ÌâÄ¿ÄÚÈÝ
7£®¹ýÑõ»¯Ã¾MgO2Ò×ÈÜÓÚÏ¡ËᣬÈÜÓÚËáºó²úÉú¹ýÑõ»¯Ç⣬ÔÚҽѧÉÏ¿É×÷Ϊ½âËá¼ÁµÈ£®¹ýÑõ»¯Ã¾²úÆ·Öг£»ìÓÐÉÙÁ¿MgO£¬ÊµÑéÊÒ¿Éͨ¹ý¶àÖÖ·½°¸²â¶¨ÑùÆ·¹ýÑõ»¯Ã¾µÄº¬Á¿£®£¨1£©Ä³Ñо¿Ð¡×éÄâÓÃÈçͼװÖòⶨһ¶¨ÖÊÁ¿µÄÑùÆ·ÖйýÑõ»¯Ã¾µÄº¬Á¿£®
¢ÙÏ¡ÑÎËáÖмÓÈëÉÙÁ¿FeCl3ÈÜÒºµÄ×÷ÓÃÊÇÓÃ×÷´ß»¯¼Á£¨»ò´ß»¯H2O2µÄ·Ö½â£©£¬MgO2µÄµç×ÓʽΪ£®
¢Ú·ÖҺ©¶·É϶ËÁ¬½ÓµÄÏðƤ¹ÜµÄ×÷ÓÃÓÐÁ½¸ö£º
Ò»£ºÊ¹·ÖҺ©¶·ÖеÄÈÜҺ˳ÀûµÎÏ£»
¶þ£ºÏû³ýµÎÈëÈÜÒºµÄÌå»ý¶ÔËù²âÆøÌåÌå»ýµÄÓ°Ï죮
£¨2£©ÊµÑéÊÒ»¹¿Éͨ¹ýÏÂÁÐÁ½ÖÖ·½°¸²â¶¨ÑùÆ·ÖйýÑõ»¯Ã¾µÄº¬Á¿£º
·½°¸I£ºÈ¡mgÑùÆ·£¬¼ÓÈë×ãÁ¿Ï¡ÑÎËᣬ³ä·Ö·´Ó¦ºóÔÙ¼ÓÈëNaOHÈÜÒºÖÁMg2+³ÁµíÍêÈ«£¬¹ýÂË¡¢Ï´µÓºó£¬½«ÂËÔü³ä·Ö×ÆÉÕ£¬×îÖյõ½ng¹ÌÌ壮
¢ÙÒÑÖª³£ÎÂÏÂKsp[Mg£¨OH£©2]=1.0¡Á10-11£®ÎªÊ¹·½°¸IÖÐMg2+ÍêÈ«³Áµí[¼´ÈÜÒºÖÐc £¨Mg2+£©¡Ü10-5mol•L-1]£¬ÈÜÒºµÄpHÖÁÉÙÓ¦µ÷ÖÁ11£®·½°¸IÖйýÑõ»¯Ã¾µÄÖÊÁ¿·ÖÊýΪ$\frac{7£¨m-n£©}{2m}$£¨Óú¬m¡¢nµÄ±í´ïʽ±íʾ£©£®
·½°¸II£º³ÆÈ¡0.56gÑùÆ·ÖÃÓÚµâÁ¿Æ¿ÖУ¬¼ÓÈëKIÈÜÒººÍ×ãÁ¿ÑÎËᣬҡÔȺóÔÚ°µ´¦¾²ÖÃ5min£¬È»ºóÓÃ0.1mol•L-11Na2S2O3ÈÜÒºµÎ¶¨£¬µÎ¶¨µ½ÖÕµãʱ¹²ÏûºÄVmL Na2S2O3ÈÜÒº£®
£¨ÒÑÖª£ºI2+2Na2S2O3=Na2S4O6+2NaI£©
¢Ú·½°¸IIÖмÓÈë0.1mol•L-1KIÈÜÒºµÄÌå»ýÖÁÉÙΪ200mL£¬µÎ¶¨Ç°Ðè¼ÓÈëÉÙÁ¿µí·ÛÈÜÒº×÷ָʾ¼Á£¬ÑùÆ·ÖйýÑõ»¯Ã¾µÄÖÊÁ¿·ÖÊýΪ0.005V£¨Óú¬VµÄ±í´ïʽ±íʾ£©£®
·ÖÎö £¨1£©¢Ùͨ¹ý²âÁ¿Ë«ÑõË®·Ö½âÉú³ÉµÄÑõÆø£¬¿É¼ì²â¹ýÑõ»¯Ã¾µÄº¬Á¿£¬¶øÂÈ»¯ÌúÄÜ×÷´ß»¯¼Á¼Ó¿ìË«ÑõË®µÄ·Ö½â£¬¼Ó¿ì·´Ó¦ËÙÂÊ£¬¹ýÑõ»¯Ã¾ÊÇÀë×Ó»¯ºÏÎ¾Ý´Ëд³öµç×Óʽ£»
¢ÚÆøÌåµÄÌå»ýÊÜζȺÍѹǿӰÏì´ó£¬·ÖҺ©¶·É϶ËÁ¬½ÓµÄÏðƤ¹ÜµÄ×÷ÓÃÊDZ£³Öºãѹ£¬Óúãѹ·ÖҺ©¶·µÄÓŵ㻹ÓÐÏû³ýµÎÈëÈÜÒºµÄÌå»ý¶ÔËù²âÆøÌåÌå»ýµÄÓ°Ï죬ʹ·ÖҺ©¶·ÖеÄÈÜҺ˳ÀûµÎÏ£»
£¨2£©·½°¸I¸ù¾ÝÇâÑõ»¯Ã¾µÄÈܶȻý£¬¼ÆËã³öÈÜÒºÖÐÇâÑõ¸ùÀë×ÓµÄŨ¶È£¬Çó³öÈÜÒºµÄPH£»ÖÊÁ¿nÊÇÑõ»¯Ã¾£¬ÖÊÁ¿mÊǹýÑõ»¯Ã¾ºÍÑõ»¯Ã¾ÔÓÖÊÖÊÁ¿£¬Éè³ö¶þÕßÎïÖʵÄÁ¿£¬ÁÐʽ¼ÆËã³ö¸÷×ÔÎïÖʵÄÁ¿£¬×îºó¸ù¾Ý¹ýÑõ»¯Ã¾µÄÎïÖʵÄÁ¿£¬Ëã³öÆäÖÊÁ¿·ÖÊý£»
·½°¸¢ò¼ÓÈë0.1mol•L-1KIÈÜÒºµÄҪʹ¹ýÑõ»¯Ã¾ÍêÈ«·´Ó¦£¬¸ù¾Ýµç×ÓÊغã¿ÉÖª¹ØϵʽMgO2¡«2KI£¬¾Ý´Ë¼ÆËãKIÈÜÒºµÄÌå»ý£¬ÓÉÓÚµâÓöµí·ÛÏÔÀ¶É«£¬¼ÓÈëÉÙÁ¿µí·ÛÈÜÒº×÷ָʾ¼Á£»¸ù¾Ýµç×ÓÊغãÕÒ³ö¹ØϵʽMgO2¡«2KI¡«I2¡«2Na2S2O3£¬¼ÆËã³ö¹ýÑõ»¯Ã¾µÄÖÊÁ¿·ÖÊý£®
½â´ð ½â£º£¨1£©¢ÙʵÑé×°ÖÃÁ¬½ÓºÃÒÔºó£¬ÊµÑéÇ°Ðè½øÐеIJÙ×÷ÊǼì²é×°ÖõÄÆøÃÜÐÔ£»¹ýÑõ»¯Ã¾ÈÜÓÚË®Éú³ÉË«ÑõË®£¬Ë«ÑõË®Ò׷ֽ⣬ͨ¹ý²âÁ¿Ë«ÑõË®·Ö½âÉú³ÉµÄÑõÆø£¬¿É¼ì²â¹ýÑõ»¯Ã¾µÄº¬Á¿£¬¶øÂÈ»¯ÌúÄÜ×÷´ß»¯¼Á¼Ó¿ìË«ÑõË®µÄ·Ö½â£¬¼Ó¿ì·´Ó¦ËÙÂÊ£¬¹ýÑõ»¯Ã¾ÊÇÀë×Ó»¯ºÏÎ¾Ý´Ëд³öµç×ÓʽΪ£»
¹Ê´ð°¸Îª£ºÓÃ×÷´ß»¯¼Á£¨»ò´ß»¯H2O2µÄ·Ö½â£©£»£»
¢ÚÓÉÓÚÆøÌåµÄÌå»ýÊÜζȺÍѹǿӰÏì´ó£¬ËùÒÔÓúãѹ·ÖҺ©¶·µÄÓŵ㻹ÓÐÏû³ýµÎÈëÈÜÒºµÄÌå»ý¶ÔËù²âÆøÌåÌå»ýµÄÓ°Ï죬ʹ·ÖҺ©¶·ÖеÄÈÜҺ˳ÀûµÎÏ£¬
¹Ê´ð°¸Îª£ºÏû³ýµÎÈëÈÜÒºµÄÌå»ý¶ÔËù²âÆøÌåÌå»ýµÄÓ°Ï죻
£¨2£©·½°¸I¸ù¾ÝÇâÑõ»¯Ã¾µÄÈܶȻý³£Êý¿ÉÖª£¬µ±ÈÜÒºÖÐc£¨Mg2+£©=l¡Á10-5mol/Lʱ£¬Ksp[Mg£¨OH£©2]=1¡Á10-11=c£¨Mg2+£©•c2£¨OH-£©£¬ÈÜÒºÖÐOH-Ũ¶ÈµÈÓÚ1¡Á10-3mol/L£¬ËùÒÔÈÜÒºµÄpH=11£¬Éè¹ýÑõ»¯Ã¾µÄÎïÖʵÄÁ¿ÊÇxmol£¬Ñõ»¯Ã¾ÎïÖʵÄÁ¿ÊÇymol£¬Ôò56x+40y=m£¬£¨x+y£©¡Á40=n£¬½âµÃx=$\frac{m-n}{16}$£¬·½°¸IÖйýÑõ»¯Ã¾µÄÖÊÁ¿·ÖÊýΪ£º$\frac{\frac{m-n}{16}¡Á56}{m}$=$\frac{7£¨m-n£©}{2m}$£¬
¹Ê´ð°¸ÊÇ£º11£»$\frac{7£¨m-n£©}{2m}$£»
·½°¸¢ò0.56gÑùÆ·ÖÃÓÚµâÁ¿Æ¿ÖУ¬¼ÙÉèÑùÆ·Öж¼ÊÇMgO2£¬ÔòÆäÎïÖʵÄÁ¿Îª0.01mol£¬¼ÓÈë0.1mol•L-1KIÈÜÒºµÄҪʹ¹ýÑõ»¯Ã¾ÍêÈ«·´Ó¦£¬¸ù¾Ýµç×ÓÊغã¿ÉÖª¹ØϵʽMgO2¡«2KI£¬ËùÒÔKIÈÜÒºµÄÌå»ýΪ$\frac{0.01mol¡Á2}{0.1mol•{L}^{-1}}$=0.2L=200mL£¬ÓÉÓÚµâÓöµí·ÛÏÔÀ¶É«£¬¼ÓÈëÉÙÁ¿µí·ÛÈÜÒº×÷ָʾ¼Á£¬ÓÃ0.1mol•L-11Na2S2O3ÈÜÒºµÎ¶¨£¬µÎ¶¨µ½ÖÕµãʱ¹²ÏûºÄVmL£¬ÔòµÄÎïÖʵÄÁ¿ÎªV¡Á10-4mol£¬¸ù¾Ýµç×ÓÊغãÕÒ³ö¹ØϵʽMgO2¡«2KI¡«I2¡«2Na2S2O3£¬¼ÆËã³ö¹ýÑõ»¯Ã¾µÄÎïÖʵÄÁ¿Îª$\frac{1}{2}$¡ÁV¡Á10-4mol£¬ËùÒÔ¹ýÑõ»¯Ã¾µÄÖÊÁ¿·ÖÊýΪ$\frac{\frac{1}{2}¡ÁV¡Á1{0}^{-4}¡Á56}{0.56}$=0.005V£¬
¹Ê´ð°¸Îª£º200£»µí·ÛÈÜÒº£»0.005V£®
µãÆÀ ¸ÃÌâÊǸ߿¼Öеij£¼ûÌâÐÍ£¬ÊôÓÚÖеÈÄѶȵÄÊÔÌ⣬ÊÔÌâ×ÛºÏÐÔÇ¿£¬²àÖضÔѧÉúÄÜÁ¦µÄÅàÑøºÍѵÁ·£¬ÓÐÀûÓÚÅàÑøѧÉú¹æ·¶ÑϽ÷µÄʵÑéÉè¼Æ¡¢²Ù×÷ÄÜÁ¦£®
A£® | ͼ4ΪÓÃŨÏõËáÈÜÒºÓëÍ·´Ó¦ÖƱ¸²¢ÊÕ¼¯ÉÙÁ¿NO2 | |
B£® | ͼ2ΪÖƱ¸ÉÙÁ¿ÑõÆø | |
C£® | ͼ3ΪÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÁòËáÈÜÒº | |
D£® | ͼ1Ϊ֤Ã÷·Ç½ðÊôÐÔÇ¿Èõ£ºS£¾C£¾Si |
A£® | m+n£¾p+q£¬¡÷H£¼0 | B£® | m+n£¾p+q£¬¡÷H£¾0 | C£® | m+n£¼p+q£¬¡÷H£¾0 | D£® | m+n£¼p+q£¬¡÷H£¼0 |
A£® | Éú³É0.3 mol Cl2ʱ£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª0.6 mol | |
B£® | ¸Ã·´Ó¦ÖУ¬Éú³ÉµÄCl2µÄĦ¶ûÖÊÁ¿Ô¼Îª70.7 g/mol | |
C£® | ±»0.1 mol K37ClO3Ñõ»¯µÄH35ClµÄÎïÖʵÄÁ¿Îª0.6 mol | |
D£® | ¸Ã·´Ó¦ÖУ¬H35Cl·¢ÉúÑõ»¯·´Ó¦£¬KClÊÇ»¹Ô²úÎï |
A£® | ȼÉճס¢¼¯ÆøÆ¿ | B£® | Õô·¢Ãó¡¢ÊÔ¹Ü | C£® | ÛáÛö¡¢ÉÕ± | D£® | ÉÕÆ¿¡¢ÉÕ± |
A£® | SO2¡¢O2¡¢H2S | B£® | H2¡¢O2¡¢F2 | C£® | O2¡¢C12¡¢HBr | D£® | H2S¡¢CO2¡¢H2 |
A£® | ÁòËáÌú | B£® | ̼ËáÇâÄÆ | C£® | ¸ßÃÌËá¼Ø | D£® | ÂÈ»¯ÂÁ |