ÓÐÒ»·Ýº¬1 mol HClºÍ1 mol MgSO4µÄÈÜÒº£¬µ±ÏòÆäÖеμÓ1 mol/L µÄBa(OH)2 ÈÜҺʱ£¬¼ÆË㣺
£¨1£©µ±Ba(OH)2 ÈÜÒºµÄµÎ¼ÓÁ¿µ½´ï Lʱ£¬¿ªÊ¼ÓÐÇâÑõ»¯Ã¾³ÁµíÎö³ö£¿
£¨2£©µ±µÎ¼ÓµÄBa(OH)2 ÈÜҺΪ1Lʱ£¬³ÁµíÎïµÄ»¯Ñ§Ê½Îª
ÿÖÖ³ÁµíÎïµÄÎïÖʵÄÁ¿Îª ¡£
£¨3£©µ±Éú³É³Áµí×ÜÖÊÁ¿×î´óʱ£¬Ba(OH)2 ÈÜÒºµÄµÎ¼ÓÁ¿Îª L£¬´Ëʱ³ÁµíµÄ×ÜÖÊÁ¿Îª g¡£
£¨4£©ÔÚÒÔÏÂ×ø±êͼÖлæÖƳöÕû¸ö¹ý³ÌÖвúÉú³ÁµíµÄ×ÜÎïÖʵÄÁ¿£¨n£©Óë¼ÓÈë Ba(OH)2 ÈÜÒºÌå»ý£¨V£©Ö®¼äµÄ¹Øϵͼ¡£
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
ÓÐÒ»·Ý¹ÌÌ壬¿ÉÄÜÓÉNaOH¡¢NaHCO3ºÍNa2CO3ÖеÄÒ»ÖÖ»òÁ½ÖÖ×é³É£¬¿ÉÓ÷ֲ½µÎ¶¨·¨£¨Ë«Ö¸Ê¾¼Á·¨£©²â¶¨Æä×é³ÉºÍ¸÷×é·ÖµÄÖÊÁ¿·ÖÊý£¬Æä²½ÖèÈçÏ£º
£¨1£©³ÆÈ¡w g¹ÌÌå·ÅÈë׶ÐÎÆ¿ÖУ¬¼ÓÈë30 mLÕôÁóˮʹ֮ÍêÈ«Èܽ⣬ÏȼÓÈë2¡«3µÎ·Ó̪×÷ָʾ¼Á¡£
£¨2£©ÓÃ0.1 mol¡¤L-1µÄ±ê×¼ÑÎËá½øÐе樣¬ÖÁ׶ÐÎÆ¿ÖÐÈÜÒºÓÉ__________É«±ä³É__________ɫʱ£¬Í£Ö¹µÎ¶¨£¬´ËʱÏûºÄ0.1 mol¡¤L-1µÄ±ê×¼ÑÎËáV1 mL¡£ÔÚÕâ²½²Ù×÷ÖУ¬Ó¦ÓÃ__________ÊÖ¿ØÖÆ»îÈû£¬__________ÊÖ³Ö׶ÐÎÆ¿²¢²»¶Ï__________£¬Á½ÑÛ×¢ÊÓ______________________________¡£
£¨3£©ÔÙÏò׶ÐÎÆ¿ÖмÓ2¡«3µÎ¼×»ù³È×÷ָʾ¼Á£¬¼ÌÐøÓÃ0.1 mol¡¤L-1±ê×¼ÑÎËáµÎ¶¨£¬ÖÁÈÜÒºÓÉ__________É«±äΪ__________ɫʱֹͣµÎ¶¨£¬´ËʱÏûºÄ0.1 mol¡¤L-1±ê×¼ÑÎËáV2 mL¡£
£¨4£©ÈôV2£¾V1ʱ£¬ÔòÔ¹ÌÌå×é³É¿ÉÄÜΪ__________£¨Ìî±àºÅ£©¡£
A.Ö»º¬NaOH
B.Ö»º¬Na2CO3
C.Na2CO3ºÍNaHCO3»ìºÏÎï
D.NaOHºÍNa2CO3µÄ»ìºÏÎï
E.Ö»º¬ÓÐNaHCO3
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêÁÉÄþÊ¡µ¤¶«Êпíµé¶þÖиßÒ»ÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÊµÑéÌâ
£¨10·Ö£©ÓÐÒ»·Ýï§ÑλìºÏÎȡÆäÖÊÁ¿Îª10.00gʱ£¬º¬ÓÐxmolµÄ(NH4)2SO4ºÍymolµÄNH4HSO4¡£ÒÔÏÂʵÑéÈ¡Óò»Í¬ÖÊÁ¿µÄï§ÑλìºÏÎï¾ù¼ÓÈëͬŨ¶ÈµÄ50.00mLµÄNaOHÈÜÒº, ¼ÓÈȳä·Ö·´Ó¦£¬½«²úÉúµÄÆøÌåA¾¸ÉÔïÖ®ºóÓÃŨÁòËáÎüÊÕ¡£
²â¶¨½á¹ûÈçÏ£»
ï§ÑλìºÏÎïÖÊÁ¿Îª10.00gºÍ20.00gʱ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿Ïàͬ£»
ï§ÑλìºÏÎïÖÊÁ¿Îª30.00gʱ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿Îª0.68g£»
ï§ÑλìºÏÎïÖÊÁ¿Îª40.00gʱ£¬Å¨ÁòËáµÄÖÊÁ¿²»±ä¡£
£¨1£©AÆøÌåµÄ»¯Ñ§Ê½Îª______________
£¨2£©10.00gï§ÑλìºÏÎïÖк¬ÓÐNH4+µÄÎïÖʵÄÁ¿Îª_____mol(Óú¬x£¬yµÄ¹Øϵʽ±íʾ)£¬10.00gï§ÑλìºÏÎïÈÜÓÚË®ºó¿ÉµçÀë³öH+µÄÎïÖʵÄÁ¿Îª___mol(Óú¬x£¬yµÄ¹Øϵʽ±íʾ)¡£
£¨3£©ÔÚº¬ÓÐ2 mol NH4+Óë1 mol H+µÄÈÜÒºÖмÓÈë2 molOH¡ª£¬¼ÓÈȳä·Ö·´Ó¦£¬¿É²úÉúÆøÌåA__________ mol
£¨4£©¼ÆË㣺50.00mLµÄNaOHÈÜÒºµÄŨ¶ÈΪ__________mol/L,¸Ãï§ÑÎÖеªÔªËصÄÖÊÁ¿·ÖÊýÊÇ %£»£¨¼ÆËã½á¹û¾ù±£ÁôÁ½Î»Ð¡Êý£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015½ìÁÉÄþÊ¡µ¤¶«ÊиßÒ»ÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÊµÑéÌâ
£¨10·Ö£©ÓÐÒ»·Ýï§ÑλìºÏÎȡÆäÖÊÁ¿Îª10.00gʱ£¬º¬ÓÐxmolµÄ(NH4)2SO4ºÍymolµÄNH4HSO4¡£ÒÔÏÂʵÑéÈ¡Óò»Í¬ÖÊÁ¿µÄï§ÑλìºÏÎï¾ù¼ÓÈëͬŨ¶ÈµÄ50.00mLµÄNaOHÈÜÒº, ¼ÓÈȳä·Ö·´Ó¦£¬½«²úÉúµÄÆøÌåA¾¸ÉÔïÖ®ºóÓÃŨÁòËáÎüÊÕ¡£
²â¶¨½á¹ûÈçÏ£»
ï§ÑλìºÏÎïÖÊÁ¿Îª10.00gºÍ20.00gʱ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿Ïàͬ£»
ï§ÑλìºÏÎïÖÊÁ¿Îª30.00gʱ£¬Å¨ÁòËáÔö¼ÓµÄÖÊÁ¿Îª0.68g£»
ï§ÑλìºÏÎïÖÊÁ¿Îª40.00gʱ£¬Å¨ÁòËáµÄÖÊÁ¿²»±ä¡£
£¨1£©AÆøÌåµÄ»¯Ñ§Ê½Îª______________
£¨2£©10.00gï§ÑλìºÏÎïÖк¬ÓÐNH4+µÄÎïÖʵÄÁ¿Îª_____mol(Óú¬x£¬yµÄ¹Øϵʽ±íʾ)£¬10.00gï§ÑλìºÏÎïÈÜÓÚË®ºó¿ÉµçÀë³öH+µÄÎïÖʵÄÁ¿Îª___mol(Óú¬x£¬yµÄ¹Øϵʽ±íʾ)¡£
£¨3£©ÔÚº¬ÓÐ2 mol NH4+Óë1 mol H+µÄÈÜÒºÖмÓÈë2 molOH¡ª£¬¼ÓÈȳä·Ö·´Ó¦£¬¿É²úÉúÆøÌåA__________ mol
£¨4£©¼ÆË㣺50.00mLµÄNaOHÈÜÒºµÄŨ¶ÈΪ__________mol/L,¸Ãï§ÑÎÖеªÔªËصÄÖÊÁ¿·ÖÊýÊÇ %£»£¨¼ÆËã½á¹û¾ù±£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