·ÖÎö £¨1£©ÏÈÇó³öagÑõÆøº¬ÓеķÖ×ÓÊý£¬ÔÙÇó³öÆäÎïÖʵÄÁ¿£¬×îºó¸ù¾ÝÆøÌåÌå»ý¹«Ê½Çó³öÌå»ý£»
£¨2£©¸ù¾Ýn=cV¼ÆËã³öÁ½ÁòËáÈÜÒºÖÐÁòËáµÄÎïÖʵÄÁ¿£¬´Ó¶øµÃ³ö»ìºÏÒºÖÐÁòËáµÄ×ÜÎïÖʵÄÁ¿£¬ÔÙ¸ù¾Ýc=$\frac{n}{V}$¼ÆËã³ö»ìºÏÒºÖÐÁòËáµÄÎïÖʵÄÁ¿Å¨¶ÈºÍÇâÀë×ÓŨ¶È£®
£¨3£©¸ù¾Ýn=$\frac{m}{M}$¼ÆËãmgAl3+µÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾Ý$\frac{V}{5}$mLÈÜÒºÖк¬ÓеÄAl3+µÄÎïÖʵÄÁ¿£¬¸ù¾Ý»¯Ñ§Ê½¿ÉÖªn£¨SO42-£©=$\frac{3}{2}$n£¨Al3+£©£¬ÔÙ¸ù¾Ýc=$\frac{n}{V}$¼ÆËãÏ¡ÊͺóÈÜÒºÖÐÁòËá¸ùÀë×ÓµÄŨ¶È£®
£¨4£©¸ù¾Ýn=$\frac{V}{{V}_{m}}$¼ÆËãÂÈ»¯ÇâÆøÌåµÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾Ým=nM¼ÆËãÂÈ»¯ÇâÆøÌåµÄÖÊÁ¿£¬¸ù¾Ým=¦ÑV¼ÆËãË®µÄÖÊÁ¿£¬ÈÜÒºÖÊÁ¿=ÆøÌåÖÊÁ¿+Ë®µÄÖÊÁ¿£¬ÀûÓÃV=$\frac{m}{V}$À´¼ÆËãÈÜÒºµÄÌå»ý£¬¸ù¾Ýc=$\frac{n}{V}$¼ÆËã¸ÃÆøÌåµÄÎïÖʵÄÁ¿Å¨¶È£®
½â´ð ½â£º£¨1£©¸ù¾ÝÆä×é·Ö¶¼ÎªÑõÆøÏàͬ£¬ËùÒÔÆäÖÊÁ¿Óë·Ö×ÓÊý³ÉÕý±È£¬agÑõÆøº¬ÓеÄÑõÔ×ÓÊýΪb£¬º¬ÓеķÖ×ÓÊýΪ£º$\frac{b}{2}$¸ö£¬ÖÊÁ¿Óë·Ö×ÓÊýµÄ±ÈÁÐʽΪ£ºag£º$\frac{b}{2}$¸ö=cg£ºx¸ö£¬x=$\frac{bc}{2a}$¸ö£»
cgÑõÆøº¬ÓеÄÎïÖʵÄÁ¿Îª£ºn=$\frac{bc}{2a{N}_{A}}$molÆäÌå»ýΪ£ºV=n¡ÁVm =$\frac{bc}{2a{N}_{A}}$mol¡ÁVm =$\frac{11.2bc}{a•{N}_{A}}$L
¹Ê´ð°¸Îª£º$\frac{11.2bc}{a•{N}_{A}}$L£»
£¨2£©100mL 0.3mol•L-1 ºÍ300mL 0.25mol•L-1µÄÁòËá»ìºÏÒºÖÐÁòËáµÄÎïÖʵÄÁ¿Îª£º0.3mol/L¡Á0.1L+0.25mol/L¡Á0.3L=0.105mol£¬
ÅäÖƵÄÈÜÒºµÄÌå»ýΪ0.5L£¬Ôò¸Ã»ìºÏÈÜÒºÖÐÁòËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ£º$\frac{0.105mol}{0.5L}$=0.21mol/L£¬
ËùÒÔÇâÀë×ÓŨ¶ÈΪ0.42mol/L£»
¹Ê´ð°¸Îª£º0.42mol/L£»
£¨3£©mgAl3+µÄÎïÖʵÄÁ¿Îª$\frac{mg}{27g/mol}$=$\frac{m}{27}$mol£¬
È¡$\frac{V}{5}$mlÈÜÒºÖк¬ÓÐAl3+µÄÎïÖʵÄÁ¿Îª$\frac{1}{5}$¡Á$\frac{m}{27}$mol£¬
¹Ên£¨SO42-£©=$\frac{3}{2}$n£¨Al3+£©=$\frac{1}{5}$¡Á$\frac{m}{27}$mol¡Á$\frac{3}{2}$=$\frac{m}{90}$mol£¬
¹ÊÏ¡ÊÍΪ5VmLºóÈÜÒºµÄc£¨SO42-£©=$\frac{\frac{m}{90}mol}{5V¡Á1{0}^{-3}L}$=$\frac{20m}{9V}$mol/L£¬
¹Ê´ð°¸Îª£º$\frac{20m}{9V}$mol/L£»
£¨4£©±ê×¼×´¿öÏ£¬½«VLÂÈ»¯ÇâµÄÎïÖʵÄÁ¿Îª$\frac{VL}{22.4L/mol}$=$\frac{V}{22.4}$mol£¬
ËùÒÔÂÈ»¯ÇâÆøÌåµÄÖÊÁ¿Îª$\frac{V}{22.4}$mol¡Á36.5g/mol=$\frac{36.5V}{22.4}$g£¬
ËùÒÔÈÜÒºµÄÖÊÁ¿Îª1000mL¡Á1g/mL+$\frac{36.5V}{22.4}$g=£¨1000+$\frac{36.5V}{22.4}$£©g£¬
ËùÒÔÈÜÒºµÄÌå»ýΪ$\frac{£¨1000+\frac{36.5V}{22.4}£©}{dg/mL}$¡Á10-3L=$\frac{22400+36.5V}{22400d}$L£¬
ËùÒÔÎïÖʵÄÁ¿Å¨¶ÈΪ$\frac{\frac{V}{22.4}mol}{\frac{22400+36.5V}{22400d}L}$=$\frac{1000dV}{22400+36.5V}$mol/L£»
¹Ê´ð°¸Îª£ºA£®
µãÆÀ ±¾Ì⿼²éÁË»ìºÏÒºÖÐÎïÖʵÄÁ¿Å¨¶ÈµÄ¼ÆË㣬ÌâÄ¿ÄѶÈÖеȣ¬Ã÷È·ÎïÖʵÄÁ¿Å¨¶ÈµÄ¸ÅÄî¼°±í´ïʽ¼´¿É½â´ð£¬ÊÔÌâ²àÖØ»ù´¡ÖªÊ¶µÄ¿¼²é£¬ÅàÑøѧÉúÁé»îÓ¦ÓÃËùѧ֪ʶµÄÄÜÁ¦£®
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£ºÑ¡ÔñÌâ
A£® | CH3£¨ CH2£©2 CH3 ºÍ CH3 CH=CH CH3 | |
B£® | CH3£¨ CH2£©2 CH3 ºÍ CH3 CH2 CH£¨ CH3£© CH2 CH3 | |
C£® | CH3 C¡ÔC CH3 ºÍCH2=CH CH=CH2 | |
D£® | CH3 C¡ÔC CH3 ºÍ CH3 CH=CH CH3 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ±ê×¼×´¿öÏ£¬2.24 L CCl4Ëùº¬µÄÔ×ÓÊýΪ0.5 NA | |
B£® | ±ê×¼×´¿öÏÂ0.5mol NOºÍ0.5mol O2×é³ÉµÄ»ìºÏÆøÌåµÄÌå»ýԼΪ22.4L | |
C£® | 78 g Na2O2 Óë×ãÁ¿Ë®·´Ó¦×ªÒƵç×ÓÊýΪNA | |
D£® | ÎïÖʵÄÁ¿Å¨¶ÈΪ0.5mol/LµÄMgCl2ÈÜÒºÖУ¬º¬ÓÐCl-¸öÊýΪ1 NA |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | 0.25mol | B£® | 2.5mol | C£® | 0.15mol | D£® | 1.5mol |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | Ïòä廯ÑÇÌúÈÜÒºÖÐͨÈëÉÙÁ¿ÂÈÆø£º2Fe2++Cl2¨T2Fe3++2Cl- | |
B£® | ±½·ÓÄÆÈÜÒºÖÐͨÈëÉÙÁ¿CO2£º2C6H5O-+CO2+H2O-¡ú2C6H5OH+CO32- | |
C£® | ÏòÁòËáÇâÄÆÈÜÒºÖеÎÈëÇâÑõ»¯±µÈÜÒºÖÁÖÐÐÔ£ºH++SO42-+Ba2++OH-¨TBaSO4¡ý+H2O | |
D£® | ½«ÇâÑõ»¯Ìú·ÛÄ©¼ÓÈëÇâµâËáÖУºFe£¨OH£©3+3H+¨TFe3++3H2O |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | Cl2¡¢HClO¡¢HCl¡¢H2O | |
B£® | Cl¡¢Cl-¡¢Cl2¡¢H2O | |
C£® | Cl2¡¢HCl¡¢H2O¡¢Cl-¡¢H+¡¢ClO-ºÍ¼«ÉÙÁ¿OH- | |
D£® | Cl2¡¢HClO¡¢H2O¡¢Cl-¡¢H+¡¢ClO-ºÍ¼«ÉÙÁ¿OH- |
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