ʵÑéÊÒÓмס¢ÒÒÁ½Æ¿ÎÞÉ«ÈÜÒº£¬ÆäÖÐһƿÊÇÏ¡ÑÎËᣬÁíһƿÊÇ̼ËáÄÆÈÜÒº¡£Îª²â¶¨¼×¡¢ÒÒÁ½Æ¿ÈÜÒºµÄ³É·Ö¼°ÎïÖʵÄÁ¿Å¨¶È£¬½øÐÐÒÔÏÂʵÑ飺
¢ÙÁ¿È¡25.00 mL¼×ÈÜÒº£¬ÏòÆäÖлºÂýµÎÈëÒÒÈÜÒº15.00 mL£¬¹²ÊÕ¼¯µ½224 mL(±ê×¼×´¿ö)ÆøÌå¡£
¢ÚÁ¿È¡15.00 mLÒÒÈÜÒº£¬ÏòÆäÖлºÂýµÎÈë¼×ÈÜÒº25.00 mL£¬¹²ÊÕ¼¯µ½112 mL(±ê×¼×´¿ö)ÆøÌå¡£
(1)Åжϣº¼×ÊÇ________ÈÜÒº£¬ÒÒÊÇ_________ÈÜÒº¡£
(2)ʵÑé¢ÚÖÐËù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________¡£
(3)¼×ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ £¬ÒÒÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ__________¡£
(4)½«n mL¼×ÈÜÒºÓëµÈÌå»ýÒÒÈÜÒº°´ÉÏÊöÁ½ÖÖʵÑ鷽ʽ½øÐз´Ó¦£¬Ëù²úÉúµÄÆøÌåµÄÌå»ýΪV mL(±ê×¼×´¿ö)£¬ÔòVµÄȡֵ·¶Î§ÊÇ_____________________________¡£
(1)HCl Na2CO3
(2)CO
+H+ ==== HCO
HCO
+H+ ==== CO2¡ü+H2O
(3)0.8 mol¡¤L£1 1 mol¡¤L£1
(4)0¡ÜV¡Ü8.96n
(1)µ±ÏòÑÎËáÖмÓÈëNa2CO3ÈÜҺʱ£¬·¢Éú·´Ó¦2H++CO
==== CO2¡ü+H2O,ÑÎËáÍêÈ«·´Ó¦ºó£¬ÔÙ¼ÓÈëµÄNa2CO3²»ÔٷųöCO2£»ÏòNa2CO3ÈÜÒºÖмÓÈëÑÎËáʱ£¬ÏÈ·¢Éú·´Ó¦CO
+H+ ==== HCO
,µ±ËùÓÐCO
¾ùת»¯ÎªHCO
ʱ£¬ÔÙ·¢Éú·´Ó¦HCO
+H+ ==== CO2¡ü+H2O¡£ÓÉʵÑé¢Ù¢ÚÖÐÊý¾Ý¿ÉÖªµÈÁ¿µÄ¼×ÓëµÈÁ¿µÄÒÒ·´Ó¦£¬ÊµÑé¢Ù²úÉúµÄCO2ÆøÌå¶à£¬ËµÃ÷¼×ΪHCl¡¢ÒÒΪNa2CO3¡£
(3)ÉèÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪc(HCl)£¬Na2CO3ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪc(Na2CO3)£¬ÓÉʵÑé¢ÙµÃ£º25.00¡Á10 ¨C3 L¡¤ c(HCl) ¡¤
=
£¬c(HCl)=0.8 mol¡¤L£1;
ÓÉʵÑé¢ÚµÃ£º 15.00¡Á10 £3 L¡¤c(Na2CO3)=25.00¡Á10£3 L¡Á0.8 mol¡¤L£1-
,
c(Na2CO3)=1 mol¡¤L£1¡£
(4)µÈÌå»ýµÄ¼×¡¢ÒÒ°´ÊµÑé¢ÙµÄ·½Ê½½øÐÐʱ²úÉúÆøÌåµÄÌå»ýΪ£º
¡Án¡Á10£3 L¡Á0.8 mol¡¤L£1¡Á22.4 l¡¤mol£1 =8.96¡Á10£3 nL,Èô°´ÊµÑé¢ÚµÄ·½Ê½½øÐУ¬ÓÉ·´Ó¦CO
+H+ ==== HCO-3¿ÉÖªÑÎËá²»×㣬²»²úÉúÆøÌ壬¹Ê0¡ÜV¡Ü8.96 n¡£
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
| ʵÑéÄ¿µÄ | ÊÔÑé·½·¨ |
| ¼ìÑéÂÈ»¯ÑÇÌúÊÇ·ñ±äÖÊ | D D |
| ³ýȥʳÑÎÖÐÉÙÁ¿Ï¸É³ | C C |
| ³ýȥ̼ËáÄÆ¹ÌÌåÖÐÉÙÁ¿Ì¼ËáÇâÄÆ | A A |
| ³ýȥþ·ÛÖлìÓеÄÉÙÁ¿ÂÁ·Û | B B |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨8·Ö£©
£¨1£©ÊµÑéÊÒÓмס¢ÒÒÁ½Æ¿ÎÞÉ«ÈÜÒº£¬ÆäÖÐһƿÊÇÏ¡ÑÎËᣬÁíһƿÊÇ̼ËáÄÆÈÜÒº¡£Îª²â¶¨¼×¡¢ÒÒÁ½Æ¿ÈÜÒºµÄ³É·Ö£¬½øÐÐÒÔÏÂʵÑ飺ȡ¼×ÈÜÒº£¬ÏòÆäÖлºÂýµÎÈëÒÒÈÜÒº£¬²¢±ßµÎ¼Ó±ßÕñµ´£¬¹Û²ìµ½¿ªÊ¼ÎÞÃ÷ÏÔÏÖÏ󣬺óÀ´ÓдóÁ¿ÆøÌåÉú³É¡£ÊµÑé¹ý³ÌÖÐËù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º £»¼×ÊÇ ÈÜÒº£¬ÒÒÊÇ ÈÜÒº£»
(2)ΪÁ˴ﵽϱíËùÁеÄʵÑéÄ¿µÄ£¬ÇëÑ¡ÔñºÏÊʵÄʵÑé·½·¨£¬½«Æä±êºÅÌîÔÚÏàÓ¦µÄ¿Õ¸ñÖÐ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêºÚÁú½Ê¡Äµµ¤½Ò»ÖиßÒ»ÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÊµÑéÌâ
£¨8·Ö£©
£¨1£©ÊµÑéÊÒÓмס¢ÒÒÁ½Æ¿ÎÞÉ«ÈÜÒº£¬ÆäÖÐһƿÊÇÏ¡ÑÎËᣬÁíһƿÊÇ̼ËáÄÆÈÜÒº¡£Îª²â¶¨¼×¡¢ÒÒÁ½Æ¿ÈÜÒºµÄ³É·Ö£¬½øÐÐÒÔÏÂʵÑ飺ȡ¼×ÈÜÒº£¬ÏòÆäÖлºÂýµÎÈëÒÒÈÜÒº£¬²¢±ßµÎ¼Ó±ßÕñµ´£¬¹Û²ìµ½¿ªÊ¼ÎÞÃ÷ÏÔÏÖÏ󣬺óÀ´ÓдóÁ¿ÆøÌåÉú³É¡£ÊµÑé¹ý³ÌÖÐËù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º £»¼×ÊÇ ÈÜÒº£¬ÒÒÊÇ ÈÜÒº£»
(2)ΪÁ˴ﵽϱíËùÁеÄʵÑéÄ¿µÄ£¬ÇëÑ¡ÔñºÏÊʵÄʵÑé·½·¨£¬½«Æä±êºÅÌîÔÚÏàÓ¦µÄ¿Õ¸ñÖÐ![]()
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìºÚÁú½Ê¡¸ßÒ»ÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÊµÑéÌâ
£¨8·Ö£©
£¨1£©ÊµÑéÊÒÓмס¢ÒÒÁ½Æ¿ÎÞÉ«ÈÜÒº£¬ÆäÖÐһƿÊÇÏ¡ÑÎËᣬÁíһƿÊÇ̼ËáÄÆÈÜÒº¡£Îª²â¶¨¼×¡¢ÒÒÁ½Æ¿ÈÜÒºµÄ³É·Ö£¬½øÐÐÒÔÏÂʵÑ飺ȡ¼×ÈÜÒº£¬ÏòÆäÖлºÂýµÎÈëÒÒÈÜÒº£¬²¢±ßµÎ¼Ó±ßÕñµ´£¬¹Û²ìµ½¿ªÊ¼ÎÞÃ÷ÏÔÏÖÏ󣬺óÀ´ÓдóÁ¿ÆøÌåÉú³É¡£ÊµÑé¹ý³ÌÖÐËù·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º £»¼×ÊÇ ÈÜÒº£¬ÒÒÊÇ ÈÜÒº£»
(2)ΪÁ˴ﵽϱíËùÁеÄʵÑéÄ¿µÄ£¬ÇëÑ¡ÔñºÏÊʵÄʵÑé·½·¨£¬½«Æä±êºÅÌîÔÚÏàÓ¦µÄ¿Õ¸ñÖÐ
![]()
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ʵÑéÊÒÓмס¢ÒÒÁ½Æ¿ÎÞÉ«ÈÜÒº£¬Ò»Æ¿ÊÇÏ¡ÑÎËᣬÁíһƿÊÇNaAlO2
ÈÜÒº£¬Îª²â¶¨¼×¡¢ÒÒÁ½Æ¿ÈÜÒºµÄ³É·Ö¼°ÎïÖʵÄÁ¿Å¨¶È£¬½øÐÐÒÔÏÂʵÑ飺
¢ÙÏò¼×ÈÜÒºÖлºÂýµÎ¼ÓÒÒÈÜÒº£¬Á¢¼´²úÉú³Áµí£»¶øÏòÒÒÈÜÒºÖлºÂýµÎ¼Ó¼×ÈÜÒº£¬¿ªÊ¼²¢ÎÞ³Áµí£¬Ò»¶Îʱ¼äºó²úÉú³Áµí¡£
¢ÚÈ¡40mLÒÒÈÜÒº£¬ÏòÆäÖлºÂýµÎÈë¼×ÈÜÒº20mL£¬¹²²úÉú³Áµí3.12g¡£
£¨1£©¼×Ê¢×°µÄÊÇ ¡£
£¨2£©ÈôÏò¼×ÈÜÒºÖлºÂýµÎ¼ÓÒÒÈÜÒºÖÁ¹ýÁ¿£¬ÏȺó·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
¡¢ ¡£
£¨3£©¢ÚÖгÁµíµÄÎïÖʵÄÁ¿Îª£º__________mol£¬¼×ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_____
mol/L£¬ÒÒÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_________mol/L¡£
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - 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