ÌâÄ¿ÄÚÈÝ

£¨1£©ÊµÑéÊÒÓÃÂÈ»¯ï§¹ÌÌåÖÆÈ¡°±ÆøµÄ»¯Ñ§·½³ÌʽÊÇ          ¡£
£¨2£©½«4.48L£¨±ê×¼×´¿ö£©°±ÆøͨÈëË®Öеõ½0.05LÈÜÒº£¬ËùµÃÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ      ¡£
£¨3£©ÏÖÓÐ100mL AlCl3ÓëMgSO4µÄ»ìºÏÈÜÒº£¬·Ö³ÉÁ½µÈ·Ý¡£
¢ÙÏòÆäÖÐÒ»·ÝÖмÓÈë10mL 4mol/LµÄ°±Ë®£¬Ç¡ºÃÍêÈ«·´Ó¦£¬ÆäÖÐAlCl3Ó백ˮ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ          ¡£¼ÌÐø¼ÓÈël mol/L NaOHÈÜÒºÖÁ10mLʱ£¬³Áµí²»ÔÙ¼õÉÙ£¬³Áµí¼õÉÙµÄÀë×Ó·½³ÌʽÊÇ          £¬¼õÉٵijÁµíµÄÎïÖʵÄÁ¿ÊÇ          ¡£
¢ÚÏòÁíÒ»·ÝÖмÓÈëa mL 1mol/LBaCl2ÈÜÒºÄÜʹSO42£­³ÁµíÍêÈ«£¬a=          ¡£

£¨1£©Ca£¨OH£©2£«2NH4ClCaCl2£«2H2O£«2NH3¡ü
£¨2£©4 mol/L £¨1·Ö£©
£¨3£©¢ÙA13+£«3NH3¡¤H2O=Al£¨OH£©3¡ý£«3NH4+
Al£¨OH£©3£«OH£­=AlO2£­£«2H2O
0.01 mol
¢Ú 5

½âÎöÊÔÌâ·ÖÎö£º¢ÅʵÑéÊÒÓÃÂÈ»¯ï§¹ÌÌåÖÆÈ¡°±ÆøµÄ»¯Ñ§·½³ÌʽÊÇCa£¨OH£©2£«2NH4ClCaCl2£«2H2O£«2NH3¡ü
¢Æ¸ù¾Ýc=n/V,µÃ°±ÆøµÄÎïÖʵÄÁ¿ÊÇ4.48L/22.4L/mol=0.2mol£¬ËùÒÔËùµÃÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈΪ£º0.2mol/0.05L="4" mol/L
¢Ç¢ÙAlCl3Ó백ˮ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇA13+£«3NH3¡¤H2O=Al£¨OH£©3¡ý£«3NH4+£¬³Áµí¼õÉÙ¼´ÇâÑõ»¯ÂÁÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£¬Àë×Ó·½³ÌʽΪAl£¨OH£©3£«OH£­=AlO2£­£«2H2O£¬³Áµí¼õÉÙµÄÎïÖʵÄÁ¿¼´ÎªÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿0.01mol
¢ÚÓÉ¢ÙÖªAl(OH)3µÄÎïÖʵÄÁ¿Îª0.01mol£¬ÏûºÄNH3¡¤H2O0.03mol£¬ËùÒÔMg2+ÏûºÄNH3¡¤H2O£¨0.04mol-0.03mol£©=0.01mol,Ôòn£¨Mg2+£©=0.005mol=n(SO42-£©£¬ËùÒÔÐèÒª1mol/LBaCl2ÈÜÒº5mlʹSO42-³ÁµíÍêÈ«¡£
¿¼µã£ºÀë×Ó·´Ó¦¼°Ïà¹Ø¼ÆËã

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÁòËáµÄ¹¤ÒµÖƱ¸ÊÇÒ»¸öÖØÒªµÄ»¯¹¤Éú²ú¹ý³Ì£¬µ«ÔÚÉú²ú¹ý³ÌÖлá²úÉú´óÁ¿ÎÛȾ£¬ÐèÒªÔÚÉú²ú¹¤ÒÕÖп¼Âǵ½ÂÌÉ«¹¤ÒÕ¡£
IβÆøµÄÎüÊÕºÍ×ÛºÏÀûÓá£
ÒÔ¹¤ÒµÖÆÁòËáµÄβÆø¡¢°±Ë®¡¢Ê¯»Òʯ¡¢½¹Ì¿¡¢Ì¼ËáÂÈ狀ÍKCIΪԭÁÏ¿ÉÒԺϳÉÁò»¯¸Æ¡¢ÁòËá¼Ø¡¢ÑÇÁòËá淋ÈÎïÖÊ¡£ºÏ³É·ÏßÈçÏ£º

£¨1£©·´Ó¦IIIÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ            ¡£
£¨2£©·´Ó¦¢ôµÄ»¯Ñ§·½³ÌʽΪ                      ¡£
£¨3£©·´Ó¦VÔÚ25¡æ¡¢40%µÄÒÒ¶þ´¼ÈÜÒºÖнøÐУ¬¸Ã·´Ó¦ÄÜ˳Àû½øÐеÄÔ­ÒòΪ           ¡£
¢ò´ß»¯¼ÁµÄ»ØÊÕÀûÓá£
SO2µÄ´ß»¯Ñõ»¯ËùʹÓõĴ߻¯¼ÁΪV2O5£¬Êµ¼ÊÉú²úÖУ¬´ß»¯¼ÁÔÚʹÓÃÒ»¶Îʱ¼äºó£¬»áº¬ÓÐV2O5¡¢VOSO4ºÍSiO2µÈ£¬ÆäÖÐVOSO4¡£ÄÜÈÜÓÚË®¡£»ØÊÕV2O5£¬µÄÖ÷ÒªÁ÷³ÌÈçÏ£º

£¨4£©Èô·´ÝÍȡʹÓõÄÁòËáÓÃÁ¿¹ý´ó£¬½øÒ»²½´¦Àíʱ»áÔö¼Ó____       µÄÓÃÁ¿¡£
£¨5£©½þÈ¡»¹Ô­¹ý³ÌµÄ²úÎïÖ®Ò»ÊÇVOSO4£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                  ¡£
Ñõ»¯¹ý³ÌµÄ»¯Ñ§·½³ÌʽΪKClO3+6VOSO4+3H2SO4= 2(VO)2(SO4)3+KCl+3H2O£»ÈôÁ½²½ËùÓÃÊÔ¼ÁNa2SO3ÓëKC1O3µÄÎïÖʵÄÁ¿Ö®±ÈΪ12£º7£¬Ôò¸Ã´ß»¯¼ÁÖÐV2O5¡¢VOSO4µÄÎïÖʵÄÁ¿Ö®±ÈΪ               ¡£

ijͬѧ¹ºÂòÁËһƿ¡Á¡ÁÅÆ¡°84Ïû¶¾Òº¡±£¬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¸Ã¡°84Ïû¶¾Òº¡±Ï¡ÊͺóÓÃÓÚÏû¶¾£¬Ï¡ÊͺóµÄÈÜÒºÖÐc(Na£«)£½________ mol¡¤L£­1(¼ÙÉèÏ¡ÊͺóÈÜÒºÃܶÈΪ1.0 g¡¤cm£­3)¡£
(3)ijʵÑéÐèÓÃ480 mLº¬25%NaClOµÄÏû¶¾Òº¡£¸Ãͬѧ²ÎÔĸá°84Ïû¶¾Òº¡±µÄÅä·½£¬ÓûÓÃNaClO¹ÌÌåÅäÖƸÃÏû¶¾Òº¡£
¢ÙÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________¡£

A£®ÈçÉÏͼËùʾµÄÒÇÆ÷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
¢ÚÔÚÅäÖƹý³ÌÖУ¬ÏÂÁвÙ×÷¿ÉÄÜʹÅäÖƵÄÈÜÒºµÄŨ¶ÈÆ«´óµÄÊÇ________¡£
A£®ÉÕ±­ÖÐÈÜҺתÒƵ½ÈÝÁ¿Æ¿ÖÐʱ£¬Î´Ï´µÓÉÕ±­
B£®¶¨ÈÝʱ£¬¸©Êӿ̶ÈÏß
C£®¶¨ÈÝʱ£¬ÑöÊӿ̶ÈÏß
D£®ÒÆҺʱ£¬ÓÐÉÙÁ¿ÒºÌ彦³ö

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

¾«Ó¢¼Ò½ÌÍø