A | B | C | |
ÆðʼʱºÏ½ð·ÛÄ©µÄÖÊÁ¿/mg | 204 | 399 | 561 |
·´Ó¦½áÊøʱÉú³ÉÆøÌåµÄÌå»ý/mL | 224 | 336 | 336 |
A£® | ÆðʼʱÑÎËáµÄŨ¶ÈΪ1mol•L-1 | B£® | ¼ÓÈëµÄNaOHÈÜҺΪ40mL | ||
C£® | ºÏ½ðÖÐþ¡¢ÂÁµÄÎïÖʵÄÁ¿ÏàµÈ | D£® | ËùµÃµÄÂËÒºÖк¬ÈýÖÖÒõÀë×Ó |
·ÖÎö ÑÎËáŨ¶È¡¢Ìå»ýÒ»¶¨£¬AÖкϽðÖÊÁ¿Ð¡ÓÚBÖкϽðÖÊÁ¿£¬ÇÒAÖÐÉú³ÉÆøÌåÌå»ýСÓÚBÖÐÆøÌåÌå»ý£¬ËµÃ÷AÖÐÑÎËá¹ýÁ¿¡¢½ðÊôÍêÈ«·´Ó¦£¬BÖкϽðÖÊÁ¿Ð¡ÓÚCÖкϽðÖÊÁ¿£¬ÇÒB¡¢CÉú³ÉÆøÌåÌå»ýÏàµÈ£¬ËµÃ÷B¡¢CÖÐÑÎËáÍêÈ«·´Ó¦£¬Éú³É336mLÇâÆøÐèÒª½ðÊôµÄÖÊÁ¿Îª£º204mg¡Á$\frac{336mL}{224mL}$=306mg£¼399mg£¬¹ÊBÖнðÊôÊ£Ó࣬ÑÎËá²»×㣮
A£®B¡¢CÖÐÑÎËáÍêÈ«£¬¸ù¾Ýn=$\frac{V}{{V}_{m}}$¼ÆËãÇâÆøµÄÎïÖʵÄÁ¿£¬¸ù¾ÝÇâÔªËØÊغã¿ÉÖªn£¨HCl£©=2n£¨H2£©£¬½ø¶ø¼ÆËãÑÎËáµÄÎïÖʵÄÁ¿Å¨¶È£¬
B£®AÖÐÑÎËáÓÐÊ£Ó࣬½ðÊôÍêÈ«·´Ó¦£¬´ËʱÉú³ÉÇâÆø2224mL£¬¹Ê¿ÉÒÔ¸ù¾ÝA×éÊý¾Ý¼ÆËã½ðÊôµÄÎïÖʵÄÁ¿£¬Áîþ¡¢ÂÁµÄÎïÖʵÄÁ¿·Ö±ðΪxmol¡¢ymol£¬¸ù¾Ý¶þÕßÖÊÁ¿Ö®ºÍÓëµç×ÓתÒÆÊغãÁз½³Ì¼ÆËãx¡¢yµÄÖµ£¬½ø¶ø¼ÆËãCÖÐAlµÄÎïÖʵÄÁ¿£»
ÏòCÉÕ±ÖмÓÈëÒ»¶¨Á¿µÄ1mol/LµÄNaOHÈÜÒº£¬Ê¹ºÏ½ðÖеÄÂÁ·ÛÇ¡ºÃÍêÈ«Èܽ⣬ȴ²»Äܺ¬ÂÁµÄ³Áµí£¬²¢Ê¹Mg2+¸ÕºÃ³ÁµíÍêÈ«£¬ÈÜÒºÖÐÈÜÖÊΪNaCl¡¢NaAlO2£¬¸ù¾ÝÄÆÀë×ÓÊغã¿ÉÖªn£¨NaOH£©=n£¨NaCl£©+n£¨NaAlO2£©£¬¾Ý´Ë¼ÆËãÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾ÝV=$\frac{n}{c}$¼ÆËãÇâÑõ»¯ÄƵÄÌå»ý£»
C£®¸ù¾ÝBÖмÆËã¿ÉÖªAl¡¢MgÎïÖʵÄÁ¿¹Øϵ£»
D£®ÈÜÒºÖÐÈÜÖÊΪNaCl¡¢NaAlO2£®
½â´ð ½â£ºÑÎËáŨ¶È¡¢Ìå»ýÒ»¶¨£¬AÖкϽðÖÊÁ¿Ð¡ÓÚBÖкϽðÖÊÁ¿£¬ÇÒAÖÐÉú³ÉÆøÌåÌå»ýСÓÚBÖÐÆøÌåÌå»ý£¬ËµÃ÷AÖÐÑÎËá¹ýÁ¿¡¢½ðÊôÍêÈ«·´Ó¦£¬BÖкϽðÖÊÁ¿Ð¡ÓÚCÖкϽðÖÊÁ¿£¬ÇÒB¡¢CÉú³ÉÆøÌåÌå»ýÏàµÈ£¬ËµÃ÷B¡¢CÖÐÑÎËáÍêÈ«·´Ó¦£¬Éú³É336mLÇâÆøÐèÒª½ðÊôµÄÖÊÁ¿Îª£º204mg¡Á$\frac{336mL}{224mL}$=306mg£¼399mg£¬¹ÊBÖнðÊôÊ£Ó࣬ÑÎËá²»×㣮
A£®B¡¢CÖÐÑÎËáÍêÈ«£¬Éú³ÉÇâÆøµÄÎïÖʵÄÁ¿Îª$\frac{0.336L}{22.4L/mol}$=0.015mol£¬¸ù¾ÝÇâÔªËØÊغã¿ÉÖªn£¨HCl£©=2n£¨H2£©=0.03mol£¬ÔòÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ$\frac{0.03mol}{0.03L}$=1mol/L£¬¹ÊAÕýÈ·£»
B£®ÉèAÖÐMgµÄÎïÖʵÄÁ¿Îªx£¬AlµÄÎïÖʵÄÁ¿Îªy£¬Éú³ÉÆøÌåµÄÎïÖʵÄÁ¿Îª£º$\frac{0.224L}{22.4L/mol}$=0.01mol£¬
ÓɺϽðµÄÖÊÁ¿¿ÉµÃ£º24x+27y=0.204£¬
¸ù¾Ýµç×ÓתÒÆÊغã¿ÉµÃ£º2x+3y=0.01¡Á2
ÁªÁ¢·½³Ì£¬½âµÃ£ºx=0.004mol£»y=0.004mol£¬
¿ÉÖªCÖÐAl¡¢MgµÄÎïÖʵÄÁ¿¾ùΪ0.004mol¡Á$\frac{561g}{204g}$=0.011mol£¬
ÏòCÉÕ±ÖмÓÈëÒ»¶¨Á¿µÄ1mol/LµÄNaOHÈÜÒº£¬Ê¹ºÏ½ðÖеÄÂÁ·ÛÇ¡ºÃÍêÈ«Èܽ⣬ȴ²»Äܺ¬ÂÁµÄ³Áµí£¬²¢Ê¹Mg2+¸ÕºÃ³ÁµíÍêÈ«£¬ÈÜÒºÖÐÈÜÖÊΪNaCl¡¢NaAlO2£¬¸ù¾ÝÄÆÀë×ÓÊغã¿ÉÖªn£¨NaOH£©=n£¨NaCl£©+n£¨NaAlO2£©=
0.03mol+0.011mol=0.041mol£¬Ôò¼ÓÈëNaOHÈÜÒºµÄÌå»ýΪ$\frac{0.041mol}{1mol/L}$=0.041L=41mL£¬¹ÊB´íÎó£»
C£®¸ù¾ÝBÖмÆËã¿ÉÖªAl¡¢MgÎïÖʵÄÁ¿ÏàµÈ£¬¹ÊCÕýÈ·£»
D£®ÈÜÒºÖÐÈÜÖÊΪNaCl¡¢NaAlO2£¬ÂËÒºÖк¬ÓÐÒõÀë×ÓΪCl-¡¢AlO2-¡¢OH-£¬¹ÊDÕýÈ·£¬
¹ÊÑ¡£ºB£®
µãÆÀ ±¾Ì⿼²é»ìºÏÎïµÄ¼ÆË㣬ÌâÄ¿ÄѶȽϴ󣬸ù¾Ý±íÖÐÊý¾ÝÅжÏÎïÖʹýÁ¿Óë·ñÊǽâÌâµÄ¹Ø¼ü£¬²àÖØ¿¼²éѧÉú¶ÔÊý¾ÝµÄ·ÖÎöÄÜÁ¦Óë¼ÆËãÄÜÁ¦£®
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£® | ÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶ÈÏàͬ¡¢ÈÜÖʵÄÖÊÁ¿·ÖÊý²»Í¬ | |
B£® | ÈÜÖʵÄÖÊÁ¿·ÖÊýÏàͬ¡¢ÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶È²»Í¬ | |
C£® | ÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶ÈºÍÈÜÖʵÄÖÊÁ¿·ÖÊý¶¼²»Í¬ | |
D£® | ÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶ÈºÍÈÜÖʵÄÖÊÁ¿·ÖÊý¶¼Ïàͬ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ÊÇ·ñÊÇ´óÁ¿·Ö×Ó»òÀë×ӵļ¯ºÏÌå | B£® | ·ÖÉ¢ÖÊ΢Á£Ö±¾¶µÄ´óС²»Í¬ | ||
C£® | ÊÇ·ñÓж¡´ï¶ûЧӦ | D£® | ÊÇ·ñ¾ùÒ»¡¢Îȶ¨¡¢Í¸Ã÷ |
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£º½â´ðÌâ
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£® | CH3CH2 CH2CH3 | B£® | CH3COOCH2CH3 | C£® | CH3CH2COOH | D£® | CH3CH3 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | Íƹ㡰µÍ̼¾¼Ã¡±£¬¼õÉÙÎÂÊÒÆøÌåµÄÅÅ·Å | |
B£® | ´óÁ¦·¢Õ¹ÐÂÄÜÔ´£¬¼õÉÙÎÂÊÒÆøÌåµÄÅÅ·Å | |
C£® | Íƹ㡰ÂÌÉ«×ÔÓÉ¡±¼Æ»®£¬ÎüÊÕ¿ÕÆøÖеÄCO2ÓÃÁ®¼ÛÄÜÔ´ºÏ³ÉÆûÓÍ | |
D£® | ÍƹãС»ðÁ¦·¢µçÕ¾µÄÐ˽¨£¬»º½âµØ·½ÓõçÀ§ÄÑ£¬´Ù½øµØ·½¾¼ÃµÄ¿ìËÙ·¢Õ¹ |
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