ÒÑÖªH3FeO4+µÄµçÀë°ëºâ³£Êý·Ö±ðΪ£ºK1=2.5¡Á10-2£¬K2=4.8¡Á10-4£¬K3=5.0¡Á10-8£¬µ±pII=4…¼£¬ÈÜÒºÖÐ
B. Ϊ»ñµÃ¾¡¿ÉÄÜ´¿¾»µÄ¸ßÌúËáÑΣ¬Ó¦¿ØÖÆpH¡Ý9
C. ÏòpH=5µÄ¸ßÌúËáÑÎÈÜÒºÖмÓÈëKOHÈÜÒº£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪHFeO4¡¥+OH¡¥=FeO42-+H2O
D. pH=2ʱ£¬ÈÜÒºÖÐÖ÷Òªº¬ÌúÐÍÌåŨ¶ÈµÄ´óС¹ØϵΪc(H2FeO4)>c(H3FeO4+)> c (HFeO4¡¥)
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¹¤ÒµÉÏ¿Éͨ¹ýúµÄÒº»¯ºÏ³É¼×´¼£¬Ö÷·´Ó¦Îª:
CO(g)+2H2(g)CH3OH(l) ¡÷H=x¡£
(1)ÒÑÖª³£ÎÂÏÂCH3OH¡¢H2ºÍCO µÄȼÉÕÈÈ·Ö±ðΪ726.5 kJ/mol¡¢285.5 kJ/mol¡¢283.0 k J/mol£¬Ôòx=____£»ÎªÌá¸ßºÏ³É¼×´¼·´Ó¦µÄÑ¡ÔñÐÔ£¬¹Ø¼üÒòËØÊÇ__________________________¡£
(2)TKÏ£¬ÔÚÈÝ»ýΪ1.00 LµÄijÃܱÕÈÝÆ÷ÖнøÐÐÉÏÊö·´Ó¦,Ïà¹ØÊý¾ÝÈçͼһ¡£
¢Ù¸Ã»¯Ñ§·´Ó¦0~10 minµÄƽ¾ùËÙÂÊv(H2)=_______£»MºÍNµãµÄÄæ·´Ó¦ËÙÂʽϴóµÄÊÇ_____(Ìî¡°vÄæ(M)¡±¡¢¡°vÄæ(N)¡±»ò¡°²»ÄÜÈ·¶¨¡±)¡£
¢Ú10 minʱÈÝÆ÷ÄÚCOµÄÌå»ý·ÖÊýΪ______¡£
¢Û¶ÔÓÚÆøÏà·´Ó¦£¬³£ÓÃij×é·Ö(B)µÄƽºâѹǿ(pB)´úÌæÎïÖʵÄÁ¿Å¨¶È(cB)±íʾƽºâ³£Êý(ÒÔKP±íʾ)£¬ÆäÖУ¬pB=p×Ü¡ÁBµÄÌå»ý·ÖÊý£»ÈôÔÚTK ÏÂƽºâÆøÌå×ÜѹǿΪx atm£¬Ôò¸Ã·´Ó¦KP=____(¼ÆËã±í´ïʽ)¡£ÊµÑé²âµÃ²»Í¬Î¶ÈϵÄlnK(»¯Ñ§Æ½ºâ³£ÊýK µÄ×ÔÈ»¶ÔÊý)Èçͼ¶þ£¬Çë·ÖÎö1nK ËæT³ÊÏÖÉÏÊö±ä»¯Ç÷ÊƵÄÔÒòÊÇ________________¡£
(3)¸ÉÔïµÄ¼×´¼¿ÉÓÃÓÚÖÆÔìȼÁϵç³Ø¡£
¢ÙÑо¿ÈËÔ±·¢ÏÖÀûÓÃNaOH ¸ÉÔï¼×´¼Ê±£¬Î¶ȿØÖƲ»µ±»áÓм×ËáÑκÍH2Éú³É£¬Æä·´Ó¦·½³ÌʽΪ______________________________£»
¢Úij¸ßУÌá³öÓÃCH3OH-O2ȼÁϵç³Ø×÷µçÔ´µç½â´¦ÀíË®Ä೧²úÉúµÄCO2(ÒÔÈÛÈÚ̼ËáÑÎΪ½éÖÊ)£¬²úÎïΪC ºÍO2¡£ÆäÑô¼«µç¼«·´Ó¦Ê½Îª___________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿½«N2¡¢H2µÄ»ìºÏÆøÌå·Ö±ð³äÈë¼×¡¢ÒÒ¡¢±ûÈý¸öÈÝÆ÷ÖУ¬½øÐкϳɰ±·´Ó¦£¬¾¹ýÏàͬµÄÒ»¶Îʱ¼äºó£¬²âµÃ·´Ó¦ËÙÂÊ·Ö±ðΪ£º¼×£ºv£¨H2£©£½3 mol¡¤L£1¡¤min£1£»
ÒÒ£ºv£¨N2£©£½2 mol¡¤L£1¡¤min£1£»±û£ºv£¨NH3£©£½ 1 mol¡¤L£1¡¤min£1¡£
ÔòÈý¸öÈÝÆ÷Öкϳɰ±µÄ·´Ó¦ËÙÂÊ£¨ £©
A. v(¼×)£¾v(ÒÒ)£¾v(±û£© B. v(ÒÒ)£¾v(±û)£¾v(¼×£©
C. v(±û)£¾v(¼×)£¾v(ÒÒ) D. v(ÒÒ)£¾v(¼×)£¾v(±û£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¹ý̼ËáÄÆ(Na2CO4)ÊÇÒ»ÖֺܺõĹ©Ñõ¼Á£¬ÆäÓëÏ¡ÑÎËá·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2Na2CO4+4HCl=4NaCl+2CO2¡ü+O2¡ü+2H2O¡£ÊÐÊÛ¹ý̼ËáÄÆÒ»°ã¶¼º¬ÓÐ̼ËáÄÆ£¬Îª²â¶¨Ä³¹ý̼ËáÄÆÑùÆ·(Ö»º¬Na2CO4ºÍNa2CO3)µÄ´¿¶È£¬Ä³»¯Ñ§ÐËȤС×é²ÉÓÃÒÔÏÂÁ½ÖÖ·½°¸ÊµÊ©£º
·½°¸Ò»£º
(1)²Ù×÷¢ÙºÍ¢ÛµÄÃû³Æ·Ö±ðΪ_______¡¢________¡£
(2)ÉÏÊö²Ù×÷ÖУ¬Ê¹Óõ½²£Á§°ôµÄÓÐ________(Ìî²Ù×÷ÐòºÅ)¡£
(3)Çë¼òÊö²Ù×÷¢ÛµÄ²Ù×÷¹ý³Ì___________¡£
·½°¸¶þ£º°´ÏÂͼ×é×°ºÃʵÑé×°Öã¬QΪһ¿É¹ÄÕ͵ÄËÜÁÏÆø´ü£¬È¡ÊÊÁ¿ÑùÆ·ÓÚÆäÖУ¬´ò¿ª·ÖҺ©¶·»îÈû£¬½«Ï¡ÑÎËáµÎÈëÆø´üÖÐÖÁ³ä·Ö·´Ó¦
(4)Ϊ²â¶¨·´Ó¦Éú³ÉÆøÌåµÄ×ÜÌå»ý£¬µÎÏ¡ÑÎËáÇ°±ØÐë¹Ø±Õ____´ò¿ª____ (¾ùÌî¡°K1¡±¡¢¡°K2¡±»ò¡°K3¡±)£»µ¼¹ÜAµÄ×÷ÓÃÊÇ________¡£
(5)µ±ÉÏÊö·´Ó¦Í£Ö¹ºó£¬Ê¹K1¡¢K3´¦ÓڹرÕ״̬£¬K2´¦ÓÚ´ò¿ª×´Ì¬£¬ÔÙ»º»º´ò¿ªK1¡£BÖÐ×°µÄ¹ÌÌåÊÔ¼ÁÊÇ_________£¬ÎªÊ²Ã´Òª»º»º´ò¿ªK1?_______________¡£
(6)ʵÑé½áÊøʱ£¬Á¿Í²1ÖÐÓÐxmLË®£¬Á¿Í²IIÖÐÊÕ¼¯µ½ÁËymLÆøÌ壬ÔòÑùÆ·Öйý̼ËáÄƵÄÖÊÁ¿·ÖÊýÊÇ________(Óú¬ÓÐx¡¢yµÄ´úÊýʽ±íʾ)
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿·´Ó¦4NH3£¨g£©£«5O2£¨g£©4NO£¨g£©£«6H2O£¨g£©ÔÚ5LÃܱÕÈÝÆ÷ÖнøÐУ¬°ë·ÖÖÓºóNOµÄÎïÖʵÄÁ¿Ôö¼ÓÁË0.3mol£¬Ôò´Ë·´Ó¦µÄƽ¾ùËÙÂÊvΪ
A. v£¨O2£©£½0.01mol/£¨L¡¤s£© B. v£¨NO£©£½0.08mol/£¨L¡¤s£©
C. v£¨H2O£©£½0.003mol/£¨L¡¤s£© D. v£¨NH3£©£½0.001mol/£¨L¡¤s£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿·´Ó¦A£¨g£©+3B£¨g£©2C£¨g£©+2D£¨g£©£¬ÔÚËÄÖÖ²»Í¬Çé¿öÏÂÓò»Í¬ÎïÖʱíʾµÄ·´Ó¦ËÙÂÊ·Ö±ðÈçÏ£¬ÆäÖз´Ó¦ËÙÂÊ×î´óµÄÊÇ£¨ £©
A. v£¨A£©=0.15mol/£¨Lmin£©
B. v£¨B£©=0.04mol/£¨Ls£©
C. v£¨C£©=0.03mol/£¨Ls£©
D. v£¨D£©=0.4mol/£¨Lmin£©
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