ÇâÑõ»¯ÄÆÈÜÒº ·Ö×Óʽ£ºNaOH Ïà¶Ô·Ö×ÓÖÊÁ¿£º40 Ãܶȣº1.2g?cm-3 ÖÊÁ¿·ÖÊý£º20% |
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
ÇâÑõ»¯ÄÆÈÜÒº ·Ö×Óʽ£ºNaOH Ïà¶Ô·Ö×ÓÖÊÁ¿£º40 Ãܶȣº1.2g?cm-3 ÖÊÁ¿·ÖÊý£º20%£¨1£©¸ÃNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ 6 6 mol/L£®£¨2£©ÏÖÔÚÒªÅäÖƸÃŨ¶ÈµÄNaOHÈÜÒº100ml£¬Ðè³ÆÁ¿ 24.0 24.0 g¹ÌÌåÇâÑõ»¯ÄÆ£®ÈÜÒºÅäÖƵÄËùÐèµÄ»ù±¾²½ÖèÈçÏ£º£¨3£©½«ÉÏÊöʵÑé²½ÖèAµ½F°´ÊµÑé¹ý³ÌÏȺó´ÎÐòÅÅÁÐ CBDFAE CBDFAE £®£¨4£©ÉÏÊöʵÑé²½ÖèA¡¢B¡¢E¡¢F¶¼Óõ½µÄÒÇÆ÷Ãû³ÆΪ 100mlÈÝÁ¿Æ¿ 100mlÈÝÁ¿Æ¿ £®£¨5£©ÏÂÁвÙ×÷¶ÔNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£® ¢ÙÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏßÔÙ¼ÓË® Æ«µÍ Æ«µÍ £»¢ÚÈÝÁ¿Æ¿ÖÐÔÓÐÉÙÁ¿ÕôÁóË® ÎÞÓ°Ïì ÎÞÓ°Ïì £»¢Û¶¨ÈÝʱ¸©ÊÓ¹Û²ìÒºÃæ Æ«¸ß Æ«¸ß £®
²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â ʵÑ飺 £¨1£©ÈçͼΪʵÑéÊÒijŨÑÎËáÊÔ¼ÁÆ¿±êÇ©ÉϵÄÓйØÊý¾Ý£¬ÊÔ¸ù¾Ý±êÇ©ÉϵÄÓйØÊý¾Ý»Ø´ðÏÂÁÐÎÊÌ⣺ ¢Ù¸ÃŨÑÎËáÖÐHClµÄÎïÖʵÄÁ¿Å¨¶ÈΪ 11.9 11.9 mol?L-1£®¢ÚÈ¡ÓÃÈÎÒâÌå»ýµÄ¸ÃÑÎËáÈÜҺʱ£¬ÏÂÁÐÎïÀíÁ¿Öв»ËæËùÈ¡Ìå»ýµÄ¶àÉÙ¶ø±ä»¯µÄÊÇBD BD £®A£®ÈÜÒºÖÐHClµÄÎïÖʵÄÁ¿ B£®ÈÜÒºµÄŨ¶È C£®ÈÜÒºÖÐCl-µÄÊýÄ¿ DÈÜÒºµÄÃÜ¶È £¨2£©ÊµÑéÊÒÅäÖÆ480mL0.08mol/LNa2CO3ÈÜÒº»Ø´ðÏÂÁÐÎÊÌâ ¢ÙÓ¦ÓÃÍÐÅÌÌìƽ³Æȡʮˮ̼ËáÄƾ§Ìå 11.4 11.4 g¢ÚÈôÔÚ³ÆÁ¿ÑùƷʱ£¬Ò©Æ··ÅÔÚÌìƽÓÒÅÌÉÏ£¬íÀÂë·ÅÔÚÌìƽ×óÅÌÉÏ£¬Ììƽƽºâʱ£¬Ôòʵ¼Ê³ÆÁ¿µÄ̼ËáÄƾ§ÌåÊÇ 10.6 10.6 g£¨1gÒÔÏÂÓÃÓÎÂ룩¢ÛÓÃÈÝÁ¿Æ¿ÅäÖÆÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº£¬¸ÃÈÝÁ¿Æ¿±ØÐëÊÇ B B A¡¢¸ÉÔïµÄ¡¡¡¡B¡¢Æ¿Èû²»Â©Ë®¡¡C¡¢ÓÃÓûÅäÖƵÄÈÜÒºÈóÏ´¹ý¡¡¡¡D¡¢ÒÔÉÏÈýÏҪÇó ¢ÜÈôʵÑéÓöÏÂÁÐÇé¿ö£¬ÈÜÒºµÄŨ¶ÈÊÇ¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»¹ÊÇ¡°²»±ä¡±£¿ A£®¼ÓˮʱԽ¹ý¿Ì¶ÈÏß Æ«µÍ Æ«µÍ £»B£®Íü¼Ç½«Ï´µÓÒº¼ÓÈëÈÝÁ¿Æ¿ Æ«µÍ Æ«µÍ £»C£®ÈÝÁ¿Æ¿ÄÚ±Ú¸½ÓÐË®Öé¶øδ¸ÉÔï´¦Àí ²»±ä ²»±ä £»D£®ÈܽâºóûÓÐÀäÈ´±ã½øÐж¨ÈÝ Æ«¸ß Æ«¸ß £®£¨3£©¢ÙÈ¡ÉÙÁ¿Fe2O3·ÛÄ©£¨ºì×ØÉ«£©¼ÓÈëÊÊÁ¿ÑÎËᣬ·´Ó¦µÄÀë×Ó·½³ÌʽΪ Fe2O3+6H+=2Fe3++3H2O Fe2O3+6H+=2Fe3++3H2O £¬·´Ó¦ºóµÃµ½»ÆÉ«µÄFeCl3ÈÜÒº£®ÓôËÈÜÒº×öÒÔÏÂʵÑ飺¢ÚÈ¡ÉÙÁ¿ÈÜÒºÖÃÓÚÊÔ¹ÜÖУ¬µÎÈëNaOHÈÜÒº£¬¿´µ½ÓкìºÖÉ«³ÁµíÉú³É£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ Fe3++3O H-=Fe£¨OH£©3¡ý Fe3++3O H-=Fe£¨OH£©3¡ý £®¢ÛÔÚСÉÕ±ÖмÓÈë25mLÕôÁóË®£¬¼ÓÈÈÖÁ·ÐÌÚºó£¬Ïò·ÐË®ÖмÓÈë2mL FeCl3±¥ºÍÈÜÒº£¬¼ÌÐøÖó·ÐÖÁÈÜÒº³Ê ºìºÖ ºìºÖ É«£¬¼´¿ÉÖƵÃFe£¨OH£©3½ºÌ壮¢ÜÁíȡһСÉÕ±¼ÓÈë25mLÕôÁóË®ºó£¬ÏòÉÕ±ÖÐÔÙ¼ÓÈë2mL FeCl3±¥ºÍÈÜÒº£¬Õñµ´¾ùÔȺ󣬽«´ËÉÕ±£¨±àºÅ¼×£©ÓëÊ¢ÓÐFe£¨OH£©3½ºÌåµÄÉÕ±£¨±àºÅÒÒ£©Ò»Æð·ÅÖðµ´¦£¬·Ö±ðÓü¤¹â±ÊÕÕÉäÉÕ±ÖеÄÒºÌ壬¿ÉÒÔ¿´µ½ ÒÒ ÒÒ £¨Ìî¼×»òÒÒ£©ÉÕ±Öлá²úÉú¶¡´ï¶ûЧӦ£®´ËʵÑé¿ÉÒÔÇø±ðÈÜÒººÍ½ºÌå ÈÜÒººÍ½ºÌå £®²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ ÈçͼΪʵÑéÊÒijÊÔ¼ÁÆ¿±êÇ©ÉϵÄÓйØÊý¾Ý£¬ÊÔ¸ù¾Ý±êÇ©ÉϵÄÓйØÊý¾Ý»Ø´ðÏÂÁÐÎÊÌ⣺
£¨2£©ÏÖÔÚÒªÅäÖƸÃŨ¶ÈµÄNaOHÈÜÒº100ml£¬Ðè³ÆÁ¿______g¹ÌÌåÇâÑõ»¯ÄÆ£®ÈÜÒºÅäÖƵÄËùÐèµÄ»ù±¾²½ÖèÈçÏ£º £¨3£©½«ÉÏÊöʵÑé²½ÖèAµ½F°´ÊµÑé¹ý³ÌÏȺó´ÎÐòÅÅÁÐ______£® £¨4£©ÉÏÊöʵÑé²½ÖèA¡¢B¡¢E¡¢F¶¼Óõ½µÄÒÇÆ÷Ãû³ÆΪ______£® £¨5£©ÏÂÁвÙ×÷¶ÔNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£® ¢ÙÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏßÔÙ¼ÓË®______£» ¢ÚÈÝÁ¿Æ¿ÖÐÔÓÐÉÙÁ¿ÕôÁóË®______£» ¢Û¶¨ÈÝʱ¸©ÊÓ¹Û²ìÒºÃæ______£® ²é¿´´ð°¸ºÍ½âÎö>> ¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ ÈçͼΪʵÑéÊÒijÊÔ¼ÁÆ¿±êÇ©ÉϵÄÓйØÊý¾Ý£¬ÊÔ¸ù¾Ý±êÇ©ÉϵÄÓйØÊý¾Ý»Ø´ðÏÂÁÐÎÊÌ⣺
£¨2£©ÏÖÔÚÒªÅäÖƸÃŨ¶ÈµÄNaOHÈÜÒº100ml£¬Ðè³ÆÁ¿______g¹ÌÌåÇâÑõ»¯ÄÆ£®ÈÜÒºÅäÖƵÄËùÐèµÄ»ù±¾²½ÖèÈçÏ£º £¨3£©½«ÉÏÊöʵÑé²½ÖèAµ½F°´ÊµÑé¹ý³ÌÏȺó´ÎÐòÅÅÁÐ______£® £¨4£©ÉÏÊöʵÑé²½ÖèA¡¢B¡¢E¡¢F¶¼Óõ½µÄÒÇÆ÷Ãû³ÆΪ______£® £¨5£©ÏÂÁвÙ×÷¶ÔNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÓкÎÓ°Ï죨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£® ¢ÙÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏßÔÙ¼ÓË®______£» ¢ÚÈÝÁ¿Æ¿ÖÐÔÓÐÉÙÁ¿ÕôÁóË®______£» ¢Û¶¨ÈÝʱ¸©ÊÓ¹Û²ìÒºÃæ______£® ²é¿´´ð°¸ºÍ½âÎö>> ͬ²½Á·Ï°²á´ð°¸ °Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁÐ±í 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°æȨÉùÃ÷£º±¾Õ¾ËùÓÐÎÄÕ£¬Í¼Æ¬À´Ô´ÓÚÍøÂ磬Öø×÷Ȩ¼°°æȨ¹éÔ×÷ÕßËùÓУ¬×ªÔØÎÞÒâÇÖ·¸°æȨ£¬ÈçÓÐÇÖȨ£¬Çë×÷ÕßËÙÀ´º¯¸æÖª£¬ÎÒÃǽ«¾¡¿ì´¦Àí£¬ÁªÏµqq£º3310059649¡£ ICP±¸°¸ÐòºÅ: »¦ICP±¸07509807ºÅ-10 ¶õ¹«Íø°²±¸42018502000812ºÅ |