ÌâÄ¿ÄÚÈÝ

10£®´Óº£Ë®ÖÐÌáÈ¡äåµÄÖ÷Òª¹¤ÒÕÁ÷³ÌÓÐ
£¨1£©ÏòËữµÄº£Ë®ÖÐͨÈëÊÊÁ¿µÄÂÈÆø£¬Ê¹äåÀë×Óת»¯Îªµ¥ÖÊä壺2NaBr+Cl2¨TBr2+2NaCl
£¨2£©ÏòÎüÊÕËþÄÚµÄÈÜÒºÖÐͨÈëÊÊÁ¿µÄÂÈÆø£º2HBr+Cl2¨T2HCl+Br2
£¨3£©ÕôÁ󷨽«º£Ë®Å¨Ëõ£¬ÓÃÁòËὫŨËõµÄº£Ë®Ëữ
£¨4£©Ïòº¬ÓÐäåµ¥ÖʵÄË®ÈÜÒºÖÐͨÈë¿ÕÆøºÍË®ÕôÆø£¬½«äåµ¥ÖÊ´µÈëÊ¢ÓжþÑõ»¯ÁòÈÜÒºµÄÎüÊÕËþÄÚÒÔ´ïµ½¸»¼¯µÄÄ¿µÄ£ºBr2+SO2+2H2O¨T2HBr+H2SO4
£¨5£©ÓÃËÄÂÈ»¯Ì¼£¨»ò±½£©ÝÍÈ¡ÎüÊÕËþÄÚµÄÈÜÒºÖеÄäåµ¥ÖÊ£®
ÕýÈ·µÄ˳ÐòÊÇ£¨¡¡¡¡£©
A£®£¨1£©£¨2£©£¨3£©£¨4£©£¨5£©B£®£¨3£©£¨1£©£¨4£©£¨2£©£¨5£©C£®£¨5£©£¨1£©£¨3£©£¨2£©£¨4£©D£®£¨5£©£¨4£©£¨3£©£¨2£©£¨1£©

·ÖÎö º£Ë®ÓÃÕôÁó·½·¨Å¨Ëõº£Ë®£¬¼ÓÈëÁòËὫŨËõµÄº£Ë®Ëữ£»ËữµÄº£Ë®ÖÐͨÈëÊÊÁ¿µÄÂÈÆø£¬ºÍäåÀë×Ó·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³Éäåµ¥ÖÊ£»Ïòº¬äåµ¥ÖʵÄË®ÈÜÒºÖÐͨÈë¿ÕÆøºÍË®ÕôÆø£¬½«äåµ¥ÖÊ´µÈëÊ¢¶þÑõ»¯ÁòµÄÈÜÒºÖб»»¹Ô­Îªä廯Ç⣬ÏòÎüÊÕËþÄÚÖÐÈÜÒºÄÚͨÈëÊÊÁ¿µÄÂÈÆøÑõ»¯äåÀë×ÓΪäåµ¥ÖÊ£»ÀûÓÃÝÍÈ¡·ÖÒº·½·¨ÝÍÈ¡äåµ¥ÖÊ£®

½â´ð ½â£º´Óº£Ë®ÖÐÌáÈ¡äåµÄÖ÷Òª¹¤ÒÕÁ÷³ÌµÄ˳ÐòΪ£º
£¨3£©º£Ë®ÓÃÕôÁó·½·¨Å¨Ëõº£Ë®£¬¼ÓÈëÁòËὫŨËõµÄº£Ë®Ëữ£»
£¨1£©ËữµÄº£Ë®ÖÐͨÈëÊÊÁ¿µÄÂÈÆø£¬ºÍäåÀë×Ó·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³Éäåµ¥ÖÊ£¬2Br-+Cl2=Br2+2Cl-£»
£¨4£©Ïòº¬äåµ¥ÖʵÄË®ÈÜÒºÖÐͨÈë¿ÕÆøºÍË®ÕôÆø£¬½«äåµ¥ÖÊ´µÈëÊ¢¶þÑõ»¯ÁòµÄÈÜÒºÖб»»¹Ô­Îªä廯Ç⣬Br2+SO2+2H2O=H2SO4+2HBr£¬
£¨2£©ÏòÎüÊÕËþÄÚÖÐÈÜÒºÄÚͨÈëÊÊÁ¿µÄÂÈÆøÑõ»¯äåÀë×ÓΪäåµ¥ÖÊ£¬2HBr+Cl2=Br2+2HCl£»
£¨5£©ÀûÓÃÝÍÈ¡·ÖÒº·½·¨ÝÍÈ¡äåµ¥ÖÊ£¬
¹ÊÁ÷³Ì˳ÐòΪ£º£¨3£©£¨1£©£¨4£©£¨2£©£¨5£©£¬
¹ÊÑ¡B£®

µãÆÀ ±¾Ì⿼²éÁ˺£Ë®ÌáäåµÄ¹¤ÒµÉú²ú¹ý³ÌºÍÊÔ¼ÁÑ¡Ôñ£¬×¢ÒâÔ­ÀíµÄ·ÖÎöÅжϣ¬»¯Ñ§·½³ÌʽÊéдÊǹؼü£¬ÌâÄ¿½Ï¼òµ¥£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
20£®Ä³ÐËȤС×éÉè¼ÆÁËÈçͼËùʾµÄʵÑé×°Ö㬼ȿÉÓÃÓÚÖÆÈ¡ÆøÌ壬ÓÖ¿ÉÓÃÓÚÑéÖ¤ÎïÖʵÄÐÔÖÊ£®
£¨1£©ÀûÓÃI¡¢¢ò×°ÖÿÉÖÆÈ¡ÊÕ¼¯ÆøÌ壨K2¹Ø±Õ£¬K1´ò¿ª£©
¢ÙÏÂÁи÷×éÎïÖÊÖÐÄÜÖ±½ÓÀûÓÃÕâÌ××°ÖýøÐÐʵÑéµÄÊÇAC £¨ÌîÕýÈ·Ñ¡ÏîµÄ±êºÅ£©
A£®Ð¿ºÍÏ¡ÁòËáB£®Ë«ÑõË®ºÍ¶þÑõ»¯ÃÌC£®Å¨°±Ë®ºÍÑõ»¯¸ÆD£®Å¨ÑÎËáºÍ¶þÑõ»¯ÃÌ
¢ÚÔÚ²»¸Ä±äI¡¢II×°Öü°Î»ÖõÄÇ°ÌáÏ£¬ÈôÒªÖÆÈ¡ÊÕ¼¯NOÆøÌ壬¿É²ÉÈ¡µÄ¸Ä½ø·½·¨ÊÇ×°ÖÃIIÄÚ³äÂúË®£»ÈôÒªÖÆÈ¡ÊÕ¼¯NO2ÆøÌ壬¿É²ÉÈ¡µÄ¸Ä½ø·½·¨ÊÇ£º½«¢ò×°ÖüÓÂú£¨Äѻӷ¢µÄ£©ÓлúÈܼÁ£®
£¨2£©ÀûÓÃI¡¢III×°ÖÿÉÑéÖ¤ÎïÖʵÄÐÔÖÊ£¨K2´ò¿ª£¬K1¹Ø±Õ£©
¢ÙÈôÒªÖ¤Ã÷Ñõ»¯ÐÔ£ºKMnO4£¾C12£¾Br2£¬ÔòAÖмÓŨÑÎËᣬBÖмÓKMnO4£¬CÖз¢ÉúµÄÀë×Ó·´Ó¦·½³ÌʽCl2+2Br-=2Cl-+Br2£®
¢ÚÈôÒªÖ¤Ã÷ÒÒȲΪ²»±¥ºÍÌþ£¬ÔòIÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ£ºCaC2+2H2O¡úCa£¨OH£©2+C2H2¡ü£®CÖеÄÏÖÏóΪäåË®ÍÊÉ«»ò¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£®
£¨3£©×°ÖâóÖÐʹÓÃÇòÐθÉÔï¹ÜD¿É·ÀÖ¹ÒºÌåµ¹Îü£¬ÆäÔ­ÒòÊǵ±ÆøÌå±»ÎüÊÕʱ£¬ÒºÌåÉÏÉýµ½ÇòÐιÜÖУ¬ÓÉÓÚÇòÐιÜÈÝ»ý½Ï´ó£¬µ¼ÖÂÊÔ¹ÜÖÐÒºÃæÃ÷ÏÔϽµ£¬Ê¹ÇòÐιܿÚÍÑÀëÒºÃ棬ÇòÐιÜÖеÄÒºÌåÓÖÁ÷»ØÊÔ¹ÜÖУ¬´Ó¶ø·ÀÖ¹Á˵¹Îü£®
2£®ÏÖÓв¿·Ö¶ÌÖÜÆÚÔªËصÄÐÔÖÊ»òÔ­×ӽṹÈçÏÂ±í£º
ÔªËرàºÅÔªËØÐÔÖÊ»òÔ­×ӽṹ
AÓÐÈý¸öµç×Ӳ㣬K¡¢M²ãµç×ÓÊýÖ®ºÍµÈÓÚL²ãµç×ÓÊý
B¶ÌÖÜÆÚÖнðÊôÐÔ×îÇ¿
C³£ÎÂϵ¥ÖÊΪ˫ԭ×Ó·Ö×Ó£¬Ç⻯ÎïµÄË®ÈÜÒº³Ê¼îÐÔ
DÔªËØ×î¸ßÕý¼ÛÊÇ+7¼Û
£¨1£©Óû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙAÔªËØÔÚÖÜÆÚ±íÖеÄλÖõÚÈýÖÜÆÚ¢öA×壻
¢ÚBÔªËØÔ­×ӽṹʾÒâͼ£»
¢ÛCµ¥ÖÊ·Ö×ӵĵç×Óʽ£»Óõç×Óʽ±íʾAºÍBÔªËØ×é³ÉµÄ»¯ºÏÎïµÄÐγɹý³Ì£»
¢ÜDÔªËØÔÚ×ÔÈ»½çÓÐÁ½ÖÖºËËØ£¬ÓÃÔ­×Ó·ûºÅ±íʾÆäÖÐ×ÓÊýΪ20µÄºËËØ£®
£¨2£©ÔªËØDÓëÔªËØAÏà±È£¬·Ç½ðÊôÐÔ½ÏÇ¿µÄÊÇCl£¨ÓÃÔªËØ·ûºÅ±íʾ£©£¬ÏÂÁбíÊöÖÐÄÜÖ¤Ã÷ÕâÒ»ÊÂʵµÄÊÇbde£¨ÌîÑ¡ÏîÐòºÅ£©£®
a£®³£ÎÂÏÂDµÄµ¥ÖʺÍAµÄµ¥ÖÊ״̬²»Í¬
b£®DµÄÇ⻯Îï±ÈAµÄÇ⻯ÎïÎȶ¨
c£®Ò»¶¨Ìõ¼þÏÂDºÍAµÄµ¥Öʶ¼ÄÜÓëÄÆ·´Ó¦
d£®AµÄ×î¸ß¼Ûº¬ÑõËáËáÐÔÈõÓÚDµÄ×î¸ß¼Ûº¬ÑõËá
e£®Dµ¥ÖÊÄÜÓëAµÄÇ⻯Îï·´Ó¦Éú³ÉAµ¥ÖÊ
£¨3£©Ì½Ñ°ÎïÖʵÄÐÔÖʲîÒìÐÔÊÇѧϰµÄÖØÒª·½·¨Ö®Ò»£®A¡¢B¡¢C¡¢DËÄÖÖÔªËصÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïÖл¯Ñ§ÐÔÖÊÃ÷ÏÔ²»Í¬ÓÚÆäËûÈýÖÖµÄÊÇNaOH£¨Ð´»¯Ñ§Ê½£©£®
£¨4£©XÊÇÓÉA¡¢B¡¢C¡¢DËÄÖÖÔªËØÖеÄijÖÖÔªËØ×é³ÉµÄµ¥ÖÊ£¬Äܾ­ÏÂͼËùʾµÄ¹ý³Ìת»¯ÎªW£¨ÆäËûÌõ¼þÂÔÈ¥£©£®
X$\stackrel{O_{2}}{¡ú}$Y$\stackrel{O_{2}}{¡ú}$Z$\stackrel{H_{2}O}{¡ú}$W
¢ÙÈôYÊÇÓд̼¤ÐÔÆøζµÄÎÞÉ«ÆøÌ壬°ÑYͨÈëBaCl2ÈÜÒºÖУ¬È»ºóµÎ¼ÓÊÊÁ¿H2O2ÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬´Ë°×É«³ÁµíµÄ»¯Ñ§Ê½Îª£ºBaSO4£»Éú³É¸Ã°×É«³ÁµíµÄ»¯Ñ§·½³ÌʽΪBaCl2+SO2+H2O2=BaSO4¡ý+2HCl£¨»òBa2++SO2+H2O2=BaSO4¡ý+2H+£©£»
¢ÚÈôZÊǺì×ØÉ«ÆøÌ壬ÔòZ¡úWµÄ·´Ó¦ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁÎïÖʵÄÁ¿Ö®±ÈΪ£º1£º2£®
19£®£¨1£©ÔÚʵÑéÊÒÀïÖÆÈ¡ÒÒÏ©³£Òòζȹý¸ß¶øʹÒÒ´¼ºÍŨH2SO4·´Ó¦Éú³ÉÉÙÁ¿µÄSO2£¬ÓÐÈËÉè¼ÆÈçͼËùʾʵÑéÒÔÈ·ÈÏÉÏÊö»ìºÍÆøÌåÖÐÓÐC2H4ºÍSO2£® £¨ÒÒÏ©µÄÖÆȡװÖÃÂÔ£©
¢Ù¢ñ¡¢¢ò¡¢¢ó¡¢¢ô×°ÖÿÉÊ¢ÈëµÄÊÔ¼ÁÊÇ¢ñA¡¢¢òB¡¢¢óA¡¢¢ôD£®£¨½«ÏÂÁÐÓйØÊÔ¼ÁµÄÐòºÅÌîÈë¿Õ¸ñÄÚ£©
A£®Æ·ºìÈÜÒº           B£®NaOHÈÜÒº       C£®Å¨H2SO4        D£®KMnO4ËáÐÔÈÜÒº
¢ÚÒÒ´¼ÖÆÈ¡ÒÒÏ©µÄ·´Ó¦×°Öô˴¦ÂÔÈ¥£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºCH3-CH2-OH $¡ú_{170¡æ}^{ŨÁòËá}$CH2=CH2¡ü+H2O
¢ÛÄÜ˵Ã÷SO2ÆøÌå´æÔÚµÄÏÖÏóÊÇ×°ÖâñÖÐÆ·ºìÈÜÒºÍÊÉ«£®
¢ÜʹÓÃ×°ÖâóµÄÄ¿µÄÊǼì²é¶þÑõ»¯ÁòÊÇ·ñ³ý¾¡£®
¢ÝÈ·¶¨º¬ÓÐÒÒÏ©µÄÏÖÏóÊÇ×°ÖâóÖеÄÆ·ºìÈÜÒº²»ÍÊÉ«£¬×°ÖâôÖеÄËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£®
£¨2£©1£¬2-¶þäåÒÒÍé¿É×÷ÆûÓÍ¿¹±¬¼ÁµÄÌí¼Ó¼Á£¬³£ÎÂÏÂÊÇÎÞÉ«ÒºÌ壬ÃܶÈ2.18g/cm3£¬·Ðµã131.4¡æ£¬ÈÛ
µã9.79¡æ£¬²»ÈÜÓÚË®¡¢Ò×ÈÜÓÚ´¼¡¢ÃÑ¡¢±ûͪµÈÓлúÈܼÁ£®ÔÚʵÑéÖпÉÒÔÓÃÒÒÏ©À´ÖƱ¸1£¬2-¶þäåÒÒÍ飮 ÇëÌîдÏÂÁпհףº
¢Ùд³öÓÃÒÒÏ©ÖƱ¸1£¬2-¶þäåÒÒÍéµÄ»¯Ñ§·½³Ìʽ£ºCH2=CH2+Br2¡úCH2BrCH2Br£®
¢ÚÒª¼ìÑéijäåÒÒÍéÖеÄäåÔªËØ£¬ÕýÈ·µÄʵÑé·½·¨ÊÇD
A£®¼ÓÈëÂÈË®Õñµ´£¬¹Û²ìË®²ãÊÇ·ñÓÐ×غìÉ«³öÏÖ
B£®µÎÈëAgNO3ÈÜÒº£¬ÔÙ¼ÓÈëÏ¡HNO3£¬¹Û²ìÓÐÎÞdz»ÆÉ«³ÁµíÉú³É
C£®¼ÓÈëNaOHÈÜÒº¹²ÈÈ£¬ÀäÈ´ºó¼ÓÈëAgNO3ÈÜÒº£¬¹Û²ìÓÐÎÞdz»ÆÉ«³ÁµíÉú³É
D£®¼ÓÈëNaOHÈÜÒº¹²ÈÈ£¬È»ºó¼ÓÈëÏ¡HNO3ʹÈÜÒº³ÊËáÐÔ£¬ÔÙµÎÈëAgNO3ÈÜÒº£¬¹Û²ìÓÐÎÞdz»ÆÉ«³ÁµíÉú³É£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø