ÓÐÏÂÁл¯Ñ§ÒÇÆ÷£º
¢ÙÍÐÅÌÌìƽ¢Ú²£Á§°ô¢ÛÒ©³×¢ÜÉÕ±¢ÝÁ¿Í²¢ÞÈÝÁ¿Æ¿¢ß½ºÍ·µÎ¹Ü¢àϸ¿ÚÊÔ¼ÁÆ¿¢á±êÇ©Ö½
£¨1£©ÏÖÐèÒªÅäÖÆ500m L 1mol/L H2SO4ÈÜÒº£¬ÐèÒªÖÊÁ¿·ÖÊýΪ98%£¬ÃܶÈΪ1.84g/cm3µÄŨH2SO4 £»
£¨2£©´ÓÉÏÊöÒÇÆ÷ÖУ¬°´ÊµÑéÒªÇóʹÓõÄÏȺó˳Ðò£¬Æä±àºÅÅÅÁÐÊÇ £»
£¨3£©ÈÝÁ¿Æ¿Ê¹ÓÃÇ°¼ìÑé©ˮµÄ·½·¨ÊÇ £»
£¨4£©ÈôʵÑéÓöµ½ÏÂÁÐÇé¿ö£¬¶ÔÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©
¢ÙÓÃÒÔÏ¡ÊÍŨÁòËáµÄÉձδϴµÓ£¬ £»
¢Úδ¾ÀäÈ´½«ÈÜҺעÈëÈÝÁ¿Æ¿ÖУ¬ £»
¢ÛÒ¡ÔȺó·¢ÏÖÒºÃæϽµÔÙ¼ÓË®£¬ £»
¢Ü¶¨ÈÝʱ¹Û²ìÒºÃ温ÊÓ£¬ ¡£
£¨1£©2.72 mL
£¨2£©¢Ý¢Ü¢Ú¢Þ¢ß¢à¢á
£¨3£©ÍùÈÝÁ¿Æ¿ÄÚ¼ÓÊÊÁ¿Ë®,ÈûºÃÆ¿Èû£¬ÓÃʳָ¶¥×¡Æ¿Èû£¬ÓÃÁíÒ»Ö»ÊÖµÄÎåÖ¸ÍÐסƿµ×£¬°ÑÆ¿µ¹Á¢¹ýÀ´£¬È粻©ˮ£¬°ÑÆ¿ÈûÐýת180??ºóÈû½ô£¬ÔÙ°ÑÆ¿µ¹Á¢¹ýÀ´£¬Èô²»Â©Ë®£¬²ÅÄÜʹÓá£
£¨4£©¢ÙÆ«µÍ¢ÚÆ«¸ß¢ÛÆ«µÍ¢ÜÆ«¸ß
½âÌâ̽¾¿£º±¾ÌâÖ÷Òª¿¼²éÁËÎïÖʵÄÁ¿Å¨¶È¡¢ÈÜÒºµÄÅäÖƼ°Îó²î·ÖÎö¡£±¾ÌâÄѵãÔÚÓÚÎó²î·ÖÎö£¬Àí½âÒýÆðÎó²îµÄÔÒòÊǽâÌâµÄ¹Ø¼ü¡£
£¨1£©V£¨Å¨H2SO4£©¡Á1.84g/mL¡Á98%=0.500L¡Á1 mol/L¡Á98g/mol
£¨2£©ÅäÖÆ˳ÐòÊÇ£º¼ÆËã¡úÁ¿È¡¡úÏ¡ÊÍ¡úÒÆÒº¡ú¶¨ÈÝ
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
£¨1£©ÏÖÐèÒªÅäÖÆ500 mL 1 mol¡¤L-1ÁòËáÈÜÒº£¬ÐèÓÃÖÊÁ¿·ÖÊýΪ98%¡¢ÃܶÈΪ1.84 g¡¤cm-3µÄŨÁòËá ________mL¡£
£¨2£©´ÓÉÏÊöÒÇÆ÷ÖУ¬°´ÊµÑéʹÓõÄÏȺó˳Ðò£¬Æä±àºÅÅÅÁÐÊÇ___________________¡£
£¨3£©ÈÝÁ¿Æ¿Ê¹ÓÃÇ°¼ìÑé©ˮµÄ·½·¨ÊÇ___________¡£
£¨4£©ÈôʵÑéÓöµ½ÏÂÁÐÇé¿ö£¬¶ÔÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죨Ìîд¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©£¿
¢ÙÓÃÒÔÏ¡ÊÍÁòËáµÄÉձδϴµÓ£¬_________________________________¡£
¢Úδ¾ÀäÈ´³ÃÈȽ«ÈÜҺעÈëÈÝÁ¿Æ¿ÖУ¬___________________________¡£
¢ÛÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏßÔÙ¼ÓË®£¬___________________________¡£
¢ÜÈÝÁ¿Æ¿ÖÐÔÓÐÉÙÁ¿ÕôÁóË®£¬___________________________________¡£
¢Ý¶¨ÈÝʱ¹Û²ìÒºÃ温ÊÓ£¬_______________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÁÉÄþÊ¡¶«±±Óý²ÅѧУ¸ßÈýÉÏѧÆÚµÚÒ»´ÎÄ£Ä⿼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÊµÑéÌâ
£¨10·Ö£©ÓÐÏÂÁл¯Ñ§ÒÇÆ÷£º
¢ÙÍÐÅÌÌìƽ¢Ú²£Á§°ô¢ÛÒ©³×¢ÜÉÕ±¢ÝÁ¿Í²¢ÞÈÝÁ¿Æ¿¢ß½ºÍ·µÎ¹Ü¢àϸ¿ÚÊÔ¼ÁÆ¿¢á±êÇ©Ö½
£¨1£©ÏÖÐèÒªÅäÖÆ500m L 1mol/L H2SO4ÈÜÒº£¬ÐèÒªÖÊÁ¿·ÖÊýΪ98%£¬ÃܶÈΪ1.84g/cm3µÄŨH2SO4 £»
£¨2£©´ÓÉÏÊöÒÇÆ÷ÖУ¬°´ÊµÑéÒªÇóʹÓõÄÏȺó˳Ðò£¬Æä±àºÅÅÅÁÐÊÇ £»
£¨3£©ÈÝÁ¿Æ¿Ê¹ÓÃÇ°¼ìÑé©ˮµÄ·½·¨ÊÇ £»
£¨4£©ÈôʵÑéÓöµ½ÏÂÁÐÇé¿ö£¬¶ÔÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©
¢ÙÓÃÒÔÏ¡ÊÍŨÁòËáµÄÉձδϴµÓ£¬ £»
¢Úδ¾ÀäÈ´½«ÈÜҺעÈëÈÝÁ¿Æ¿ÖУ¬ £»
¢ÛÒ¡ÔȺó·¢ÏÖÒºÃæϽµÔÙ¼ÓË®£¬ £»
¢Ü¶¨ÈÝʱ¹Û²ìÒºÃ温ÊÓ£¬ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêÁÉÄþÊ¡¸ßÈýÉÏѧÆÚµÚÒ»´ÎÄ£Ä⿼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÊµÑéÌâ
£¨10·Ö£©ÓÐÏÂÁл¯Ñ§ÒÇÆ÷£º
¢ÙÍÐÅÌÌìƽ¢Ú²£Á§°ô¢ÛÒ©³×¢ÜÉÕ±¢ÝÁ¿Í²¢ÞÈÝÁ¿Æ¿¢ß½ºÍ·µÎ¹Ü¢àϸ¿ÚÊÔ¼ÁÆ¿¢á±êÇ©Ö½
£¨1£©ÏÖÐèÒªÅäÖÆ500m L 1mol/L H2SO4ÈÜÒº£¬ÐèÒªÖÊÁ¿·ÖÊýΪ98%£¬ÃܶÈΪ1.84g/cm3µÄŨH2SO4 £»
£¨2£©´ÓÉÏÊöÒÇÆ÷ÖУ¬°´ÊµÑéÒªÇóʹÓõÄÏȺó˳Ðò£¬Æä±àºÅÅÅÁÐÊÇ £»
£¨3£©ÈÝÁ¿Æ¿Ê¹ÓÃÇ°¼ìÑé©ˮµÄ·½·¨ÊÇ £»
£¨4£©ÈôʵÑéÓöµ½ÏÂÁÐÇé¿ö£¬¶ÔÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©
¢ÙÓÃÒÔÏ¡ÊÍŨÁòËáµÄÉձδϴµÓ£¬ £»
¢Úδ¾ÀäÈ´½«ÈÜҺעÈëÈÝÁ¿Æ¿ÖУ¬ £»
¢ÛÒ¡ÔȺó·¢ÏÖÒºÃæϽµÔÙ¼ÓË®£¬ £»
¢Ü¶¨ÈÝʱ¹Û²ìÒºÃ温ÊÓ£¬ ¡£
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