ÒÑÖª£ºH2(g)£«F2(g)2HF(g)¡¡¦¤H£½£270 kJ/mol£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ | |
[¡¡¡¡] | |
A£® |
1¸öÇâÆø·Ö×ÓÓë1¸ö·úÆø·Ö×Ó·´Ó¦Éú³É2¸ö·ú»¯Çâ·Ö×ӷųö270 kJ |
B£® |
1 molÇâÆøÓë1 mol·úÆø·´Ó¦Éú³É2 molҺ̬·ú»¯Çâ·Å³öµÄÈÈÁ¿Ð¡ÓÚ270 kJ |
C£® |
ÔÚÏàͬÌõ¼þÏ£¬1 molÇâÆøÓë1 mol·úÆøµÄÄÜÁ¿×ܺʹóÓÚ2 mol·ú»¯ÇâÆøÌåµÄÄÜÁ¿ |
D£® |
2 mol·ú»¯ÇâÆøÌå·Ö½â³É1 molµÄÇâÆøºÍ1 molµÄ·úÆø·Å³ö270 kJÈÈÁ¿ |
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â
(12·Ö)£¨1£©¡°æ϶ðÒ»ºÅ¡±µÇÔ³ɹ¦£¬ÊµÏÖÁËÖйúÈË¡°±¼Ô¡±µÄÃÎÏ룮¡°æ϶ðÒ»ºÅ¡±Ê¹ÓõÄÍƽø¼ÁÊÇÒºÇâºÍÒºÑõ£¬ÕâÖÖÍƽø¼ÁµÄÓŵãÊÇÏàͬÖÊÁ¿Ê±£¬ÇâÆø·Å³öµÄÈÈÁ¿¶à£¬²úÎïΪˮÎÞÎÛȾ¡£
ÒÑÖª£ºH2(g)£«O2(g)=H2O(l)¡¡¦¤H=£285.8kJ/mol ; H2(g)=H2(l)¡¡¦¤H=£0.92 kJ/mol
O2(g)=O2(l)¡¡¦¤H£½£6.84kJ/mol ; H2O(l)=H2O(g)¡¡¦¤H£½£«44.0kJ/mol
Çëд³öÒºÇâºÍÒºÑõÉú³ÉÆø̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º_______________________________.
£¨2£©ÒÑÖª£º¢ÙNH3(g)£«HCl(g)===NH4Cl(s) ¦¤H£½£176 kJ/mol
¢ÚNH3(g)£«H2O(l)===NH3¡¤H2O(aq) ¦¤H £½£35.1 kJ/mol
¢ÛHCl(g)===HCl(aq)¡¡ ¦¤H £½ £72.3 kJ/mol
¢ÜNH3¡¤H2O(aq)£«HCl(aq)===NH4Cl(aq)£«H2O(l) ¦¤H£½£52.3 kJ/mol
ÔòNH4Cl(s)===NH4Cl(aq)µÄ¦¤H£½______
£¨3£©·Ö±ðÈ¡40 mLµÄ0.50mol/LÑÎËáÓë0.55 mol/LÇâÑõ»¯ÄÆÈÜÒº½øÐÐÖкͷ´Ó¦£®Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣮
¢Ù ÀíÂÛÉÏÏ¡Ç¿Ëᡢϡǿ¼î·´Ó¦Éú³É1 molˮʱ·Å³ö57.3 kJµÄÈÈÁ¿£¬Ð´³ö±íʾϡÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ __
¢Ú ¼ÙÉèÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒºµÄÃܶȶ¼ÊÇ1 g/cm3£¬ÓÖÖªÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝ
c£½4.18 J/(g¡¤¡æ)£®ÎªÁ˼ÆËãÖкÍÈÈ£¬ÊµÑéʱ»¹Ðè²âÁ¿µÄÊý¾ÝÓÐ(ÌîÐòºÅ)__________£»
A£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÎÂ¶È B£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÖÊÁ¿
C£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄÎÂ¶È D£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿
E£®·´Ó¦ºó»ìºÏÈÜÒºµÄ×î¸ßÎÂ¶È F£®·´Ó¦ºó»ìºÏÈÜÒºµÄÖÊÁ¿
¢Û ijѧÉúʵÑé¼Ç¼Êý¾ÝÈçÏ£º
ʵÑé ÐòºÅ | ÆðʼζÈt1/¡æ | ÖÕֹζÈt2/¡æ | |
ÑÎËá | ÇâÑõ»¯ÄÆ | »ìºÏÈÜÒº | |
1 | 20.0 | 20.1 | 23.2 |
2 | 20.2 | 20.4 | 23.4 |
3 | 20.5 | 20.6 | 23.6 |
ÒÀ¾Ý¸ÃѧÉúµÄʵÑéÊý¾Ý¼ÆË㣬¸ÃʵÑé²âµÃµÄÖкÍÈȦ¤H£½__________ __£»
¢Ü¼Ù¶¨¸ÃѧÉúµÄ²Ù×÷ÍêȫͬÉÏ£¬ÊµÑéÖиÄÓÃ100 mL 0.5 mol/LÑÎËá¸ú100 mL 0.55 mol/LÇâÑõ»¯ÄÆÈÜÒº½øÐз´Ó¦£¬ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿________(Ìî¡°ÏàµÈ¡±»ò¡°²»ÏàµÈ¡±)£¬ËùÇóÖкÍÈÈ__________(Ìî¡°ÏàµÈ¡±»ò¡°²»ÏàµÈ¡±)£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÄÚÃɹŰÍÑåÄ׶ûÊÐÖÐѧ°ÍÊÐÒ»Öи߶þÉÏѧÆÚ12ÔÂÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºµ¥Ñ¡Ìâ
ÒÑÖª£ºH2(g)+F2(g)="===2HF(g)" ¦¤H="-270" kJ/mol£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©
A£®44.8 L HFÆøÌå·Ö½â³É22.4 LµÄH2ºÍ22.4 LµÄF2ÎüÊÕ270 kJÈÈÁ¿ |
B£®1 mol H2Óë1 mol F2·´Ó¦Éú³É2 molҺ̬HF·Å³öµÄÈÈÁ¿´óÓÚ270 kJ |
C£®ÏàͬÌõ¼þÏ£¬1 molH2Óë1 molF2µÄÄÜÁ¿×ܺ͵ÍÓÚ2 molHFÆøÌåµÄÄÜÁ¿ |
D£®¶Ï¿ª2 mol H¡ªF¼üËùÐèÒªµÄÄÜÁ¿±È¶Ï¿ª1 mol H¡ªH¼üºÍ1 mol F¡ªF¼üËùÐèÒªµÄÄÜÁ¿Ö®ºÍ´ó270 kJ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêºÓ±±Ê¡ºâË®ÖÐѧ¸ßÒ»ÏÂѧÆÚÈýµ÷¿¼ÊÔ»¯Ñ§ÊÔ¾í £¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ
(12·Ö)£¨1£©¡°æ϶ðÒ»ºÅ¡±µÇÔ³ɹ¦£¬ÊµÏÖÁËÖйúÈË¡°±¼Ô¡±µÄÃÎÏ룮¡°æ϶ðÒ»ºÅ¡±Ê¹ÓõÄÍƽø¼ÁÊÇÒºÇâºÍÒºÑõ£¬ÕâÖÖÍƽø¼ÁµÄÓŵãÊÇÏàͬÖÊÁ¿Ê±£¬ÇâÆø·Å³öµÄÈÈÁ¿¶à£¬²úÎïΪˮÎÞÎÛȾ¡£
ÒÑÖª£ºH2(g)£«O2(g)=H2O(l)¡¡¦¤H=£285.8 kJ/mol ; H2(g)=H2(l)¡¡¦¤H=£0.92 kJ/mol
O2(g)=O2(l)¡¡¦¤H£½£6.84 kJ/mol ; H2O(l)=H2O(g)¡¡¦¤H£½£«44.0 kJ/mol
Çëд³öÒºÇâºÍÒºÑõÉú³ÉÆø̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º_______________________________.
£¨2£©ÒÑÖª£º¢ÙNH3(g)£«HCl(g)===NH4Cl(s) ¦¤H£½£176 kJ/mol
¢ÚNH3(g)£«H2O(l)===NH3¡¤H2O(aq) ¦¤H£½£35.1 kJ/mol
¢ÛHCl(g)===HCl(aq)¡¡ ¦¤H£½ £72.3 kJ/mol
¢ÜNH3¡¤H2O(aq)£«HCl(aq)===NH4Cl(aq)£«H2O(l) ¦¤H£½£52.3 kJ/mol
ÔòNH4Cl(s)===NH4Cl(aq)µÄ¦¤H£½______
£¨3£©·Ö±ðÈ¡40 mLµÄ0.50 mol/LÑÎËáÓë0.55 mol/LÇâÑõ»¯ÄÆÈÜÒº½øÐÐÖкͷ´Ó¦£®Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣮
¢Ù ÀíÂÛÉÏÏ¡Ç¿Ëᡢϡǿ¼î·´Ó¦Éú³É1 molˮʱ·Å³ö57.3 kJµÄÈÈÁ¿£¬Ð´³ö±íʾϡÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ __
¢Ú ¼ÙÉèÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒºµÄÃܶȶ¼ÊÇ1 g/cm3£¬ÓÖÖªÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝ
c£½4.18 J/(g¡¤¡æ)£®ÎªÁ˼ÆËãÖкÍÈÈ£¬ÊµÑéʱ»¹Ðè²âÁ¿µÄÊý¾ÝÓÐ(ÌîÐòºÅ)__________£»
A£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÎÂ¶È | B£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÖÊÁ¿ |
C£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄÎÂ¶È | D£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿ |
ʵÑé ÐòºÅ | ÆðʼζÈt1/¡æ | ÖÕֹζÈt2/¡æ | |
ÑÎËá | ÇâÑõ»¯ÄÆ | »ìºÏÈÜÒº | |
1 | 20.0 | 20.1 | 23.2 |
2 | 20.2 | 20.4 | 23.4 |
3 | 20.5 | 20.6 | 23.6 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìºÓ±±Ê¡¸ßÒ»ÏÂѧÆÚÈýµ÷¿¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
(12·Ö)£¨1£©¡°æ϶ðÒ»ºÅ¡±µÇÔ³ɹ¦£¬ÊµÏÖÁËÖйúÈË¡°±¼Ô¡±µÄÃÎÏ룮¡°æ϶ðÒ»ºÅ¡±Ê¹ÓõÄÍƽø¼ÁÊÇÒºÇâºÍÒºÑõ£¬ÕâÖÖÍƽø¼ÁµÄÓŵãÊÇÏàͬÖÊÁ¿Ê±£¬ÇâÆø·Å³öµÄÈÈÁ¿¶à£¬²úÎïΪˮÎÞÎÛȾ¡£
ÒÑÖª£ºH2(g)£«O2(g)=H2O(l)¡¡¦¤H=£285.8 kJ/mol ; H2(g)=H2(l)¡¡¦¤H=£0.92 kJ/mol
O2(g)=O2(l)¡¡¦¤H£½£6.84 kJ/mol ; H2O(l)=H2O(g)¡¡¦¤H£½£«44.0 kJ/mol
Çëд³öÒºÇâºÍÒºÑõÉú³ÉÆø̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º_______________________________.
£¨2£©ÒÑÖª£º¢ÙNH3(g)£«HCl(g)===NH4Cl(s) ¦¤H£½£176 kJ/mol
¢ÚNH3(g)£«H2O(l)===NH3¡¤H2O(aq) ¦¤H £½£35.1 kJ/mol
¢ÛHCl(g)===HCl(aq)¡¡ ¦¤H £½ £72.3 kJ/mol
¢ÜNH3¡¤H2O(aq)£«HCl(aq)===NH4Cl(aq)£«H2O(l) ¦¤H£½£52.3 kJ/mol
ÔòNH4Cl(s)===NH4Cl(aq)µÄ¦¤H£½______
£¨3£©·Ö±ðÈ¡40 mLµÄ0.50 mol/LÑÎËáÓë0.55 mol/LÇâÑõ»¯ÄÆÈÜÒº½øÐÐÖкͷ´Ó¦£®Í¨¹ý²â¶¨·´Ó¦¹ý³ÌÖÐËù·Å³öµÄÈÈÁ¿¿É¼ÆËãÖкÍÈÈ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣮
¢Ù ÀíÂÛÉÏÏ¡Ç¿Ëᡢϡǿ¼î·´Ó¦Éú³É1 molˮʱ·Å³ö57.3 kJµÄÈÈÁ¿£¬Ð´³ö±íʾϡÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ __
¢Ú ¼ÙÉèÑÎËáºÍÇâÑõ»¯ÄÆÈÜÒºµÄÃܶȶ¼ÊÇ1 g/cm3£¬ÓÖÖªÖкͺóÉú³ÉÈÜÒºµÄ±ÈÈÈÈÝ
c£½4.18 J/(g¡¤¡æ)£®ÎªÁ˼ÆËãÖкÍÈÈ£¬ÊµÑéʱ»¹Ðè²âÁ¿µÄÊý¾ÝÓÐ(ÌîÐòºÅ)__________£»
A£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÎÂ¶È B£®·´Ó¦Ç°ÑÎËáÈÜÒºµÄÖÊÁ¿
C£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄÎÂ¶È D£®·´Ó¦Ç°ÇâÑõ»¯ÄÆÈÜÒºµÄÖÊÁ¿
E£®·´Ó¦ºó»ìºÏÈÜÒºµÄ×î¸ßÎÂ¶È F£®·´Ó¦ºó»ìºÏÈÜÒºµÄÖÊÁ¿
¢Û ijѧÉúʵÑé¼Ç¼Êý¾ÝÈçÏ£º
ʵÑé ÐòºÅ |
ÆðʼζÈt1/¡æ |
ÖÕֹζÈt2/¡æ |
|
ÑÎËá |
ÇâÑõ»¯ÄÆ |
»ìºÏÈÜÒº |
|
1 |
20.0 |
20.1 |
23.2 |
2 |
20.2 |
20.4 |
23.4 |
3 |
20.5 |
20.6 |
23.6 |
ÒÀ¾Ý¸ÃѧÉúµÄʵÑéÊý¾Ý¼ÆË㣬¸ÃʵÑé²âµÃµÄÖкÍÈȦ¤H£½__________ __£»
¢Ü¼Ù¶¨¸ÃѧÉúµÄ²Ù×÷ÍêȫͬÉÏ£¬ÊµÑéÖиÄÓÃ100 mL 0.5 mol/LÑÎËá¸ú100 mL 0.55 mol/LÇâÑõ»¯ÄÆÈÜÒº½øÐз´Ó¦£¬ÓëÉÏÊöʵÑéÏà±È£¬Ëù·Å³öµÄÈÈÁ¿________(Ìî¡°ÏàµÈ¡±»ò¡°²»ÏàµÈ¡±)£¬ËùÇóÖкÍÈÈ__________(Ìî¡°ÏàµÈ¡±»ò¡°²»ÏàµÈ¡±)£®
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÄÚÃɹŰÍÑåÄ׶ûÊаÍÊÐÒ»Öи߶þÉÏѧÆÚ12ÔÂÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÑ¡ÔñÌâ
ÒÑÖª£ºH2(g)+F2(g)====2HF(g) ¦¤H=-270 kJ/mol£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©
A.44.8 L HFÆøÌå·Ö½â³É22.4 LµÄH2ºÍ22.4 LµÄF2ÎüÊÕ270 kJÈÈÁ¿
B.1 mol H2Óë1 mol F2·´Ó¦Éú³É2 molҺ̬HF·Å³öµÄÈÈÁ¿´óÓÚ270 kJ
C.ÏàͬÌõ¼þÏ£¬1 molH2Óë1 molF2µÄÄÜÁ¿×ܺ͵ÍÓÚ2 molHFÆøÌåµÄÄÜÁ¿
D.¶Ï¿ª2 mol H¡ªF¼üËùÐèÒªµÄÄÜÁ¿±È¶Ï¿ª1 mol H¡ªH¼üºÍ1 mol F¡ªF¼üËùÐèÒªµÄÄÜÁ¿Ö®ºÍ´ó270 kJ
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - 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