ÌâÄ¿ÄÚÈÝ
14£®Áò´úÁòËáÄƾ§Ì壨Na2S2O3•5H2O£¬Ê½Á¿248£©Ë׳ƺ£²¨»ò´óËÕ´ò£¬ËüÒ×ÈÜÓÚË®£¬ÇÒÈܽâ¶ÈËæζÈÉý¸ß¶øÏÔÖøÔö´ó£¬ÄÑÈÜÓÚÒÒ´¼£¬¼ÓÈÈʱÒ׷ֽ⣬¿ÉÓÃÓÚÕÕÏàÐÐÒµµÄ¶¨Ó°¼Á£®ÊµÑéÊÒÄ£Ä⹤ҵÖƱ¸Áò´úÁòËáÄƾ§Ìåͨ³£ÓÐÒÔÏ·½·¨£¬Çë»Ø´ðÓйØÎÊÌ⣮ÑÇÁòËáÄÆ·¨£ºNa2SO3+S+5H2O=Na2S2O3•5H2O£¬¼òÒ×ʵÑéÁ÷³ÌÈçÏ£º
£¨1£©Áò·ÛÓÃÒÒ´¼ÈóʪµÄÄ¿µÄÊÇÓÐÀûÓÚÁò·ÛÓëÑÇÁòËáÄÆÈÜÒº³ä·Ö½Ó´¥£¬¼Ó¿ì·´Ó¦ËÙÂÊ£®
£¨2£©²Ù×÷Öв»Äܽ«ÈÜÒºÕô·¢ÖÁ¸ÉµÄÔÒòÊÇÕô¸É»áʹÁò´úÁòËáÄƾ§ÌåÍÑË®²¢·Ö½â£®
£¨3£©ËùµÃ´Ö²úÆ·Ò»°ãͨ¹ýÖؽᾧ·½·¨Ìá´¿£®
Áò»¯¼î·¨£º2Na2S+Na2CO3+4SO2=3Na2S2O3+CO2£¬Ö÷ҪʵÑé×°ÖÃÈçÏ£º
£¨4£©×°ÖÃCµÄ×÷ÓÃÊÇÎüÊÕ·´Ó¦Éú³ÉµÄCO2ºÍ¶àÓàµÄSO2£¬·ÀÖ¹ÎÛȾ´óÆø£®
£¨5£©Îª³ä·ÖÀûÓÃSO2£¬¶Ô×°ÖÃB½øÐиĽø£¨ÈçÉÏÓÒͼËùʾ£©£ºµ±AÖз´Ó¦·¢Éúºó£¬¹Ø±ÕÐýÈûb¡¢e£¬´ò¿ªa¡¢c¡¢d£¬ÍùBÖÐÈÜҺͨÈëSO2£¬Î´·´Ó¦µÄSO2±»ÊÕ¼¯µ½ÆøÄÒfÖУ®´ýfÊÕ¼¯µ½½Ï¶àÆøÌåʱ£¨¼Ù¶¨´ËʱװÖÃAÖз´Ó¦ÒÑÍ£Ö¹£©£¬¹Ø±ÕÐýÈûac£¬´ò¿ªÐýÈûbde£¬ÇáÇἷѹf£¬Ê¹SO2»º»ºµØѹÈëBÖÐÈÜÒºÔٴη´Ó¦£¬Î´·´Ó¦µÄSO2ÓÖ±»ÊÕ¼¯ÔÚÆøÄÒgÖУ®ÔÙ½«gÖеÄÆøÌ強ѹÈëfÖУ¬Èç´Ë·´¸´£¬Ö±ÖÁÍêÈ«·´Ó¦£®
£¨6£©Îª²â¶¨Áò´úÁòËáÄƾ§Ìå´Ö²úÆ·µÄ´¿¶È£®Ä³ÐËȤС×é³ÆÈ¡5.0¿Ë´Ö²úÆ·Åä³É250mLÈÜÒº£¬²¢Óüä½ÓµâÁ¿·¨±ê¶¨¸ÃÈÜÒºµÄŨ¶È£ºÔÚ׶ÐÎÆ¿ÖмÓÈë25mL 0.01mol/L KIO3ÈÜÒº£¬²¢¼ÓÈë¹ýÁ¿µÄKI²¢Ëữ£¬·¢ÉúÏÂÁз´Ó¦£º5I-+IO3-+6H+=3I2+3H2O£¬ÔÙ¼ÓÈ뼸µÎµí·ÛÈÜÒº£¬Á¢¼´ÓÃËùÅäNa2S2O3ÈÜÒºµÎ¶¨£¬·¢Éú·´Ó¦£ºI2+2S2O32-=2I-+S4O62-£¬µ±À¶É«ÍÊÈ¥ÇÒ°ë·ÖÖÓ²»±äɫʱµ½´ïµÎ¶¨Öյ㣮ʵÑéÊý¾ÝÈçÏÂ±í£º
µÎ¶¨´ÎÊý | 1 | 2 | 3 |
ÏûºÄNa2S2O3ÈÜÒº£¨mL£© | 19.98 | 21.18 | 20.02 |
¢Ú¿ÉÄÜÔì³ÉʵÑé½á¹ûÆ«µÍµÄÓÐBD£¨Ìî±àºÅ£©£®
A£®×¶ÐÎÆ¿ÓÃÕôÁóË®ÈóÏ´
B£®µÎ¶¨¹ÜδÓÃNa2S2O3ÈÜÒºÈóÏ´
C£®µÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý
D£®ÈôµÎ¶¨Ç°µÎ¶¨¹Ü¼â×ì´¦ÓÐÆøÅÝ£¬µÎ¶¨ºóÏûʧ£®
·ÖÎö £¨1£©ÒÒ´¼Ò×ÈÜÓÚË®£¬Áò·ÛÓÃÒÒ´¼Èóʪ£¬ÓÐÀûÓÚÁò·ÛÓëÑÇÁòËáÄÆÈÜÒº³ä·Ö½Ó´¥£»
£¨2£©ÒÑÖªÁò´úÁòËáÄƾ§Ìå¼ÓÈÈÒ׷ֽ⣻
£¨3£©Áò´úÁòËáÄƵÄÈܽâ¶ÈËæζÈÉý¸ß¶øÏÔÖøÔö´ó£¬½µµÍζȣ¬²ÉÓÃÖؽᾧ·¨ÖÆÈ¡´Ö²úÆ·£»
£¨4£©×°ÖÃCÖ÷ÒªÊÇβÆø´¦Àí£¬ÎüÊÕ·´Ó¦Éú³ÉµÄCO2ºÍ¶àÓàµÄSO2£¬·ÀÖ¹ÎÛȾ´óÆø£¬
£¨5£©¹Ø±ÕÐýÈûa£¬´ò¿ªÐýÈûc£¬ÇáÇἷѹf£¬Ê¹SO2»º»ºµØѹÈëBÖÐÈÜÒºÔٴη´Ó¦£¬Î´·´Ó¦µÄSO2ÓÖ±»ÊÕ¼¯ÔÚÆøÄÒgÖУ»
£¨6£©ÔÚ׶ÐÎÆ¿ÖмÓÈë25mL 0.0lmol•L-1KIO3ÈÜÒº£¬²¢¼ÓÈë¹ýÁ¿µÄKI²¢Ëữ£¬·¢ÉúÏÂÁз´Ó¦£º5I-+IO3-+6H+¨T3I2+3H2O£¬ÔÙ¼ÓÈ뼸µÎµí·ÛÈÜÒº£¬Á¢¼´ÓÃËùÅäNa2S2O3ÈÜÒºµÎ¶¨£¬·¢Éú·´Ó¦£ºI2+2S2O32-¨T2I-+S4O62-£¬
Ôò¿ÉµÃ¹Øϵʽ£ºIO3-¡«6S2O32-£¬¾Ý´Ë¼ÆË㣻
A£®×¶ÐÎÆ¿ÓÃÕôÁóË®ÈóÏ´£¬¶ÔʵÑé½á¹ûûӰÏ죻
B£®µÎ¶¨¹ÜÄ©ÓÃNa2S2O3ÈÜÒºÈóÏ´£¬ÔòNa2S2O3ÈÜÒº»á±»Ï¡ÊÍ£»
C£®µÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý£¬Ê¹Na2S2O3ÈÜÒºÌå»ýƫС£»
D£®µÎ¶¨Ç°µÎ¶¨¹Ü¼â×ì´¦ÓÐÆøÅÝ£¬µÎ¶¨ºóÏûʧ£¬Ê¹¶Á³öµÄNa2S2O3µÄÌå»ý±ä´ó£®
½â´ð ½â£º£¨1£©ÒÒ´¼Ò×ÈÜÓÚË®£¬Áò·ÛÓÃÒÒ´¼Èóʪ£¬ÓÐÀûÓÚÁò·ÛÓëÑÇÁòËáÄÆÈÜÒº³ä·Ö½Ó´¥£¬¼Ó¿ì·´Ó¦ËÙÂÊ£»
¹Ê´ð°¸Îª£ºÓÐÀûÓÚÁò·ÛÓëÑÇÁòËáÄÆÈÜÒº³ä·Ö½Ó´¥£¬¼Ó¿ì·´Ó¦ËÙÂÊ£»
£¨2£©ÒÑÖªÁò´úÁòËáÄƾ§Ìå¼ÓÈÈÒ׷ֽ⣬¹ÊÕô¸É»áʹÁò´úÁòËáÄÆÍÑË®²¢·Ö½â£¬ËùÒÔ´ÓÁò´úÁòËáÄÆÈÜÒºÖзÖÀëÈÜÖÊʱ²»Äܽ«ÈÜÒºÕô·¢ÖÁ¸É£»
¹Ê´ð°¸Îª£ºÕô¸É»áʹÁò´úÁòËáÄÆÍÑË®²¢·Ö½â£»
£¨3£©Áò´úÁòËáÄƵÄÈܽâ¶ÈËæζÈÉý¸ß¶øÏÔÖøÔö´ó£¬½µµÍζȣ¬²ÉÓÃÖؽᾧ·¨ÖÆÈ¡´Ö²úÆ·£»
¹Ê´ð°¸Îª£ºÖؽᾧ£»
£¨4£©×°ÖÃCÖ÷ÒªÊÇβÆø´¦Àí£¬ÎüÊÕ·´Ó¦Éú³ÉµÄCO2ºÍ¶àÓàµÄSO2£¬·ÀÖ¹ÎÛȾ´óÆø£¬
¹Ê´ð°¸Îª£ºÎüÊÕ·´Ó¦Éú³ÉµÄCO2ºÍ¶àÓàµÄSO2£¬·ÀÖ¹ÎÛȾ´óÆø£»
£¨5£©¹Ø±ÕÐýÈûac£¬´ò¿ªÐýÈûbde£¬ÇáÇἷѹf£¬Ê¹SO2»º»ºµØѹÈëBÖÐÈÜÒºÔٴη´Ó¦£¬Î´·´Ó¦µÄSO2ÓÖ±»ÊÕ¼¯ÔÚÆøÄÒgÖУ»
¹Ê´ð°¸Îª£ºac£»bde£»
£¨6£©ÔÚ׶ÐÎÆ¿ÖмÓÈë25mL 0.0lmol•L-1KIO3ÈÜÒº£¬²¢¼ÓÈë¹ýÁ¿µÄKI²¢Ëữ£¬·¢ÉúÏÂÁз´Ó¦£º5I-+IO3-+6H+¨T3I2+3H2O£¬ÔÙ¼ÓÈ뼸µÎµí·ÛÈÜÒº£¬Á¢¼´ÓÃËùÅäNa2S2O3ÈÜÒºµÎ¶¨£¬·¢Éú·´Ó¦£ºI2+2S2O32-¨T2I-+S4O62-£¬
Ôò¿ÉµÃ¹Øϵʽ£ºIO3-¡«6S2O32-£¬
1mol 6mol
0.025L¡Á0.0lmol•L-1 n£¨S2O32-£©
Ôòn£¨S2O32-£©=0.0015mol£¬
µÚÈý´ÎʵÑéµÄÊý¾ÝÎó²î½Ï´ó£¬ÉáÈ¥£¬
ËùÒÔ250mLÁò´úÁòËáÄÆÈÜÒºÖÐÁò´úÁòËáÄƵÄÎïÖʵÄÁ¿Îª0.0015mol¡Á$\frac{250}{\frac{1}{2}£¨19.98+20.02£©}$=0.01875mol£¬
ÔòÁò´úÁòËáÄƵÄÖÊÁ¿Îª0.01875mol¡Á248g/mol=4.65g£¬
Ôò¸Ã²úÆ·µÄ´¿¶ÈÊÇ$\frac{4.65g}{5.0g}$¡Á100%=93%£»
A£®×¶ÐÎÆ¿ÓÃÕôÁóË®ÈóÏ´£¬¶ÔʵÑé½á¹ûûӰÏ죬´¿¶È²»±ä£¬¹ÊA²»Ñ¡£»
B£®µÎ¶¨¹ÜÄ©ÓÃNa2S2O3ÈÜÒºÈóÏ´£¬ÔòNa2S2O3ÈÜÒº»á±»Ï¡ÊÍ£¬ËùÒÔ²â³öÁò´úÁòËáÄƵÄÖÊÁ¿Æ«Ð¡£¬¹Ê´¿¶ÈƫС£¬¹ÊBÑ¡£»
C£®µÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý£¬Ê¹Na2S2O3ÈÜÒºÌå»ýƫС£¬¼ÆËã³öµÄÁò´úÁòËáÄƵÄÖÊÁ¿Æ«´ó£¬¹Ê´¿¶ÈÆ«´ó£¬¹ÊC²»Ñ¡£»
D£®µÎ¶¨Ç°µÎ¶¨¹Ü¼â×ì´¦ÓÐÆøÅÝ£¬µÎ¶¨ºóÏûʧ£¬Ê¹¶Á³öµÄNa2S2O3µÄÌå»ý±ä´ó£¬¼ÆËã³öÔÈÜÒºÖеÄÁò´úÁòËáÄƵÄÖÊÁ¿Æ«Ð¡£¬Ôò´¿¶ÈƫС£¬¹ÊDÑ¡£®
¹Ê´ð°¸Îª£º93%£»BD£®
µãÆÀ ±¾Ìâͨ¹ýÖÆÈ¡Na2S2O3•5H2OµÄʵÑé²Ù×÷£¬¿¼²éÁËÎïÖÊÖƱ¸·½°¸µÄÉè¼Æ¡¢»ù±¾ÊµÑé²Ù×÷¡¢Àë×Ó·½³ÌʽµÄÊéд¡¢ÎïÖÊ´¿¶ÈµÄ¼ÆËã¡¢µÎ¶¨Îó²î·ÖÎöµÈ£¬ÌâÄ¿ÄѶÈÖеȣ¬Ã÷ȷʵÑé²Ù×÷ÓëÉè¼Æ¼°Ïà¹ØÎïÖʵÄÐÔÖÊÊǽâ´ð±¾ÌâµÄ¹Ø¼ü£¬ÊÔÌâ³ä·Ö¿¼²éÁËѧÉúµÄ·ÖÎö¡¢Àí½âÄÜÁ¦¼°Áé»îÓ¦ÓÃËùѧ֪ʶµÄÄÜÁ¦£®
A£® | Z¡¢NµÄ¼òµ¥Àë×ӵĻ¹ÔÐÔ£ºZ2-£¾N- | |
B£® | Y2XºÍY2X2¶¼ÊǼîÐÔÑõ»¯Îï | |
C£® | ZÓëXÐγɵĻ¯ºÏÎï¶ÔÓ¦µÄË®»¯ÎïÒ»¶¨ÊÇÇ¿Ëá | |
D£® | ZÔªËصķǽðÊôÐÔ±ÈNÔªËصķǽðÊôÐÔÇ¿ |
·Ö×Óʽ | Íâ¹Û | ÈÈ·Ö½âÎÂ¶È | ÈÛµã | ÈܽâÐÔ |
CO£¨NH2£©2•H2O2 | °×É«¾§Ìå | 45¡æ | 75-85¡æ | Ò×ÈÜÓÚË®¡¢ÓлúÈܼÁ |
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©·´Ó¦Æ÷µÄ¼ÓÈÈ·½Ê½ÊÇˮԡ¼ÓÈÈ£»·´Ó¦Î¶ÈÐè¿ØÖÆÔÚÊʵ±Î¶ÈϽøÐУ¬Î¶Ȳ»Äܹý¸ßµÄÔÒòÊÇζȹý¸ß£¬²úÆ·Ò׷ֽ⣬ÖÂʹ»îÐÔÑõº¬Á¿½µµÍ£¬Î¶ÈÒ²²»ÄܹýµÍµÄÔÒòÊÇζȹýµÍ£¬·´Ó¦ËÙÂÊÌ«Âý£¬ÇÒ·´Ó¦ÌåϵÖÆÀäÐèÒªÏûºÄ´óÁ¿ÄÜÁ¿£®
£¨2£©ÈôÓû´ÓĸҺÖзÖÀë³öH2O2ºÍÄòËØ£¬¿É²ÉÓõIJÙ×÷ÊǼõѹÕôÁ󡢽ᾧ£®
£¨3£©¸É·¨¹¤ÒÕÖÆÈ¡¹ýÑõ»¯ÄòËصķ½·¨ÊÇ£º²ÉÓøßŨ¶ÈË«ÑõˮˮÈÜÒºÅçÎíµ½ÎÞË®ÄòËعÌÌåÉϽøÐз´Ó¦£¬Ë®ºÍ·´Ó¦ÈÈͨ¹ýÁ÷̬´²ÒÆÈ¥¶øµÃµ½¸ÉÔïµÄ¹ýÑõ»¯ÄòËزúÆ·£®
±È½Ï¸É·¨Óëʪ·¨Á½ÖÖ¹¤ÒÕ£¬ÄãÈÏΪ¸É·¨¹¤ÒÕµÄÓŵãÊÇ£ºÁ÷³Ì¶Ì£¬¹¤ÒÕ¼òµ¥£¨´ð³öÒ»µã¼´¿É£©£¬¸É·¨¹¤ÒÕµÄȱµãÊÇ£ºË«ÑõˮŨ¶È¸ß¾¼ÃЧÒæµÍ£¬É豸¸´Ôӵȣ¨´ð³öÁ½µã¼´¿É£©£¬Êª·¨¹¤ÒÕµÄÓŵãÊÇ£ºµÍŨ¶ÈË«ÑõË®£¬¾¼ÃЧÒæ¸ß£¬É豸¼òµ¥Ò×ÓÚ´ïµ½£¬Ä¸Òº¿ÉÑ»·Ê¹Óõȣ¨´ð³öÁ½µã¼´¿É£©£®
£¨4£©×¼È·³ÆÈ¡0.6000g²úÆ·ÓÚ250mL׶ÐÎÆ¿ÖУ¬¼ÓÊÊÁ¿ÕôÁóË®Èܽ⣬ÔÙ¼Ó1mL 6mol•L-1H2SO4£¬ÓÃ0.1000mol•L-1KMnO4±ê×¼ÈÜÒºµÎ¶¨ÖÁÖÕµãʱÏûºÄ20.00mL£¨ÄòËØÓëKMnO4ÈÜÒº²»·´Ó¦£©£¬Ôò²úÆ·ÖÐCO£¨NH2£©2•H2O2µÄÖÊÁ¿·ÖÊýΪ78.3%£¨½á¹û±£Áôµ½Ð¡Êýµãºóһ룩£®
ÒÑÖª£º¢ÙKsp£¨CaF2£©=1.46¡Á10-10£¬Ksp£¨CaC2O4£©=2.34¡Á10-9£®
½ðÊôÀë×Ó | ¿ªÊ¼³ÁµíµÄpH | ³ÁµíÍêÈ«µÄpH |
Fe3+ | 1.1 | 3.2 |
Fe2+ | 5.8 | 8.8 |
Al3+ | 3.0 | 5.0 |
Ni2+ | 6.7 | 9.5 |
£¨1£©¡°·ÛË顱µÄÄ¿µÄÊÇÔö´ó½Ó´¥Ãæ»ý£¬¼Ó¿ì·´Ó¦ËÙÂÊ£»Ìá¸ßÄøµÄ½þ³öÂÊ£®
£¨2£©±£³ÖÆäËûÌõ¼þÏàͬ£¬ÔÚ²»Í¬Î¶È϶ԷÏÄø´ß»¯¼Á½øÐС°Ëá½þ¡±£¬Äø½þ³öÂÊËæʱ¼ä±ä»¯Èçͼ£®¡°Ëá½þ¡±µÄÊÊÒËζÈÓëʱ¼ä·Ö±ðΪc£¨Ìî×Öĸ£©£®
a.30¡æ¡¢30min b.90¡æ¡¢150min c.70¡æ¡¢120min d.90¡æ¡¢120min
£¨3£©Ö¤Ã÷¡°³ÁÄø¡±¹¤ÐòÖÐNi2+ÒѾ³ÁµíÍêÈ«µÄʵÑé²½Öè¼°ÏÖÏóÊǾ²Öã¬ÔÚÉϲãÇåÒºÖмÌÐøµÎ¼Ó£¨NH4£©2C2O4ÈÜÒº£¬Èô²»ÔÙ²úÉú³Áµí£¬Ôò¡°³ÁÄø¡±¹¤ÐòÒѾÍê³É£®½«¡°³ÁÄø¡±¹¤ÐòµÃµ½µÄ»ìºÏÎï¹ýÂË£¬ËùµÃ¹ÌÌåÓÃ75%ÒÒ´¼ÈÜҺϴµÓ¡¢110¡æϺæ¸É£¬µÃ²ÝËáÄø¾§Ì壮ÓÃ75%ÒÒ´¼ÈÜҺϴµÓµÄÄ¿µÄÊÇÏ´È¥£¨NH4£©2SO4ÔÓÖÊ¡¢±ãÓÚºæ¸É¡¢¼õÉÙ²ÝËáÄø¾§ÌåËðʧ£®
£¨4£©ÔÚ³ýÌúºÍÂÁ¹¤ÐòÖУ¬Ó¦ÏȼÓÈëH2O2Ñõ»¯£¬ÔÙ¼ÓÇâÑõ»¯Äøµ÷½ÚpHÖµµÄ·¶Î§Îª5.0¡ÜpH£¼6.7£®µÚ2²½ÖмÓÈëÊÊÁ¿NH4FÈÜÒºµÄ×÷ÓÃÊdzýÈ¥ÔÓÖÊCa2+£®
£¨5£©½«µÃµ½µÄ²ÝËáÄø¾§ÌåÔÚÕæ¿ÕÖмÓÈÈÖÁ320¡æ·Ö½â¿ÉÖØÐÂÖƵõ¥ÖÊÄø´ß»¯¼Á£¬Ð´³ö¸ÃÖƱ¸¹ý³ÌµÄ»¯Ñ§·½³Ìʽ£ºNiC2O4•2H2O$\frac{\underline{\;320¡æ\;}}{\;}$Ni+2CO2¡ü+2H2O£®
£¨6£©ÒÑÖª·ÏÄø´ß»¯¼ÁÖÐÄøµÄÖÊÁ¿·ÖÊýΪ5.9%£¬Ôò100kg·ÏÄø´ß»¯¼Á×î¶à¿ÉÖƵÃ18.3kg²ÝËáÄø¾§Ì壨Ni£º59£¬C£º12£¬H£º1£¬O£º16£©£®
A£® | $\frac{1000V¦Ñ}{22400+36.5V}$ mol/L | B£® | $\frac{V¦Ñ}{22400}$ mol/L | ||
C£® | $\frac{V¦Ñ}{22400+36.5}$ mol/L | D£® | $\frac{V}{22.4}$ mol/L |
A£® | 1mol¸ÃÓлúÎïÒ»¶¨Ìõ¼þÏ¿ÉÒÔºÍ5moläåË®·´Ó¦ | |
B£® | ¸ÃÓлúÎïµÄ·Ö×ÓʽΪC15H8O7R | |
C£® | 1mol¸ÃÓлúÎïÒ»¶¨Ìõ¼þÏÂ×î¶àÏûºÄ8 mol H2 | |
D£® | 1molάÉúËØP¿ÉÒÔºÍ4molNaOH·´Ó¦ |