ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÇúÏßA±íÃ÷£¬ÔÈÜÒºÖÐͨÈëCO2ºó£¬ËùµÃÈÜÒºÖеÄÈÜÖÊΪ£¨Ð´»¯Ñ§Ê½£©_____________£¬¸÷ÖÖÈÜÖʵÄÎïÖʵÄÁ¿Ö®±ÈΪ_____________£»ÏòÏ¡ÊͺóµÄÈÜÒºÖÐÖðµÎ¼ÓÈë0.1 mol¡¤L-1µÄÑÎËᣬ²úÉúCO2Ìå»ýµÄ×î´óֵΪ_____________mL¡£
£¨2£©ÇúÏßB±íÃ÷£¬ÔÈÜÒºÖÐͨÈëCO2ºó£¬ËùµÃÈÜÒºÖеÄÈÜÖÊΪ£¨Ð´»¯Ñ§Ê½£©_____________£»ÏòÏ¡ÊͺóµÄÈÜÒºÖÐÖðµÎ¼ÓÈë0.1 mol¡¤L-1µÄÑÎËᣬ²úÉúCO2Ìå»ý×î´óÖµ_____________mL¡£
£¨3£©ÔNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_____________mol¡¤L-1¡£
£¨1£©NaOH¡¢Na2CO3 3¡Ã1 33.6£¨2£©Na2CO3¡¢NaHCO3 112 £¨3£©0.75
½âÎö£º£¨1£©AÇúÏß²úÉúCO2ÓÃn(HCl)=£¨75 mL-60 mL£©¡Á10-3L¡¤mL-1¡Á0.1 mol¡¤L-1=0.001 5 mol£¬²úÉúCO2±ÈBÉÙ£¬¹ÊAÖÐͨÈëCO2²»×㣬Òò²úÉúÆøÌå·´Ó¦¾ùΪ
NaHCO3+HCl====NaCl+CO2¡ü+H2O£¬¹ÊAΪNaOHºÍNa2CO3µÄ»ìºÏÎï¡£ÓÉ
Na2CO3 + HCl ==== NaHCO3 +H2O
0.001 5 mol 0.001 5 mol 0.001 5 mol
NaHCO3 + HCl ==== NaCl+H2O
0.001 5 mol 0.001 5 mol
¹ÊÖкÍNaOHÏûºÄHClΪ75 mL¡Á10-3 L¡¤mL-1¡Á0.1 mol¡¤L-1-(0.001 5 mol¡Á2)=0.004 5 mol
=3¡Ã1,V(CO2)=33.6 mL
(2)BÇúÏß·¢Éú·´Ó¦NaHCO3+HCl====NaCl+CO2¡ü+H2O¡£ÏûºÄn(HCl)=50 mL¡Á10-3L¡¤mL-1¡Á0.1 mol¡¤L-1=0.005 mol£¬¹ÊBΪNa2CO3ºÍNaHCO3µÄ»ìºÏÎÓÉ
Na2CO3 + HCl ==== NaCl+NaHCO3
0.0025 mol 0.0025 mol
Ôòn£¨NaHCO3£©=0.002 5 mol,V(CO2)=112 mL¡£¾ÝAΪ10 mL£¬n£¨Na+£©=0.004 5 mol+0.001 5 mol¡Á2=0.007 5 mol
c==0.75 mol¡¤L-1
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£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¢ÙÁ¿È¡25.00 mL¼×ÈÜÒº£¬ÏòÆäÖлºÂýµÎ¼ÓÒÒÈÜÒº15.00 mL£¬¹²ÊÕ¼¯µ½¶þÑõ»¯Ì¼ÆøÌå224 mL¡£
¢ÚÁíÈ¡15.00 mLÒÒÈÜÒº£¬ÏòÆäÖлºÂýµÎ¼Ó¼×ÈÜÒº25.00 mL£¬¹²ÊÕ¼¯µ½¶þÑõ»¯Ì¼ÆøÌå112 mL¡£
ÉÏÊöÆøÌåÌå»ý¾ùÒÑ»»ËãΪ±ê×¼×´¿ö£¬Ôò¸ù¾ÝÉÏÊö²ÙÊö¼°ÊµÑéÊý¾ÝÌî¿Õ£º
£¨1£©Ð´³öÉÏÊö¹ý³ÌÖÐËùÉæ¼°·´Ó¦µÄÀë×Ó·½³Ìʽ£º¢Ù_______________£»¢Ú______________£»¢Û______________¡£
£¨2£©¼×ÊÇ______________£¬¼×ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______________£¬ÒÒÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______________¡££¨¶þÑõ»¯Ì¼ÔÚÈÜÒºÖеÄÉÙÁ¿ÈܽâºöÂÔ²»¼Æ£©
£¨3£©½«n mLµÄ¼×ÈÜÒºÓëµÈÌå»ýµÄÒÒÈÜÒº°´¸÷ÖÖ¿ÉÄܵķ½Ê½»ìºÏ£¬²úÉúµÄÆøÌåÌå»ýΪV mL£¨±ê×¼×´¿ö£©£¬ÔòVµÄÈ¡Öµ·¶Î§Îª____________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ìÉϺ£ÊзîÏÍÇø¸ßÈý4Ôµ÷ÑвâÊÔ£¨¶þÄ££©»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
¸ßÖл¯Ñ§Ñ§Ï°½×¶Î£¬°±µÄ´ß»¯Ñõ»¯·´Ó¦Ò²¿ÉÓÃij½ðÊôM£¨Ïà¶ÔÔ×ÓÖÊÁ¿£¼100£©µÄÑõ»¯Îï×ö´ß»¯¼Á¡£MÔªËصÄÀë×ÓÔÚÒ»¶¨Ìõ¼þÏÂÄÜÓëNH3¡¢Â±Àë×ӵȰ´Ä³¹Ì¶¨±ÈÀýÀÎÀνáºÏ³ÉÎȶ¨µÄÅäºÏÎïÀë×Ó£¨ÀàËÆÓÚÔ×ÓÍÅ£©¡£
£¨1£©150¡æ¡¢³£Ñ¹Ï£¬6.08g¸ÃÑõ»¯ÎïÓë0.6mol HClÆøÌå³ä·Ö·¢Éú¸´·Ö½â·´Ó¦ºó£¬Ñõ»¯ÎïÎÞÊ£Ó࣬ÆøÌå±äΪ0.48mol¡£¸Ã½ðÊôMµÄÏà¶ÔÔ×ÓÖÊÁ¿Îª___________¡£
½«ÉÏÊö·´Ó¦ËùµÃ¹ÌÌåÈÜÓÚË®Åä³É50.0mLÈÜÒº£¬¸ÃÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶ÈΪ
_______mol¡¤L-1¡£
£¨2£©È¡£¨1£©ÖÐËùÅäÈÜÒº12.5mL£¬Ï¡ÊÍÖÁ25mL£¬»º»ºÍ¨Èë2688mL°±Æø£¨±ê¿öÏ£©£¬ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Ç¡ºÃÍêÈ«£¬µÃµ½²úÎïB£¨Ä¦¶ûÖÊÁ¿Îª260.5g/mol£©£¬¼ÌÐøÓÃ1.5mol/LµÄAgNO3ÈÜÒºµÎ¶¨£¬´ïµ½ÖÕµãʱ£¬ºÄÈ¥AgNO3ÈÜÒº40.0mL¡£½«BͶÈë¹ýÁ¿ÉÕ¼îÈÜÒºÖУ¬Î´¼ì²â³öNH3µÄÒݳö¡£ÔòBµÄ»¯Ñ§Ê½¿É±íʾΪ ¡£
£¨3£©ÒÑÖªÔÚÏÂͼÖУ¬LµÄλÖÃÍêÈ«Ïàͬ¡£ÏÖÓÐÁíÒ»ÖÖÅäÀë×Ó[M(NH3)6-xClx]n¡À£¨1¡Üx¡Ü5£¬ÇÒxΪÕûÊý£©£¬½á¹¹ÀàËÆÏÂͼ¡£
Èç¹û¸ÃÅäÀë×Ó¹²ÓÐ2ÖÖ²»Í¬½á¹¹£¬¸ÃÅäÀë×ÓµÄʽÁ¿Îª ¡£
£¨4£©Ò»¶¨Ìõ¼þÏ£¬3.04g¸Ã½ðÊôÑõ»¯ÎïÇ¡ºÃ±»Ä³Ñõ»¯¼Á£¨Ð§¹ûÏ൱ÓÚ0.03mol O2£©Ñõ»¯£¬ÔÙ¼ÓÈë0.05mol KCl£¬¾Ò»Ð©ÌØÊ⹤ÒÕ´¦Àí£¬K¡¢MÁ½ÔªËØÇ¡ºÃÍêÈ«Ðγɺ¬ÑõËáÑΣ¨Ê½Á¿£¼360£©£¬ÇÒÑÎÄÚ²¿ÒõÑôÀë×Ó¸öÊý±ÈΪ1:2¡£Ôò¸ÃÑεÄʽÁ¿ÒÔ¼°ÎïÖʵÄÁ¿·Ö±ðΪ_______________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêÉϺ£ÊÐËɽÇø¸ßÈýÉÏѧÆÚÆÚÄ©£¨Ò»Ä££©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
ÂÌ·¯£¨FeSO4¡¤7H2O£©ÔÚ»¯Ñ§ºÏ³ÉÉÏÓÃ×÷»¹Ô¼Á¼°´ß»¯¼Á£¬¹¤ÒµÉϳ£Ó÷ÏÌúмÈÜÓÚÒ»¶¨Å¨¶ÈµÄÁòËáÈÜÒºÖƱ¸ÂÌ·¯¡£
£¨1£©98% 1.84 g/cm3µÄŨÁòËáÔÚÏ¡Ê͹ý³ÌÖУ¬ÃܶÈϽµ£¬µ±Ï¡ÊÍÖÁ50%ʱ£¬ÃܶÈΪ1.4g/cm3£¬50%µÄÁòËáÎïÖʵÄÁ¿Å¨¶ÈΪ (±£ÁôÁ½Î»Ð¡Êý)£¬50%µÄÁòËáÓë30%µÄÁòËáµÈÌå»ý»ìºÏ£¬»ìºÏËáµÄŨ¶ÈΪ £¨Ìî>¡¢<¡¢=£©40%¡£
£¨2£©Êµ¼ÊÉú²úÓÃ20%·¢ÑÌÁòËᣨ100¿Ë·¢ÑÌÁòËẬSO3 20¿Ë£©ÅäÖÆÏ¡ÁòËᣬÈôÓÃSO3¡¤nH2O±íʾ20%µÄ·¢ÑÌÁòËᣬÔòn=____________(±£ÁôÁ½Î»Ð¡Êý)¡£
£¨3£©ÂÌ·¯ÔÚ¿ÕÆøÖÐÈÝÒ×±»²¿·ÖÑõ»¯ÎªÁòËáÌú£¬ÏÖÈ¡7.32¿Ë¾§ÌåÈÜÓÚÏ¡ÑÎËáºó£¬¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬¹ýÂ˵óÁµí9.32¿Ë£»ÔÙͨÈë112mL£¨±ê×¼×´¿ö£©ÂÈÆøÇ¡ºÃ½«Fe2£«ÍêÈ«Ñõ»¯£¬ÍƲ⾧ÌåµÄ»¯Ñ§Ê½Îª ¡£
£¨4£©ÁòËáÑÇÌú泥Û(NH4)2SO4¡¤FeSO4¡¤6H2O£Ý(Ë׳ÆĪ¶ûÑÎ)£¬½ÏÂÌ·¯Îȶ¨£¬ÔÚ·ÖÎö»¯Ñ§Öг£ÓÃÀ´ÅäÖÆFe2+µÄ±ê×¼ÈÜÒº£¬ÓôËFe2+µÄ±ê×¼ÈÜÒº¿ÉÒԲⶨʣÓàÏ¡ÏõËáµÄÁ¿¡£ÏÖÈ¡8.64¿ËCu2SºÍCuSµÄ»ìºÏÎïÓÃ200mL2mol/LÏ¡ÏõËáÈÜÒº´¦Àí£¬·¢Éú·´Ó¦ÈçÏ£º
10NO3-£«3Cu2S£«16H£«¡ú6Cu2£«£«10NO¡ü£«3SO42-£«8H2O
8NO3-£«3CuS£«8H£«¡ú 3Cu2£«£«3 SO42-£«8NO¡ü+ 4H2O
Ê£ÓàµÄÏ¡ÏõËáÇ¡ºÃÓëV mL 2 mol/L (NH4)2Fe(SO4)2ÈÜÒºÍêÈ«·´Ó¦¡£
ÒÑÖª£ºNO3-£«3Fe2£«£«4H£«¡ú NO¡ü£«3Fe3+£«2H2O
¢Ù VÖµ·¶Î§ £»
¢Ú ÈôV=48£¬ÊÔ¼ÆËã»ìºÏÎïÖÐCuSµÄÖÊÁ¿·ÖÊý£¨±£ÁôÁ½Î»Ð¡Êý£©¡£
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