ÌâÄ¿ÄÚÈÝ

¡¾»¯Ñ§¡ª¡ªÑ¡ÐÞ2£º»¯Ñ§Óë¼¼Êõ¡¿£¨15·Ö£©

£¨1£©ÎÒ¹úijµØÇøÒÑ̽Ã÷Ô̲ØÓзḻµÄ³àÌú¿ó(Ö÷Òª³É·ÖΪFe2O3£¬»¹º¬ÓÐSiO2µÈÔÓÖÊ)¡¢Ãº¿ó¡¢Ê¯»ÒʯºÍð¤ÍÁ¡£ÄâÔڸõØÇø½¨Éè´óÐÍÁ¶Ìú³§¡£

¢ÙËæ×ÅÌú¿óµÄ¿ª·¢ºÍÁ¶Ìú³§µÄ½¨Á¢£¬ÐèÒªÔڸõØÇøÏàÓ¦½¨Á¢½¹»¯³§¡¢·¢µç³§¡¢Ë®Äà³§µÈ£¬ÐγɹæÄ£µÄ¹¤ÒµÌåϵ¡£¾Ý´ËÈ·¶¨Í¼ÖÐÏàÓ¦¹¤³§µÄÃû³Æ£ºA£®________£¬B.________£¬C.________£¬D.________£»

¢ÚÒÔ³àÌú¿óΪԭÁÏ£¬Ð´³ö¸ß¯Á¶ÌúÖеõ½ÉúÌúºÍ²úÉú¯ÔüµÄ»¯Ñ§·½³Ìʽ£º______________¡£

£¨2£©²£Á§¸Ö¿ÉÓÉ·ÓÈ©Ê÷Ö¬ºÍ²£Á§ÏËÎ¬ÖÆ³É¡£

¢Ù·ÓÈ©Ê÷Ö¬ÓÉ·ÓÈ©ºÍ¼×È©Ëõ¾Û¶ø³É£¬·´Ó¦ÓдóÁ¿Èȷųö£¬Îª·Àֹζȹý¸ß£¬Ó¦ÏòÓб½·ÓµÄ·´Ó¦¸ª_______µØ¼ÓÈë¼×È©£¬ÇÒ·´Ó¦¸ªÓ¦×°ÓÐ________×°Öá£

¢Ú²£Á§¸ÖÖв£Á§ÏËάµÄ×÷ÓÃÊÇ______¡£²£Á§¸Ö¾ßÓÐ µÈÓÅÒìÐÔÄÜ£¨Ð´³öÁ½µã¼´¿É£©¡£

¢ÛÏÂÁд¦Àí·Ï¾ÉÈȹÌÐÔ·ÓÈ©ËÜÁϵÄ×ö·¨ºÏÀíµÄÊÇ________¡£

a£®ÉîÂñ b£®·ÛËéºóÓÃ×÷Ê÷Ö¬ÌîÁÏ

c£®ÓÃ×÷ȼÁÏ d£®ÓÃÓлúÈܼÁ½«ÆäÈܽ⣬»ØÊÕÊ÷Ö¬

£¨3£©¹¤ÒµÉÏÖ÷Òª²ÉÓð±Ñõ»¯·¨Éú²úÏõËᣬÈçͼÊǰ±Ñõ»¯ÂÊÓë°±£­¿ÕÆø»ìºÏÆøÖÐÑõ°±±ÈµÄ¹ØÏµ¡£ÆäÖÐÖ±Ïß±íʾ·´Ó¦µÄÀíÂÛÖµ£»ÇúÏß±íʾÉú²úʵ¼ÊÇé¿ö¡£µ±°±Ñõ»¯ÂÊ´ïµ½100%£¬ÀíÂÛÉÏr[n(O2)/n(NH3)]£½_____£¬Êµ¼ÊÉú²úÒª½«rֵά³ÖÔÚ1.7¡«2.2Ö®¼ä£¬Ô­ÒòÊÇ__________________________________________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨16·Ö£©Ä³»¯Ñ§ÐËȤС×é½øÐÐÓйصç½âʳÑÎË®µÄ̽¾¿ÊµÑ飬µç½â×°ÖÃÈçÓÒͼËùʾ¡£

ʵÑéÒ»£ºµç½â±¥ºÍʳÑÎË®¡£

£¨1£©¼òÊöÅäÖÆ±¥ºÍʳÑÎË®µÄ²Ù×÷£º

£¨2£©µç½â±¥ºÍʳÑÎË®µÄÀë×Ó·½³ÌʽΪ

ʵÑé¶þ£ºµç½â²»±¥ºÍʳÑÎË®¼°²úÎï·ÖÎö¡£

ÏàͬÌõ¼þÏ£¬µç½â1 mol¡¤LÒ»1NaClÈÜÒº²¢ÊÕ¼¯Á½¼«²úÉúµÄÆøÌå¡£ÔÚX´¦ÊÕ¼¯µ½V1mLÆøÌ壬ͬʱ£¬ÔÚY´¦ÊÕ¼¯µ½V2mLÆøÌ壬ֹͣµç½â¡£½á¹û·¢ÏÖV2£¼V1£¬ÇÒÓëµç½â±¥ºÍʳÑÎË®Ïà±È£¬Y´¦ÊÕ¼¯µ½µÄÆøÌåÑÕÉ«Ã÷ÏÔ½Ïdz¡£¾­ÌÖÂÛ·ÖÎö£¬µ¼ÖÂÉÏÊöÏÖÏóµÄÔ­ÒòÓУº

i£®Óв¿·ÖC12ÈܽâÓÚNaClÈÜÒºÖУ»ii£®ÓÐ02Éú³É¡£

£¨3£©Éè¼ÆÊµÑéÖ¤Ã÷Óв¿·ÖC12ÈܽâÓÚNaClÈÜÒºÖС£ÊµÑé·½°¸Îª ¡£

£¨4£©Ö¤Ã÷ÓÐO2Éú³É²¢²â¶¨O2µÄÌå»ý¡£

°´ÈçͼËùʾװÖýøÐÐʵÑ顣ͨ¹ý×¢ÉäÆ÷»º»ºµØ½«ÔÚY´¦ÊÕ¼¯µ½µÄV2mLÆøÌåÈ«²¿ÍÆÈë×°ÖÃA(Ê¢ÓÐ×ãÁ¿ÊÔ¼Á£©ÖУ¬×îÖÕ£¬Á¿Æø¹ÜÖÐÊÕ¼¯µ½V3mLÆøÌ壨Éè¾ùÔÚÏàͬÌõ¼þϲâµÃ£©¡£

¢Ù×°ÖÃAµÄ×÷ÓÃÊÇ ¡£

¢Ú±¾ÊµÑéÖУ¬¹Û²ìµ½ µÄÏÖÏó£¬ËµÃ÷ʯīµç¼«ÉÏÓÐ02Éú³É¡£

¢ÛʵÑéÖÐÊÇ·ñÐèÒªÔ¤Ïȳý¾»×°ÖÃÖÐµÄ¿ÕÆø£¿ £¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©¡£

£¨5£©ÊµÑé¶þÖУ¬ÔÚʯīµç¼«ÉÏÉú³ÉCl2µÄ×ÜÌå»ýΪ mL£¨ÓôúÊýʽ±íʾ£©¡£

ʵÑ鷴˼£º

£¨6£©ÓÉÒÔÉÏʵÑéÍÆÖª£¬Óûͨ¹ýµç½âʳÑÎË®³ÖÐøµØ»ñµÃ½Ï´¿¾»µÄÂÈÆø£¬µç½âʱӦ¿ØÖƵÄÌõ¼þ£º

¢Ù £»¢Ú ¡£Òª½øÒ»²½Ö¤Ã÷¸ÃÍÆÂÛ£¬»¹Ðè½øÐеç½â²»Í¬Å¨¶ÈʳÑÎË®µÄƽÐÐʵÑé¡£

£¨15·Ö£©2Na2CO3¡¤3H2O2ÊÇÒ»ÖÖÐÂÐ͵ÄÑõϵƯ°×¼Á¡£Ä³ÊµÑéÐËȤС×é½øÐÐÁËÈçÏÂʵÑé¡£

¢ñ£®ÊµÑéÖÆ±¸

ʵÑéÔ­Àí£º2Na2CO3£«3H2O2£½2Na2CO3¡¤3H2O2

ʵÑé²½Ö裺ȡ3£®5 g Na2CO3ÈÜÓÚ10 mL H2O£¬¼ÓÈë0£®1 gÎȶ¨¼Á£¬ÓôÅÁ¦½Á°èÆ÷½Á°èÍêÈ«Èܽâºó£¬½«6£®0 mL 30£¥H2O2ÔÚ15 minÄÚ»ºÂý¼ÓÈëµ½Èý¾±ÉÕÆ¿ÖУ¬ÊµÑé×°ÖÃÈçͼ¡£

·´Ó¦1Сʱºó£¬¼ÓÈë1gÂÈ»¯Äƺ󣬾²Öýᾧ£¬È»ºó³éÂË£¬¸ÉÔïÒ»Öܺ󣬳ÆÖØ¡£

£¨1£©×°ÖÃÖÐÇòÐÎÀäÄý¹ÜµÄ×÷ÓÃÊÇ______________¡£

£¨2£©Ê¹ÓÃÀäˮԡµÄ×÷ÓÃÊÇ______________________________________________¡£

£¨3£©¼ÓÈëÊÊÁ¿NaCl¹ÌÌåµÄÔ­ÒòÊÇ_______________________________________¡£

£¨4£©2Na2CO3¡¤3H2O2¼«Ò׷ֽ⣬Æä·´Ó¦·½³Ìʽ¿É±íʾΪ_________________________________¡£

¢ò£®»îÐÔÑõº¬Á¿²â¶¨

׼ȷ³ÆÈ¡ÊÔÑù0£®2000 g£¬ÖÃÓÚ250 mL×¶ÐÎÆ¿ÖУ¬¼Ó100 mLŨ¶ÈΪ6£¥µÄÁòËáÈÜÒº£¬ÓÃ0£®0200 mol£¯L¸ßÃÌËá¼Ø±ê×¼ÈÜÒºµÎ¶¨£¬¼Ç¼¸ßÃÌËá¼Ø±ê×¼ÈÜÒºÏûºÄµÄÌå»ýΪ

32.70 mL¡£

»îÐÔÑõº¬Á¿¼ÆË㹫ʽ£ºQ£¥£½£¨40cV£¯m£©¡Á100£¥[cKMnO4±ê×¼ÈÜҺŨ¶È£¨mol£¯L£©£»VÏûºÄµÄKMnO4±ê×¼ÈÜÒºÌå»ý£¨L£©£»mÊÔÑùÖÊÁ¿£¨g£©]

£¨5£©µÎ¶¨ÖÕµãµÄÅжÏÒÀ¾ÝΪ______________________________________________¡£

£¨6£©µÎ¶¨¹ý³ÌÖÐÉæ¼°µÄ»¯Ñ§·½³Ìʽ_______________________________________¡£

£¨7£©»îÐÔÑõº¬Á¿Îª__________________________¡£

¢ó£®²úÆ·´¿¶ÈµÄ²â¶¨

£¨8£©ÎªÁ˲ⶨ²úÆ·ÖÐ2Na2CO3¡¤3H2O2µÄÖÊÁ¿·ÖÊý,Éè¼ÆÁ˼¸ÖÖ·½°¸,Éæ¼°²»Í¬µÄ·´Ó¦Ô­Àí¡£

·½°¸Ò» ½«ÊÔÑùÓëMnO2»ìºÏ¾ùÔÈ£¬Ïò»ìºÏÎïÖеμÓË®£¬²âÉú³ÉÆøÌåµÄÌå»ý£¬½ø¶ø½øÐмÆËã¡£

·½°¸¶þ _________________________________________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø