°×Áס¢ºìÁ×ÊÇÁ×µÄÁ½ÖÖͬËØÒìÐÎÌ壬ÔÚ¿ÕÆøÖÐȼÉյõ½Á×µÄÑõ»¯Î¿ÕÆø²»×ãʱÉú³ÉP4O6£¬¿ÕÆø³ä×ãʱÉú³ÉP4O10¡£
£¨1£©ÒÑÖª298 Kʱ°×Áס¢ºìÁ×ÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪP4£¨s£¬°×Á×£©£«5O2£¨g£©=P4O10£¨s£©¦¤H1£½£2 983.2 kJ¡¤mol£1£»P£¨s£¬ºìÁ×£©£«O2£¨g£©=P4O10£¨s£©¦¤H2£½£738.5 kJ¡¤mol£1£¬Ôò¸ÃζÈÏ°×Á×ת»¯ÎªºìÁ×µÄÈÈ»¯Ñ§·½³ÌʽΪ________________¡£
£¨2£©ÒÑÖª298 Kʱ°×Áײ»ÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪP4£¨s£¬°×Á×£©£«3O2£¨g£©=P4O6£¨s£©¦¤H£½£1 638 kJ¡¤mol£1¡£ÔÚijÃܱÕÈÝÆ÷ÖмÓÈë62 g°×Á׺Í50.4 LÑõÆø£¨±ê×¼×´¿ö£©£¬¿ØÖÆÌõ¼þʹ֮ǡºÃÍêÈ«·´Ó¦¡£ÔòËùµÃµ½µÄP4O10ÓëP4O6µÄÎïÖʵÄÁ¿Ö®±ÈΪ________£¬·´Ó¦¹ý³ÌÖзųöµÄÈÈÁ¿Îª________¡£
£¨3£©ÒÑÖª°×Á׺ÍPCl3µÄ·Ö×ӽṹÈçͼËùʾ£¬ÏÖÌṩÒÔÏ»¯Ñ§¼üµÄ¼üÄÜ£¨kJ¡¤mol£1£©£ºP¡ªP 198£¬Cl¡ªCl 243£¬P¡ªCl 331¡£
Ôò·´Ó¦P4£¨s£¬°×Á×£©£«6Cl2£¨g£©=4PCl3£¨s£©µÄ·´Ó¦ÈȦ¤H£½________¡£
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£ºÌî¿ÕÌâ
¼×Íé×÷ΪһÖÖÐÂÄÜÔ´ÔÚ»¯Ñ§ÁìÓòÓ¦Óù㷺£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¸ß¯ұÌú¹ý³ÌÖУ¬¼×ÍéÔÚ´ß»¯·´Ó¦ÊÒÖвúÉúˮúÆø(COºÍH2)»¹ÔÑõ»¯Ìú£¬Óйط´Ó¦Îª£ºCH4(g)£«CO2(g)=2CO(g)£«2H2(g)¡¡¦¤H£½260 kJ¡¤mol£1
ÒÑÖª£º2CO(g)£«O2(g)=2CO2(g)¡¡¦¤H£½£566 kJ¡¤mol£1
ÔòCH4ÓëO2·´Ó¦Éú³ÉCOºÍH2µÄÈÈ»¯Ñ§·½³ÌʽΪ_____________________£»
(2)ÈçÏÂͼËùʾ£¬×°ÖâñΪ¼×ÍéȼÁϵç³Ø(µç½âÖÊÈÜҺΪKOHÈÜÒº)£¬Í¨¹ý×°ÖâòʵÏÖÌú°ôÉ϶ÆÍ¡£
¢Ùa´¦Ó¦Í¨Èë________(Ìî¡°CH4¡±»ò¡°O2¡±)£¬b´¦µç¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½ÊÇ________£»
¢Úµç¶Æ½áÊøºó£¬×°ÖâñÖÐÈÜÒºµÄpH________(Ìîд¡°±ä´ó¡±¡°±äС¡±»ò¡°²»±ä¡±£¬ÏÂͬ)£¬×°ÖâòÖÐCu2£«µÄÎïÖʵÄÁ¿Å¨¶È________£»
¢Ûµç¶Æ½áÊøºó£¬×°ÖâñÈÜÒºÖеÄÒõÀë×Ó³ýÁËOH£ÒÔÍ⻹º¬ÓÐ________(ºöÂÔË®½â)£»
¢ÜÔڴ˹ý³ÌÖÐÈôÍêÈ«·´Ó¦£¬×°ÖâòÖÐÒõ¼«ÖÊÁ¿±ä»¯12.8 g£¬Ôò×°ÖâñÖÐÀíÂÛÉÏÏûºÄ¼×Íé________L(±ê×¼×´¿öÏÂ)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
úÆø»¯ºÍÒº»¯ÊÇÏÖ´úÄÜÔ´¹¤ÒµÖÐÖص㿼ÂǵÄÄÜÔ´×ÛºÏÀûÓ÷½°¸¡£×î³£¼ûµÄÆø»¯·½·¨ÎªÓÃúÉú²úˮúÆø£¬¶øµ±Ç°±È½ÏÁ÷ÐеÄÒº»¯·½·¨ÎªÓÃúÉú²úCH3OH¡£
£¨1£©ÒÑÖª£ºCO2(g)£«3H2(g)=CH3OH(g)£«H2O(g)¡¡¦¤H1
2CO(g)£«O2(g)=2CO2(g)¡¡¦¤H2
2H2(g)£«O2(g)=2H2O(g)¡¡¦¤H3
Ôò·´Ó¦CO(g)£«2H2(g)=CH3OH(g)µÄ¦¤H£½______¡£
£¨2£©ÈçͼÊǸ÷´Ó¦ÔÚ²»Í¬Î¶ÈÏÂCOµÄת»¯ÂÊËæʱ¼ä±ä»¯µÄÇúÏß¡£
¢ÙT1ºÍT2ζÈϵÄƽºâ³£Êý´óС¹ØϵÊÇK1________K2(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±)¡£
¢ÚÓÉCOºÏ³É¼×´¼Ê±£¬COÔÚ250 ¡æ¡¢300 ¡æ¡¢350 ¡æÏ´ﵽƽºâʱת»¯ÂÊÓëѹǿµÄ¹ØϵÇúÏßÈçÏÂͼËùʾ£¬ÔòÇúÏßcËù±íʾµÄζÈΪ________ ¡æ¡£Êµ¼ÊÉú²úÌõ¼þ¿ØÖÆÔÚ250 ¡æ¡¢1.3¡Á104 kPa×óÓÒ£¬Ñ¡Ôñ´ËѹǿµÄÀíÓÉÊÇ____________¡£
¢ÛÒÔÏÂÓйظ÷´Ó¦µÄ˵·¨ÕýÈ·µÄÊÇ________(ÌîÐòºÅ)¡£
A£®ºãΡ¢ºãÈÝÌõ¼þÏ£¬ÈôÈÝÆ÷ÄÚµÄѹǿ²»·¢Éú±ä»¯£¬Ôò¿ÉÄæ·´Ó¦´ïµ½Æ½ºâ
B£®Ò»¶¨Ìõ¼þÏ£¬H2µÄÏûºÄËÙÂÊÊÇCOµÄÏûºÄËÙÂʵÄ2±¶Ê±£¬¿ÉÄæ·´Ó¦´ïµ½Æ½ºâ
C£®Ê¹ÓúÏÊʵĴ߻¯¼ÁÄÜËõ¶Ì´ïµ½Æ½ºâµÄʱ¼ä²¢Ìá¸ßCH3OHµÄ²úÂÊ
D£®Ä³Î¶ÈÏ£¬½«2 mol COºÍ6 mol H2³äÈë2 LÃܱÕÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦£¬´ïµ½Æ½ºâºó£¬²âµÃc(CO)£½0.2 mol¡¤L£1£¬ÔòCOµÄת»¯ÂÊΪ80%
£¨3£©Ò»¶¨Î¶ÈÏ£¬Ïò2 L¹Ì¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖмÓÈë1 mol CH3OH(g)£¬·¢Éú·´Ó¦£ºCH3OH(g)??CO(g)£«2H2(g)£¬H2µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯µÄÇúÏßÈçͼËùʾ¡£
0¡«2 minÄÚµÄƽ¾ù·´Ó¦ËÙÂÊv(CH3OH)£½__________¡£¸ÃζÈÏ£¬·´Ó¦CO(g)£«2H2(g)??CH3OH(g)µÄƽºâ³£ÊýK£½__________¡£ÏàͬζÈÏ£¬Èô¿ªÊ¼Ê±¼ÓÈëCH3OH(g)µÄÎïÖʵÄÁ¿ÊÇÔÀ´µÄ2±¶£¬Ôò__________(ÌîÐòºÅ)ÊÇÔÀ´µÄ2±¶¡£
A£®Æ½ºâ³£Êý B£®CH3OHµÄƽºâŨ¶È
C£®´ïµ½Æ½ºâµÄʱ¼ä D£®Æ½ºâʱÆøÌåµÄÃܶÈ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
¸ù¾ÝÏÂÁÐÌõ¼þ¼ÆËãÓйط´Ó¦µÄìʱ䣺
£¨1£©ÒÑÖª£ºTi(s)£«2Cl2(g)=TiCl4(l)¦¤H£½£804£®2 kJ¡¤mol£1
2Na(s)£«Cl2(g)=2NaCl(s)¡¡¦¤H£½£882£®0 kJ¡¤mol£1
Na(s)=Na(l)¡¡¦¤H£½£«2£®6 kJ¡¤mol£1
Ôò·´Ó¦TiCl4(l)£«4Na(l)=Ti(s)£«4NaCl(s)µÄ¦¤H£½________ kJ¡¤mol£1¡£
£¨2£©ÒÑÖªÏÂÁз´Ó¦ÊýÖµ£º
·´Ó¦ÐòºÅ | »¯Ñ§·´Ó¦ | ·´Ó¦ÈÈ |
¢Ù | Fe2O3(s)£«3CO(g)= 2Fe(s)£«3CO2(g) | ¦¤H1£½£26£®7 kJ¡¤mol£1 |
¢Ú | 3Fe2O3(s)£«CO(g)=2Fe3O4(s)£«CO2(g) | ¦¤H2£½£50£®8 kJ¡¤mol£ |
¢Û | Fe3O4(s)£«CO(g)=3FeO(s)£«CO2(g) | ¦¤H3£½£36£®5 kJ¡¤mol£1 |
¢Ü | FeO(s)£«CO(g)=Fe(s)£«CO2(g) | ¦¤H4 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
£¨1£©Æû³µ·¢¶¯»ú¹¤×÷ʱ»áÒý·¢N2ºÍO2·´Ó¦£¬ÆäÄÜÁ¿±ä»¯Ê¾ÒâͼÈçÏ£º
д³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º________________¡£
£¨2£©¶þ¼×ÃÑ(CH3OCH3)ÊÇÎÞÉ«ÆøÌ壬¿É×÷ΪһÖÖÐÂÐÍÄÜÔ´¡£ÓɺϳÉÆø(×é³ÉΪH2¡¢COºÍÉÙÁ¿µÄCO2)Ö±½ÓÖƱ¸¶þ¼×ÃÑ£¬ÆäÖеÄÖ÷Òª¹ý³Ì°üÀ¨ÒÔÏÂËĸö·´Ó¦£º
¼×´¼ºÏ³É·´Ó¦£º
(¢¡)CO(g)£«2H2(g)=CH3OH(g)¦¤H1£½£90.1 kJ¡¤mol£1
(¢¢)CO2(g)£«3H2(g)=CH3OH(g)£«H2O(g)¦¤H2£½£49.0 kJ¡¤mol£1
ˮúÆø±ä»»·´Ó¦£º
(¢£)CO(g)£«H2O(g)=CO2(g)£«H2(g)¦¤H3£½£41.1 kJ¡¤mol£1
¶þ¼×ÃѺϳɷ´Ó¦£º
(¢¤)2CH3OH(g)=CH3OCH3(g)£«H2O(g)¦¤H4£½£24.5 kJ¡¤mol£1
ÓÉH2ºÍCOÖ±½ÓÖƱ¸¶þ¼×ÃÑ(ÁíÒ»²úÎïΪˮÕôÆø)µÄÈÈ»¯Ñ§·½³ÌʽΪ_______________¡£
¸ù¾Ý»¯Ñ§·´Ó¦ÔÀí£¬·ÖÎöÔö¼Óѹǿ¶ÔÖ±½ÓÖƱ¸¶þ¼×ÃÑ·´Ó¦µÄÓ°Ïì_________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÁòÔڵؿÇÖÐÖ÷ÒªÒÔÁò»¯Îï¡¢ÁòËáÑεÈÐÎʽ´æÔÚ£¬Æäµ¥Öʺͻ¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Óá£
£¨1£©ÒÑÖª£ºÖؾ§Ê¯£¨BaSO4£©¸ßÎÂìÑÉÕ¿É·¢ÉúһϵÁз´Ó¦£¬ÆäÖв¿·Ö·´Ó¦ÈçÏ£º
BaSO4(s)+4C(s)=BaS(s)+4CO(g) ¡÷H=" +" 571.2 kJ?mol¡ª1
BaS(s)= Ba(s)+S(s) ¡÷H=" +460" kJ?mol¡ª1
ÒÑÖª£º2C(s)+O2(g)=2CO(g) ¡÷H=" -221" kJ?mol¡ª1
Ôò£ºBa(s)+S(s)+2O2(g)=BaSO4(s) ¡÷H= ¡£
£¨2£©ÐÛ»Æ(As4S4)ºÍ´Æ»Æ(As2S3)ÊÇÌáÈ¡ÉéµÄÖ÷Òª¿óÎïÔÁÏ¡£ÒÑÖªAs2S3ºÍHNO3ÓÐÈçÏ·´Ó¦£º
As2S3+10H++ 10NO3-=2H3AsO4+3S+10NO2¡ü+ 2H2O
µ±·´Ó¦ÖÐתÒƵç×ÓµÄÊýĿΪ2molʱ£¬Éú³ÉH3AsO4µÄÎïÖʵÄÁ¿Îª ¡£
£¨3£©ÏòµÈÎïÖʵÄÁ¿Å¨¶ÈNa2S¡¢NaOH»ìºÏÈÜÒºÖеμÓÏ¡ÑÎËáÖÁ¹ýÁ¿¡£ÆäÖÐÖ÷Òªº¬Áò¸÷ÎïÖÖ£¨H2S¡¢HS¡ª¡¢S2¡ª£©µÄ·Ö²¼·ÖÊý£¨Æ½ºâʱijÎïÖÖµÄŨ¶ÈÕ¼¸÷ÎïÖÖŨ¶ÈÖ®ºÍµÄ·ÖÊý£©ÓëµÎ¼ÓÑÎËáÌå»ýµÄ¹ØϵÈçÏÂͼËùʾ£¨ºöÂԵμӹý³ÌH2SÆøÌåµÄÒݳö£©¡£
¢Ùº¬ÁòÎïÖÖB±íʾ ¡£ÔڵμÓÑÎËá¹ý³ÌÖУ¬ÈÜÒºÖÐc(Na+)Ó뺬Áò¸÷ÎïÖÖŨ¶ÈµÄ´óС¹ØϵΪ (Ìî×Öĸ)¡£
a£®c(Na+)= c(H2S)+c(HS¡ª)+2c(S2¡ª)
b£®2c(Na+)=c(H2S)+c(HS¡ª)+c(S2¡ª)
c£®c(Na+)=3[c(H2S)+c(HS¡ª)+c(S2¡ª)]
¢ÚNaHSÈÜÒº³Ê¼îÐÔ£¬ÈôÏòÈÜÒºÖмÓÈëCuSO4ÈÜÒº£¬Ç¡ºÃÍêÈ«·´Ó¦£¬ËùµÃÈÜÒº³ÊÇ¿ËáÐÔ£¬ÆäÔÒòÊÇ £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨4£©ÁòµÄÓлúÎ£©Óë¼×È©¡¢ÂÈ»¯ÇâÒÔÎïÖʵÄÁ¿Ö®±È1:1:1·´Ó¦£¬¿É»ñµÃÒ»ÖÖɱ³æ¼ÁÖмäÌåXºÍH2O¡£
¼°XµÄºË´Å¹²ÕñÇâÆ×ÈçÏÂͼ£¬ÆäÖÐ £¨Ìî¡°¢ñ¡±»ò¡°¢ò¡±£©ÎªµÄºË´Å¹²ÕñÇâÆ×ͼ¡£Ð´³öXµÄ½á¹¹¼òʽ£º ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
Ò»¶¨Ìõ¼þÏ£¬ÔÚÌå»ýΪ3 LµÄÃܱÕÈÝÆ÷Öз´Ó¦£ºCO£¨g£©+ 2H2£¨g£©CH3OH£¨g£©´ïµ½»¯Ñ§Æ½ºâ״̬¡£
£¨1£©¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽK= £»¸ù¾ÝÏÂͼ£¬Éý¸ßζȣ¬KÖµ½« £¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©¡£
£¨2£©500¡æʱ£¬´Ó·´Ó¦¿ªÊ¼µ½´ïµ½»¯Ñ§Æ½ºâ£¬ÒÔH2µÄŨ¶È±ä»¯±íʾµÄ»¯Ñ§·´Ó¦ËÙÂÊÊÇ £¨ÓÃnB¡¢tB±íʾ£©¡£
£¨3£©ÅжϸÿÉÄæ·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄ±êÖ¾ÊÇ £¨Ìî×Öĸ£©¡£
a¡¢CO¡¢H2¡¢CH3OHµÄŨ¶È¾ù²»Ôٱ仯
b¡¢»ìºÏÆøÌåµÄÃܶȲ»Ôٸıä
c¡¢»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»Ôٸıä
d¡¢vÉú³É£¨CH3OH£©= vÏûºÄ£¨CO£©
£¨4£©300¡æʱ£¬½«ÈÝÆ÷µÄÈÝ»ýѹËõµ½ÔÀ´µÄ1/2£¬ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¶ÔƽºâÌåϵ²úÉúµÄÓ°ÏìÊÇ £¨Ìî×Öĸ£©¡£
a¡¢c£¨H2£©¼õÉÙ
b¡¢Õý·´Ó¦ËÙÂʼӿ죬Äæ·´Ó¦ËÙÂʼõÂý
c¡¢CH3OH µÄÎïÖʵÄÁ¿Ôö¼Ó
d¡¢ÖØÐÂƽºâʱc£¨H2£©/ c£¨CH3OH£©¼õС
£¨5£©¸ù¾ÝÌâÄ¿ÓйØÐÅÏ¢£¬ÇëÔÚÓÒÏÂ×ø±êͼÖбêʾ³ö¸Ã»¯Ñ§·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯£¨±êÃ÷ÐÅÏ¢£©¡£
£¨6£©ÒÔ¼×´¼¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪÔÁÏ£¬Ê¯Ä«Îªµç¼«¿É¹¹³ÉȼÁϵç³Ø¡£ÒÑÖª¸ÃȼÁϵç³ØµÄ×Ü·´Ó¦Ê½Îª£º2CH3OH +3O2+4OH- = 2CO32- + 6H2O£¬¸Ãµç³ØÖиº¼«Éϵĵ缫·´Ó¦Ê½ÊÇ£º2CH3OH¨C12e£+16OH££½ 2CO32£+ 12H2O £¬ÔòÕý¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Îª£º ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÄÜԴΣ»úÊǵ±Ç°È«ÇòÎÊÌ⣬¿ªÔ´½ÚÁ÷ÊÇÓ¦¶ÔÄÜԴΣ»úµÄÖØÒª¾Ù´ë¡£
(1)ÏÂÁÐ×ö·¨ÓÐÖúÓÚÄÜÔ´¡°¿ªÔ´½ÚÁ÷¡±µÄÊÇ________(ÌîÐòºÅ)¡£
a£®´óÁ¦·¢Õ¹Å©´åÕÓÆø£¬½«·ÏÆúµÄ½Õ¸Ñת»¯ÎªÇå½à¸ßЧµÄÄÜÔ´
b£®´óÁ¦¿ª²Éú¡¢Ê¯ÓͺÍÌìÈ»ÆøÒÔÂú×ãÈËÃÇÈÕÒæÔö³¤µÄÄÜÔ´ÐèÇó
c£®¿ª·¢Ì«ÑôÄÜ¡¢Ë®ÄÜ¡¢·çÄÜ¡¢µØÈÈÄܵÈÐÂÄÜÔ´¡¢¼õÉÙʹÓÃú¡¢Ê¯Ó͵Ȼ¯Ê¯È¼ÁÏ
d£®¼õÉÙ×ÊÔ´ÏûºÄ£¬Ôö¼Ó×ÊÔ´µÄÖظ´Ê¹Óá¢×ÊÔ´µÄÑ»·ÔÙÉú
(2)½ð¸ÕʯºÍʯī¾ùΪ̼µÄͬËØÒìÐÎÌ壬ËüÃÇȼÉÕÑõÆø²»×ãʱÉú³ÉÒ»Ñõ»¯Ì¼£¬³ä·ÖȼÉÕÉú³É¶þÑõ»¯Ì¼£¬·´Ó¦ÖзųöµÄÈÈÁ¿ÈçͼËùʾ¡£
¢ÙÔÚͨ³£×´¿öÏ£¬½ð¸ÕʯºÍʯīÖÐ________(Ìî¡°½ð¸Õʯ¡±»ò¡°Ê¯Ä«¡±)¸üÎȶ¨£¬Ê¯Ä«µÄȼÉÕÈÈΪ________¡£
¢Ú12 gʯīÔÚÒ»¶¨Á¿¿ÕÆøÖÐȼÉÕ£¬Éú³ÉÆøÌå36 g£¬¸Ã¹ý³Ì·Å³öµÄÈÈÁ¿________¡£
(3)ÒÑÖª£ºN2¡¢O2·Ö×ÓÖл¯Ñ§¼üµÄ¼üÄÜ·Ö±ðÊÇ946 kJ¡¤mol£1¡¢497 kJ¡¤mol£1¡£
N2(g)£«O2(g)=2NO(g)¡¡¦¤H£½180.0 kJ¡¤mol£1¡£
NO·Ö×ÓÖл¯Ñ§¼üµÄ¼üÄÜΪ________kJ¡¤mol£1¡£
(4)×ÛºÏÉÏÊöÓйØÐÅÏ¢£¬Çëд³öCOºÍNO·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÇëÓÃÊʵ±µÄ»¯Ñ§ÓÃÓïÌî¿Õ¡£
£¨1£©Na2CO3Ë®½âµÄÀë×Ó·½³Ìʽ£º £»
£¨2£©H2SµçÀë·½³Ìʽ£º £»
£¨3£©AlCl3Ë®½âµÄÀë×Ó·½³Ìʽ£º £»
£¨4£©ÔÚ25¡æ¡¢101 kPaÏ£¬l g¼×ÍéÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ55£®6 kJÈÈÁ¿£¬Ð´³ö±íʾ¼×ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º £»
£¨5£©¼îÐÔÇâÑõȼÁϵç³ØµÄÁ½¼«µç¼«·½³Ìʽ
¸º¼«£º £»
Õý¼«£º ¡£
£¨6£©Ð´³öNaHCO3ÈÜÒºÖеÄÀë×ÓŨ¶È¹Øϵ
c(H+)+c(Na+)= £»
c(Na+)= ¡£
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