ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÅðÊǵÚIIIA×åÔªËØ£¬µ¥ÖÊÅðÔÚ¼ÓÈÈÌõ¼þÏÂÄÜÓë¶àÖַǽðÊô·´Ó¦¡£Ä³Í¬Ñ§ÓûÀûÓÃÂÈÆøºÍµ¥ÖÊÅð·´Ó¦ÖƱ¸ÈýÂÈ»¯Åð¡£ÒÑÖªBCl3µÄ·ÐµãΪ12.5¡æ£¬ÈÛµãΪ-107.3¡æ£¬ÓöË®¾çÁÒ·´Ó¦¡£

£¨1£©Ñ¡ÓÃÏÂͼËùʾµÄ×°Ö㨿ÉÒÔÖظ´Ñ¡Ó㩽øÐÐʵÑ飬װÖÃÒÀ´ÎÁ¬½ÓµÄºÏÀí˳ÐòΪ________¡£

£¨2£©Í¼ÖÐg¹ÜµÄ×÷ÓÃÊÇ______£¬×°ÖÃEµÄ×÷ÓÃÊÇ_______¡£

£¨3£©¿ªÊ¼ÊµÑéʱ£¬Ïȵãȼ____(Ìî¡°A¡±»ò¡°B¡±)´¦µÄ¾Æ¾«µÆ¡£

£¨4£©Çëд³öBCl3ÓöË®±äÖʵĻ¯Ñ§·½³Ìʽ___________¡£

£¨5£©ÅðËáÊÇÒ»ÔªÈõËᣬÆäÄÆÑλ¯Ñ§Ê½ÎªNa[B(OH)4]£¬ÔòÅðËáÔÚË®ÖеçÀë·½³ÌʽÊÇ______¡£

£¨6£©ÊµÑéÍê³Éºó£¬Ä³Í¬Ñ§ÏòF (ÈÜÒºÖк¬ÓÐ0.05 mol/LNaC10¡¢0.05 mol/LNaCl¡¢0.1 mol/L NaOH£©ÖеμÓÆ·ºìÈÜÒº£¬·¢ÏÖÈÜÒºÍÊÉ«¡£ÏÖÉè¼ÆʵÑé̽¾¿ÈÜÒºÍÊÉ«µÄÔ­Òò£¬Ç뽫±íÖÐÊý¾Ý²¹³äÍêÕû£¬Íê³ÉʵÑé·½°¸¡£

ʵÑéÐòºÅ

0.1mol/LNaClOÈÜÒº/mL

0.1mol/LNaClÈÜÒº/mL

0.2mol/LNaOHÈÜÒº/mL

H2O /mL

Æ·ºìÈÜÒº

ÏÖÏó

¢Ù

4.0

0

0

x

3µÎ

½Ï¿ìÍÊÉ«

¢Ú

0

4.0

4.0

0

3µÎ

²»ÍÊÉ«

¢Û

4.0

0

4.0

0

3µÎ

»ºÂýÍÊÉ«

x=_______£¬½áÂÛ£º_______________¡£

¡¾´ð°¸¡¿ ABDCEDF ±£³ÖÆøѹƽºâ£¬±ãÓÚ·ÖҺ©¶·ÖеÄÒºÌåÁ÷ÈëÕôÁóÉÕÆ¿ ÀäÄý²¢ÊÕ¼¯BCl3 A BCl3£«3H2O=H3BO3£«3HCl H3BO3£«H2O[B(OH)4]-£«H+ 4.0 NaClOʹƷºìÈÜÒºÍÊÉ«£¬ÈÜÒº¼îÐÔԽǿÍÊÉ«Ô½Âý¡£

¡¾½âÎö¡¿£¨1£©ÓûÀûÓÃÂÈÆøºÍµ¥ÖÊÅð·´Ó¦ÖƱ¸ÈýÂÈ»¯Åð£¬ÒÑÖªµ¥ÖÊÅðÔÚ¼ÓÈÈÌõ¼þÏÂÄÜÓë¶àÖַǽðÊô·´Ó¦£¬BCl3µÄ·ÐµãΪ12.5¡æ£¬ÈÛµãΪ-107.3¡æ£¬ÓöË®¾çÁÒ·´Ó¦¡£Ôò±ØÐëÏÈÖƱ¸¸ÉÔï½à¾»µÄÂÈÆø£¬¶øÇÒ·´Ó¦¹ý³ÌÖбØÓзÀֹˮÕôÆû½øÐУ¬¹ÊAΪÖÆÈ¡ÂÈÆø×°Öã¬È»ºó½«ËùµÃÂÈÆøͨ¹ýB×°ÖÃÒÔ³ýÈ¥ÂÈÆøÖеÄÂÈ»¯Ç⣬ÔÙͨ¹ýD×°ÖøÉÔïºó½«ÂÈÆøͨÈëC×°ÖÃÖмÓÈÈÌõ¼þÏÂÓëÅð·´Ó¦£¬²úÉúµÄÈýÂÈ»¯ÅðÆøÌå½øÈëE×°ÖÃÀäÈ´ÊÕ¼¯ÆðÀ´£¬ÔÙÁ¬½ÓD×°ÖøÉÔï·ÀÖ¹Íâ½çË®ÕôÆû½øÐУ¬×îºóÁ¬½ÓF½øÐÐβÆø´¦Àí¡£Òò´Ë½ÓÁ¬µÄ˳ÐòΪ£ºABDCEDF£»£¨2£©Í¼ÖÐg¹ÜµÄ×÷ÓÃÊÇ£º±£³ÖÆøѹƽºâ£¬±ãÓÚ·ÖҺ©¶·ÖеÄÒºÌåÁ÷ÈëÕôÁóÉÕÆ¿£»×°ÖÃEµÄ×÷ÓÃÊÇ£ºÀäÄý²¢ÊÕ¼¯BCl3£»£¨3£©¿ªÊ¼ÊµÑéʱ£¬ÏȵãȼA´¦µÄ¾Æ¾«µÆÖÆÈ¡ÂÈÆø£¬Ê¹ºóÃ淴ӦװÖÃÖгäÂú¸ÉÔïµÄÂÈÆø£»£¨4£©Çëд³öBCl3Óöˮˮ½â±äÖÊÉú³ÉÅðËá¼°ÑÎËᣬÆ仯ѧ·½³Ìʽ£ºBCl3£«3H2O=H3BO3£«3HCl£»£¨5£©ÅðËáÊÇÒ»ÔªÈõËᣬÆäÄÆÑλ¯Ñ§Ê½ÎªNa[B(OH)4]£¬ÔòÅðËáÔÚË®ÖеçÀë·½³ÌʽÊÇH3BO3£«H2O[B(OH)4]-£«H+£»£¨6£©Éè¼ÆʵÑé̽¾¿²»Í¬ÈÜÒº¶ÔÆ·ºìÍÊÉ«µÄÔ­Òò£¬Í¨³£ÈÜÒºµÄŨ¶È²»Í¬µ«ÈÜÒºµÄ×ÜÌå»ýÏàͬ£¬¸ù¾Ý¢Ú¢Û¿ÉÖª£¬ÈÜÒº×ÜÌå»ýΪ8.0mL£¬Ôòx=4.0mL£¬ÊµÑé¢Ù¢ÛÓÐNaClOÔòÍÊÉ«£¬ÊµÑé¢ÚûÓÐNaClOÔò²»ÍÊÉ«£¬ÊµÑé¢ÙÓëʵÑé¢Û¶Ô±È£¬¼îÐÔÌõ¼þÏ£¬ÍÊÉ«Âý£¬Ôò½áÂÛΪ£ºNaClOʹƷºìÈÜÒºÍÊÉ«£¬ÈÜÒº¼îÐÔԽǿÍÊÉ«Ô½Âý¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ïÈ£¨Sr£©ÎªµÚÎåÖÜÆÚIIA×åÔªËØ£¬Æ仯ºÏÎïÁùË®ÂÈ»¯ïÈ£¨SrCl2¡¤6H2O£©ÊÇʵÑéÊÒÖØÒªµÄ·ÖÎöÊÔ¼Á£¬¹¤ÒµÉϳ£ÒÔÌìÇàʯ£¨Ö÷Òª³É·ÖΪSrSO4£©ÎªÔ­ÁÏÖƱ¸£¬Éú²úÁ÷³ÌÈçÏ£º

£¨1£©SrCl2µÄµç×ÓʽΪ_____________¡£

£¨2£©¹¤ÒµÉÏÌìÇàʯ¸ô¾ø¿ÕÆø¸ßαºÉÕ£¬Èô0.5 mol SrSO4ÖÐÖ»ÓÐS±»»¹Ô­£¬×ªÒÆÁË4 molµç×Ó¡£Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_______________________¡£

£¨3£©¼ÓÈëÁòËáµÄÄ¿µÄÊÇ______________________¡£ÎªÁËÌá¸ßÔ­ÁϵÄÀûÓÃÂÊ£¬ÂËÒºÖÐSr2+µÄŨ¶ÈÓ¦²»¸ßÓÚ_________ mol/L£¨×¢£º´ËʱÂËÒºÖÐBa2+Ũ¶ÈΪ1¡Á10-5mol/L£»SrSO4ºÍBaSO4ÈܶȻý³£ÊýÒÀ´ÎΪ3¡Á10-7¡¢1¡Á10-10¡££©

£¨4£©²úÆ·´¿¶È¼ì²â£º³ÆÈ¡1.000 g²úÆ·ÈܽâÓÚÊÊÁ¿Ë®ÖУ¬ÏòÆäÖмÓÈ뺬AgNO31.100¡Á10-2molµÄAgNO3ÈÜÒº£¨ÈÜÒºÖгýCl-Í⣬²»º¬ÆäËüÓëAg+·´Ó¦µÄÀë×Ó£©£¬´ýCl-ÍêÈ«³Áµíºó£¬Óú¬Fe3+µÄÈÜÒº×÷ָʾ¼Á£¬ÓÃ0.2000 mol/LµÄNH4SCN±ê×¼ÈÜÒºµÎ¶¨Ê£ÓàµÄAgNO3£¬Ê¹Ê£ÓàµÄAg+ÒÔAgSCN°×É«³ÁµíµÄÐÎʽÎö³ö¡£

¢ÙµÎ¶¨·´Ó¦´ïµ½ÖÕµãµÄÏÖÏóÊÇ_______________________¡£

¢ÚÈôµÎ¶¨¹ý³ÌÓÃÈ¥ÉÏÊöŨ¶ÈµÄNH4SCNÈÜÒº20.00 mL£¬Ôò²úÆ·ÖÐSrCl2¡¤6H2OµÄÖÊÁ¿°Ù·Öº¬Á¿Îª______________£¨±£Áô4λÓÐЧÊý×Ö£©¡£

£¨5£©¹¤ÒµÉϳ£Óõç½âÈÛÈÚSrCl2ÖÆïȵ¥ÖÊ¡£ÓÉSrCl2¡¤6H2OÖÆÈ¡ÎÞË®ÂÈ»¯ïȵķ½·¨ÊÇ_________¡£

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

¾«Ó¢¼Ò½ÌÍø