ÁòËáÍË®ÈÜÒº³ÊËáÐÔ£¬Êô±£»¤ÐÔÎÞ»úɱ¾ú¼Á£¬¶ÔÈËÐó±È½Ï°²È«£¬Æäͬʯ»ÒÈé»ìºÏ¿ÉµÃ¡°²¨¶û¶à¡±ÈÜÒº¡£ÊµÑéÊÒÀïÐèÓÃ480 mL 0.10 mol/LµÄÁòËáÍÈÜÒº£¬ÔòӦѡÓõÄÈÝÁ¿Æ¿¹æ¸ñºÍ³ÆÈ¡ÈÜÖʵÄÖÊÁ¿·Ö±ðΪ£¨ £©
A.480 mLÈÝÁ¿Æ¿£¬³ÆÈ¡7.68 gÁòËáÍ
B.480 mLÈÝÁ¿Æ¿£¬³ÆÈ¡12.0 gµ¨·¯
C.500 mLÈÝÁ¿Æ¿£¬³ÆÈ¡8.00 gÁòËáÍ
D.500 mLÈÝÁ¿Æ¿£¬³ÆÈ¡12.5 gµ¨·¯
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´£º2014Ä껯ѧ¸ß¿¼×ܸ´Ï°¿ÎʱÑÝÁ· 2-1ÎïÖʵķÖÀàÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
ʵÑéÊÒÀï³£ÓõĸÉÔï¼ÁÓУº¢ÙŨÁòËᣨ98%£©£¬¢ÚÎÞË®ÂÈ»¯¸Æ£¬¢Û±äÉ«¹è½º£Û¹è½ºµÄÖ÷Òª³É·ÖÊǶþÑõ»¯¹è£¬ÔÚÆäÖвôÈëÉÙÁ¿µÄÎÞË®ÂÈ»¯îÜ£¨CoCl2£©×÷ָʾ¼Á£¬ÎÞË®ÂÈ»¯îܳÊÀ¶É«£¬ÎüË®ºó±äΪCoCl2¡¤6H2O³Ê·ÛºìÉ«£Ý£¬¢ÜÎåÑõ»¯¶þÁ×£¬¢Ý¼îʯ»Ò£¨Ö÷Òª³É·ÖÊÇÇâÑõ»¯ÄÆ¡¢Ñõ»¯¸Æ£¬ÖÆ·¨ÊÇ£º°ÑÉúʯ»Ò¼Óµ½Å¨µÄÉÕ¼îÈÜÒºÖУ¬ÔÙ¼ÓÇ¿ÈÈÕô¸É£©µÈ¡£
£¨1£©Ð´³öÖÆÈ¡¸ÉÔï¼Á¼îʯ»Ò¹ý³ÌÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º__________________________________¡£
£¨2£©ÉÏÊöÎïÖÊÖУ¬ÊôÓÚ´¿¾»ÎïµÄÊÇ______¡£
A£®¢Ù¢Ú¢Ü B£®¢Ú¢Ü
C£®¢Ù¢Ú¢Ü¢Ý D£®È«²¿
£¨3£©ÉÏÊö¸ÉÔï¼ÁÖУ¬²»ÒËÓÃÓÚ¸ÉÔïÂÈ»¯ÇâÆøÌåµÄÊÇ______¡£
£¨4£©ÉÏÊö¢Ù¡«¢Ü£¬ÆäÖ÷Òª»¯Ñ§³É·ÖÒÀ´ÎÊôÓÚ______¡¢______¡¢______¡¢______ £¨Ìîд¸÷ÎïÖÊËùÊôµÄÀà±ð£©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Ä껯ѧ¸ß¿¼×ܸ´Ï°¿ÎʱÑÝÁ· 11-2ÎïÖʵļìÑé¡¢·ÖÀëºÍÌá´¿Á·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ijÎÞÉ«ÈÜÒºº¬ÓТÙNa+¡¢¢ÚBa2+¡¢¢ÛCl-¡¢¢ÜBr-¡¢¢ÝSO32-¡¢¢ÞSO42-ÖеÄÈô¸ÉÖÖ£¬ÒÀ´Î½øÐÐÏÂÁÐʵÑ飬ÇÒÿ²½Ëù¼ÓÊÔ¼Á¾ù¹ýÁ¿£¬¹Û²ìµ½µÄÏÖÏóÈçÏ£º
²½Öè | ²Ù×÷ | ÏÖÏó |
(1) | ÓÃpHÊÔÖ½¼ìÑé | ÈÜÒºµÄpH´óÓÚ7 |
(2) | ÏòÈÜÒºÖеμÓÂÈË®£¬ÔÙ¼ÓÈëCCl4Õñµ´£¬¾²Öà | CCl4²ã³Ê³ÈÉ« |
(3) | ÏòËùµÃË®ÈÜÒºÖмÓÈëBa(NO3)2ÈÜÒººÍÏ¡ÏõËá | Óа×É«³Áµí²úÉú |
(4) | ¹ýÂË£¬ÏòÂËÒºÖмÓÈëAgNO3ÈÜÒººÍÏ¡ÏõËá | Óа×É«³Áµí²úÉú |
ÏÂÁнáÂÛÕýÈ·µÄÊÇ( )
A.¿Ï¶¨º¬ÓеÄÀë×ÓÊǢ٢ܢÝ
B.¿Ï¶¨Ã»ÓеÄÀë×ÓÊÇ¢Ú¢Þ
C.²»ÄÜÈ·¶¨µÄÀë×ÓÊÇ¢Ù
D.²»ÄÜÈ·¶¨µÄÀë×ÓÊÇ¢Û¢Ý
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Ä껯ѧ¸ß¿¼×ܸ´Ï°¿ÎʱÑÝÁ· 11-1»¯Ñ§ÊµÑéÒÇÆ÷ºÍ»ù±¾²Ù×÷Á·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÇëÕÒ³öÏÂÁÐͼʾÖÐÕýÈ·µÄʵÑé²Ù×÷( )
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Ä껯ѧ¸ß¿¼×ܸ´Ï°¿ÎʱÑÝÁ· 1-2ÎïÖʵÄÁ¿ÔÚʵÑéÖеÄÓ¦ÓÃÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
¡°84Ïû¶¾Òº¡±ÄÜÓÐЧɱÃð¼×ÐÍH1N1²¡¶¾£¬Ä³Í¬Ñ§¹ºÂòÁËһƿijƷÅÆ¡°84Ïû¶¾Òº¡±£¬²¢²éÔÄÏà¹Ø×ÊÁϺÍÏû¶¾Òº°üװ˵Ã÷µÃµ½ÈçÏÂÐÅÏ¢£º
¡°84Ïû¶¾Òº¡±£ºº¬25% NaClO¡¢1 000 mL¡¢ÃܶÈ1.19 g¡¤cm-3£¬Ï¡ÊÍ100±¶£¨Ìå»ý±È£©ºóʹÓá£
Çë¸ù¾ÝÒÔÉÏÐÅÏ¢ºÍÏà¹Ø֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸Ã¡°84Ïû¶¾Òº¡±µÄÎïÖʵÄÁ¿Å¨¶ÈΪ_____mol¡¤L-1¡£
£¨2£©¸Ãͬѧȡ100 mL¸ÃÆ·ÅÆ¡°84Ïû¶¾Òº¡±Ï¡ÊͺóÓÃÓÚÏû¶¾£¬Ï¡ÊͺóµÄÈÜÒºÖÐ
c£¨Na£«£©¡Ö_____mol¡¤L-1£¨¼ÙÉèÏ¡ÊͺóÈÜÒºÃܶÈΪ1.0 g¡¤cm-3£©¡£
£¨3£©¸Ãͬѧ²ÎÔĸÃÆ·ÅÆ¡°84Ïû¶¾Òº¡±µÄÅä·½£¬ÓûÓÃNaClO¹ÌÌåÅäÖÆ480 mLº¬25% NaClOµÄÏû¶¾Òº¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_____¡£
A£®ÈçͼËùʾµÄÒÇÆ÷ÖУ¬ÓÐËÄÖÖÊDz»ÐèÒªµÄ£¬»¹ÐèÒ»ÖÖ²£Á§ÒÇÆ÷
B£®ÈÝÁ¿Æ¿ÓÃÕôÁóˮϴ¾»ºó£¬Ó¦ºæ¸É²ÅÄÜÓÃÓÚÈÜÒºÅäÖÆ
C£®ÀûÓùºÂòµÄÉÌÆ·NaClOÀ´ÅäÖÆ¿ÉÄܵ¼Ö½á¹ûÆ«µÍ
D£®ÐèÒª³ÆÁ¿µÄNaClO¹ÌÌåÖÊÁ¿Îª143 g
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Ä껯ѧ¸ß¿¼×ܸ´Ï°¿ÎʱÑÝÁ· 1-2ÎïÖʵÄÁ¿ÔÚʵÑéÖеÄÓ¦ÓÃÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÓÐÁòËáþÈÜÒº500 mL£¬ËüµÄÃܶÈÊÇ1.20 g/cm3£¬ÆäÖÐþÀë×ÓµÄÖÊÁ¿·ÖÊýÊÇ4.8%£¬ÔòÓйظÃÈÜÒºµÄ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©
A£®ÈÜÖʵÄÖÊÁ¿·ÖÊýÊÇ24.0%
B£®ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ2.4 mol/L
C£®ÈÜÖʺÍÈܼÁµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ1¡Ã40
D£®ÁòËá¸ùÀë×ÓµÄÖÊÁ¿·ÖÊýÊÇ19.2%
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Ä껯ѧ¸ß¿¼×ܸ´Ï°¿ÎʱÑÝÁ· 1-1ÎïÖʵÄÁ¿ ÆøÌåĦ¶ûÌå»ýÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ͬÎÂͬѹÏ£¬ÖÊÁ¿ºöÂÔ²»¼ÆµÄÁ½ÆøÇòAºÍB£¬·Ö±ð³äÈëXÆøÌåºÍYÆøÌ壬ÇÒ³äÆøºóÁ½ÆøÇòµÄÌå»ýÏàͬ¡£ÈôÏàͬÌõ¼þÏ£¬AÆøÇò·ÅÔÚCOÖо²Ö¹²»¶¯£¬BÆøÇò·ÅÔÚO2ÖÐÉϸ¡¡£ÏÂÁÐÐðÊö»ò±íʾÕýÈ·µÄÊÇ£¨ £©
A£®XÆøÌåµÄÏà¶Ô·Ö×ÓÖÊÁ¿±ÈYÆøÌåµÄÏà¶Ô·Ö×ÓÖÊÁ¿´ó
B£®X¿ÉÄÜÊÇC2H4£¬Y¿ÉÄÜÊÇCH4
C£®XÆøÌåµÄÃܶÈСÓÚYÆøÌåµÄÃܶÈ
D£®³äÆøºóAÆøÇòÖÊÁ¿±ÈBÆøÇòÖÊÁ¿´ó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Ä껯ѧ¸ß¿¼×ܸ´Ï°µ¥Ôª¼ì²â µÚËÄÕ·ǽðÊô¼°Æ仯ºÏÎïÁ·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
ÏÂÁÐʵÑé×°ÖÃÖÐÄܴﵽʵÑéÄ¿µÄµÄÊÇ( )
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014Ä껯ѧ¸ß¿¼×ܸ´Ï°µ¥Ôª¼ì²â µÚÁù¡¢ÆßÕ»¯Ñ§·´Ó¦ÓëÄÜÁ¿ËÙÂÊƽºâ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÑ¡ÔñÌâ
Éè·´Ó¦¢ÙFe(s)+CO2(g)FeO(s)+CO(g) ¦¤H=a kJ/mol£¬·´Ó¦¢ÚFe(s)+H2O(g)FeO(s)+H2(g) ¦¤H=b kJ/mol£¬ÒÔÉÏÁ½·´Ó¦µÄƽºâ³£Êý·Ö±ðΪK1ºÍK2¡£ÔÚ²»Í¬Î¶ÈÏ£¬K1¡¢K2µÄÖµÈçÏ£º
ÏÂÁÐÓйØÐðÊöÕýÈ·µÄÊÇ( )
A£®b£¾0B£®ÔÚ973KÏÂÔö´óѹǿ£¬K2Ôö´ó
C£®a£¾bD£®ÔÚ³£ÎÂÏ·´Ó¦¢ÙÒ»¶¨ÄÜ×Ô·¢½øÐÐ
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