ʵÑ飨1£©50mL0.50mol/LÑÎËáÓë50mL0.55mol/LNaOHÈÜÒºÔÚÈçͼËùʾµÄ×°ÖÃÖнøÐÐÖкͷ´Ó¦¡£Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ¡£
¢Ù´óÉÕ±ÉÏÈç²»¸ÇÓ²Ö½°å£¬ÇóµÃµÄÖкÍÈÈÊýÖµ £¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°ÎÞÓ°Ï족£©¡£
¢ÚʵÑéÖÐÄÜ·ñÓû·ÐÎÍË¿½Á°è°ô´úÌæ»·Ðβ£Á§½Á°è°ô £¨Ìî¡°ÄÜ¡±¡¢¡°²»ÄÜ¡±£©¡£
¢ÛijͬѧÓÃ0.25mol/LµÄÏ¡ÁòËá´úÌæÑÎËá½øÐÐÉÏÊöʵÑ飬ʵÑéÊý¾ÝÈçϱí
i£©ÇëÌîдϱíÖеĿհףº
| ÎÂ¶È ÊµÑé´ÎÊý | ÆðʼζÈt1/¡æ | ÖÕֹΠ¶Èt2/¡æ | ƽ¾ùζȲî (t2£t1)/¡æ | ||
| H2SO4ÈÜÒº | NaOHÈÜÒº | 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 |
ii£©½üËÆÈÏΪ0.55 mol/L NaOHÈÜÒººÍ0.25mol/L H2SO4ÈÜÒºµÄÃܶȶ¼ÊÇ1 g/cm3£¬ÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝc£½4.18 J/(g¡¤¡æ)¡£ÔòÖкÍÈȦ¤H£½______(±£ÁôСÊýµãºóһλ)¡£
iii£©ÉÏÊöʵÑé½á¹ûµÄÊýÖµÓë57.3 kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔÒò¿ÉÄÜÊÇ(Ìî×Öĸ)____¡£
a£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î b£®ÔÚÁ¿È¡NaOHÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý
c£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÏ¡ÁòËáµÄСÉÕ±ÖÐ
d£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζÈ
£¨2£©ÈçͼËùʾ£¬ÉÕÆ¿A¡¢BÖÐ×°ÓÐÏàͬŨ¶ÈµÄNO2ºÍN2O4µÄ»ìºÏÆøÌ壬ÖмäÓÃֹˮ¼ÐK¼Ð½ô£¬ÉÕ±¼×ÖÐÊ¢·Å100ml¡¢6 mol /LµÄHClÈÜÒº£¬ÉÕ±ÒÒÖÐÊ¢·Å100mlË®£¬£¨ÑÎËáºÍË®µÄÆðʼζÈÏàͬ£©ÏÖÏòÉÕ±¼×µÄÈÜÒºÖмÓÈë25gNaOH¹ÌÌ壬ͬʱÏòÉÕ±ÒÒÖмÓÈë25gNH4NO3¹ÌÌå½Á°èʹ֮Èܽâ:
¢ÙAÉÕÆ¿ÖÐÆøÌåÑÕÉ« £¬»ìºÏÆøÌåµÄƽ¾ù·Ö×ÓÁ¿ £¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±¡¢¡°²»±ä¡±£©¡£
¢ÚBÉÕÆ¿ÖÐÆøÌåÑÕÉ« £¬ÀíÓÉÊÇ ¡£
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ʵÑéÊÒÓûÓõ¨·¯ÅäÖÆ250 mL0.2 mol¡¤L-1 µÄCuSO4ÈÜÒº¡£
(1)ÅäÖÆÈÜҺʱ£¬Ò»°ã¿ÉÒÔ·ÖΪÒÔϼ¸¸ö²½Ö裺(Íê³ÉÏÂÁпոñ)
A.¼ÆËã¡¡B. ³ÆÁ¿¡¡C.Èܽ⡡¡¡D. E. F. ¶¨ÈÝ¡¡G. Ò¡ÔÈ¡¢×°Æ¿¡¡
(2)±¾ÊµÑ鱨ÐëÓõ½µÄÒÇÆ÷ÓÐÌìÆ½¡¢Ò©³×¡¢²£Á§°ô¡¢ÉÕ±¡¢Á¿Í²»¹ÓÐ______
ºÍ £¨ÌîÒÇÆ÷Ãû³Æ£©
(3) ¾¼ÆËãÐèÒª³ÆÈ¡CuSO4¡¤5H2OµÄÖÊÁ¿Îª ¡£
(4) Èô³ÆÈ¡µ¨·¯Ê±íÀÂë±»·ÅÔÚ×óÅÌ(10gÒÔÏÂÓÃÓÎÂë)£¬ÔòÅäÖÆµÄCuSO4ÈÜÒºµÄŨ¶È (Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±¡¢¡°ÎÞÓ°Ï족)£»Èô¶¨ÈÝʱÑöÊӿ̶ÈÏߣ¬ÔòÅäÖÆµÄCuSO4ÈÜÒºµÄŨ¶È (Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±¡¢¡°ÎÞÓ°Ï족)¡£
(5)´Ó׼ȷÅäÖÆºÃµÄCuSO4ÈÜÒºÖÐÈ¡³ö50mL £¬ÔòÕâ50mL CuSO4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______________£¬Ëùº¬Cu2+µÄÖÊÁ¿Îª ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
Ñо¿ÈËÔ±ÑÐÖÆ³öÒ»ÖÖï®Ë®µç³Ø£¬¿É×÷ΪÓãÀ׺ÍDZͧµÄ´¢±¸µçÔ´¡£¸Ãµç³ØÒÔ½ðÊô﮺͸ְåΪµç¼«²ÄÁÏ£¬ÒÔLiOHΪµç½âÖÊ£¬Ê¹ÓÃʱ¼ÓÈëË®¼´¿É·Åµç¡£¹ØÓÚ¸Ãµç³ØµÄÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©
A£®Ë®¼ÈÊÇÑõ»¯¼ÁÓÖÊÇÈܼÁ
B£®·ÅµçʱÕý¼«ÉÏÓÐÇâÆøÉú³É
C£®·ÅµçʱOH£ÏòÕý¼«Òƶ¯ D£®×Ü·´Ó¦Îª£º2Li£«2H2O=== 2LiOH£«H2¡ü
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁÐ˵·¨»ò±íʾ·¨ÕýÈ·µÄÊÇ£¨ £©
A£®ÓÉ¡°C£¨Ê¯£©¡úC£¨½ð£©£»¡÷H= +1.9 kJ/mol ¡±¿ÉÖª½ð¸Õʯ±ÈʯīÎȶ¨
B£®ÔÚ101KPaʱ£¬1mol̼ȼÉÕËù·Å³öµÄÈÈÁ¿ÎªÌ¼µÄȼÉÕÈÈ
C£®ÔÚ101kPaʱ£¬2gH2ÍêȫȼÉÕÉú³ÉҺ̬ˮ£¬·Å³ö285.8kJÈÈÁ¿£¬ÇâÆøÈ¼ÉÕµÄÈÈ»¯Ñ§·½³Ìʽ±íʾΪ£º2H2(g)+O2(g) = 2H2O(l)£»¡÷H= -571.6kJ/mol
D£®HClºÍNaOH·´Ó¦µÄÖкÍÈÈ¡÷H= -57.3kJ/mol£¬ÔòH2SO4ºÍCa(OH)2·´Ó¦µÄÖкÍÈÈΪ
¡÷H= £(2¡Á57.3)kJ/mol
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
Ò»¶¨Ìõ¼þÏ£¬½øÐÐÏÂÁз´Ó¦£ºMgSO4(s) + CO(g)
MgO(s) + CO2(g) +SO2(g) ¡÷H>0¡£¸Ã·´Ó¦ÔÚºãÈݵÄÃܱÕÈÝÆ÷Öдﵽƽºâºó£¬Èô½ö¸Ä±äͼÖкá×ø±êxµÄÖµ£¬ÖØÐ´ﵽƽºâºó£¬×Ý×ø±êyËæx±ä»¯Ç÷ÊÆºÏÀíµÄÊÇ£¨ £©
| Ñ¡Ïî | x | y |
|
| A | ÎÂ¶È | ÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄÃÜ¶È | |
| B | COµÄÎïÖʵÄÁ¿ | CO2ÓëCOµÄÎïÖʵÄÁ¿Ö®±È | |
| C | SO2µÄŨ¶È | ƽºâ³£ÊýK | |
| D | MgSO4µÄÖÊÁ¿£¨ºöÂÔÌå»ý£© | COµÄת»¯ÂÊ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÓÃ0.1026mol¡¤L-1µÄÑÎËáµÎ¶¨25.00mLδ֪Ũ¶ÈµÄÇâÑõ»¯ÄÆÈÜÒº£¬µÎ¶¨´ïÖÕµãʱ£¬µÎ¶¨¹ÜÖеÄÒºÃæÈçÏÂͼËùʾ£¬ÕýÈ·µÄ¶ÁÊýΪ(¡¡¡¡)¡£
A. 22.30mL B. 22.35mL
C. 23.65mL D. 23.70 mL
![]()
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁйØÓÚ0.10 mol¡¤L£1 NaHCO3ÈÜÒºµÄ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)¡£
A£®ÈÜÖʵĵçÀë·½³ÌʽΪNaHCO3£½Na£«£« H£«£« CO32 £
B£®25 ¡æÊ±£¬¼ÓˮϡÊͺó£¬n(H£«)Óën(OH£)µÄ³Ë»ý±ä´ó
C£®Àë×ÓŨ¶È¹ØÏµ£ºc(Na£«)£«c(H£«)£½c(OH£)£«c(HCO3£ )£«c(CO32 £)
D£®Î¶ÈÉý¸ß£¬c(HCO3£ )Ôö´ó
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÒÑÖª¢Ù±ûÍé¢ÚÕý¶¡Íé¢ÛÒì¶¡Íé¢ÜÎìÍé¢Ý¼ºÍ飬ÉÏÊöÎïÖʵķе㰴Óɵ͵½¸ßµÄ˳ÐòÅÅÁеÄÊÇ(¡¡¡¡)¡£
A£®¢Ù¢Û¢Ú¢Ü¢Ý¡¡¡¡£Â£®¢Ý¢Ü¢Û¢Ú¢Ù¡¡£Ã£®¢Ù¢Ú¢Û¢Ü¢Ý¡¡£Ä£®¢Ý¢Ù¢Ú¢Ü¢Û
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡£©
£Á.CaCl2ÔÚÈÛÈÚ״̬Ï¿ɵ¼µç£¬ËüÊÇÇ¿µç½âÖÊ
£Â.ûÓе¥ÖʲμӵϝºÏ·´Ó¦Ò»¶¨ÊÇ·ÇÑõ»¯»¹Ô·´Ó¦
£Ã.ÔªËØ´¦ÓÚ×î¸ß¼Û̬ʱһ¶¨¾ßÓÐÇ¿Ñõ»¯ÐÔ
D£®ÔÚ2KClO3+4HCl(Ũ)=2KCl+2ClO2+Cl2+2H2OÖУ¬ÂÈÆø¼ÈÊÇÑõ»¯²úÎïÓÖÊÇ»¹Ô²úÎï
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com