ÏÂÁÐÊÂʵ£º¢ÙÃ÷·¯¿É×ö¾»Ë®¼Á£»¢ÚNaHSO4Ë®ÈÜÒº³ÊËáÐÔ£»¢ÛNa2SiO3¡¢Na2CO3¡¢NaAlO2µÈÈÜÒº²»ÄÜÖü´æÔÚÄ¥¿Ú²£Á§Æ¿ÈûµÄÊÔ¼ÁÆ¿ÖУ»¢Üï§Ì¬µª·Ê²»ÄÜÓë²Ýľ»Ò»ìºÏÊ©Ó㻢ݼÓÈÈÄÜʹ´¿¼îÈÜҺȥÎÛÄÜÁ¦ÔöÇ¿£»¢ÞÅäÖÆFeCl3ÈÜÒº£¬ÐèÓÃŨÑÎËáÈܽâFeCl3¹ÌÌ壻¢ßNH4FÈÜÒº²»ÄÜÓò£Á§Æ¿Ê¢·Å£®ÆäÖÐÓëÑÎÀàË®½âÓйصÄÊÇ£¨ £©
A£®È«²¿
B£®³ý¢ßÒÔÍâ
C£®³ý¢ÚÒÔÍâ
D£®³ý¢Ü¡¢¢ÞÒÔÍâ
¡¾´ð°¸¡¿·ÖÎö£ºÑÎÀàË®½âµÄʵÖÊÊÇÑÎÖеÄijЩÀë×ÓÓëË®µçÀë³öµÄH+»òOH-Éú³ÉÈõµç½âÖÊ£¬´Ù½øÁËË®µÄµçÀ룬Ôö´óÁËË®µÄµçÀë¶È£¬Ç¿ËáÈõ¼îÇ¿¼îÈõËáÑÎÒÔ¼°ÈõËáÈõ¼îÑοɷ¢ÉúË®½â£®
½â´ð£º½â£º¢ÙÃ÷·¯¿É×ö¾»Ë®¼ÁÊÇÓÉÓÚÂÁÀë×ÓË®½âÉú³ÉÇâÑõ»¯ÂÁ½ºÌå¾ßÓÐÎü¸½ÐÔ¶ø¾»Ë®£¬ÓëÑÎÀàË®½âÓйأ¬¹Ê¢Ù×÷Ϊ£»
¢ÚNaHSO4Ë®ÈÜÒº³ÊËáÐÔÊÇÓÉÓÚÁòËáÇâ¸ùÀë×ÓµçÀë³öÇâÀë×ÓµÄÔµ¹Ê£¬ÓëÑÎÀàË®½âÎ޹أ¬¹Ê¢Ú´íÎó£»
¢ÛNa2SiO3¡¢Na2CO3¡¢NaAlO2µÈ¶¼ÊÇÇ¿¼îÈõËáÑΣ¬Ë®½â³Ê¼îÐÔ£¬ÔòÈÜÒº²»ÄÜÖü´æÔÚÄ¥¿Ú²£Á§Æ¿ÈûµÄÊÔ¼ÁÆ¿ÖУ¬ÓëÑÎÀàË®½âÓйأ¬¹Ê¢ÛÕýÈ·£»
¢Üï§Ì¬µª·Ê²»ÄÜÓë²Ýľ»Ò»ìºÏÊ©Óã¬ÊÇÓÉÓÚ笠ùÀë×ÓÓë̼Ëá¸ùÀë×Ó·¢Éú»¥´ÙË®½âµÄÔ­Òò£¬¹Ê¢ÜÕýÈ·£»
¢Ý¼ÓÈÈÄÜʹ´¿¼îÈÜҺȥÎÛÄÜÁ¦ÔöÇ¿ÊÇÓÉÓÚ¼ÓÈÈ´Ù½øÑÎÀàµÄË®½â£¬¹Ê¢ÝÕýÈ·£»
¢ÞÅäÖÆFeCl3ÈÜÒº£¬ÐèÓÃŨÑÎËáÈܽâFeCl3¹ÌÌ壬ĿµÄÊÇÒÖÖÆÌúÀë×ÓµÄË®½â£¬·ÀÖ¹ÈÜÒº»ë×Ç£¬ÓëÑÎÀàË®½âÓйأ¬¹Ê¢ÞÕýÈ·£»
¢ßNH4FÈÜÒº²»ÄÜÓò£Á§Æ¿Ê¢·Å£¬ÊÇÓÉÓÚË®½âÉú³ÉHFÄܸ¯Ê´²£Á§£¬¹Ê¢ßÕýÈ·£®
¹ÊÑ¡C£®
µãÆÀ£º±¾Ì⿼²éÑÎÀàË®½âµÄÓ¦Óã¬ÌâÄ¿ÄѶÈÖеȣ¬×¢Òâ³£¼ûÄÜË®½âµÄÑÎÀ࣬°ÑÎÕË®½âÔ­Àí£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÊÂʵ£º¢ÙÃ÷·¯¿É×ö¾»Ë®¼Á£»¢ÚNaHSO4Ë®ÈÜÒº³ÊËáÐÔ£»¢ÛNa2SiO3¡¢Na2CO3¡¢NaAlO2µÈÈÜÒº²»ÄÜÖü´æÔÚÄ¥¿Ú²£Á§Æ¿ÈûµÄÊÔ¼ÁÆ¿ÖУ»¢Üï§Ì¬µª·Ê²»ÄÜÓë²Ýľ»Ò»ìºÏÊ©Ó㻢ݼÓÈÈÄÜʹ´¿¼îÈÜҺȥÎÛÄÜÁ¦ÔöÇ¿£»¢ÞÅäÖÆFeCl3ÈÜÒº£¬ÐèÓÃŨÑÎËáÈܽâFeCl3¹ÌÌ壻¢ßNH4FÈÜÒº²»ÄÜÓò£Á§Æ¿Ê¢·Å£®ÆäÖÐÓëÑÎÀàË®½âÓйصÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨1£©298K¡¢100kPaʱ£¬C£¨s£¬Ê¯Ä«£©+O2£¨g£©¨TCO2£¨g£©¡÷H1=-393.5kJ?mol-1    2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H2=-571.6kJ?mol-1  2C2H2£¨g£©+5O2£¨g£©=4CO2£¨g£©+2H2O£¨l£©¡÷H3=-2599kJ?mol-1£¬Çëд³ö298KʱÓÉC£¨s£¬Ê¯Ä«£©ºÍH2£¨g£©Éú³É1molC2H2£¨g£©µÄÈÈ»¯Ñ§·½³Ìʽ
C£¨s£¬Ê¯Ä«£©+H2£¨g£©=C2H2£¨g£©¡÷H=226.7kJ?mol-1
C£¨s£¬Ê¯Ä«£©+H2£¨g£©=C2H2£¨g£©¡÷H=226.7kJ?mol-1
£»
£¨2£©Ë®ÊÇÉúÃüÖ®Ô´£¬Ò²ÊÇ»¯Ñ§·´Ó¦ÖеÄÖ÷½Ç£®ÊԻشðÏÂÁÐÎÊÌ⣺A£®B£®CÊÇÖÐѧ»¯Ñ§³£¼ûµÄÈýÖÖÓÐÉ«ÎïÖÊ£¨Æä×é³ÉµÄÔªËؾùÊô¶ÌÖÜÆÚÔªËØ£©£¬ËüÃǾùÄÜÓëË®·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬µ«Ë®¼È²»ÊÇÑõ»¯¼ÁÒ²²»ÊÇ»¹Ô­¼Á£¬Çëд³öA£®B£®CÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
3NO2+H2O=2HNO3+NO¡¢Cl2+H2O=HCl+HClO¡¢2Na2O2+2H2O=4NaOH+O2¡ü
3NO2+H2O=2HNO3+NO¡¢Cl2+H2O=HCl+HClO¡¢2Na2O2+2H2O=4NaOH+O2¡ü
£»
£¨3£©Ð´³öÁò»¯ÄÆÔÚË®ÈÜÒºÖÐË®½âµÄÀë×Ó·½³Ìʽ
S2-+H2OHS-+OH-
S2-+H2OHS-+OH-
ÔÚÅäÖÆÁò»¯ÄÆÈÜҺʱ£¬ÎªÁË·ÀÖ¹·¢ÉúË®½â£¬¿ÉÒÔ¼ÓÈëÉÙÁ¿µÄ
NaOH
NaOH
£»
£¨4£©Ã÷·¯¿É×ö¾»Ë®¼ÁÊÇÒòΪ
Al3+Ë®½â²úÉúµÄ½º×´µÄAl£¨OH£©3¾ßÓÐÎü¸½ÐÔ£¬¿ÉÒÔÎü¸½Ë®ÖеÄÔÓÖÊ
Al3+Ë®½â²úÉúµÄ½º×´µÄAl£¨OH£©3¾ßÓÐÎü¸½ÐÔ£¬¿ÉÒÔÎü¸½Ë®ÖеÄÔÓÖÊ
£¬ÓйصÄÀë×Ó·½³ÌʽΪ
Al3++3H2OAl£¨OH£©3+3H+ÓÐÎÞÉ«ÆøÌåºÍ°×É«Ðõ×´³Áµí²úÉú
Al3++3H2OAl£¨OH£©3+3H+ÓÐÎÞÉ«ÆøÌåºÍ°×É«Ðõ×´³Áµí²úÉú
£»ÏòÃ÷·¯µÄË®ÈÜÒºÖмÓÈë±¥ºÍµÄСËÕ´òÈÜÒº£¬Ôò¹Û²ìµ½µÄÏÖÏóÊÇ
ÓÐÎÞÉ«ÆøÌåºÍ°×É«Ðõ×´³Áµí²úÉú
ÓÐÎÞÉ«ÆøÌåºÍ°×É«Ðõ×´³Áµí²úÉú
£¬ÓйصÄÀë×Ó·½³Ìʽ
3HCO3-+Al3+=Al£¨OH£©3¡ý+CO2¡ü
3HCO3-+Al3+=Al£¨OH£©3¡ý+CO2¡ü
£»
£¨5£©ÏÂÁÐÄÄЩÊÂʵÄÜ˵Ã÷´×ËáÊÇÈõËá
¢Ú¢Ü¢Ý¢Þ
¢Ú¢Ü¢Ý¢Þ

¢Ù´×Ëá²»Ò׸¯Ê´Ò·þ£»
¢Ú0.1mol/LµÄCH3COONaÈÜÒºµÄPHԼΪ9£»
¢Û½øÐÐÖк͵ζ¨Ê±£¬µÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄH2SO4ÈÜÒº±ÈµÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄCH3COOHÈÜÒºÏûºÄµÄNaOHÈÜÒº¶à£»
¢Ü0.1mol/LµÄCH3COOHÈÜÒºPHԼΪ2.9£»
¢ÝÏàͬÌå»ýµÄPH¾ùµÈÓÚ4µÄÑÎËáºÍCH3COOHÈÜÒº£¬±»Í¬Ò»ÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒºÖкͣ¬CH3COOHÈÜÒºÏûºÄµÄNaOHÈÜÒº¶à£»
¢Þþ·ÛÓëÒ»¶¨Á¿Ï¡ÁòËá·´Ó¦£¬Èç¹ûÏòÆäÖмÓÈëÉÙÁ¿´×ËáÄÆ¿ÉÒÔ½µµÍ·´Ó¦ËÙÂʵ«²»¸Ä±ä²úÉúÆøÌåµÄ×ÜÁ¿£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÊÂʵ£º¢ÙÃ÷·¯¿É×ö¾»Ë®¼Á£»¢ÚNaHSO4Ë®ÈÜÒº³ÊËáÐÔ£»¢ÛNa2SiO3¡¢Na2CO3¡¢NaAlO2µÈÈÜÒº²»ÄÜÖü´æÔÚÄ¥¿Ú²£Á§Æ¿ÈûµÄÊÔ¼ÁÆ¿ÖУ»¢Üï§Ì¬µª·Ê²»ÄÜÓë²Ýľ»Ò»ìºÏÊ©Ó㻢ݼÓÈÈÄÜʹ´¿¼îÈÜҺȥÎÛÄÜÁ¦ÔöÇ¿£»¢ÞÅäÖÆFeCl3ÈÜÒº£¬ÐèÓÃŨÑÎËáÈܽâFeCl3¹ÌÌ壻¢ßNH4FÈÜÒº²»ÄÜÓò£Á§Æ¿Ê¢·Å£®ÆäÖÐÓëÑÎÀàË®½âÓйصÄÊÇ£¨¡¡¡¡£©
A£®È«²¿B£®³ý¢ßÒÔÍâC£®³ý¢ÚÒÔÍâD£®³ý¢Ü¡¢¢ÞÒÔÍâ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äêɽ¶«Ê¡¼ÃÄÏÍâ¹úÓïѧУ¸ß¿¼»¯Ñ§ÖʼìÊÔ¾í£¨Èý£©£¨½âÎö°æ£© ÌâÐÍ£º½â´ðÌâ

£¨1£©298K¡¢100kPaʱ£¬C£¨s£¬Ê¯Ä«£©+O2£¨g£©¨TCO2£¨g£©¡÷H1=-393.5kJ?mol-1    2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H2=-571.6kJ?mol-1  2C2H2£¨g£©+5O2£¨g£©=4CO2£¨g£©+2H2O£¨l£©¡÷H3=-2599kJ?mol-1£¬Çëд³ö298KʱÓÉC£¨s£¬Ê¯Ä«£©ºÍH2£¨g£©Éú³É1molC2H2£¨g£©µÄÈÈ»¯Ñ§·½³Ìʽ______£»
£¨2£©Ë®ÊÇÉúÃüÖ®Ô´£¬Ò²ÊÇ»¯Ñ§·´Ó¦ÖеÄÖ÷½Ç£®ÊԻشðÏÂÁÐÎÊÌ⣺A£®B£®CÊÇÖÐѧ»¯Ñ§³£¼ûµÄÈýÖÖÓÐÉ«ÎïÖÊ£¨Æä×é³ÉµÄÔªËؾùÊô¶ÌÖÜÆÚÔªËØ£©£¬ËüÃǾùÄÜÓëË®·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬µ«Ë®¼È²»ÊÇÑõ»¯¼ÁÒ²²»ÊÇ»¹Ô­¼Á£¬Çëд³öA£®B£®CÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______£»
£¨3£©Ð´³öÁò»¯ÄÆÔÚË®ÈÜÒºÖÐË®½âµÄÀë×Ó·½³Ìʽ______ÔÚÅäÖÆÁò»¯ÄÆÈÜҺʱ£¬ÎªÁË·ÀÖ¹·¢ÉúË®½â£¬¿ÉÒÔ¼ÓÈëÉÙÁ¿µÄ______£»
£¨4£©Ã÷·¯¿É×ö¾»Ë®¼ÁÊÇÒòΪ______£¬ÓйصÄÀë×Ó·½³ÌʽΪ______£»ÏòÃ÷·¯µÄË®ÈÜÒºÖмÓÈë±¥ºÍµÄСËÕ´òÈÜÒº£¬Ôò¹Û²ìµ½µÄÏÖÏóÊÇ______£¬ÓйصÄÀë×Ó·½³Ìʽ______£»
£¨5£©ÏÂÁÐÄÄЩÊÂʵÄÜ˵Ã÷´×ËáÊÇÈõËá______
¢Ù´×Ëá²»Ò׸¯Ê´Ò·þ£»
¢Ú0.1mol/LµÄCH3COONaÈÜÒºµÄPHԼΪ9£»
¢Û½øÐÐÖк͵ζ¨Ê±£¬µÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄH2SO4ÈÜÒº±ÈµÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄCH3COOHÈÜÒºÏûºÄµÄNaOHÈÜÒº¶à£»
¢Ü0.1mol/LµÄCH3COOHÈÜÒºPHԼΪ2.9£»
¢ÝÏàͬÌå»ýµÄPH¾ùµÈÓÚ4µÄÑÎËáºÍCH3COOHÈÜÒº£¬±»Í¬Ò»ÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒºÖкͣ¬CH3COOHÈÜÒºÏûºÄµÄNaOHÈÜÒº¶à£»
¢Þþ·ÛÓëÒ»¶¨Á¿Ï¡ÁòËá·´Ó¦£¬Èç¹ûÏòÆäÖмÓÈëÉÙÁ¿´×ËáÄÆ¿ÉÒÔ½µµÍ·´Ó¦ËÙÂʵ«²»¸Ä±ä²úÉúÆøÌåµÄ×ÜÁ¿£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸