ÁòËáÊÇ»¯Ñ§¹¤ÒµÖеÄÖØÒªÔÁÏ£¬ÖÁ2010ÄêÎÒ¹úÒѳÉΪȫÇòÁòËá²úÄÜ×î¸ß¡¢²úÁ¿×î´óµÄ¹ú¼Ò¡£
(1)18.4mol/L£¨ÖÊÁ¿·ÖÊý0.98£¬ÃܶÈ1.84g/cm3£©Å¨ÁòËáÊdz£ÓõĸÉÔï¼Á£¬ÓÃÓÚÎüÊÕ³±ÊªÆøÌåÖеÄË®ÕôÆø¡£µ±Å¨ÁòËáŨ¶È½µµ½16 mol/L£¨ÃܶÈ1.8g/cm3£©ÒÔÏÂʱ£¬Ôòʧȥ¸ÉÔïÄÜÁ¦¡£
¢Ù16 mol/LµÄÁòËáµÄÖÊÁ¿·ÖÊýΪ £¨±£ÁôÁ½Î»Ð¡Êý£¬ÏÂͬ£©¡£
¢Ú50mLÖÊÁ¿·ÖÊýΪ0.98µÄŨÁòËá×÷Ϊ¸ÉÔï¼Áʱ£¬×î¶à¿ÉÎüË® g¡£
(2)½«Ìú·ÛÓëÁò·ÛÔÚ¸ô¾ø¿ÕÆøÌõ¼þÏ·´Ó¦ËùµÃµÄ¹ÌÌåM 9.920 g£¬Óë×ãÁ¿Ï¡ÁòËá·´Ó¦£¬ÊÕ¼¯µ½ÆøÌå2.688 L£¨»»Ëãµ½±ê×¼×´¿ö£©£¬ÖÊÁ¿Îª3.440 g¡£Ôò¹ÌÌåMµÄ³É·ÖΪ
£¨Ð´»¯Ñ§Ê½£©£¬ÆäÖÐÌúÔªËØÓëÁòÔªËصÄÖÊÁ¿±ÈΪ ¡£
(3)µ±´úÁòËṤҵ´ó¶àÓýӴ¥·¨ÖÆÁòËᣨÉè¿ÕÆøÖÐÑõÆøµÄÌå»ý·ÖÊýΪ0.20£©¡£
¢ÙΪʹ»ÆÌú¿óìÑÉÕ³ä·Ö£¬³£Í¨Èë¹ýÁ¿40%µÄ¿ÕÆø£¬ÔòìÑÉÕºó¯ÆøÖÐSO2µÄÌå»ý·ÖÊýΪ ¡£
¢Ú½«¢ÙÖеįÆø¾¾»»¯³ý³¾ºóÖ±½ÓËÍÈë½Ó´¥ÊÒ£¬Á÷Á¿Îª1.00m3/s£¬´Ó½Ó´¥ÊÒµ¼³öÆøÌåµÄÁ÷Á¿Îª0.95m3/s£¨Í¬ÎÂͬѹϲⶨ£©£¬ÔòSO2µÄת»¯ÂÊΪ %¡£
(4)½Ó´¥·¨ÖÆÁòËáÅŷŵÄβÆøÏÈÓð±Ë®ÎüÊÕ£¬ÔÙÓÃŨÁòËá´¦Àí£¬µÃµ½½Ï¸ßŨ¶ÈµÄSO2£¨Ñ»·ÀûÓ㩺ͻìºÏï§ÑΡ£Îª²â¶¨´Ëï§ÑÎÖеªÔªËصÄÖÊÁ¿·ÖÊý£¬½«²»Í¬ÖÊÁ¿µÄï§Ñηֱð¼ÓÈëµ½50.00mLÏàͬŨ¶ÈµÄNaOHÈÜÒºÖУ¬·Ðˮԡ¼ÓÈÈÖÁÆøÌåÈ«²¿Òݳö£¨´ËζÈÏÂï§Ñβ»·Ö½â£©¡£¸ÃÆøÌ徸ÉÔïºóÓÃŨÁòËáÎüÊÕÍêÈ«£¬²â¶¨Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿¡£
²¿·Ö²â¶¨½á¹ûÈçÏ£º
ï§ÑÎÖÊÁ¿Îª10.000 gºÍ20.000 gʱ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿Ïàͬ£»
ï§ÑÎÖÊÁ¿Îª30.000 gʱ£¬Å¨ÁòËáÖÊÁ¿ÔöÖØ0.680 g£»
ï§ÑÎÖÊÁ¿Îª40.000 gʱ£¬Å¨ÁòËáµÄÖÊÁ¿²»±ä¡£
¢Ù¼ÆËã¸Ã»ìºÏÑÎÖеªÔªËصÄÖÊÁ¿·ÖÊý¡£
¢Ú¼ÆËãÉÏÊöÇâÑõ»¯ÄÆÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£
(1)£¨4·Ö£©¢Ù 0.87£»¢Ú11.63£»£¨Ã¿Ð¡Ìâ2·Ö£©
(2)£¨3·Ö£©FeS¡¢Fe£¨1·Ö£©£¬21£º10£»£¨2 ·Ö£©
(3)£¨4·Ö£©¢Ù 0.11£»¢Ú 92.5%£»£¨Ã¿Ð¡Ìâ2·Ö£©
(4)£¨5·Ö£©
¢ÙÉ裺NaOHΪamol,NH4HSO4Ϊxmol, £¨NH4£©2SO4Ϊymol¡£
ÔòÓУº
½âµÃ£ºa=0.232
x=0.064
y=0.02
£¨3·Ö£©
¢Ú £¨2·Ö£©
½âÎö(1)£¨4·Ö£©¢Ù
¢ÚÓÉÈÜÖÊÖÊÁ¿Êغã
50ml¡Á1.84g/ml¡Á0.98=16mol/l¡Á(50ml¡Á1.84g/ml+mg)/1.8g/ml¡Á98g¡¤mol-1
m=11.63g
(2)£¨3·Ö£©FeS¡¢Fe£¨1·Ö£©£¬21£º10£»£¨2 ·Ö£©
Fe£«SFeS
FeS£«H2SO4=FeSO4£«H2S¡ü
Fe£«H2SO4=FeSO4£«H2¡ü,
n(Æø)= 2.688 L/22.4l¡¤mol-1=0.12mol
n(H2S)+n(H2)=0.12mol
n(H2S)¡Á34g¡¤mol-1+n(H2)¡Á2 g¡¤mol-1= 3.440 g
n(H2S)=n(FeS)=0.1mol,n(H2)=0.02mol,
m(Fe):m(S)=(9.920g-0.1mol¡Á32g¡¤mol-1):( 0.1mol¡Á32g¡¤mol-1)=21:10
(3)£¨4·Ö£©¢Ù 0.11£»¢Ú 92.5%£»£¨Ã¿Ð¡Ìâ2·Ö£©
¢Ù4FeS2£«11O22Fe2O3£«8SO2,
ͨÈë¹ýÁ¿40%µÄ¿ÕÆø£¬ÔòìÑÉÕºó¯ÆøÖÐSO2µÄÌå»ý·ÖÊýΪ
8¡Â£¨11¡Á4+11¡Á40%¡Á5+8£©¡Á100%=8¡Â74¡Á100 %=11%
¢Ú 2 SO2 £« O2 2SO3 ¡÷V
2 1 1
0.1m3/s 0.05m3/s 1.00m3/s¨D0.95m3/s
SO2µÄ×ÜÌå»ý£º1.00m3/s¡Á8¡Â74¡Á100 %="0.108" m3/s£¬
SO2µÄת»¯ÂÊΪ0.1m3/s¡Â0.108 m3/s¡Á100%=92.5%
(4)£¨5·Ö£©
¢ÙÉ裺NaOHΪamol,NH4HSO4Ϊxmol, £¨NH4£©2SO4Ϊymol¡£
ÔòÓУº
½âµÃ£ºa=0.232
x=0.064
y=0.02
£¨3·Ö£©
¢Ú £¨2·Ö£©
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
| ||
| ||
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â
ÁòËáÊÇ»¯Ñ§¹¤ÒµÖеÄÖØÒªÔÁÏ£¬ÖÁ2010ÄêÎÒ¹úÒѳÉΪȫÇòÁòËá²úÄÜ×î¸ß¡¢²úÁ¿×î´óµÄ¹ú¼Ò¡£
(1)18.4mol/L£¨ÖÊÁ¿·ÖÊý0.98£¬ÃܶÈ1.84g/cm3£©Å¨ÁòËáÊdz£ÓõĸÉÔï¼Á£¬ÓÃÓÚÎüÊÕ³±ÊªÆøÌåÖеÄË®ÕôÆø¡£µ±Å¨ÁòËáŨ¶È½µµ½16 mol/L£¨ÃܶÈ1.8g/cm3£©ÒÔÏÂʱ£¬Ôòʧȥ¸ÉÔïÄÜÁ¦¡£
¢Ù16 mol/LµÄÁòËáµÄÖÊÁ¿·ÖÊýΪ £¨±£ÁôÁ½Î»Ð¡Êý£¬ÏÂͬ£©¡£
¢Ú50mLÖÊÁ¿·ÖÊýΪ0.98µÄŨÁòËá×÷Ϊ¸ÉÔï¼Áʱ£¬×î¶à¿ÉÎüË® g¡£
(2)½«Ìú·ÛÓëÁò·ÛÔÚ¸ô¾ø¿ÕÆøÌõ¼þÏ·´Ó¦ËùµÃµÄ¹ÌÌåM 9.920 g£¬Óë×ãÁ¿Ï¡ÁòËá·´Ó¦£¬ÊÕ¼¯µ½ÆøÌå2.688 L£¨»»Ëãµ½±ê×¼×´¿ö£©£¬ÖÊÁ¿Îª3.440 g¡£Ôò¹ÌÌåMµÄ³É·ÖΪ
£¨Ð´»¯Ñ§Ê½£©£¬ÆäÖÐÌúÔªËØÓëÁòÔªËصÄÖÊÁ¿±ÈΪ ¡£
(3)µ±´úÁòËṤҵ´ó¶àÓýӴ¥·¨ÖÆÁòËᣨÉè¿ÕÆøÖÐÑõÆøµÄÌå»ý·ÖÊýΪ0.20£©¡£
¢ÙΪʹ»ÆÌú¿óìÑÉÕ³ä·Ö£¬³£Í¨Èë¹ýÁ¿40%µÄ¿ÕÆø£¬ÔòìÑÉÕºó¯ÆøÖÐSO2µÄÌå»ý·ÖÊýΪ ¡£
¢Ú½«¢ÙÖеįÆø¾¾»»¯³ý³¾ºóÖ±½ÓËÍÈë½Ó´¥ÊÒ£¬Á÷Á¿Îª1.00m3/s£¬´Ó½Ó´¥ÊÒµ¼³öÆøÌåµÄÁ÷Á¿Îª0.95m3/s£¨Í¬ÎÂͬѹϲⶨ£©£¬ÔòSO2µÄת»¯ÂÊΪ %¡£
(4)½Ó´¥·¨ÖÆÁòËáÅŷŵÄβÆøÏÈÓð±Ë®ÎüÊÕ£¬ÔÙÓÃŨÁòËá´¦Àí£¬µÃµ½½Ï¸ßŨ¶ÈµÄSO2£¨Ñ»·ÀûÓ㩺ͻìºÏï§ÑΡ£Îª²â¶¨´Ëï§ÑÎÖеªÔªËصÄÖÊÁ¿·ÖÊý£¬½«²»Í¬ÖÊÁ¿µÄï§Ñηֱð¼ÓÈëµ½50.00mLÏàͬŨ¶ÈµÄNaOHÈÜÒºÖУ¬·Ðˮԡ¼ÓÈÈÖÁÆøÌåÈ«²¿Òݳö£¨´ËζÈÏÂï§Ñβ»·Ö½â£©¡£¸ÃÆøÌ徸ÉÔïºóÓÃŨÁòËáÎüÊÕÍêÈ«£¬²â¶¨Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿¡£
²¿·Ö²â¶¨½á¹ûÈçÏ£º
ï§ÑÎÖÊÁ¿Îª10.000 gºÍ20.000 gʱ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿Ïàͬ£»
ï§ÑÎÖÊÁ¿Îª30.000 gʱ£¬Å¨ÁòËáÖÊÁ¿ÔöÖØ0.680 g£»
ï§ÑÎÖÊÁ¿Îª40.000 gʱ£¬Å¨ÁòËáµÄÖÊÁ¿²»±ä¡£
¢Ù¼ÆËã¸Ã»ìºÏÑÎÖеªÔªËصÄÖÊÁ¿·ÖÊý¡£
¢Ú¼ÆËãÉÏÊöÇâÑõ»¯ÄÆÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÉϺ£ÊÐãÉÐÐÇø¸ßÈýÉÏѧÆÚÆÚÄ©ÖÊÁ¿³é²â»¯Ñ§ÊÔ¾í ÌâÐÍ£º¼ÆËãÌâ
ÁòËáÊÇ»¯Ñ§¹¤ÒµÖеÄÖØÒªÔÁÏ£¬ÖÁ2010ÄêÎÒ¹úÒѳÉΪȫÇòÁòËá²úÄÜ×î¸ß¡¢²úÁ¿×î´óµÄ¹ú¼Ò¡£
(1)18.4mol/L£¨ÖÊÁ¿·ÖÊý0.98£¬ÃܶÈ1.84g/cm3£©Å¨ÁòËáÊdz£ÓõĸÉÔï¼Á£¬ÓÃÓÚÎüÊÕ³±ÊªÆøÌåÖеÄË®ÕôÆø¡£µ±Å¨ÁòËáŨ¶È½µµ½16 mol/L£¨ÃܶÈ1.8g/cm3£©ÒÔÏÂʱ£¬Ôòʧȥ¸ÉÔïÄÜÁ¦¡£
¢Ù16 mol/LµÄÁòËáµÄÖÊÁ¿·ÖÊýΪ £¨±£ÁôÁ½Î»Ð¡Êý£¬ÏÂͬ£©¡£
¢Ú50mLÖÊÁ¿·ÖÊýΪ0.98µÄŨÁòËá×÷Ϊ¸ÉÔï¼Áʱ£¬×î¶à¿ÉÎüË® g¡£
(2)½«Ìú·ÛÓëÁò·ÛÔÚ¸ô¾ø¿ÕÆøÌõ¼þÏ·´Ó¦ËùµÃµÄ¹ÌÌåM 9.920 g£¬Óë×ãÁ¿Ï¡ÁòËá·´Ó¦£¬ÊÕ¼¯µ½ÆøÌå2.688 L£¨»»Ëãµ½±ê×¼×´¿ö£©£¬ÖÊÁ¿Îª3.440 g¡£Ôò¹ÌÌåMµÄ³É·ÖΪ
£¨Ð´»¯Ñ§Ê½£©£¬ÆäÖÐÌúÔªËØÓëÁòÔªËصÄÖÊÁ¿±ÈΪ ¡£
(3)µ±´úÁòËṤҵ´ó¶àÓýӴ¥·¨ÖÆÁòËᣨÉè¿ÕÆøÖÐÑõÆøµÄÌå»ý·ÖÊýΪ0.20£©¡£
¢ÙΪʹ»ÆÌú¿óìÑÉÕ³ä·Ö£¬³£Í¨Èë¹ýÁ¿40%µÄ¿ÕÆø£¬ÔòìÑÉÕºó¯ÆøÖÐSO2µÄÌå»ý·ÖÊýΪ ¡£
¢Ú½«¢ÙÖеįÆø¾¾»»¯³ý³¾ºóÖ±½ÓËÍÈë½Ó´¥ÊÒ£¬Á÷Á¿Îª1.00m3/s£¬´Ó½Ó´¥ÊÒµ¼³öÆøÌåµÄÁ÷Á¿Îª0.95m3/s£¨Í¬ÎÂͬѹϲⶨ£©£¬ÔòSO2µÄת»¯ÂÊΪ %¡£
(4)½Ó´¥·¨ÖÆÁòËáÅŷŵÄβÆøÏÈÓð±Ë®ÎüÊÕ£¬ÔÙÓÃŨÁòËá´¦Àí£¬µÃµ½½Ï¸ßŨ¶ÈµÄSO2£¨Ñ»·ÀûÓ㩺ͻìºÏï§ÑΡ£Îª²â¶¨´Ëï§ÑÎÖеªÔªËصÄÖÊÁ¿·ÖÊý£¬½«²»Í¬ÖÊÁ¿µÄï§Ñηֱð¼ÓÈëµ½50.00mLÏàͬŨ¶ÈµÄNaOHÈÜÒºÖУ¬·Ðˮԡ¼ÓÈÈÖÁÆøÌåÈ«²¿Òݳö£¨´ËζÈÏÂï§Ñβ»·Ö½â£©¡£¸ÃÆøÌ徸ÉÔïºóÓÃŨÁòËáÎüÊÕÍêÈ«£¬²â¶¨Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿¡£
²¿·Ö²â¶¨½á¹ûÈçÏ£º
ï§ÑÎÖÊÁ¿Îª10.000 gºÍ20.000 gʱ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿Ïàͬ£»
ï§ÑÎÖÊÁ¿Îª30.000 gʱ£¬Å¨ÁòËáÖÊÁ¿ÔöÖØ0.680 g£»
ï§ÑÎÖÊÁ¿Îª40.000 gʱ£¬Å¨ÁòËáµÄÖÊÁ¿²»±ä¡£
¢Ù¼ÆËã¸Ã»ìºÏÑÎÖеªÔªËصÄÖÊÁ¿·ÖÊý¡£
¢Ú¼ÆËãÉÏÊöÇâÑõ»¯ÄÆÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê½ËÕÊ¡ÎÞÎýÒ»Öи߶þ£¨Ï£©ÆÚÖл¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
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