ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿»¯Ñ§ÊµÑéÊÇѧϰÀí½â»¯Ñ§ÖªÊ¶µÄ»ù´¡,ËüÒÔÆäÉú¶¯µÄ÷ÈÁ¦ºÍ·á¸»µÄÄÚº­ÔÚ»¯Ñ§Ñ§Ï°Öз¢»Ó×ŶÀÌصŦÄܺÍ×÷Óá£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

I .ʵÑéÊÒÓûÓùÌÌåNaOHÅäÖÆ100 mL 1mol¡¤L-1µÄNaOHÈÜÒº¡£

£¨1£©ÅäÖÆÉÏÊöÈÜÒº£¬ÏÂÁв»ÄÜÓõ½µÄÒÇÆ÷ÊÇ_______________(ÓÃÐòºÅ»Ø´ð)¡£

A.ÉÕ±­ B.´óÊÔ¹Ü C.²£Á§°ô D.100 mLÈÝÁ¿Æ¿

£¨2£©³ýÉÏÊöÒÇÆ÷Í⣬»¹Ò»¶¨ÒªµÄ²£Á§ÒÇÆ÷ÓÐ______________¡£

£¨3£©ÓÃÍÐÅÌÌìƽ³ÆÈ¡NaOH¹ÌÌåµÄÖÊÁ¿Îª_________g£¬´ÓϱíÖÐÑ¡Ôñ³ÆÁ¿NaOH¹ÌÌåËùÐèÒªµÄÒÇÆ÷ __________________£¨ÌîÐòºÅ£©¡£

£¨4£©ÏÂÁÐÇé¿öʹËùÅäµÃÈÜÒºµÄŨ¶ÈÈçºÎ±ä»¯£¿£¨Ìî¡°Æ«¸ß¡±¡° Æ«µÍ¡±»ò¡°²»±ä¡±£©

A.δϴµÓÈܽâÇâÑõ»¯ÄƵÄÉÕ±­_________¡£

B.ÈÝÁ¿Æ¿Ê¹ÓÃÇ°ÓÃÕôÁóˮϴ¹ý£¬ÄÚ±Ú¸½ÓÐË®Öé¶øδ¸ÉÔï´¦Àí___________¡£

C.¶¨ÈÝʱ¼ÓË®³¬¹ýÁ˿̶ÈÏߣ¬½«¶à³öµÄÒºÌåÎü³ö___________¡£

II£®ÒÑÖª»¯ºÏÎï¼×ºÍÒÒ¶¼²»ÈÜÓÚË®£¬¼×¿ÉÈÜÓÚÖÊ×î·ÖÊý´óÓÚ»òµÈÓÚ98%µÄŨÁòËᣬ¶øÒÒ²»ÈÜ¡£ÏÖÓÐÒ»·Ý¼×ºÍÒҵĻìºÏÎïÑùÆ·£¬Í¨¹ýʵÑé½øÐзÖÀ룬¿ÉµÃµ½¹ÌÌå¼×£¨ÊµÑéÖÐʹÓõĹýÂËÆ÷ÊÇÓÃÓÚ¹ýÂËÇ¿ËáÐÔÒºÌåµÄÄÍËá¹ýÂËÆ÷£¬¼×ÓëÁòËá²»·¢Éú»¯Ñ§·´Ó¦£©¡£

ÇëÌîдÏÂÁпհף¬Íê³ÉÓÉÉÏÊö»ìºÏÎïµÃµ½¹ÌÌå¼×µÄʵÑéÉè¼Æ¡£

ÐòºÅ

ʵÑé²½Öè

¼òÊöʵÑé²Ù×÷£¨²»±ØÐðÊöÈçºÎ×éװʵÑé×°Öã©

¢Ù

Èܽâ

½«»ìºÏÎï·ÅÈëÉÕ±­ÖУ¬¼ÓÈë98% H2SO4____¡£

¢Ú

_____________

¢Û

Ï¡ÊÍ£¨³Áµí£©

____________

¢Ü

¹ýÂË

¢Ý

_________

Ïò¢ÜµÄ¹ýÂËÆ÷ÖÐ×¢ÈëÉÙÁ¿ÕôÁóË®£¬Ê¹Ë®Ãæ½þ¹ý³ÁµíÎ´ýË®Â˳öºó£¬ÔٴμÓˮϴµÓ£¬Á¬Ï´¼¸´Î¡£

¢Þ

¼ìÑé³ÁµíÊÇ·ñÏ´¾»

____________________¡£

¡¾´ð°¸¡¿ B ½ºÍ·µÎ¹Ü 4.0 abe Æ«µÍ ²»±ä Æ«µÍ ³ä·Ö½Á°è£¬Ö±ÖÁ¹ÌÌå²»ÔÙÈܽ⠹ýÂË ½«¢ÚÖÐËùµÃÂËÒºÑØÉÕ±­ÄÚ±ÚÂýÂý×¢Èë×ãÁ¿Ë®ÖУ¬²¢²»¶Ï½Á°è£¬Ö±ÖÁÎö³öÈ«²¿¹ÌÌå Ï´µÓ³Áµí ÓÃСÊԹܴӢݵÄ©¶·Ï¿ÚÈ¡ÉÙÁ¿Ï´³öÒº£¬µÎÈëBaCl2ÈÜÒº¡£ÈçûÓа×É«³Áµí£¬ËµÃ÷³ÁµíÒÑÏ´¾»

¡¾½âÎö¡¿I .£¨1£©´óÊÔ¹ÜÓÃÓÚ·¢Éú»¯Ñ§·´Ó¦£»£¨2£©¶¨ÈÝʱҪÓýºÍ·µÎ¹Ü£»£¨3£©m(NaOH)=0.1L¡Á1mol¡¤L-1¡Á40g¡¤mol£­1=4.0g£»½«ÇâÑõ»¯ÄÆ·ÅÔÚÉÕ±­ÖÐÓÃÍÐÅÌÌìƽ³ÆÁ¿£»£¨4£©¸ù¾Ýc=n/V·ÖÎöÎó²î£»II£®·ÖÀë»ìºÏÎÁ½ÖÖ»ìºÏÎïÒªÒÔÔ­ÓеÄ×é³É·Ö¿ª£®½â´ËÌâµÄ˼·Ӧ´Ó¡°ÒÑÖª»¯ºÏÎï¼×ºÍÒÒ¶¼²»ÈÜÓÚË®£¬¼×¿ÉÈÜÓÚÖÊÁ¿·ÖÊý´óÓÚ»òµÈÓÚ98%µÄÁòËá¶øÒÒ²»ÈÜ¡±²úÉú£®¼ÈÈ»ÓÐÒ»ÖÖ¿ÉÈÜÓÚÖÊÁ¿·ÖÊý´óÓÚ»òµÈÓÚ98%µÄÁòËᣬÕâÒѾ­¸æËßÎÒÃǶþÕßÔÚŨÁòËáÖеÄ״̬²»Í¬£¬¿É²ÉÓùýÂ˵ķ½·¨½«¶þÕß·Ö¿ª£®µ±È»£¬ÓÉÓÚÈÜҺΪŨÁòËᣬËùÒÔʹÓõIJ»ÊÇÆÕͨµÄÂËÖ½£¬Ó¦ÎªÓÃÓÚÇ¿ËáÐÔÒºÌåµÄÄÍËáÐÔ¹ýÂËÆ÷£®ÔÙÓÃˮϴµÓ£¬¼´¿ÉµÃµ½²»ÈÜÓÚŨÁòËáµÄÄÑÈÜÐÔ¹ÌÌ壮ÁíÒ»ÖÖ¹ÌÌåÈܽâÔÚŨÁòËáÖУ¬ÒªµÃµ½Ëü£¬ÐèÒªÔÙ´ÎÑо¿¡°ÒÑÖª»¯ºÏÎï¼×ºÍÒÒ¶¼²»ÈÜÓÚË®£¬¼×¿ÉÈÜÓÚÖÊÁ¿·ÖÊý´óÓÚ»òµÈÓÚ98%µÄÁòËᡱ£¬Õâ˵Ã÷¼×ÔÚÏ¡ÁòËáÖв»Èܽ⣬Ҳ¾ÍÊÇ˵£¬Ö»ÒªÎÒÃǼÓˮϡÊͼ׵ÄŨÁòËáÈÜÒº£¬¾Í¿ÉÈü״ÓÈÜÒºÖÐÖØÐÂÓÎÀë³öÀ´£®

I .ʵÑéÊÒÓûÓùÌÌåNaOHÅäÖÆ100 mL 1mol¡¤L-1µÄNaOHÈÜÒº¡££¨1£©ÅäÖÆÉÏÊöÈÜÒº£¬ÎÞÐèÓõ½µÄÒÇÆ÷ÊÇB´óÊԹܣ»A.ÉÕ±­ÓÃÓÚÈܽ⣻ C.²£Á§°ôÓÃÓÚ½Á°è¡¢ÒýÁ÷£»D.100 mLÈÝÁ¿Æ¿ÓÃÓÚÅäÖÆÒ»¶¨Ìå»ýµÄÈÜÒº£»¹ÊÑ¡B¡££¨2£©³ýÉÏÊöÒÇÆ÷Í⣬¶¨ÈÝʱҪÓýºÍ·µÎ¹ÜµÎ¼ÓÉÙÁ¿ÒºÌ壬¹Ê»¹Ò»¶¨ÒªµÄ²£Á§ÒÇÆ÷ÓнºÍ·µÎ¹Ü¡££¨3£©m(NaOH)=0.1L¡Á1mol¡¤L-1¡Á40g¡¤mol£­1=4.0g£¬ÓÃÍÐÅÌÌìƽ³ÆÈ¡NaOH¹ÌÌåµÄÖÊÁ¿Îª4.0g£¬½«ÇâÑõ»¯ÄÆ·ÅÔÚÉÕ±­ÖÐÓÃÍÐÅÌÌìƽ³ÆÁ¿£¬³ÆÁ¿NaOH¹ÌÌåËùÐèÒªµÄÒÇÆ÷ abe£¨ÌîÐòºÅ£©¡££¨4£©A.δϴµÓÈܽâÇâÑõ»¯ÄƵÄÉÕ±­£¬ÒýÆðÈÜÖʱäÉÙ£¬¹ÊŨ¶ÈÆ«µÍ ¡£B.ÈÝÁ¿Æ¿Ê¹ÓÃÇ°ÓÃÕôÁóˮϴ¹ý£¬ÄÚ±Ú¸½ÓÐË®Öé¶øδ¸ÉÔï´¦Àí£¬¶¨ÈÝʱҪ¼ÓË®£¬¹Ê¶ÔŨ¶ÈûÓÐÓ°Ï죬¼´²»±ä£»C.¶¨ÈÝʱ¼ÓË®³¬¹ýÁ˿̶ÈÏߣ¬½«¶à³öµÄÒºÌåÎü³ö£¬ÒýÆðÈÜÖʱäÉÙ£¬¹ÊŨ¶ÈÆ«µÍ¡£

II£®¢ÙÈܽ⣬½«»ìºÏÎï·ÅÈëÉÕ±­ÖУ¬¼ÓÈë98% H2SO4£¬³ä·Ö½Á°è£¬Ö±ÖÁ¹ÌÌå²»ÔÙÈܽ⡣¢Ú¹ýÂË·ÖÀë²»ÈÜÎïºÍ¿ÉÈÜÎ½«ÉÕ±­ÖеĻìºÏÒºÑز£Á§°ô»ºÂýµ¹Èë¹ýÂËÆ÷ÖйýÂË£¬¢ÛÏ¡ÊÍ£¨³Áµí£©£º¹Ê½«¢ÚÖÐËùµÃÂËÒºÑØÉÕ±­ÄÚ±ÚÂýÂý×¢Èë×ãÁ¿Ë®ÖУ¬²¢²»¶Ï½Á°è£¬Ö±ÖÁÎö³öÈ«²¿¹ÌÌ壻¢Ü¹ýÂË¢ÝÖ±½ÓÏò¹ýÂËÆ÷ÖмÓÈëÕôÁóˮϴµÓ³Áµí£¬Öظ´2-3´Î£¬¹ÊÏ´µÓ³Áµí£»Ïò¢ÜµÄ¹ýÂËÆ÷ÖÐ×¢ÈëÉÙÁ¿ÕôÁóË®£¬Ê¹Ë®Ãæ½þ¹ý³ÁµíÎ´ýË®Â˳öºó£¬ÔٴμÓˮϴµÓ£¬Á¬Ï´¼¸´Î¡£¢Þ¼ìÑé³ÁµíÊÇ·ñÏ´¾»£ºÓÃСÊԹܴӢݵÄ©¶·Ï¿ÚÈ¡ÉÙÁ¿Ï´³öÒº£¬µÎÈëBaCl2ÈÜÒº¡£ÈçûÓа×É«³Áµí£¬ËµÃ÷³ÁµíÒÑÏ´¾»¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿µªÊÇÓëÈËÀàÉú»îÃÜÇÐÏà¹ØµÄÒ»ÖÖÖØÒªÔªËØ¡£¸ù¾ÝÒªÇó»Ø´ð£º

(1)д³öN2µÄµç×Óʽ_________£»½«¿ÕÆøÖеĵªÆøת»¯Îªº¬µª»¯ºÏÎïµÄ¹ý³Ì³ÆΪ¹Ìµª£¬µ±½ñ´ó¹æÄ£È˹¤¹ÌµªµÄÖ÷Òª·½·¨ÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©______________________________¡£

(2)ijÎÞÉ«»ìºÏÆøÌåÖпÉÄܺ¬ÓÐNH3¡¢O2¡¢H2¡¢NO¡¢HClµÈÆøÌ壬½«Ëüͨ¹ýŨH2SO4 ºó·¢ÏÖÆøÌåÌå»ý¼õÉÙ£¬½«Ê£ÓàÆøÌåÓë¿ÕÆø½Ó´¥ºó³Êºì×ØÉ«£¬´Ë»ìºÏÆøÖÐÒ»¶¨²»º¬ÓÐ_________¡£

(3)ʵÑéÊÒÓжàÖÖÖÆÈ¡°±ÆøµÄ·½·¨£¬ÆäÖмÓÈÈÏûʯ»ÒÓëÂÈ»¯ï§µÄ»ìºÏ¹ÌÌå¾ÍÊÇ·½·¨Ö®Ò»¡£

¢Ùд³ö´Ë·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____________________________________________¡£

¢Ú°±Æø¿ÉʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÔ­Òò£¨ÓÃÏàÓ¦µÄ·½³Ìʽ˵Ã÷£©_______________¡£

¢ÛÓÃÒÒ×°Ö㨼ûͼ£©ÎüÊÕÒ»¶Îʱ¼ä°±Æøºó£¬ÔÙͨÈë¿ÕÆø£¬Í¬Ê±½«¼ÓÈȵIJ¬Ë¿²åÈëÒÒ×°ÖõÄ׶ÐÎÆ¿ÄÚ£¬×¶ÐÎÆ¿Öв»¿ÉÄÜÉú³ÉµÄÎïÖÊÊÇ___£¨Ñ¡ÌîÐòºÅ£©¡£

a H2 b NO2 c HNO3 d NH4NO3

д³öÒÒ×°ÖÃÖа±´ß»¯Ñõ»¯µÄ»¯Ñ§·½³Ìʽ£º__________________________¡£

(4)ÒÑÖª3Cl2+2NH3==N2 + 6HCl£¬ÈôNH3¹ýÁ¿Ôò»¹Óз´Ó¦£ºNH3+ HCl== NH4Cl ¡£³£Î³£Ñ¹Ï£¬ÈôÔÚÒ»ÃܱÕÈÝÆ÷Öн«15mLCl2ºÍ40mL NH3³ä·Ö»ìºÏ·´Ó¦ºó£¬Ê£ÓàÆøÌåµÄÌå»ýΪ____mL¡£

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

¾«Ó¢¼Ò½ÌÍø