ijÈÜÒº¼×ÖпÉÄܺ¬ÓÐÏÂÁÐÀë×ÓÖеļ¸ÖÖ£ºK+¡¢NO¡¢SO¡¢NH¡¢CO£¨²»¿¼ÂÇÈÜÒºÖÐÓÉË®µçÀëµÄÉÙÁ¿µÄH+ºÍOH££©£¬È¡200mL¸ÃÈÜÒº£¬·ÖΪÁ½µÈ·Ý½øÐÐÏÂÁÐʵÑ飺
ʵÑé1£ºµÚÒ»·Ý¼ÓÈë×ãÁ¿µÄÉռ¼ÓÈÈ£¬²úÉúµÄÆøÌåÔÚ±ê×¼×´¿öÏÂΪ672mL£»
ʵÑé2£ºµÚ¶þ·ÝÏȼÓÈë×ãÁ¿µÄÑÎËᣬÎÞÏÖÏó£¬ÔÙ¼ÓÈë×ãÁ¿µÄBaCl2ÈÜÒº£¬µÃ¹ÌÌå2.33g¡£
£¨1£©¼×ÈÜÒºÖÐÒ»¶¨´æÔÚµÄÀë×ÓÊÇ £»
£¨2£©¼×ÈÜÒºÖÐÒ»¶¨²»´æÔÚµÄÀë×ÓÊÇ £»
£¨3£©¼×ÈÜÒºÖпÉÄÜ´æÔÚµÄÀë×ÓÊÇ £»ÄãµÃ³ö´Ë½áÂÛµÄÀíÓÉÊÇ ¡£
£¨1£©NO¡¢SO¡¢NH £¨2£©CO
£¨3£©K+ ÓÉÌâÒâÖª£ºÈÜÒºÖк¬NH0.03mol£¬SO0.01mol£¬¾ÝµçÖÐÐÔÔÀí£¬±ØÓÐNO£¬ÈôNO Ϊ0.01mol£¬ÔòÎÞK+£¬ÈôNO´óÓÚ0.01mol£¬ÔòÓÐK+¡£
½âÎöÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝʵÑé1£ºµÚÒ»·Ý¼ÓÈë×ãÁ¿µÄÉռ¼ÓÈÈ£¬ÒòNH4++OH-=NH3¡ü+H2O£¬»á²úÉú±ê×¼×´¿öÏÂΪ224mLÆøÌ壬֤Ã÷º¬ÓÐNH4+£¬ÇÒÎïÖʵÄÁ¿Îª0.01mol£»
ʵÑé2£ºµÚ¶þ·ÝÏȼÓÈë×ãÁ¿µÄÑÎËᣬÎÞÏÖÏó£¬ÔòÒ»¶¨²»º¬ÓÐCO32-£¬ÔÙ¼Ó×ãÁ¿µÄBaCl2ÈÜÒº£¬µÃ¹ÌÌå2.33g£¬Ö¤Ã÷Ò»¶¨º¬ÓÐSO42-£¬ÇÒÎïÖʵÄÁ¿Îª£ºn="m/M" ="2.33g/233g/mol" =0.01mol£»¸ù¾ÝÈÜÒºÖеĵçºÉÊغ㣬ÈÜÒº³ÊµçÖÐÐÔ£¬ÔòÒ»¶¨º¬ÓмØÀë×Ó£¬ÇÒ¼ØÀë×ÓµÄŨ¶È¡Ý0.01mol¡Á2?0.01mol¡Á 0.1L ¨T0.1mol/L£¬ËùÒÔ¸ÃÈÜÒºÖп϶¨º¬ÓÐNH4+¡¢S042-¡¢K+£¬
¹Ê´ð°¸Îª£ºK+¡¢NH4+¡¢S042-£»
£¨2£©ÊµÑé2£ºµÚ¶þ·ÝÏȼÓÈë×ãÁ¿µÄÑÎËᣬÎÞÏÖÏó£¬Ôò²»º¬ÓÐCO32-£¬Èç¹ûÓÐCO32-£¬CO32-+2H+=H2O+CO2¡ü£¬»áÓжþÑõ»¯Ì¼ÆøÌå²úÉú£¬¹Ê´ð°¸Îª£ºCO32-£»
£¨3£©¸ù¾Ý£¨1£©Öª£º¼ØÀë×ÓµÄŨ¶È¡Ý0.01mol¡Á2?0.01mol ¡Á0.1L =0.1mol/L£¬Èç¹ûK+µÄÎïÖʵÄÁ¿µÈÓÚ0.01mol£¬Ôò²»º¬NO3-£¬Èç¹ûK+µÄÎïÖʵÄÁ¿´óÓÚ0.01mol£¬Ôò»¹Ó¦º¬ÓÐNO3-£¬
¹Ê´ð°¸Îª£ºNO3-£»ÓÉÌâÒâÖª£¬NH4+ÎïÖʵÄÁ¿Îª0.01mol£¬SO42-ÎïÖʵÄÁ¿Îª0.01mol£¬¸ù¾ÝÈÜÒº³ÊµçÖÐÐÔÔÀí£¬Ó¦¸Ãº¬ÓÐK+£¬Èç¹ûK+µÄÎïÖʵÄÁ¿µÈÓÚ0.01mol£¬Ôò²»º¬NO3-£¬Èç¹ûK+µÄÎïÖʵÄÁ¿´óÓÚ0.01mol£¬Ôò»¹Ó¦º¬ÓÐNO3-£®
¿¼µã£º±¾Ì⿼²éÁËÈÜÒºÖгɷֵļø±ð
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£ºÌî¿ÕÌâ
£¨1£©º¬ÂÈÏû¶¾¼Á¿É·À¼×ÐÍH1N1Á÷¸Ð¡£¶þÑõ»¯ÂÈÊÇÄ¿Ç°¹ú¼ÊÉϹ«ÈϵĵÚËÄ´ú¸ßЧ¡¢ÎÞ¶¾µÄ¹ãÆ×Ïû¶¾¼Á£¬Ëü¿ÉÓÉKClO3ÔÚH2SO4´æÔÚÏÂÓëNa2SO3·´Ó¦ÖƵá£Çëд³ö·´Ó¦µÄÀë×Ó·½³Ìʽ_____________________________________________________________________________¡£
£¨2£©Ä³ÎÞÉ«ÈÜÒºÖ»º¬ÓÐÏÂÁÐ8ÖÖÀë×ÓÖеÄij¼¸ÖÖ£ºNa+¡¢H+¡¢Mg2+¡¢Ag+¡¢Cl-¡¢OH-¡¢¡£ÒÑÖª¸ÃÈÜÒº¿ÉÓëAl2O3·´Ó¦£¬Ôò£º
¢Ù¸ÃÈÜÒºÓëAl2O3·´Ó¦ºóÓÐAl3+Éú³É£¬ÔòÔÈÜÒºÖÐÒ»¶¨º¬ÓÐ_______£¬Ò»¶¨²»»áº¬ÓдóÁ¿µÄ_______¡£
¢Ú¸ÃÈÜÒºÓëAl2O3·´Ó¦ºóÓÐÉú³É£¬ÔòÔÈÜÒºÖÐÒ»¶¨º¬ÓÐ_______£¬¿ÉÄܺ¬ÓдóÁ¿µÄ_______¡£
¢Ûд³ö¸ÃÈÜÒºÓëAl2O3·´Ó¦Éú³ÉµÄÀë×Ó·½³Ìʽ____________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ij»ìºÏÈÜÒºÖпÉÄÜ´óÁ¿º¬ÓеÄÀë×ÓÈçϱíËùʾ£º
ÑôÀë×Ó | H£«¡¢K£«¡¢Al3£«¡¢NH4+¡¢Mg2£« |
ÒõÀë×Ó | Cl£¡¢Br£¡¢OH£¡¢CO32-¡¢AlO2- |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÓÐÏÂÁÐÎïÖÊ£º¢ÙÇâÑõ»¯ÄƹÌÌå ¢ÚÍË¿ ¢ÛÑÎËá ¢ÜÈÛÈÚÂÈ»¯ÄÆ ¢Ý¶þÑõ»¯Ì¼ÆøÌå ¢Þ°±Ë® ¢ßÕáÌǾ§Ìå¡£ÇëÓÃÐòºÅÌî¿Õ£º
(1)ÉÏÊö״̬Ï¿ɵ¼µçµÄÊÇ________¡£ (2)ÊôÓÚµç½âÖʵÄÊÇ________¡£
(3)ÊôÓڷǵç½âÖʵÄÊÇ________¡£ (4)ÉÏÊö״̬ϵĵç½âÖʲ»Äܵ¼µçµÄÊÇ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÔÚNa£«Å¨¶ÈΪ0£®5mol£¯LµÄij³ÎÇåÈÜÒºÖУ¬»¹¿ÉÄܺ¬ÓÐϱíÖеÄÈô¸ÉÖÖÀë×Ó£º
È¡¸ÃÈÜÒº100mL½øÐÐÈçÏÂʵÑ飨ÆøÌåÌå»ýÔÚ±ê×¼×´¿öϲⶨ£©£º
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑé¢ñÄÜÈ·¶¨Ò»¶¨²»´æÔÚµÄÑôÀë×ÓÊÇ____________¡£
£¨2£©Í¨¹ýʵÑé¢ñ¡¢¢ò¡¢¢óºÍ±ØÒª¼ÆË㣬ÌîдϱíÖÐÒõÀë×ÓµÄŨ¶È£¨ÄܼÆËã³öµÄ£¬Ìîд¼ÆËã½á¹û£¬Ò»¶¨²»´æÔÚµÄÀë×ÓÌî¡°0¡±£¬²»ÄÜÈ·¶¨ÊÇ·ñ´æÔÚµÄÀë×ÓÌî¡°?¡±£©
£¨3£©ÅжÏK£«ÊÇ·ñ´æÔÚ£¬Èô´æÔÚÇóÆä×îСŨ¶È£¬Èô²»´æÔÚ˵Ã÷ÀíÓÉ_____________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
£¨1£©¸ù¾Ý»¯Ñ§·½³Ìʽ£¬Ð´³ö¶ÔÓ¦µÄÀë×Ó·½³Ìʽ£º
¢ÙAgNO3+KCl=AgCl¡ý£«KNO3
¢ÚCuO+H2SO4=CuSO4£«H2O
£¨2£©Ð´³öÒ»¸öÄÜʵÏÖÏÂÁÐÀë×Ó·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢ÙFe+Cu2+=Fe2++Cu
¢ÚCO32-+2H+=CO2¡ü£«H2O
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
I£®£¨1£©ÏòBa(OH)2ÈÜÒºÖмÓÈëÏ¡ÁòËᣬÇëÍê³ÉÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ£º________________________________¡£
£¨2£©ÏÂÁÐÈýÖÖÇé¿öÏ£¬Àë×Ó·½³ÌʽÓ루1£©ÏàͬµÄÊÇ_________________¡£
A£®ÏòNaHSO4ÈÜÒºÖÐÖðµÎ¼ÓÈëBa(0H)2ÈÜÒºÖÁÏÔÖÐÐÔ
B£®ÏòNaHSO4ÈÜÒºÖÐÖðµÎ¼ÓÈëBa(0H)2ÈÜÒºÖÁSO42£Ç¡ºÃÍêÈ«³Áµí
C£®ÏòNaHSO4ÈÜÒºÖÐÖðµÎ¼ÓÈëBa(0H)2ÈÜÒºÖÁ¹ýÁ¿
¢ò£®ÊµÑéÊÒ¿ÉÒÔÓÃÂÈËá¼ØºÍŨÑÎËá·´Ó¦ÖÆÈ¡ÂÈÆø£¬·´Ó¦Ê½ÈçÏ£ºKClO3£«6HCl(Ũ)£½KCl£«3Cl2¡ü£«3H2O
£¨1£©ÓÃË«ÏßÇÅ·¨±íʾÉÏÊö·´Ó¦Öеç×ÓתÒƵķ½ÏòºÍÊýÄ¿¡£
£¨2£©·´Ó¦Öз¢ÉúÑõ»¯·´Ó¦µÄÎïÖÊÊÇ____________£¨Ìѧʽ£©£¬±»»¹ÔµÄÔªËØÊÇ____________________£¨ÌîÔªËØÃû³Æ£©¡£
£¨3£©Ñõ»¯¼ÁÓ뻹ԼÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ____________________¡£
£¨4£©Èô·´Ó¦Öб»Ñõ»¯µÄÑÎËáΪ1mol£¬ÔòÉú³ÉµÄÂÈÆøÌå»ýΪ_______________£¨±ê×¼×´¿öÏ£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÓÐÒ»¹ÌÌå»ìºÏÎ¿ÉÄÜÓÉNaI¡¢KCl¡¢Na2CO3¡¢Na2SO4¡¢CaCl2¡¢Cu(NO3)2ÖеÄÒ»ÖÖ»ò¼¸ÖÖ×é³É£¬ÎªÁ˼ìÑéËùº¬µÄÎïÖÊ£¬×öÁËÒÔÏÂʵÑ飺
¢ÙÈ¡ÉÙÐí¹ÌÌåÈÜÓÚË®£¬µÃµ½ÎÞɫ͸Ã÷ÈÜÒº£»
¢ÚÍù´ËÈÜÒºÖеμÓÂÈ»¯±µÈÜÒº£¬Óа×É«³ÁµíÉú³É£»
¢Û¹ýÂË£¬Íù³ÁµíÖмÓÈë×ãÁ¿µÄÏ¡ÏõËᣬ·¢ÏÖ³ÁµíûÓÐÈ«²¿ÈܽâÇÒÓÐÎÞÉ«ÎÞζµÄÆøÌåÉú³É¡£
¢ÜÍùÂËÒºÖмÓÈë×ãÁ¿µÄÐÂÖƵÄÂÈË®£¬ÔÙ¼ÓÈëÉÙÐíÆûÓÍ£¬Õñµ´£¬¾²Öã¬ÉϲãÒºÌå³Ê×ϺìÉ«¡£
£¨1£©ÊÔÅжϣº¹ÌÌå»ìºÏÎïÖп϶¨º¬ÓÐ £¬Ò»¶¨Ã»ÓÐ £¬¿ÉÄܺ¬ÓÐ________________¡£
£¨2£©¶Ô¿ÉÄܺ¬ÓеÄÎïÖÊ£¬ÈçºÎ½øÐÐʵÑéÒÔ½øÒ»²½¼ìÑé¡£ ¡£
£¨3£©ÊµÑé¢ÜÖз¢ÉúµÄ»¯Ñ§·´Ó¦ÊôÓÚ ·´Ó¦£¨Ìî·´Ó¦ÀàÐÍ£©£¬Ö÷ҪʵÑé²Ù×÷Ãû³Æ½Ð ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÔÚÏ¡ÁòËáËữµÄº¬6 mol KIÈÜÒºÖÐÖðµÎ¼ÓÈëKBrO3ÈÜÒº£¬Õû¸ö¹ý³ÌÖꬵâÎïÖÊÓëËù¼ÓÈëKBrO3ÎïÖʵÄÁ¿µÄ¹ØϵÈçͼ¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©bµãʱ£¬¶ÔÓ¦º¬µâÎïÖʵĻ¯Ñ§Ê½Îª ¡£
£¨2£©b¡úc¹ý³ÌÖУ¬½öÓÐÒ»ÖÖÔªËØ·¢Éú»¯ºÏ¼Û±ä»¯£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ²¢±ê³öµç×ÓתÒÆ·½ÏòÓëÊýÄ¿ ¡£
£¨3£©µ±n(KBrO3)=4molʱ£¬ÌåϵÖжÔÓ¦º¬µâÎïÖʵĻ¯Ñ§Ê½Îª ¡£
£¨4£©ËáÐÔÌõ¼þÏ£¬Br2¡¢IO3£¡¢BrO3£¡¢I2Ñõ»¯ÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòΪ ¡£
£¨5£©ÔÚÏ¡ÁòËáËữµÄKBrO3ÈÜÒºÖв»¶ÏµÎÈëµí·ÛKIÈÜÒº£¬±ßµÎ±ßÕñµ´¡£ÔòʵÑé¹ý³ÌÖеĿÉÄܹ۲쵽µÄÏÖÏóΪ ¡£
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