ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿²èÒ¶ÖеIJè¶à·ÓÊÇÒ»ÖÖÌìÈ»¿¹Ñõ»¯¼Á£¨Æ俹Ñõ»¯ÄÜÁ¦ÊÇ VC µÄ 5¡«10 ±¶£©£¬ËüÒ×ÈÜÓÚË®¡¢ÒÒ´¼¡¢ÒÒËáÒÒõ¥£¬ÄÑÈÜÓÚÂȷ¡£ÔÚËáÐÔ½éÖÊÖУ¬²è¶à·ÓÄܽ« Fe3+»¹Ô­Îª Fe2+£¬Fe2+Óë K3Fe(CN)6Éú³ÉµÄÉîÀ¶É«Åäλ»¯ºÏÎï KFe[Fe(CN)6]¶ÔÌض¨²¨³¤¹âµÄÎüÊճ̶ȣ¨ÓùâÃܶÈÖµ A ±íʾ£©Óë²è¶à·ÓÔÚÒ»¶¨Å¨¶È·¶Î§ÄÚ³ÉÕý±È¡£A Óë²è¶à·Ó±ê׼ҺŨ¶ÈµÄ¹ØϵÈçͼ 1 Ëùʾ£º

ijʵÑéС×éÉè¼ÆÈçÏÂʵÑéÁ÷³Ì´Ó²èÒ¶ÖÐÌáÈ¡²è¶à·Ó£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)²Ù×÷¢ÙÓÃË®½þÈ¡²è¶à·Óʱ£¬ÊµÑéС×é·¢ÏÖ½Á°èËٶȶԲè¶à·Ó½þ³öÁ¿µÄÓ°ÏìÈçͼ 2 Ëùʾ£¬Ô­ÒòÊÇ____¡£ÈôÓÃÒÒ´¼½þÈ¡²è¶à·Ó²Ù×÷ÈçÏ£º³ÆÈ¡ 10 g ²èҶĩ£¬ÓÃÂËÖ½°üºÃ£¬×°ÈëºãѹµÎҺ©¶·ÖУ¬Ô²µ×ÉÕÆ¿ÄÚ¼Ó·ÐʯºÍÊÊÁ¿ÒÒ´¼£¬Èçͼ 3 °²×°ºó£¬ ͨÀäÄýË®£¬¼ÓÈÈÌ×¼ÓÈÈ£¬µ±ÒÒ´¼±»¼ÓÈÈ·ÐÌں󣬿ØÖƼÓÈÈÌ×ζÈÔÚ 90¡æ¡£ÎªÊ¹ºãѹ©¶·ÄÚÒºÃæ¸ß³ö²èÒ¶°üÔ¼ 0.5 cm£¬²¢±£³ÖÔ¼ 1 h£¬¿ÉÐеIJÙ×÷·½·¨ÊÇ________¡£

(2)¼õѹÕô·¢Ïà¶ÔÓÚÒ»°ãÕô·¢µÄÓŵãÊÇ________£»ÂȷµÄ×÷ÓÃÊÇ________¡£

(3)ÏÂÁÐÓйØʵÑé»ù±¾²Ù×÷²»ÕýÈ·µÄÊÇ________¡£

A£®Í¼ 3 ÖÐÀäÄýË®Á÷ÏòΪ a ½ø b ³ö

B£®·ÖҺ©¶·Ê¹ÓÃÇ°Ðë¼ìÑéÊÇ·ñ©ˮ²¢Ï´¾»±¸ÓÃ

C£®²Ù×÷¢Ù¹ýÂËʱ£¬¿ÉÓò£Á§°ôÊʵ±½Á°èÒÔ¼Ó¿ì·ÖÀëËÙ¶È

D£®ÝÍÈ¡¹ý³ÌÖУ¬¾­ÕñÒ¡²¢·ÅÆøºó£¬½«·ÖҺ©¶·ÖÃÓÚÌúȦÉÏÁ¢¼´·ÖÒº

(4)³ÆÈ¡ 1.25 g ´Ö²úÆ·£¬ÓÃÕôÁóË®ÈܽⲢ¶¨ÈÝÖÁ 1000 mL£¬ÒÆÈ¡¸ÃÈÜÒº 1.00 mL£¬¼Ó¹ýÁ¿ Fe3+ºÍ K3Fe(CN)6 ËáÐÔÈÜÒº£¬ÓÃÕôÁóË®¶¨ÈÝÖÁ 100 mL ºó£¬²âµÃÈÜÒº¹âÃܶÈÖµ A=0.800£¬Ôò²úÆ·µÄ´¿¶ÈÊÇ_____ £¨ÒÔÖÊÁ¿·ÖÊý±íʾ£©¡£

¡¾´ð°¸¡¿²è¶à·ÓÒ×±»ÑõÆøÑõ»¯£¬½Á°èËÙ¶ÈÔ½¿ì£¬²è¶à·ÓÑõ»¯ËÙ¶ÈÔ½¿ì£¬½þ³öÂÊÔ½µÍ ¹Ø±Õ»îÈû£¬µ±Â©¶·ÄÚÒºÃæ¸ß³ö²èÒ¶°üÔ¼0.5cmʱ£¬µ÷½Ú»îÈûʹÒÒ´¼ÀäÄýËÙ¶ÈÓ멶··ÅÒºËÙ¶ÈÒ»Ö ½µµÍÕô·¢Î¶ȷÀÖ¹²úÎïÑõ»¯»ò·Ö½â ÝÍÈ¡£¨»ò³ýÔÓ£© CD 96%

¡¾½âÎö¡¿

²èÒ¶ÓÃË®½þÅÝʱ£¬²è¶à·Ó¡¢°±»ùËá¡¢²è¼îµÈÈܽâÔÚË®ÖУ¬¹ýÂ˺ó£¬È¥³ýÂËÔü£¬µÃµ½µÄÂËÒº¼õѹÕô·¢£¬¿É½µµÍÕô·¢Î¶ȣ¬·ÀÖ¹Ñõ»¯·´Ó¦µÄ·¢Éú£»ÍùŨËõÒºÖмÓÈëÂȷ¡¢·ÖÒº£¬¿ÉµÃº¬ÂȷµÄÓлúÈÜÒº¢Ù£¬Ë®ÈÜÒº¢ÚÖк¬Óвè¶à·Ó£¬ÓÃÒÒËáÒÒõ¥ÝÍÈ¡¡¢·ÖÒº£¬¿ÉµÃ²è¶à·ÓµÄÒÒËáÒÒõ¥ÈÜÒº£¬È»ºó¼õѹÕô·¢£¬¼´¿É»ñµÃ²è¶à·Ó´Ö²úÆ·¡£

(1)²Ù×÷¢ÙÓÃË®½þÈ¡²è¶à·Óʱ£¬´Óͼ2ÖпÉÒÔ¿´³ö£¬½Á°èËÙÂÊÔ½¿ì£¬¹âÃܶÈÖµ(A)ԽС£¬¼´²è¶à·ÓŨ¶ÈԽС£¬ÒòΪ²è¶à·ÓÒ×±»ÑõÆøÑõ»¯£¬½Á°èÔ½¿ì£¬Óë¿ÕÆøµÄ½Ó´¥Ô½¶à£¬Óɴ˵óöÔ­ÒòÊÇ£º²è¶à·ÓÒ×±»ÑõÆøÑõ»¯£¬½Á°èËÙ¶ÈÔ½¿ì£¬²è¶à·ÓÑõ»¯ËÙ¶ÈÔ½¿ì£¬½þ³öÂÊÔ½µÍ¡£ÎªÊ¹ºãѹ©¶·ÄÚÒºÃæ¸ß³ö²èÒ¶°üÔ¼ 0.5 cm£¬²¢±£³ÖÔ¼ 1 h£¬¿ÉÐеIJÙ×÷·½·¨ÊǹرջîÈû£¬µ±Â©¶·ÄÚÒºÃæ¸ß³ö²èÒ¶°üÔ¼0.5cmʱ£¬µ÷½Ú»îÈûʹÒÒ´¼ÀäÄýËÙ¶ÈÓ멶··ÅÒºËÙ¶ÈÒ»Ö¡£´ð°¸Îª£º²è¶à·ÓÒ×±»ÑõÆøÑõ»¯£¬½Á°èËÙ¶ÈÔ½¿ì£¬²è¶à·ÓÑõ»¯ËÙ¶ÈÔ½¿ì£¬½þ³öÂÊÔ½µÍ£»¹Ø±Õ»îÈû£¬µ±Â©¶·ÄÚÒºÃæ¸ß³ö²èÒ¶°üÔ¼0.5cmʱ£¬µ÷½Ú»îÈûʹÒÒ´¼ÀäÄýËÙ¶ÈÓ멶··ÅÒºËÙ¶ÈÒ»Ö£»

(2)¼õѹÕô·¢±ÈÕô·¢ËùÐèµÄζȵͣ¬³ý½ÚÊ¡ÄÜÔ´Í⣬»¹¿Éʹ»·¾³µÄζȽµµÍ£¬ÆäÓŵãÊǽµµÍÕô·¢Î¶ȷÀÖ¹²úÎïÑõ»¯»ò·Ö½â£»ÒòΪºó¼ÓÈëµÄÒÒËáÒÒõ¥ÝÍÈ¡²è¶à·Ó£¬ÔòÂÈ·ÂÝÍÈ¡µÄÓ¦ÊÇÔÓÖÊ£¬Óɴ˵óöÂȷµÄ×÷ÓÃÊÇÝÍÈ¡£¨»ò³ýÔÓ£©¡£´ð°¸Îª£º½µµÍÕô·¢Î¶ȷÀÖ¹²úÎïÑõ»¯»ò·Ö½â£»ÝÍÈ¡£¨»ò³ýÔÓ£©£»

(3)A£®Í¼ 3 ÖÐÀäÄýË®Á÷ÏòӦϽøÉϳö£¬¼´a½øb³ö£¬AÕýÈ·£»

B£®·ÖҺ©¶·Ê¹ÓÃÇ°£¬Îª·À©Һ£¬Ó¦²é©£¬È»ºóÏ´µÓ±¸Óã¬BÕýÈ·£»

C£®²Ù×÷¢Ù¹ýÂËʱ£¬²»¿ÉÓò£Á§°ô½Á°è£¬·ñÔò»áÔì³ÉÂËÖ½Ë𻵣¬C²»ÕýÈ·£»

D£®ÝÍÈ¡¹ý³ÌÖУ¬·ÖҺ©¶·Ó¦¾²Ö÷ֲãºó£¬ÔÙ½øÐзÖÒº£¬D²»ÕýÈ·£»

¹ÊÑ¡CD¡£´ð°¸Îª£ºCD£»

(4) ÈÜÒº¹âÃܶÈÖµ A=0.800£¬´Ó±íÖпɲéµÃ²è¶à·ÓµÄŨ¶ÈΪ1.2¡Á10-5g/mL£¬Óɴ˿ɵóöÔ­1000 mLÈÜÒºÖÐËùº¬²è¶à·ÓµÄÖÊÁ¿Îª1.2¡Á10-5g/mL¡Á100mL¡Á=1.2g£¬Ôò²úÆ·µÄ´¿¶ÈÊÇ=96%¡£´ð°¸Îª£º96%¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿»¯¹¤Ô­ÁÏË®ºÏ루N2H4¡¤H2O£©ÊÇÒ»ÖÖÇ¿»¹Ô­ÐԵļîÐÔÒºÌå¡£

¢ñ£®ÊµÑéÊÒÓÃÏÂͼװÖÃÖƱ¸Ë®ºÏ루N2H4¡¤H2O£©¡£

ʵÑé²½Ö裺¹Ø±ÕK2¡¢K3£¬´ò¿ªK1£¬ÖƱ¸ NaClO£»¹Ø±ÕK1¡¢K2£¬´ò¿ªK3£¬Í¨ÈëN2Ò»¶Îʱ¼ä£»¹Ø±ÕK3£¬´ò¿ªK2£¬µãȼ¾Æ¾«µÆ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ê¢·Å¼îʯ»ÒµÄÒÇÆ÷µÄÃû³ÆΪ___________________¡£

£¨2£©ÅäÖÆ30%NaOHÈÜҺʱ£¬ËùÐè²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜºÍ_________¡£

£¨3£©²½Öè2ÖÐͨÈëN2Ò»¶Îʱ¼äµÄÔ­ÒòÊÇ____________________________________¡£

£¨4£©Ë®²ÛÖзÅÈëÀäË®µÄÄ¿µÄÊÇ_____________________________________________¡£

£¨5£©NH3Óë NaClO·´Ó¦Éú³ÉN2H4¡¤H2OµÄ»¯Ñ§·½³ÌʽΪ_______________________¡£

¢ò£®ÒÑÖª£ºN2H4¡¤H2O+2I2=N2¡ü+4HI+H2O£¬²â¶¨Ë®ºÏë´ֲúÆ·µÄ´¿¶È²½ÖèÈçÏ£º

a£®³ÆÈ¡N2H4¡¤H2O´Ö²úÆ·(ÆäËüÔÓÖʲ»ÓëI2·´Ó¦)2.000g¡£

b£®¼ÓË®Åä³É250.00mLÈÜÒº¡£

c£®ÒƳö25.00mLÖÃÓÚ׶ÐÎÆ¿ÖУ¬µÎ¼Ó¼¸µÎµí·ÛÈÜÒº¡£

d£®ÓÃ0.3000mol¡¤L£­1µÄµâ±ê×¼ÈÜÒº½øÐеζ¨¡£

e£®Öظ´ÉÏÊö²Ù×÷Á½´Î¡£Èý´Î²â¶¨Êý¾ÝÈçÏÂ±í£º

ʵÑéÐòºÅ

1

2

3

ÏûºÄµâ±ê×¼ÈÜÌå»ý/mL

20.24

20.02

19.98

f£®Êý¾Ý´¦Àí¡£

£¨6£©µâ±ê×¼ÈÜҺʢ·ÅÔÚ____________£¨Ìî¡°Ëáʽ¡±»ò¡°¼îʽ¡±£©µÎ¶¨¹ÜÖС£Ôڵζ¨¹ÜÖÐ×°Èëµâ±ê×¼ÈܵÄÇ°Ò»²½£¬Ó¦½øÐеIJÙ×÷Ϊ_________£¬´ïµ½ÖÕµãµÄÏÖÏóÊÇ__________¡£

£¨7£©ÏûºÄµÄµâ±ê×¼ÈÜƽ¾ùÌå»ýΪ______mL£¬´Ö²úÆ·ÖÐË®ºÏëµÄÖÊÁ¿·ÖÊýΪ______¡£

£¨8£©ÅжÏÏÂÁвÙ×÷¶Ô²â¶¨½á¹ûµÄÓ°Ï죨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©¡£

¢ÙÈôÔÚÅäÖƵâ±ê×¼ÈÜҺʱ£¬ÉÕ±­ÖеÄÈÜÒºÓÐÉÙÁ¿½¦³ö£¬Ôò²â¶¨½á¹û___________¡£

¢ÚÈôÔڵζ¨ÖÕµã¶ÁÈ¡µÎ¶¨¹Ü¿Ì¶Èʱ£¬¸©ÊÓ±ê×¼ÒºÒºÃ棬Ôò²â¶¨½á¹û___________¡£

¡¾ÌâÄ¿¡¿ÇâÄÜÔ´ÊÇ×î¾ßÓ¦ÓÃÇ°¾°µÄÄÜÔ´Ö®Ò»£¬¸ß´¿ÇâµÄÖƱ¸ÊÇÄ¿Ç°µÄÑо¿Èȵ㡣¼×ÍéË®ÕôÆø´ß»¯ÖØÕûÊÇÖƸߴ¿ÇâµÄ·½·¨Ö®Ò»¡£

£¨1£©·´Ó¦Æ÷Öгõʼ·´Ó¦µÄÉú³ÉÎïΪH2ºÍCO2£¬ÆäÎïÖʵÄÁ¿Ö®±ÈΪ4¡Ã1£¬¼×ÍéºÍË®ÕôÆø·´Ó¦µÄ·½³ÌʽÊÇ___¡£

£¨2£©ÒÑÖª·´Ó¦Æ÷Öл¹´æÔÚÈçÏ·´Ó¦£º

i.CH4(g)+H2O(g)=CO(g)+3H2(g) ¦¤H1

ii.CO(g)+H2O(g)=CO2(g)+H2(g) ¦¤H2

iii.CH4(g)=C(s)+2H2(g) ¦¤H3

¡­¡­

·´Ó¦iiiΪ»ýÌ¿·´Ó¦£¬ÀûÓæ¤H1ºÍ¦¤H2¼ÆË㦤H3ʱ£¬»¹ÐèÒªÀûÓÃ__£¨Ð´»¯Ñ§·½³Ìʽ£©·´Ó¦µÄ¦¤H¡£

£¨3£©·´Ó¦ÎïͶÁϱȲÉÓÃn£¨H2O£©¡Ãn£¨CH4£©=4¡Ã1£¬´óÓÚ·´Ó¦µÄ¼ÆÁ¿ÊýÖ®±È£¬Ä¿µÄÊÇ__£¨Ìî×Öĸ£©¡£

a.´Ù½øCH4ת»¯ b.´Ù½øCOת»¯ÎªCO2 c.¼õÉÙ»ýÌ¿Éú³É

£¨4£©ÓÃCaO¿ÉÒÔÈ¥³ýCO2¡£H2Ìå»ý·ÖÊýºÍCaOÏûºÄÂÊËæʱ¼ä±ä»¯¹ØϵÈçͼËùʾ¡£´Ót1ʱ¿ªÊ¼£¬H2Ìå»ý·ÖÊýÏÔÖø½µµÍ£¬µ¥Î»Ê±¼äCaOÏûºÄÂÊ__£¨Ìî¡°Éý¸ß¡±¡°½µµÍ¡±»ò¡°²»±ä¡±£©¡£´ËʱCaOÏûºÄÂÊԼΪ35%£¬µ«ÒÑʧЧ£¬ÒòΪ´ËʱCaOÖ÷Òª·¢ÉúÁË__£¨Ð´»¯Ñ§·½³Ìʽ£©·´Ó¦¶øʹ£¨1£©Öз´Ó¦Æ½ºâÏò__Òƶ¯¡£

£¨5£©ÒÔ¼×´¼ÎªÈ¼ÁÏ£¬ÑõÆøΪÑõ»¯¼Á£¬KOHÈÜҺΪµç½âÖÊÈÜÒº£¬¿ÉÖƳÉȼÁϵç³Ø¡£ÒԴ˵ç³Ø×÷µçÔ´£¬ÔÚʵÑéÊÒÖÐÄ£ÄâÂÁÖÆÆ·±íÃæ¡°¶Û»¯¡±´¦Àí¹ý³Ì£¨×°ÖÃÈçͼËùʾ£©¡£ÆäÖÐÎïÖÊaÊÇ__£¬µçÔ´¸º¼«µç¼«·´Ó¦Îª___¡£¡°¶Û»¯¡±×°ÖÃÖÐÑô¼«µç¼«·´Ó¦Îª___¡£

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

¾«Ó¢¼Ò½ÌÍø