ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿µª¼°Æ仯ºÏÎïµÄÐÔÖÊÊÇÁ˽⹤ũҵÉú²úµÄÖØÒª»ù´¡¡£ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐÓйØ˵·¨´íÎóµÄÊÇ£¨ £©

A.±ê×¼×´¿öÏ£¬5.6LNOÓë5.6LO2³ä·Ö»ìºÏºóµÄ·Ö×ÓÊýΪ0.5NA

B.±ê×¼×´¿öÏ£¬22.4L15NH3º¬ÓеÄÖÊ×ÓÊýΪ10NA

C.13.8gNO2Óë×ãÁ¿Ë®·´Ó¦£¬×ªÒƵĵç×ÓÊýΪ0.2NA

D.³£ÎÂÏ£¬1L0.1mol¡¤L1NH4NO3ÈÜÒºÖк¬ÓеĵªÔ­×ÓÊýΪ0.2NA

¡¾´ð°¸¡¿A

¡¾½âÎö¡¿

A. ±ê×¼×´¿öÏ£¬5.6LNOµÄÎïÖʵÄÁ¿Îª0.25mol£¬5.6LO2µÄÎïÖʵÄÁ¿Îª0.25mol£¬Á½Õß³ä·Ö»ìºÏºó·¢Éú2NO £«O2 £½2NO2£¬¸ù¾Ý·½³Ìʽ¹Øϵ£¬ÏûºÄO2µÄÎïÖʵÄÁ¿Îª0.125mol£¬NO·´Ó¦Í꣬Éú³É0.25mol NO2£¬×ܵÄÎïÖʵÄÁ¿Îª0.25 mol£«0.125mol£½0.375mol£¬ÓÉÓÚ¶þÑõ»¯µª»¹»á×Ô·¢·¢Éú¿ÉÄæ·´Ó¦Éú³ÉËÄÑõ»¯¶þµª£¬Òò´ËÎïÖʵÄÁ¿Ð¡ÓÚ0.375mol£¬·Ö×ÓÊýСÓÚ0.375NA£¬¹ÊA´íÎó£»

B. 1¸ö15NH3º¬ÓÐ10¸öÖÊ×Ó£¬±ê×¼×´¿öÏ£¬22.4L15NH3µÄÎïÖʵÄÁ¿Îª1mol£¬Òò´Ë22.4L15NH3 º¬ÓеÄÖÊ×ÓÊýΪ10NA£¬¹ÊBÕýÈ·£»

C. NO2Óë×ãÁ¿Ë®·´Ó¦3 NO2£«H2O £½2HNO3£«NO£¬3mol NO2תÒÆ2molµç×Ó£¬13.8gNO2µÄÎïÖʵÄÁ¿£¬Òò´Ë13.8gNO2Óë×ãÁ¿Ë®·´Ó¦×ªÒƵĵç×ÓÊýΪ0.2NA£¬¹ÊCÕýÈ·£»

D. ³£ÎÂÏ£¬1L0.1mol¡¤L1NH4NO3ÈÜÒºÎïÖʵÄÁ¿Îª£¬Òò´Ëº¬ÓеĵªÔ­×ÓÊýΪ0.2 NA£¬¹ÊDÕýÈ·¡£

×ÛÉÏËùÊö£¬´ð°¸ÎªA¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿½üÄê¿Æѧ¼ÒÌá³ö¡°ÂÌÉ«×ÔÓÉ¡±¹¹Ïë¡£°Ñ¿ÕÆøÖеÄCO2½øÐÐת»¯£¬²¢Ê¹Ö®ÓëH2·´Ó¦Éú³É¿ÉÔÙÉúÄÜÔ´¼×´¼¡£

£¨1£©¹¤ÒµÉú²úÖпÉÀûÓÃH2»¹Ô­CO2ÖƱ¸Çå½àÄÜÔ´¼×´¼¡£

¢ÙÏÂÁйØÓÚÇâÄܵÄ˵·¨ÕýÈ·µÄÊÇ____¡£

A£®ÇâÄÜÊÇÇå½àÄÜÔ´ B£®ÇâÄÜÊǶþ´ÎÄÜÔ´

C£®ÇâÄÜÊDz»¿ÉÔÙÉúÄÜÔ´ D£®Æø̬ÇâÄÜÔ´¸üÈÝÒ×±£´æºÍÔËÊä

¢ÚÒÑÖªCO£¨g£©ºÍH2£¨g£©µÄȼÉÕÈÈ£¨¡÷H£©·Ö±ðΪ-283.0kJmol-1¡¢-285.8kJmol-1¡£COÓëH2ºÏ³É¼×´¼µÄÄÜÁ¿±ä»¯Èçͼ¼×Ëùʾ£º

ÔòÓÃCO2£¨g£©ºÍH2£¨g£©ÖƱ¸ÒºÌ¬¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ____¡£

¢Û½«Ò»¶¨Á¿µÄCO2ºÍH2³äÈ뵽ijºãÈÝÃܱÕÈÝÆ÷ÖУ¬²âµÃÔÚ²»Í¬´ß»¯¼Á×÷ÓÃÏ£¬Ïàͬʱ¼äÄÚCO2µÄת»¯ÂÊÓëζȵı仯ÈçͼÒÒËùʾ£º´ß»¯Ð§¹û×îºÃµÄÊÇ´ß»¯¼Á____£¨Ñ¡Ìî¡°¢ñ¡±¡°¢ò¡±»ò¡°¢ó¡±£©¡£

£¨2£©ÀûÓÃCOºÍË®ÕôÆø¿ÉÉú²úH2£¬·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºCO£¨g£©+H2O£¨g£©CO2£¨g£©+H2£¨g£©¡£½«²»Í¬Á¿µÄCO£¨g£©ºÍH2O£¨g£©·Ö±ðͨÈëµ½Ìå»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖнøÐÐÈçÏ·´Ó¦£¬µÃµ½Èý×éÊý¾ÝÈç±íËùʾ£º

ζÈ/¡æ

ÆðʼÁ¿

´ïµ½Æ½ºâ

CO/mol

H2O/mol

H2/mol

COת»¯ÂÊ

ʱ¼ä/min

650

4

2

1.6

10

900

3

2

5

900¡æʱ£¬´ïµ½Æ½ºâʱµÄ·´Ó¦ËÙÂÊv£¨H2O£©=____¡££¨±£Áô2λСÊý£©¡£

£¨3£©¿Æѧ¼Ò»¹Ñо¿ÁËÆäËûת»¯ÎÂÊÒÆøÌåµÄ·½·¨£¬ÀûÓÃÈçͼËùʾװÖÿÉÒÔ½«CO2ת»¯ÎªÆøÌåȼÁÏCO¡£¸Ã×°Öù¤×÷ʱ£¬Nµç¼«µÄµç¼«·´Ó¦Ê½Îª____¡£

¡¾ÌâÄ¿¡¿ÁòËáÑÇÌú茶§Ìå[(NH4)2Fe(SO4)2¡¤6H2O]Ò×ÈÜÓÚË®£¬ÔÚ¶¨Á¿·ÖÎöÖг£ÓÃÀ´ÅäÖÆÑÇÌúÀë×ӵıê×¼ÈÜÒº¡£Ä³»¯Ñ§¿ÎÍâС×éͬѧÉè¼Æ²¢Íê³ÉÏÂÁÐʵÑé¡£

I.(NH4)2Fe(SO4)2¡¤6H2OµÄÖÆÈ¡

ʵÑé²½Ö裺

¢Ù³ÆÈ¡4.2gËéÌúмÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë10mL30%µÄNaOHÈÜÒº£¬¼ÓÈÈÖó·Ð£­¶Îʱ¼ä¡£ÇãÈ¥¼îÒº£¬Ë®Ï´ÖÁÖÐÐÔ¡£

¢ÚÏò´¦Àí¹ýµÄÌúмÖмÓÈë25mL3mol/LH2SO4£¬Ë®Ô¡¼ÓÈÈÖÁ·´Ó¦ÍêÈ«£¬³ÃÈȹýÂË¡£

¢ÛÏòÂËÒºÖмÓÈ룭¶¨Ìå»ýµÄ±¥ºÍ(NH4)2SO4ÈÜÒº£¬¼ÓÈÈŨËõ£¬ÀäÈ´£¬³éÂË£¬Ï´µÓ£¬¸ÉÔïµÃ²úÆ·¡£

£¨1£©²½Öè¢ÙÖмÓÈëNaOHÈÜÒºµÄÄ¿µÄÊÇ___¡£

£¨2£©²½Öè¢Ú³ÃÈȹýÂ˵ÄÄ¿µÄÊÇ___¡£

£¨3£©²½Öè¢Û¼ÓÈÈŨËõÈÜҺʱ£¬ÐèÒªµÄÒÇÆ÷Óоƾ«µÆ¡¢___¡¢___£¨¹Ì¶¨¡¢¼Ð³ÖÒÇÆ÷³ýÍ⣩¡£

¢ò.(NH4)2Fe(SO4)2¡¤6H2OºÍFeSO4¡¤7H2OÎȶ¨ÐԵĶԱÈÑо¿

Èçͼ£¬ÔÚ2Ö§ÏàͬµÄ×¢ÉäÆ÷Öзֱð×°ÈëµÈÎïÖʵÄÁ¿µÄÁ½ÖÖ¾§Ì壬µ÷Õû2֧עÉäÆ÷»îÈûʹ¿ÕÆøÌå»ýÏàµÈ£¬ÓÃÕëÍ·¼°½ºÈû·âºÃ×¢ÉäÆ÷£¨ÆøÃÜÐÔÁ¼ºÃ£©¡£½Ï³¤Ê±¼äºó£¬È¡³öÁ½ÖÖ¾§Ì壬·Ö±ðµÎ¼Ó2µÎ0.01mol/LKSCNÈÜÒº¡£Õû¸ö¹ý³ÌÖÐʵÑéÏÖÏó¼Ç¼ÈçÏ£º

£¨4£©ÅжÏa___b£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©£¬³öÏִ˽á¹ûµÄÔ­Òò¿ÉÄÜÊÇ___¡£

£¨5£©ÅжϴËÌõ¼þÏÂ(NH4)2Fe(SO4)2¡¤6H2OµÄÎȶ¨ÐÔ___£¨Ìî¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©FeSO4¡¤7H2O¡£

¢ó.ÅäÖÆ(NH4)2Fe(SO4)2ÈÜÒº²¢±ê¶¨Å¨¶È

ÅäÖÆ(NH4)2Fe(SO4)2ÈÜÒº100mL£¬×¼È·Á¿È¡20.00mL£¬ÖÃÓÚ׶ÐÎÆ¿ÖУ¬¼Ó10mLH2SO4ºÍÁÚ¶þµª·Æָʾ¼Á3¡«4µÎ£¬ÓÃcmol/LK2Cr2O7±ê×¼ÒºµÎ¶¨ÖÁÈÜÒº±ä³É×غìÉ«¡£ÏûºÄK2Cr2O7ÈÜÒºµÄÌå»ýΪVmL¡£

£¨6£©¼ÆËã(NH4)2Fe(SO4)2ÈÜÒºµÄŨ¶ÈΪ___ mol/L¡£

£¨7£©ÏÂÁвÙ×÷»áʹËù²âÈÜҺŨ¶ÈÆ«´óµÄÊÇ___£¨Ìî±êºÅ£©¡£

A.׶ÐÎÆ¿Óôý²âÈÜÒºÈóÏ´

B.µÎ¶¨Ç°µÎ¶¨¹Ü¼â×ì´¦ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ

C.µÎ¶¨¹ý³ÌÖоçÁÒÒ¡»Î׶ÐÎÆ¿£¬ÈÜÒº½¦³ö

D.µÎ¶¨Ç°Æ½ÊÓ¶ÁÊý£¬µÎ¶¨ºó¸©ÊÓ¶ÁÊý

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

¾«Ó¢¼Ò½ÌÍø