ÌâÄ¿ÄÚÈÝ
£¨1£©ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÃèÊö²»ÕýÈ·µÄÓÐ £¨ÌîÐòºÅ£©¡£
A£®ÓÃÍÐÅÌÌìƽ³ÆÁ¿17.55gÂÈ»¯Äƾ§Ìå |
B£®Ì¼ËáÄÆÈÜÒº±£´æÔÚ´ø²£Á§ÈûµÄÊÔ¼ÁÆ¿ |
C£®ÓøÉÔïµÄpHÊÔÖ½²â¶¨ÐÂÖÆÂÈË®µÄpH |
D£®Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¸©Êӿ̶ÈÏ߶¨ÈݺóŨ¶ÈÆ«´ó |
F£®³ýÈ¥CO2ÆøÌåÖлìÓеÄÉÙÁ¿HCl£¬¿ÉÒÔ½«ÆøÌåͨÈë±¥ºÍ̼ËáÇâÄÆÈÜÒº
£¨2£©ÏÂͼΪÖÐѧ»¯Ñ§ÊµÑéÖг£¼ûµÄʵÑé×°ÖÃ
A B C
ʵÑéÊÒ³£ÓÃ×°ÖÃAÖƱ¸Ï±íÖÐÆøÌ壬Ç뽫·ÖҺ©¶·ºÍÔ²µ×ÉÕÆ¿ÖÐӦװµÄ»¯Ñ§ÊÔ¼ÁÌîдÍêÕû¡£
ÆøÌå | O2 | Cl2 | NH3 |
·ÖҺ©¶·ÖÐÊÔ¼Á | | | Ũ°±Ë® |
Ô²µ×ÉÕÆ¿ÖÐÊÔ¼Á | | KMnO4 | |
¿ÉÓÃB×°ÖÃÅÅÒºÊÕ¼¯ÆøÌ壬ÆøÌåÓ¦´Ó¸Ã×°ÖÃ________(Ìî¡°×󡱡°ÓÒ¡±)¹Ü¿Úµ¼½ø£¬ÌÈÈôÀûÓøÃ×°ÖÃÊÕ¼¯Cl2£¬ÊÔ¼ÁÆ¿ÖÐÊ¢·ÅµÄÊÔ¼ÁΪ ¡£
C×°ÖÃÓÃÓÚ´¦Àí¶àÓàÆøÌå¶Ô»·¾³µÄÎÛȾ£¬ÈôÀûÓøÃ×°ÖÃÎüÊÕCl2£¬´ËʱÉÕ±Öз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ ¡£ÌÈÈô¸Ã×°ÖÃÖнö½öÊ¢·ÅÏ¡ÁòËᣬͨ³£²»ÊʺÏÎüÊÕ°±ÆøµÄÔÒòÊÇ £¬ÈôÏòÉÕ±ÖжîÍâÔÙ¼ÓÈëÒ»ÖÖҺ̬ÓлúÎïÔò¿É°²È«ÎüÊÕ°±Æø£¬ÕâÖÖÓлúÎïΪ ¡£
£¨1£©ABCE£¨4·Ö ÿ¶àÑ¡1¸ö»òÒÅ©1¸ö¾ù¿Û1·Ö£©
£¨2£©¢Ù£¨¹²4·Ö£©
¢ÚÓÒ ±¥ºÍʳÑÎË®ÆøÌå O2 Cl2 NH3 ·ÖҺ©¶·ÖÐÊÔ¼Á H2O2£¨»òH2O£© ŨÑÎËá Ũ°±Ë® Ô²µ×ÉÕÆ¿ÖÐÊÔ¼Á MnO2£¨»òNa2O2£© KMnO4 NaOH£¨»òCaO£©
¢ÛCl2£«2OH£=ClO££«Cl££«H2O °±Æø¼«Ò×ÈÜÓÚË®£¨Ï¡ÁòËᣩ£¬»áÒýÆðµ¹Îü
CCl4(»òËÄÂÈ»¯Ì¼)
½âÎöÊÔÌâ·ÖÎö£º£¨1£©A£®ÓÃÍÐÅÌÌìƽ³ÆÁ¿Ö»Äܾ«È·µ½0.1g£¬´íÎó£»B£®Ì¼ËáÄÆÈÜÒº³Ê¼îÐÔ»áºÍ²£Á§ÈûÖеĶþÑõ»¯¹è·´Ó¦Éú³ÉÓÐÕ³ÐÔµÄÎïÖʹèËáÄÆ£¬»áʹ²£Á§ÈûºÍ²£Á§Æ¿Õ³ÔÚÒ»Æ𣬴íÎó£»C£®ÓøÉÔïµÄpHÊÔÖ½²â¶¨ÐÂÖÆÂÈË®µÄpH£¬ÓÉÓÚÂÈË®ÓÐƯ°×ÐÔ»áʹÊÔÖ½·¢Éú±äÉ«ÎÞ·¨¹Û²ì£¬´íÎó£»D£®Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¸©Êӿ̶ÈÏ߶¨ÈݺóŨ¶ÈÆ«´ó£¬ÕýÈ·£»E£®ÔÚÖó·ÐµÄÕôÁóË®ÖÐÂýÂý¼ÓÈë±¥ºÍFeCl3ÈÜÒºÖ±µ½Éú³ÉºìºÖÉ«ÒºÌ壬ÕâÊÇÖÆÈ¡Fe(OH)3½ºÌå µÄ³£Ó÷½·¨£¬´íÎó£»F£®³ýÈ¥CO2ÆøÌåÖлìÓеÄÉÙÁ¿HCl£¬¿ÉÒÔ½«ÆøÌåͨÈë±¥ºÍ̼ËáÇâÄÆÈÜÒº£¬ÕýÈ·£»¹ÊѡΪABCE¡££¨2£©¢ÙÏÂͼ·Ö±ðÊÇÖƱ¸ÈýÖÖÆøÌåµÄÔÁÏ
¢ÚÒªÓÃB×°ÖÃÅÅÒºÊÕ¼¯ÆøÌ壬ÆøÌåÓ¦´Ó¸Ã×°ÖÃ_Óҹܿڵ¼½ø£¬ÕâÑù²ÅÄÜ°ÑÒºÌåÅųö¡£ÊÕ¼¯Cl2£¬ÊÔ¼ÁÆ¿ÖÐÊ¢·ÅµÄÊÔ¼ÁΪ±¥ºÍʳÑÎË®¿ÉÒÔÒ»·½Ãæ³ýÈ¥ÂÈ»¯Ç⣬ÁíÒ»·½ÃæÂÈÆøµÄÈܽâÐÔ¡£¢ÛÂÈÆøºÍÇâÑõ»¯ÄƵķ´Ó¦Àë×Ó·½³ÌʽΪCl2£«2OH£=ClO££«Cl££«H2O¡£ÌÈÈô¸Ã×°ÖÃÖнö½öÊ¢·ÅÏ¡ÁòËᣬͨ³£²»ÊʺÏÎüÊÕ°±ÆøµÄÔÒòÊÇ °±Æø¼«Ò×ÈÜÓÚË®£¨Ï¡ÁòËᣩ£¬»áÒýÆðµ¹Îü£¬ÈôÏòÉÕ±ÖжîÍâÔÙ¼ÓÈëÒ»ÖÖҺ̬ÓлúÎïÔò¿É°²È«ÎüÊÕ°±Æø£¬ÕâÖÖÓлúÎïΪCCl4(»òËÄÂÈ»¯Ì¼)£¬ÎïÖʵÄÃܶȱÈË®´ó£¬°±Æø²»»áÈܽâÔÚÆäÖС£ÆøÌå O2 Cl2 NH3 ·ÖҺ©¶·ÖÐÊÔ¼Á H2O2£¨»òH2O£© ŨÑÎËá Ũ°±Ë® Ô²µ×ÉÕÆ¿ÖÐÊÔ¼Á MnO2£¨»òNa2O2£© KMnO4 NaOH£¨»òCaO£©
¿¼µã£º¿¼²é³£¹æʵÏÖÏóµÄÃèÊöºÍʵÑéÊÒÖƱ¸ÆøÌåµÄ»ù±¾ÔÀí¡£
ÀûÓ÷´Ó¦I2(s)+Cl2(g)=2ICl(l)£¬ÊµÑéÊÒ¿ÉÓÃÈçÏÂͼËùʾװÖ㨼ÓÈÈ¡¢¼Ð³ÖÒÇÆ÷ÒÑÂÔÈ¥£©ÖÆÈ¡ÉÙÁ¿IC1¡£
ÒÑÖª£ºIClµÄÈÛµãΪ13.9¡æ£¬·ÐµãΪ97.4¡æ£¬Ò×Ë®½â£¬ÇÒÄÜ·¢Éú·´Ó¦£º
ICl(l)+Cl2(g)=2ICl3(l)
£¨1£©×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ____________¡£
£¨2£©×°ÖÃBµÄ×÷ÓÃÊÇ______¡£²»ÄÜÓÃ×°ÖÃF´úÌæ×°ÖÃE£¬ÀíÓÉÊÇ____________¡£
£¨3£©ËùÖƵõÄIClÖÐÈÜÓÐÉÙÁ¿ICl3ÔÓÖÊ£¬Ìá´¿µÄ·½·¨ÊÇ______ (Ìî±êºÅ£©¡£
A£®¹ýÂË | B£®Õô·¢½á¾§ | C£®ÕôÁó | D£®·ÖÒº |
i£®
ii£®ICl+KI=I2+KCl
iii£®I2+2Na2S2O3=2NaI+Na2S4O6
ʵÑé1£º½«0.500g¸ÃÓÍÖ¬ÑùÆ·ÈÜÓÚ10mLËÄÂÈ»¯Ì¼ºó£¬¼ÓÈË20mLijIClµÄ±ù´×ËáÈÜÒº£¨¹ýÁ¿£©£¬³ä·Ö·´Ó¦ºó£¬¼ÓÈË×ãÁ¿KIÈÜÒº£¬Éú³ÉµÄµâµ¥ÖÊÓÃa mol?L¡ª1µÄNa2S2O3£¬±ê×¼ÈÜÒºµÎ¶¨¡£¾Æ½ÐÐʵÑ飬²âµÃÏûºÄµÄNa2S2O3ÈÜÒºµÄƽ¾ùÌå»ýΪV1mL¡£
ʵÑé2£¨¿Õ°×ʵÑ飩£º²»¼ÓÓÍÖ¬ÑùÆ·£¬ÆäËü²Ù×÷²½Öè¡¢ËùÓÃÊÔ¼Á¼°ÓÃÁ¿ÓëʵÑé1ÍêÈ«Ïàͬ£¬²âµÃÏûºÄµÄNa2S2O3ÈÜÒºµÄƽ¾ùÌå»ýΪV2mL¡£
¢ÙµÎ¶¨¹ý³ÌÖпÉÓÃ______×÷ָʾ¼Á¡£
¢ÚµÎ¶¨¹ý³ÌÖÐÐèÒª²»¶ÏÕñµ´£¬·ñÔò»áµ¼ÖÂV1______£¨Ìî¡°Æ«´ó¡±»ò¡°Æ«Ð¡¡±£©¡£
¢Û0.500g¸ÃÓÍÖ¬ÑùÆ·ËùÏûºÄµÄIClµÄÎïÖʵÄÁ¿Îª______mol¡£ÓÉ´ËÊý¾Ý¾»»Ëã¼´¿ÉÇóµÃ¸ÃÓÍÖ¬µÄ²»±¥ºÍ¶È¡£
ÖкÍÈȵIJⶨÊǸßÖÐÖØÒªµÄ¶¨Á¿ÊµÑ顣ȡ0.55mol/LµÄNaOHÈÜÒº50mLÓë0.25mol/LµÄÁòËá50mLÖÃÓÚͼËùʾµÄ×°ÖÃÖнøÐÐÖкÍÈȵIJⶨʵÑ飬»Ø´ðÏÂÁÐÎÊÌ⣺
(1)´ÓÉÏͼʵÑé×°Öÿ´£¬ÆäÖÐÉÐȱÉÙµÄÒ»ÖÖ²£Á§ÓÃÆ·ÊÇ_________ _£¬³ý´ËÖ®Í⣬װÖÃÖеÄÒ»¸öÃ÷ÏÔ´íÎóÊÇ ¡£
(2)Ϊ±£Ö¤¸ÃʵÑé³É¹¦¸Ãͬѧ²ÉÈ¡ÁËÐí¶à´ëÊ©£¬ÈçͼµÄËéÖ½ÌõµÄ×÷ÓÃÔÚÓÚ________ ___¡£
(3)Èô¸ÄÓÃ60mL 0.25mol¡¤L-1 H2SO4ºÍ50mL 0.55mol¡¤L-1 NaOHÈÜÒº½øÐз´Ó¦ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿ £¨Ìî¡°ÏàµÈ¡±¡¢¡°²»ÏàµÈ¡±£©£¬ÈôʵÑé²Ù×÷¾ùÕýÈ·£¬ÔòËùÇóÖкÍÈÈ Ìî¡°ÏàµÈ¡±¡°²»ÏàµÈ¡±£©
(4)µ¹ÈëNaOHÈÜÒºµÄÕýÈ·²Ù×÷ÊÇ£º________¡£ (´ÓÏÂÁÐÑ¡³ö)¡£
A£®Ñز£Á§°ô»ºÂýµ¹Èë
B£®·ÖÈý´ÎÉÙÁ¿µ¹Èë
C£®Ò»´ÎѸËÙµ¹Èë
(5)ʹÁòËáÓëNaOHÈÜÒº»ìºÏ¾ùÔȵÄÕýÈ·²Ù×÷ÊÇ£º________¡£ (´ÓÏÂÁÐÑ¡³ö)¡£
A£®ÓÃζȼÆСÐĽÁ°è
B£®½Ò¿ªÓ²Ö½Æ¬Óò£Á§°ô½Á°è
C£®ÇáÇáµØÕñµ´ÉÕ±
D£®ÓÃÌ×ÔÚζȼÆÉϵĻ·Ðβ£Á§°ôÇáÇáµØ½Á¶¯
(6)ʵÑéÊý¾ÝÈçÏÂ±í£º
¢ÙÇëÌîдϱíÖеĿհףº
ÎÂ¶È ÊµÑé´ÎÊý | ÆðʼζÈt1¡æ | ÖÕֹζÈt2/¡æ | ζȲîƽ¾ùÖµ (t2£t1)/¡æ | ||
H2SO4 | NaOH | ƽ¾ùÖµ | |||
1 | 26.2 | 26.0 | 26.1 | 29.5 | |
2 | 27.0 | 27.4 | 27.2 | 32.3 | |
3 | 25.9 | 25.9 | 25.9 | 29.2 | |
4 | 26.4 | 26.2 | 26.3 | 29.8 |
¢Ú½üËÆÈÏΪ0.55 mol/L NaOHÈÜÒººÍ0.25 mol/LÁòËáÈÜÒºµÄÃܶȶ¼ÊÇ1 g/cm3£¬ÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc£½4.18 J/(g¡¤¡æ)¡£ÔòÖкÍÈȦ¤H£½______ ____ ( ȡСÊýµãºóһλ)¡£
¢ÛÉÏÊöʵÑéÊýÖµ½á¹ûÓë57.3 kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔÒò¿ÉÄÜÊÇ(Ìî×Öĸ)____ ____¡£
a£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î
b£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζÈ
c£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±ÖÐ
ijNa2CO3ÑùÆ·ÖлìÓÐÒ»¶¨Á¿µÄNa2SO4 (Éè¾ù²»º¬½á¾§Ë®£©£¬Ä³»¯Ñ§ÐËȤС×é²ÉÓÃÁ½ÖÖ·½°¸²â¶¨¸ÃÑùÆ·ÖÐNa2CO3µÄÖÊÁ¿·ÖÊý£¬ÊԻشðÏÂÁÐÎÊÌâ¡£
·½°¸Ò»£ºÀûÓÃÏÂÁз½°¸Íê³ÉNa2CO3ÖÊÁ¿·ÖÊýµÄœy¶¨
(1)²Ù×÷¢ÛºÍ¢ÜµÄÃû³Æ·Ö±ðΪ_______¡£
(2)ÉÏÊö²Ù×÷¢Ù¡«¢ÜÖУ¬Ê¹Óõ½²£Á§°ôµÄÓÐ______(Ìî²Ù×÷ÐòºÅ)¡£
(3)ÅжϲÙ×÷¢Ú·ñÍê³ÉµÄ·½·¨ÊÇ______
·½°¸¶þ£º²ÉÓÃÏÂͼʵÑé×°Ö㨼гÖÒÇÆ÷ÒÑÊ¡ÂÔ£©.Ñ¡ÓÃÏÂÁÐÊÔ¼Á: a.ŨÁòËáb.±¥ºÍNaHCO3ÈÜÒºC.6mol/LÑÎËáD.2mol/LÁòËá, e.¼îʯ»Òf. ÎÞË®CaCl2,œy¶¨ÑùÆ·ÖÐNa2CO3,µÄÖÊÁ¿·ÖÊý£º
(4)Ìîд±íÖпոñ£º
ÒÇÆ÷ | ÊÔ¼Á | ¼ÓÈë¸ÃÊÔ¼ÁµÄÄ¿µÄ |
A | | ¹ÄÈë¿ÕÆøʱϴȥCO2 |
B | | ʹÑùÆ·³ä·Ö·´Ó¦·Å³öÆøÌå |
C | a | |
D | e | ³ä·ÖÎüÊÕCO2 |
E | e | |