ÒÀ¾ÝÊÂʵ£¬Ð´³öÏÂÁз´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£®
£¨1£©1molN2£¨g£©ÓëÊÊÁ¿O2£¨g£©Æð·´Ó¦£¬Éú³ÉNO2£¨g£©£¬ÎüÊÕ68kJÈÈÁ¿______
£¨2£©1molCu£¨s£©ÓëÊÊÁ¿O2£¨g£©Æð·´Ó¦£¬Éú³ÉCuO£¨s£©£¬·Å³ö157kJÈÈÁ¿______
£¨3£©ÎÀÐÇ·¢Éäʱ¿ÉÓÃ루N2H4£©×÷ȼÁÏ£¬1molN2H4£¨l£©ÔÚO2£¨g£©ÖÐȼÉÕ£¬Éú³ÉN2£¨g£©ºÍH2O£¨l£©£¬·Å³ö622kJÈÈÁ¿£®
______
£¨4£©½«0.3molµÄÆø̬¸ßÄÜȼÁÏÒÒÅðÍ飨B2H6£©ÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5kJÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ______£®ÓÖÒÑÖª£ºH2O£¨g£©=H2O£¨l£©£»¡÷H2=-44.0kJ/mol£¬Ôò11.2L£¨±ê×¼×´¿ö£©ÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ______kJ£®
£¨1£©1molN2£¨g£©ÓëÊÊÁ¿O2£¨g£©Æð·´Ó¦£¬Éú³ÉNO2£¨g£©£¬ÎüÊÕ68kJÈÈÁ¿£¬·´Ó¦ÎüÈÈʱìʱäֵΪÕýÖµ£¬ËùÒԸ÷´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪN2£¨g£©+2O2£¨g£©=2NO2£¨g£©¡÷H=+68kJ?mol-1£¬
¹Ê´ð°¸Îª£ºN2£¨g£©+2O2£¨g£©=2NO2£¨g£©¡÷H=+68kJ?mol-1£»
£¨2£©·ÅÈÈʱìʱäֵΪ¸ºÖµ£¬1molCu£¨s£©ÓëÊÊÁ¿O2£¨g£©Æð·´Ó¦£¬Éú³ÉCuO£¨s£©£¬·Å³ö157kJÈÈÁ¿µÄÈÈ»¯Ñ§·½³ÌʽΪCu£¨s£©+
1
2
O2£¨g£©=CuO£¨s£©£»¡÷H=-157kJ?mol-1£¬
¹Ê´ð°¸Îª£ºCu£¨s£©+
1
2
O2£¨g£©=CuO£¨s£©£»¡÷H=-157kJ?mol-1£»
£¨3£©1molN2H4£¨l£©ÔÚO2£¨g£©ÖÐȼÉÕ£¬Éú³ÉN2£¨g£©ºÍH2O£¨l£©£¬·Å³ö622kJÈÈÁ¿·ÅÈÈʱìʱäֵΪ¸ºÖµ£¬ËùÒԸ÷´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪN2H4£¨g£©+O2£¨g£©=N2£¨g£©+2H2O£¨l£©¡÷H=-622kJ?mol-l£¬
¹Ê´ð°¸Îª£ºN2H4£¨g£©+O2£¨g£©=N2£¨g£©+2H2O£¨l£©¡÷H=-622kJ?mol-l£»
£¨4£©0.3molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5KJµÄÈÈÁ¿£¬Ôò1molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö2165KJµÄÈÈÁ¿£¬·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£»
B2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol¢Ù
H2O£¨g£©=H2O£¨l£©£»¡÷H2=-44.0kJ/mol¢Ú
¢Ù-¢Ú¡Á3µÃ£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨g£©¡÷H=-2033kJ/mol£¬11.2L£¨±ê×¼×´¿ö£©¼´0.5molÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇΪ2033kJ¡Á0.5=1016.5kJ£¬
¹Ê´ð°¸Îª£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£»1016.5kJ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÎÊ´ðÌâ

£¨1£©ÒÑÖªH2£¨g£©+1/2O2£¨g£©=H2O£¨g£©£¬·´Ó¦¹ý³ÌÖÐÄÜÁ¿±ä»¯Èçͼ1£ºÇë»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ùͼ1ÖÐa£¬b·Ö±ð´ú±íʲôÒâÒ壿a£®______£»b£®______£®
¢Ú¸Ã·´Ó¦ÊÇ______·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©£¬¡÷H______£¨Ìî¡°£¼0¡±»ò¡°£¾0¡±£©£®
£¨2£©ÏÖÓпÉÄæ·´Ó¦A£¨Æø£©+B£¨Æø£©?3C£¨Æø£©¡÷H£¼0£¬Í¼2Öмס¢ÒÒ¡¢±û·Ö±ð±íʾÔÚ²»Í¬µÄÌõ¼þÏ£¬Éú³ÉÎïCÔÚ·´Ó¦»ìºÏÎïÖеİٷֺ¬Á¿£¨C%£©ºÍ·´Ó¦Ê±¼äµÄ¹Øϵ£º
¢ÙÈô¼×ͼÖÐÁ½ÌõÇúÏß·Ö±ð±íʾÓд߻¯¼ÁºÍÎÞ´ß»¯¼ÁʱµÄÇé¿ö£¬Ôò_______ÇúÏßÊDZíʾ
Óд߻¯¼ÁʱµÄÇé¿ö£®
¢ÚÈôÒÒͼÖеÄÁ½ÌõÇúÏß±íʾ100¡æ¡¢200¡æʱµÄÇé¿ö£¬Ôò______ÇúÏßÊDZíʾ100¡æµÄÇé¿ö£®
¢ÛÈô±ûͼÖÐÁ½ÌõÇúÏß·Ö±ð±íʾ²»Í¬Ñ¹Ç¿ÏµÄÇé¿ö£¬Ôò______ÇúÏßÊDZíʾѹǿ½Ï´óµÄÇé¿ö£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©ÔÚ25¡æ¡¢101kPaÏ£¬1gҺ̬¼×´¼È¼ÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22.7kJ£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽӦΪ______£®
£¨2£©ÓÉÇâÆøºÍÑõÆø·´Ó¦Éú³É1molҺ̬ˮʱ·ÅÈÈ285.8kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ______£®Èô1gË®ÕôÆøת»¯³ÉҺ̬ˮ·ÅÈÈ2.444kJ£¬Ôò·´Ó¦2H2£¨g£©+O2£¨g£©¨T2H2O£¨g£©µÄ¡÷H=______£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÓйØÈÈ»¯Ñ§·½³ÌʽµÄ±íʾ¼°Ëµ·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÒÑÖªc£¨Ê¯Ä«£¬s£©=c£¨½ð¸Õʯ£¬s£©¡÷H£¾0£¬Ôò½ð¸Õʯ±ÈʯīÎȶ¨
B£®ÒÑÖªH2£¨g£©+F2£¨g£©¨T2HF£¨g£©¡÷H=-270kJ/mol£¬Ôò2L·ú»¯ÇâÆøÌå·Ö½â³É1LÇâÆøºÍ1L·úÆøÎüÊÕ270kJÈÈÁ¿
C£®HClºÍNaOH·´Ó¦µÄÖкÍÈÈΪ-57.3kJ/mol£¬ÔòH2SO4ºÍBa£¨oH£©2·´Ó¦µÄÖкÍÈÈ¡÷H=2¡Á£¨-57.3£©kJ/mol
D£®ÒÑÖªI2£¨g£©+H2£¨g£©¨T2HI£¨g£©¡÷H1£»I2£¨s£©+H2£¨g£©¨T2HI£¨g£©¡÷H2£»Ôò¡÷H1£¼¡÷H2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÎÊ´ðÌâ

£¨1£©ÒÑÖªH2O£¨l£©=H2O£¨g£©£»¡÷H=+44KJ/mol£®ÔÚ25¡æ¡¢101kPaÏ£¬1g¼×´¼È¼ÉÕÉú³ÉCO2ºÍÆø̬ˮʱ·ÅÈÈ19.93kJ£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽӦΪ______£®
£¨2£©ÒÑÖªH2µÄȼÉÕÈÈ¡÷H=-285.8kJ?mol-1£¬COµÄȼÉÕÈÈ¡÷H=-282.8kJ?mol-1£®ÏÖÓÐCO¡¢H2¡¢CO2×é³ÉµÄ»ìºÏÆøÌå67.2L£¨±ê×¼×´¿ö£©£¬¾­³ä·ÖȼÉÕºó·Å³öµÄ×ÜÈÈÁ¿Îª710.0kJ£¬²¢Éú³É18gҺ̬ˮ£¬ÇóȼÉÕÇ°»ìºÏÆøÌåÖÐCOµÄÌå»ý·ÖÊý£®______£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

Ò»¶¨Ìõ¼þÏ£¬»¯Ñ§·´Ó¦2H2+O2¨T2H2OµÄÄÜÁ¿±ä»¯ÈçͼËùʾ£¬Ôò·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ£¨¡¡¡¡£©
A£®H2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨g£©¡÷H=£¨a+b£©kJ?mol-1
B£®2H2£¨g£©+O2£¨g£©¨T2H2O£¨g£©¡÷H=2£¨b-a£©kJ?mol-1
C£®H2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨l£©¡÷H=£¨b+c-a£©kJ?mol-1
D£®2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H=2£¨a-b-c£©kJ?mol-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

½ñÓÐÈçÏÂÈý¸öÈÈ»¯Ñ§·½³Ìʽ£º
H2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨g£©¡÷H=akJ?mol-1
H2£¨g£©+
1
2
O2£¨g£©¨TH2O£¨l£©¡÷H=bkJ?mol-1
2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H=ckJ?mol-1
¹ØÓÚËüÃǵÄÏÂÁбíÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ËüÃǶ¼ÊÇÎüÈÈ·´Ó¦B£®a¡¢bºÍc¾ùΪÕýÖµ
C£®a=b?D£®2b=c

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®·²ÊÇ·ÅÈÈ·´Ó¦¶¼ÊÇ×Ô·¢µÄ£¬ÓÉÓÚÎüÈÈ·´Ó¦¶¼ÊÇ·Ç×Ô·¢µÄ
B£®×Ô·¢·´Ó¦ÔÚÇ¡µ±Ìõ¼þϲÅÄÜʵÏÖ
C£®×Ô·¢·´Ó¦ÔÚÈκÎÌõ¼þ϶¼ÄÜʵÏÖ
D£®×Ô·¢·´Ó¦Ò»¶¨ÊÇìØÔö´ó£¬·Ç×Ô·¢·´Ó¦Ò»¶¨ÊÇìؼõС»ò²»±ä

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁи÷Ïî²»ÕýÈ·µÄÊÇ
A£®µÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏÂÁÐÈÜÒºÖТ٠NH4Al(SO4)2¢Ú NH4Cl£¬¢ÛCH3COONH4£¬¢Ü NH3¡¤H2O£» c(NH4+)£¬ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ¢Ù>¢Ú>¢Û>¢Ü
B£®ÊÒÎÂÏ£¬Ïò0£®01mol/L NH4HSO4ÈÜÒºÖеμÓNaOHÈÜÒºÖÁÖÐÐÔ
C£®25¡æʱ£¬0£®1mol/LCH3COOHÈÜÒºV1 mLºÍ0£®1mol/L NaOHÈÜÒºV2mL»ìºÏ£¬ÈôV1£¾V2£¬Ôò»ìºÏÈÜÒºµÄpHÒ»¶¨Ð¡ÓÚ7
D£®¶ÔÓÚ·´Ó¦£¬ÔÚÈκÎζÈ϶¼ÄÜ×Ô·¢½øÐÐ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸