ÒÑÖª2SO2(g)£«O2(g)2SO3(g)¡¡¦¤H<0µÄʵÑéÊý¾ÝÈçÏÂ±í£º
ÎÂ¶È | ²»Í¬Ñ¹Ç¿ÏÂSO2µÄת»¯ÂÊ(%) | ||||
1¡Á105Pa | 5¡Á105Pa | 1¡Á106Pa | 5¡Á106Pa | 1¡Á107Pa | |
450¡æ | 97.5 | 98.9 | 99.2 | 99.6 | 99.7 |
550¡æ | 85.6 | 92.9 | 94.9 | 97.7 | 98.3 |
(1)ӦѡÔñµÄζÈÊÇ________¡£
(2)Ó¦²ÉÓõÄѹǿÊÇ________£¬ÀíÓÉÊÇ
________________________________________________________________________________________________________________________________________________¡£
(3)ÔںϳÉSO3µÄ¹ý³ÌÖУ¬²»ÐèÒª·ÖÀë³öSO3µÄÔÒòÊÇ________________________________________________________________________________________________________________________________________________________________________________________________________________________¡£
½âÎö£º¸Ã·´Ó¦ÓëºÏ³É°±µÄ·´Ó¦ÏàËÆ£º¶¼ÊÇÆøÌåÌå»ýËõСµÄ·ÅÈÈ·´Ó¦£¬Î¶ÈÉý¸ßÓÐÀûÓÚ¼Ó¿ì·´Ó¦ËÙÂÊ£¬µ«½µµÍÁË·´Ó¦ÎïµÄת»¯ÂÊ£¬ÊÊÒËÌõ¼þµÄÑ¡ÔñÐèÒª×ۺϿ¼ÂÇ¡£
´ð°¸£º(1)450¡æ¡¡(2)1¡Á105Pa¡¡ÒòΪ³£Ñ¹ÏÂSO2µÄת»¯ÂÊÒѾºÜ¸ß£¬Èô²ÉÓýϴóµÄѹǿ£¬SO2µÄת»¯ÂÊÌá¸ßºÜÉÙ£¬µ«ÐèÒª¶¯Á¦¸ü´ó£¬¶ÔÉ豸µÄÒªÇó¸ü¸ß¡¡(3)ÒòΪSO2µÄת»¯ÂʱȽϸߣ¬´ïµ½Æ½ºâºóµÄ»ìºÏÆøÌåÖÐSO2µÄÓàÁ¿ºÜÉÙ£¬¹Ê²»ÐèÒª·ÖÀëSO3¡£
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£º
ij¶þÔªËá(»¯Ñ§Ê½ÓÃH2A±íʾ)ÔÚË®ÈÜÒºÖеĵçÀë·½³ÌʽÊÇH2A===H£«£«HA££»HA£
H£«£«A2£¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)Na2AÈÜÒºÏÔ________(Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±)£¬ÀíÓÉÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)________________________________________________________________________¡£
(2)Èô0.1 mol·L£1 NaHAÈÜÒºµÄpH£½2£¬Ôò0.1 mol·L£1 H2AÈÜÒºÖÐÇâÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶È¿ÉÄÜ__________0.11 mol·L£1(Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±)£¬ÀíÓÉÊÇ
________________________________________________________________________
__________________¡£
(3)0.1 mol·L£1 NaHAµÄÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
________________________________________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
Ò»ÖÖ³ÎÇå͸Ã÷µÄÈÜÒºÖУ¬¿ÉÄܺ¬ÓÐÏÂÁÐÀë×Ó£ºK£«¡¢Fe3£«¡¢Ba2£«¡¢Al3£«¡¢NH¡¢Cl£¡¢NO¡¢HCO¡¢SO¡£ÏÖ×öÒÔÏÂʵÑ飺
(1)½«ÈÜÒºµÎÔÚÀ¶É«Ê¯ÈïÊÔÖ½ÉÏ£¬ÊÔÖ½³ÊºìÉ«£»
(2)È¡ÉÙÁ¿ÈÜÒº£¬¼ÓÈëÓÃÏ¡HNO3ËữµÄBaCl2ÈÜÒº£¬²úÉú°×É«³Áµí£»
(3)½«(2)ÖеijÁµí¹ýÂË£¬ÏòÂËÒºÖмÓÈëAgNO3ÈÜÒº£¬²úÉú°×É«³Áµí£»
(4)ÁíÈ¡ÈÜÒº£¬ÖðµÎ¼ÓÈëNaOHÈÜÒºÖÁ¹ýÁ¿£¬Ö»¿´µ½ÓкìºÖÉ«³ÁµíÉú³É£¬ÇÒ³ÁµíÖÊÁ¿²»¼õÉÙ¡£
ÓÉ´Ë¿ÉÒÔÍƶϣº
ÈÜÒºÖп϶¨´æÔÚµÄÀë×ÓÓÐ_________________________________________________£»
ÈÜÒºÖп϶¨²»´æÔÚµÄÀë×ÓÓÐ________________________________________________£»
ÈÜÒºÖв»ÄÜÈ·¶¨ÊÇ·ñ´æÔÚµÄÀë×ÓÓÐ_________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁÐʵÑé²Ù×÷ÖУº¢ÙÈ¡ÒºÌåÊÔ¼Á¡¡¢ÚÈ¡¹ÌÌåÊÔ¼Á¡¡¢ÛÈܽâ
¢Ü¹ýÂË¡¡¢ÝÕô·¢£¬Ò»¶¨ÒªÓõ½²£Á§°ôµÄÊÇ(¡¡¡¡)
A£®¢Ù¢Ú¢Û¡¡¡¡¡¡¡¡¡¡¡¡¡¡ B£®¢Ú¢Û¢Ü
C£®¢Ù¢Ú¢Ý D£®¢Û¢Ü¢Ý
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÔÚºãκãÈÝÌõ¼þÏ£¬ÄÜʹA(g)£«B(g)C(g)£«D(g)Õý·´Ó¦ËÙÂÊÔö´óµÄ´ëÊ©ÊÇ(¡¡¡¡)
¢Ù¼õСC»òDµÄŨ¶È ¢ÚÔö´óDµÄŨ¶È
¢Û¼õСBµÄŨ¶È ¢ÜÔö´óA»òBµÄŨ¶È
A£®¢Ù¢Ú B£®¢Ú¢Û
C£®¢Ù¢Ü D£®¢Ú¢Ü
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
½«H2(g)ºÍBr2(g)³äÈëºãÈÝÃܱÕÈÝÆ÷£¬ºãÎÂÏ·¢Éú·´Ó¦H2(g)£«Br2(g)2HBr(g)¡¡¦¤H<0£¬Æ½ºâʱBr2(g)µÄת»¯ÂÊΪa£»Èô³õʼÌõ¼þÏàͬ£¬¾øÈÈϽøÐÐÉÏÊö·´Ó¦£¬Æ½ºâʱBr2(g)µÄת»¯ÂÊΪb¡£aÓëbµÄ¹ØϵÊÇ(¡¡¡¡)
A£®a>b B£®a£½b
C£®a<b D£®ÎÞ·¨È·¶¨
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
A£® 126C ¡¢136C ¡¢146C Ϊ̼ԪËصÄÈýÖÖºËËØ¡£ÔªËØÖÜÆÚ±íÖÐ̼µÄÏà¶ÔÔ×ÓÖÊÁ¿Îª12.01£¬ËµÃ÷×ÔÈ»½çÖеÄ̼Ö÷ÒªÒÔ126C µÄºËËØÐÎʽ´æÔÚ¡£146CΪ·ÅÉäÐÔºËËØ£¬¿ÉÓÃÓÚͬλËØʾ×Ù
B£®Æû³µÎ²Æø´ß»¯×ª»¯×°Öÿɽ«Î²ÆøÖеÄNOºÍCOµÈÓк¦ÆøÌåת»¯ÎªN2ºÍCO2£¬¸Ã×°ÖÃÖеĴ߻¯¼Á¿É½µµÍNOºÍCO·´Ó¦µÄ»î»¯ÄÜ£¬ÓÐÀûÓÚÌá¸ß¸Ã·´Ó¦µÄƽºâת»¯ÂÊ
C£®µÀ¶û¶Ù¡¢ÌÀÄ·Éú¡¢Â¬Éª¸£¡¢²£¶ûµÈ¿Æѧ¼ÒµÄÑо¿²»¶Ï¸üÐÂÈËÃǶÔÔ×ӽṹµÄÈÏʶ
D£®µØ¹µÓÍÓÉÓÚ»ìÓÐһЩ¶ÔÈËÌåÓꦵÄÔÓÖʶø²»ÄÜʳÓ㬿ɼӹ¤ÖƳÉÉúÎï²ñÓÍ£¬ÉúÎï²ñÓ͵ijɷÖÓë´ÓʯÓÍÖÐÌáÈ¡µÄ²ñÓͳɷֲ»Í¬
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨1£©¾Ý±¨µÀÒÔÅðÇ⻯ºÏÎïNaBH4£¨HµÄ»¯ºÏ¼ÛΪ-1¼Û£©ºÍH2O2×÷ÔÁϵÄȼÁϵç³Ø£¬¿ÉÓÃ×÷ͨÐÅÎÀÐǵ硣¸º¼«²ÄÁϲÉÓÃPt/C£¬Õý¼«²ÄÁϲÉÓÃMnO2£¬Æ乤×÷ÔÀíÈçͼËùʾ¡£Ð´³ö¸Ãµç³Ø·Åµçʱ¸º¼«µÄµç¼«·´Ó¦Ê½£º
¡ø ¡£
£¨2£©»ð¼ý·¢Éä³£ÒÔҺ̬루N2H4£©ÎªÈ¼ÁÏ£¬ÒºÌ¬¹ýÑõ»¯Çâ
ΪÖúȼ¼Á¡£
ÒÑÖª£º N2H4£¨l£© + O2£¨g£© £½ N2£¨g£©+ 2H2O£¨l£© ¡÷H = ¨C 534 kJ¡¤mol¡ª1
H2O2£¨l£©= H2O£¨l£© + 1/2O2£¨g£© ¡÷H = ¨C 98.6 kJ¡¤mol¡ª1
д³ö³£ÎÂÏ£¬N2H4£¨l£© Óë H2O2£¨l£©·´Ó¦Éú³ÉN2ºÍH2OµÄÈÈ»¯Ñ§·½³Ìʽ ¡ø ¡£
£¨3£©O3¿ÉÓɳôÑõ·¢ÉúÆ÷£¨ÔÀíÈçÓÒͼËùʾ£©µç½âÏ¡ÁòËáÖƵá£
¢ÙͼÖÐÒõ¼«Îª ¡ø £¨Ìî¡°A¡±»ò¡°B¡±£©¡£
¢ÚÈôC´¦Í¨ÈëO 2£¬ÔòA¼«µÄµç¼«·´Ó¦Ê½Îª£º ¡ø
£¨4£©ÏòÒ»ÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿Ò»Ñõ»¯Ì¼¸úË®ÕôÆø·¢Éú·´Ó¦ CO(g)+H2O(g)CO2(g)+H2(g)£¬ÏÂÁÐÇé¿öÏÂÄÜÅжϸ÷´Ó¦Ò»¶¨´ïµ½Æ½ºâ״̬µÄÊÇ____ ¡ø _____£¨Ñ¡Ìî±àºÅ£©¡£
A£®vÕý£¨H2O) = vÄ棨H2)
B£®ÈÝÆ÷ÖÐÆøÌåµÄѹǿ²»ÔÙ·¢Éú¸Ä±ä
C£®H2OµÄÌå»ý·ÖÊý²»Ôٸıä
D£®ÈÝÆ÷ÖÐCO2ºÍH2µÄÎïÖʵÄÁ¿Ö®±È²»ÔÙ·¢Éú¸Ä±ä
E£®ÈÝÆ÷ÖÐÆøÌåµÄÃܶȲ»ÔÙ·¢Éú¸Ä±ä
£¨5£©Î¶ÈT1ʱ£¬ÔÚÒ»Ìå»ýΪ2LµÄÃܱÕÈÝ»ýÖÐ,¼ÓÈë0.4molCO2ºÍ0.4molµÄH2,·´Ó¦ÖÐc(H2O)µÄ±ä»¯Çé¿öÈçͼËùʾ,T1ʱ·´Ó¦CO(g)+H2O(g)CO2(g)+H2(g)µÚ4·ÖÖӴﵽƽºâ¡£ÔÚµÚ5·ÖÖÓʱÏòÌåϵÖÐͬʱÔÙ³äÈë0.1molCOºÍ0.1molH2£¨ÆäËûÌõ¼þ²»±ä£©£¬ÇëÔÚÓÒͼÖл³öµÚ5·ÖÖÓµ½9·ÖÖÓc(H2O)Ũ¶È±ä»¯Ç÷ÊƵÄÇúÏß¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
½ñÓÐÏÂÁÐÆøÌ壺H2¡¢Cl2¡¢HCl¡¢NH3¡¢NO¡¢H2S¡¢SO2£¬ÓÃÈçͼËùʾµÄ×°ÖýøÐÐʵÑ飬Ìî¿ÕÏÂÁпհףº
(1)ÉÕÆ¿¸ÉÔïʱ£¬´ÓA¿Ú½øÆø¿ÉÊÕ¼¯µÄÆøÌåÊÇ________£¬´ÓB¿Ú½øÆø¿ÉÊÕ¼¯µÄÆøÌåÊÇ______________¡£
(2)ÉÕÆ¿ÖгäÂúˮʱ£¬¿ÉÓÃÀ´²âÁ¿________µÈÆøÌåµÄÌå»ý¡£
(3)µ±ÉÕÆ¿ÖÐ×°ÈëÏ´Òº£¬ÓÃÓÚÏ´Æøʱ£¬ÆøÌåÓ¦´Ó________¿Ú½øÈëÉÕÆ¿¡£
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