ÌâÄ¿ÄÚÈÝ

ijʵÑéС×éÓÃ0.50mol/LNaOHÈÜÒººÍ0.50mol/LÁòËáÈÜÒº²â¶¨ÖкÍÈÈ£®²â¶¨ÖкÍÈȵÄʵÑé×°ÖÃÈçͼ1Ëùʾ£®

£¨1£©Ð´³öÏ¡ÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ£¨ÖкÍÈÈÊýֵΪ57.3kJ/mol£©£º______£®
£¨2£©È¡50mLNaOHÈÜÒººÍ30mLÁòËáÈÜÒº½øÐÐʵÑ飬Êý¾ÝÈçϱíËùʾ£º
¢ÙÇëÌîдϱíÖеĿհףº
ζÈÆðʼζÈt1/¡æÖÕֹζÈ
t1/¡æ
ζȲî
£¨t2-t1£©t1/¡æ
ʵÑé´ÎÊýH2SO4NaOHƽ¾ùÖµ
126.226.026.130.1______
227.027.427.233.3
325.925.925.929.8
426.426.226.330.4
¢Ú½üËÆÈÏΪ0.50mol/LNaOHÈÜÒººÍ0.50mol/LÁòËáÈÜÒºµÄÃܶȶ¼ÊÇ1g/cm3£¬ÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc=4.18J/£¨g?¡æ£©£®ÔòÉú³É1molH2OµÄ¡÷H=______kJ/mol£¨È¡Ð¡Êýµãºóһ룩£®
¢ÛÉÏÊöʵÑé½á¹ûµÄÊýÖµÓë57.3kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔ­Òò¿ÉÄÜÊÇ______£¨Ìî×Öĸ£©£®
¢ÙÊÒεÍÓÚ10¡æʱ½øÐÐʵÑ飬¢ÚÔÚÁ¿È¡NaOHÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý£¬¢Û·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±­ÖУ¬¢ÜʵÑéʱÓû·ÐÎÍ­Ë¿½Á°è°ô´úÌæ»·Ðβ£Á§½Á°è°ô£®
£¨3£©½«V1mL 0.4mol/LH2SO4ÈÜÒººÍV2mLδ֪Ũ¶ÈµÄNaOHÈÜÒº»ìºÏ¾ùÔȺó²âÁ¿²¢¼Ç¼ÈÜҺζȣ¬ÊµÑé½á¹ûÈçͼ2Ëùʾ£¨ÊµÑéÖÐʼÖÕ±£³ÖV1+V2=50mL£©£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ______
A×ö¸ÃʵÑéʱ»·¾³Î¶ÈΪ22¡æ
B¸ÃʵÑé±íÃ÷»¯Ñ§ÄÜ¿Éת»¯ÎªÈÈÄÜ
CNaOHÈÜÒºµÄŨ¶ÈԼΪ1.2mol/L
D¸ÃʵÑé±íÃ÷ÓÐË®Éú³ÉµÄ·´Ó¦¶¼ÊÇ·ÅÈÈ·´Ó¦£®
£¨1£©ÒÑ֪ϡǿËᡢϡǿ¼î·´Ó¦Éú³É1molҺ̬ˮʱ·Å³ö57.3kJµÄÈÈÁ¿£¬Ï¡ÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº¶¼ÊÇÇ¿ËáºÍÇ¿¼îµÄÏ¡ÈÜÒº£¬Ôò·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºNaOH£¨aq£©+HCl £¨aq£©¨TNaCl£¨aq£©+H2O£¨l£©¡÷H=-57.3 kJ/mol£¬
¹Ê´ð°¸Îª£ºNaOH£¨aq£©+HCl£¨aq£©¨TNaCl£¨aq£©+H2O£¨l£©¡÷H=-57.3 kJ/mol£»
£¨2£©¢ÙµÚ2×éÊý¾ÝÃ÷ÏÔÓÐÎó£¬ËùÒÔɾµô£¬Èý´ÎζȲîƽ¾ùÖµ=
(30.1-26.1)+(29.8-25.9)+(30.4-26.3)
3
=4.0C£¬¹Ê´ð°¸Îª£º4.0£»
¢Ú50mL0.50mol/LÇâÑõ»¯ÄÆÓë30mL0.50mol/LÁòËáÈÜÒº½øÐÐÖкͷ´Ó¦Éú³ÉË®µÄÎïÖʵÄÁ¿Îª0.05L¡Á0.50mol/L=0.025mol£¬ÈÜÒºµÄÖÊÁ¿Îª£º80ml¡Á1g/ml=80g£¬Î¶ȱ仯µÄֵΪ¡÷T=4¡æ£¬ÔòÉú³É0.025molË®·Å³öµÄÈÈÁ¿ÎªQ=m?c?¡÷T=80g¡Á4.18J/£¨g?¡æ£©¡Á4.0¡æ=1337.6J£¬¼´1.3376KJ£¬ËùÒÔʵÑé²âµÃµÄÖкÍÈÈ¡÷H=-
1.3376KJ
0.025mol
=-53.5 kJ/mol£¬
¹Ê´ð°¸Îª£º-53.5kJ/mol£»
¢Û¢ÙÊÒεÍÓÚ10¡æʱ½øÐÐʵÑ飬ɢʧÈÈÁ¿£¬ÊµÑé½á¹ûµÄÊýֵƫС£¬¹Ê¢ÙÕýÈ·£»
¢ÚÔÚÁ¿È¡NaOHÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý£¬ÈÜÒºµÄÖÊÁ¿Æ«´ó£¬ÊµÑé½á¹ûµÄÊýֵƫС£¬¹Ê¢ÚÕýÈ·£»
¢Û·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±­ÖУ¬É¢Ê§ÈÈÁ¿£¬ÊµÑé½á¹ûµÄÊýֵƫС£¬¹Ê¢ÛÕýÈ·£»
¢ÜʵÑéʱÓû·ÐÎÍ­Ë¿½Á°è°ô´úÌæ»·Ðβ£Á§½Á°è°ô£¬É¢Ê§ÈÈÁ¿£¬ÊµÑé½á¹ûµÄÊýֵƫС£¬¹Ê¢ÜÕýÈ·£»
¹ÊÑ¡£º¢Ù¢Ú¢Û¢Ü£»
£¨3£©A£®ÊµÑéʱµÄζÈӦΪËá¼îδ»ìºÏ֮ǰµÄζȣ¬Ôò²»ÊÇ»·¾³Î¶ȣ¬¹ÊA´íÎó£»
B£®·¢ÉúÖкͷ´Ó¦£¬ÈÜҺζÈÉý¸ß£¬±íÃ÷»¯Ñ§ÄÜ¿ÉÒÔת»¯ÎªÈÈÄÜ£¬¹ÊBÕýÈ·£»
C£®Ç¡ºÃ·´Ó¦Ê±²Î¼Ó·´Ó¦µÄÁòËáÈÜÒºµÄÌå»ýÊÇ30mL£¬ÓÉV1+V2=50mL¿ÉÖª£¬ÏûºÄµÄÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ20mL£¬
Ç¡ºÃ·´Ó¦Ê±ÇâÑõ»¯ÄÆÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿ÊÇn£®
H2SO4 +2NaOH=Na2SO4+2H 2O
1 2
0.4mol/L¡Á0.03L n
Ôòn=0.4mol/L¡Á0.03L¡Á2=0.024mol£¬ËùÒÔŨ¶ÈÊÇ
0.024mol
0.02L
=1.2mol/L£¬¹ÊCÕýÈ·£»
D£®Ö»ÊǸ÷´Ó¦·ÅÈÈ£¬ÆäËûÓÐË®Éú³ÉµÄ·´Ó¦²»Ò»¶¨£¬ÈçÂÈ»¯ï§ºÍÇâÑõ»¯±µ¾§ÌåµÄ·´Ó¦£¬ËùÒÔD´íÎó£®
¹ÊÑ¡BC£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÀûÓÃÈçͼËùʾװÖòⶨÖкÍÈȵÄʵÑé²½ÖèÈçÏ£º
¢ÙÓÃÁ¿Í²Á¿È¡50mL0.50mol?L-1ÑÎËáµ¹ÈëСÉÕ±­ÖУ¬²â³öÑÎËáζȣ»¢ÚÓÃÁíÒ»Á¿Í²Á¿È¡50mL0.55mol?L-1NaOHÈÜÒº£¬²¢ÓÃÁíһζȼƲâ³öÆäζȣ»¢Û½«NaOHÈÜÒºµ¹ÈëСÉÕ±­ÖУ¬É跨ʹ֮»ìºÏ¾ùÔÈ£¬²âµÃ»ìºÏÒº×î¸ßζȣ®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÎªÊ²Ã´ËùÓÃNaOHÈÜÒºÒªÉÔ¹ýÁ¿______£®
£¨2£©µ¹ÈëNaOHÈÜÒºµÄÕýÈ·²Ù×÷ÊÇ______£¨ÌîÐòºÅ£©£®
A£®Ñز£Á§°ô»ºÂýµ¹ÈëB£®·ÖÈý´ÎÉÙÁ¿µ¹ÈëC£®Ò»´ÎѸËÙµ¹Èë
£¨3£©Ê¹ÑÎËáÓëNaOHÈÜÒº»ìºÏ¾ùÔȵÄÕýÈ·²Ù×÷ÊÇ______£¨ÌîÐòºÅ£©£®
A£®ÓÃζȼÆСÐĽÁ°èB£®½Ò¿ªÓ²Ö½Æ¬Óò£Á§°ô½Á°è
C£®ÇáÇáµØÕñµ´ÉÕ±­D£®ÓÃÌ×ÔÚζȼÆÉϵĻ·Ðβ£Á§½Á°è°ôÇáÇáµØ½Á¶¯
£¨4£©ÏÖ½«Ò»¶¨Á¿µÄÏ¡ÇâÑõ»¯ÄÆÈÜÒº¡¢Ï¡ÇâÑõ»¯¸ÆÈÜÒº¡¢Ï¡°±Ë®·Ö±ðºÍ1L1mol?L-1µÄÏ¡ÑÎËáÇ¡ºÃÍêÈ«·´Ó¦£¬Æä·´Ó¦ÈÈ·Ö±ðΪ¡÷H1¡¢¡÷H2¡¢¡÷H3£¬Ôò¡÷H1¡¢¡÷H2¡¢¡÷H3µÄ´óС¹ØϵΪ______£®
£¨5£©¼ÙÉèÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒºµÄÃܶȶ¼ÊÇ1g?cm-3£¬ÓÖÖªÖкͷ´Ó¦ºóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc=4.18J?g-1?¡æ-1£®ÎªÁ˼ÆËãÖкÍÈÈ£¬Ä³Ñ§ÉúʵÑé¼Ç¼Êý¾ÝÈçÏ£º
ʵÑéÐòºÅÆðʼζÈt1/¡æÖÕֹζÈt2/¡æ
ÑÎËáÇâÑõ»¯ÄÆÈÜÒº»ìºÏÈÜÒº
120.020.123.2
220.220.423.4
320.520.623.6
ÒÀ¾Ý¸ÃѧÉúµÄʵÑéÊý¾Ý¼ÆË㣬¸ÃʵÑé²âµÃµÄÖкÍÈÈ¡÷H=______£¨½á¹û±£ÁôһλСÊý£©£®
£¨6£©______£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©ÓÃBa£¨OH£©2ÈÜÒººÍÁòËá´úÌæÇâÑõ»¯ÄÆÈÜÒººÍÑÎËᣬÀíÓÉÊÇ______£®
ÓÐЩ»¯Ñ§·´Ó¦½øÐÐʱ¹Û²ì²»µ½Ã÷ÏÔµÄÏÖÏó£®Ä³ÐËȤС×éΪ֤Ã÷NaOHÈÜÒºÓëÏ¡ÑÎËá·¢ÉúÁËÖкͷ´Ó¦£¬´Ó²»Í¬½Ç¶ÈÉè¼ÆÈçÏ·½°¸£®
·½°¸Ò»£ºÏÈÓÃpHÊÔÖ½²â¶¨NaOHÈÜÒºµÄpH£¬ÔٵμÓÑÎËá²¢²»¶ÏÕñµ´£¬Í¬Ê±²â¶¨»ìºÏÈÜÒºµÄpH£¬Èô²âµÃpHÖð½¥±äСÇÒСÓÚ7£¬Ö¤Ã÷NaOHÓëÏ¡ÑÎËá·¢ÉúÁËÖкͷ´Ó¦£®
£¨1£©ÓÃpHÊÔÖ½²â¶¨ÈÜÒºµÄpHʱ£¬ÕýÈ·µÄ²Ù×÷ÊÇ£º______£®
£¨2£©¼òÊöÇ¿µ÷¡°²âµÃµÄpHСÓÚ7¡±µÄÀíÓÉ£º______£®
·½°¸¶þ£ºÔÚNaOHÈÜÒºÖеμӼ¸µÎ·Ó̪£¬ÈÜÒºÏÔºìÉ«£¬ÔٵμÓÏ¡ÑÎËᣬºìÉ«Öð½¥Ïûʧ£¬Ö¤Ã÷NaOHÈÜÒºÓëÏ¡ÑÎËá·¢ÉúÁËÖкͷ´Ó¦£®µ«¸Ã×éͬѧÔÚÏòNaOHÈÜÒºÖеμӷÓ̪ʱ·¢ÏÖÇâÑõ»¯ÄÆÈÜÒºÖеÎÈë·Ó̪ºó£¬ÈÜÒº±ä³ÉÁ˺ìÉ«£¬¹ýÁËÒ»»á¶ùºìÉ«Ïûʧ£®¸ÃС×é¶ÔÕâÖÖÒâÍâÏÖÏóµÄÔ­Òò×÷ÈçϲÂÏ룺
¢Ù¿ÉÄÜÊÇ·Ó̪±»¿ÕÆøÖеÄÑõÆøÑõ»¯£¬Ê¹ºìÉ«Ïûʧ£»
¢Ú¿ÉÄÜÊÇÇâÑõ»¯ÄÆÈÜÒºÓë¿ÕÆøÖеĶþÑõ»¯Ì¼·´Ó¦£¬Ê¹ºìÉ«Ïûʧ£®
£¨1£©ÎªÑéÖ¤²ÂÏë¢Ù£¬½«ÅäÖƵÄÇâÑõ»¯ÄÆÈÜÒº¼ÓÈÈ£¬ÔÚÒºÃæÉϵÎһЩֲÎïÓÍ£¬ÀäÈ´ºóÏòÈÜÒºÖÐÔÙµÎÈë·Ó̪£®¡°¼ÓÈÈ¡±ºÍ¡°µÎÈëÖ²ÎïÓÍ¡±Ä¿µÄÊÇ______£®½á¹û±íÃ÷ºìÉ«ÏûʧÓë¿ÕÆøÖеÄÑõÆøÎ޹أ®
£¨2£©ÎªÑéÖ¤²ÂÏë¢Ú£¬È¡Ò»¶¨Á¿µÄNa2CO3ÈÜÒº£¬ÏòÆäÖеÎÈë·Ó̪£¬·¢ÏÖÈÜÒº³ÊÏÖºìÉ«£¬µÃ³öÒÔϽáÂÛ£º
½áÂÛ1£ºËµÃ÷Na2CO3ÈÜÒº³Ê______ÐÔ£»
½áÂÛ2£ºËµÃ÷ÈÜÒººìÉ«ÏûʧÓë¿ÕÆøÖеĶþÑõ»¯Ì¼Î޹أ®
£¨3£©Í¨¹ý²éÔÄ×ÊÁϵÃÖª£º·Ó̪ÔÚÇâÑõ»¯ÄÆÈÜҺŨ¶ÈºÜ´óʱ£¬¿ÉÄÜÖØÐÂÍÊÖÁÎÞÉ«£®ÇëÉè¼ÆʵÑé·½°¸Ö¤Ã÷ºìÉ«ÏûʧµÄÔ­ÒòÊÇÈ¡ÓõÄNaOHÈÜҺŨ¶È¹ý´ó£ºÊµÑé·½·¨£º______£¬¹Û²ìµ½µÄÏÖÏó£º______£®
Ëá¼îÖÐÊÇÖÐѧ»¯Ñ§ÖеÄÖØҪʵÑ飮¼¸Î»Í¬Ñ§¶ÔÇâÑõ»¯ÄÆÈÜÒººÍÏ¡ÑÎËá»ìºÏºóµÄÓйØÎÊÌ⣬½øÐÐÁËÈçÏÂ̽¾¿£º
£¨1£©¼×ͬѧÉè¼ÆÍê³ÉÁËÒ»¸öʵÑ飬֤Ã÷ÇâÑõ»¯ÄÆÈÜÒºÓëÑÎËáÄܹ»·¢Éú·´Ó¦£®ÔÚÊ¢ÓÐÏ¡ÑÎËáµÄÊÔ¹ÜÖУ¬ÓýºÍ·µÎ¹ÜÂýÂýµÎÈëÇâÑõ»¯ÄÆÈÜÒº£¬²»¶ÏÕñµ´ÈÜÒº£¬Í¬Ê±²â¶¨ÈÜÒºµÄpH£¬Ö±ÖÁÇâÑõ»¯ÄƹýÁ¿£®
¢ÙÔÚʵÑéÖÐÒªÇóÓÃÊÔÖ½²âÁ¿ÈÜÒºµÄpH£¬Èçͼ1ËùʾµÄ²Ù×÷ÖÐÕýÈ·µÄÊÇ______£®£¨Ìî´úºÅ£©

¢ÚÈçͼ2ÖÐÄĸöͼÏó·ûºÏ¸ÃͬѧµÄ¼Ç¼£®______£¨ÌîÐòºÅ£©
£¨2£©ÒÒͬѧÅäÖÆ100mL1.5mol?L-1ÇâÑõ»¯ÄÆÈÜÒº£¬ÎªÁËÖ¤Ã÷ÇâÑõ»¯ÄÆÈÜÒºÓëÏ¡ÑÎËáÄÜ·¢Éú»¯Ñ§·´Ó¦£¬Éè¼Æ²¢Íê³ÉÁËÈçͼ3ËùʾµÄʵÑ飮
¢ÙXÈÜÒºÊÇ______£¬Ê¹ÓÃËáʽµÎ¶¨¹ÜÇ°Òª¼ì²é»îÈûÊÇ·ñ©ˮµÄ²Ù×÷·½·¨ÊÇ______£®
¢ÚÒÒͬѧÓÃÑÎËáÓëÇâÑõ»¯ÄÆÈÜÒºÏ໥µÎ¶¨£¬»æÖƳöÈçͼ4ÇúÏßaºÍbµÄµÎ¶¨ÇúÏߣ®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ______£®
A£®ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.1mol?L-1
B£®Pµãʱ·´Ó¦Ç¡ºÃÍêÈ«£¬ÈÜÒº³ÊÖÐÐÔ
C£®ÇúÏßaÊÇÑÎËáµÎ¶¨ÇâÑõ»¯ÄÆÈÜÒºµÄµÎ¶¨ÇúÏß
D£®Á½ÖÖ·´Ó¦¹ý³Ì¶¼ÊÇ·ÅÈȹý³Ì£¬¶¼¿ÉÒÔÉè¼Æ³ÉÔ­µç³Ø
£¨3£©±ûͬѧΪ²â¶¨±êʾÖÊÁ¿·ÖÊýΪ32%µÄÑÎËáµÄʵ¼ÊÖÊÁ¿·ÖÊý£¬ÓÃpH²â¶¨ÒÇ×é³ÉʵÑé×°Öã®ÊµÑéʱÏÈÔÚÉÕ±­ÖмÓÈë20g40%µÄÇâÑõ»¯ÄÆÈÜÒº£¬ÔÙÖðµÎ¼ÓÈë¸ÃÑÎËᣬ²â¶¨ÒÇ´òÓ¡³ö¼ÓÈëÑÎËáµÄÖÊÁ¿ÓëÉÕ±­ÖÐÈÜÒºµÄpH¹Øϵͼ£¨Èçͼ5£©£®
¢Ù¸ù¾Ý´Ë´Î²â¶¨µÄ½á¹û£¬¸ÃÑÎËáµÄʵ¼ÊÖÊÁ¿·ÖÊýÊÇ______£®
¢ÚÇë·ÖÎöÄãµÄ¼ÆËã½á¹ûÓë±êÇ©±êʾµÄÖÊÁ¿·ÖÊý²»Ò»ÖµĿÉÄÜÔ­Òò£®______£®

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

¾«Ó¢¼Ò½ÌÍø