ÌâÄ¿ÄÚÈÝ

ij¹¤Òµ·ÏË®Öнöº¬Ï±íÀë×ÓÖеÄ5ÖÖ£º

ÑôÀë×Ó

K+  Cu2+   Fe3+   Ca2+   Fe2+

ÒõÀë×Ó

Cl-   CO32-  NO3-  SO42-  SiO32-

ijͬѧÓû̽¾¿·ÏË®µÄ×é³É£¬½øÐÐÁËÈçÏÂʵÑ飺

¢ñ£®È¡·ÏË®ÉÙÐí¼ÓÈëÉÙÁ¿ÑÎËᣬÎÞ°×É«³ÁµíÎö³ö,µ«Éú³ÉÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×ǵÄÎÞÉ«ÎÞÎ¶ÆøÌå

¢ò£®Ïò¢ñÖÐËùµÃµÄÈÜÒºÖмÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³É¡£

ÏÂÁÐÍÆ¶Ï²»ÕýÈ·µÄÊÇ

A. ÈÜÒºÖÐÒ»¶¨º¬ÓеÄÑôÀë×ÓÊÇK+ ¡¢ Cl-  ¡¢CO32-  ¡¢NO3- ¡¢ SO42-

B. ¢ñÖмÓÈëÑÎËáÉú³ÉÎÞÉ«ÆøÌåµÄµÄÀë×Ó·½³ÌʽÊÇCO32-+2H+=CO2¡ü+H2O

C. Ô­ÈÜÒºÖеÄK+ ¡¢ Cl-  ¡¢NO3- ´æÔÚÓë·ñÎÞ·¨È·¶¨

D. ¢òÖвúÉú°×É«³ÁµíµÄÀë×Ó·½³ÌʽÊÇBa2++SO42-=BaSO4¡ý

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

°±ºÍ루N2H4£©ÊǵªµÄÁ½ÖÖ³£¼û»¯ºÏÎÔÚ¿ÆÑ§¼¼ÊõºÍÉú²úÖÐÓй㷺ӦÓ᣻شðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÑÖª£ºN2(g)+3H2(g) 2NH3(g) ¦¤H£½-92.4kJ¡¤mol-1

ÔÚºãΡ¢ºãÈݵÄÃܱÕÈÝÆ÷ÖУ¬ºÏ³É°±·´Ó¦µÄ¸÷ÎïÖÊŨ¶ÈµÄ±ä»¯ÇúÏßÈçͼËùʾ¡£

¢Ù ¼ÆËãÔÚ¸ÃζÈÏ·´Ó¦2NH3(g) N2(g)+3H2(g)µÄƽºâ³£ÊýK=________¡£

¢Ú ÔÚµÚ25minÄ©£¬±£³ÖÆäËüÌõ¼þ²»±ä£¬Èô½«Î¶ȽµµÍ£¬ÔÚµÚ35minÄ©Ôٴδﵽƽºâ¡£ÔÚÆ½ºâÒÆ¶¯¹ý³ÌÖÐN2Ũ¶È±ä»¯ÁË0.5mol/L£¬ÇëÔÚͼÖл­³ö25-40minNH3Ũ¶È±ä»¯ÇúÏß¡£________

¢Û ÒÑÖª£º2N2(g)+6H2O(l) 4NH3(g)+3O2£¨g£©¡÷H=+1530.0KJ/molÔòÇâÆøµÄÈÈֵΪ_____¡£

£¨2£©¢Ù N2H4ÊÇÒ»ÖÖ¸ßÄÜȼÁϾßÓл¹Ô­ÐÔ£¬Í¨³£ÓÃNaClOÓë¹ýÁ¿NH3·´Ó¦ÖƵã¬Çë½âÊÍΪʲôÓùýÁ¿°±Æø·´Ó¦µÄÔ­Òò£º__________

¢Ú ÓÃNaClOÓëNH3 ÖÆN2H4µÄ·´Ó¦ÊÇÏ൱¸´Ôӵģ¬Ö÷Òª·ÖΪÁ½²½£º

ÒÑÖªµÚÒ»²½£ºNH3+ClO-=OH-+NH2Cl

Çëд³öµÚ¶þ²½Àë×Ó·½³Ìʽ£º__________________

¢Û N2H4Ò×ÈÜÓÚË®£¬ÊÇÓë°±ÏàÀàËÆµÄÈõ¼î£¬¼ºÖªÆä³£ÎÂϵçÀë³£ÊýK1=1.0¡Á10-6£¬³£ÎÂÏ£¬½«0.2 mol/L N2H4¡¤H2OÓë0.lmol/L£¬ÑÎËáµÈÌå»ý»ìºÏ£¨ºöÂÔÌå»ý±ä»¯£©¡£Ôò´ËʱÈÜÒºµÄPHµÈÓÚ________£¨ºöÂÔN2H4µÄ¶þ¼¶µçÀ룩¡£

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

¾«Ó¢¼Ò½ÌÍø