ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿Ì¼ËáïÈÖ÷ÒªÓÃÓڲʵçÏÔÏñ¹ÜµÄÓ«ÆÁ²£Á§ºÍÌØÖÖ²£Á§µÈµÄÖÆÔ죬ͨ³£ÓÉÌìÇàʯ¿ó(Ö÷Òª³É·ÖΪSrSO4£¬»¹º¬ÓбµµÈÔÓÖÊ)ÖƱ¸£¬ÖƱ¸Á÷³ÌÈçͼËùʾ£º
ÒÑÖª£ºi£®ïÈÔÚ¡°ÂËÒº1¡±¡°ÂËÒº2¡±ÖоùÖ÷ÒªÒÔSr(HS)2¡¢Sr(OH)2ÐÎʽ´æÔÚ
ii£®SrSO4¡¢BaSO4µÄKSP·Ö±ðÊÇ3.2¡Á10-7¡¢1.0¡Á10-10
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)²½Öè¢Ù·´Ó¦£ºSrSO4+2CSrS+2CO2¡ü£¬ÈôÔÚ±ê×¼×´¿öϲúÉú3.36 LCO2£¬ÔòתÒƵç×ÓÊýΪ__________________¡£
(2)д³ö²½Öè¢ÚÖС°Ë®½þ¡±Ö÷Òª·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________¡£²½Öè¢Ú¡°¹ýÂË¡±²Ù×÷ÐèÒªµÄÖ÷Òª²£Á§ÒÇÆ÷Ϊ________________________¡£
(3)ÉÏÊöÉú²ú¹¤ÒÕµÄÓŵãÊÇÖÊÁ¿ºÃ¡¢³É±¾µÍ£¬µ«´Ó»·±£½Ç¶È¿¼ÂǸù¤ÒÕÉú²ú´æÔÚÃ÷ÏÔµÄȱµãÊÇ______________________________________________________¡£
(4)ÈôÏòº¬ÓÐSr2+¡¢Ba2+µÄ¡°ÂËÒº1¡±ÖеμÓÏ¡ÁòËᣬµ±Á½ÖÖ³Áµí¹²´æʱ£¬c(Sr2+)£ºc(Ba2+)=________________¡£
(5)д³ö¡°Ì¼»¯¡±¹ý³ÌÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_______________________________¡£
(6)ijÉú²úÆóÒµÓÃa kgµÄÌìÇàʯ¿ó(º¬SrSO4£º40£¥)ÖƱ¸£¬×îÖյõ½b kg̼ËáïȲúÆ·£¬²úÂÊΪ______________(Óú¬a¡¢bµÄʽ×Ó±íʾ)¡£
¡¾´ð°¸¡¿ 0.6NA ÉÕ±¡¢Â©¶·¡¢²£Á§°ô ²úÉúÎÛȾÎïH2SÆøÌå¡¢²úÉúÎÛȾÎïSO2ÆøÌ壨ºÏÀí¼´¿É£© 320 »ò¡¢ ¡Á100%
¡¾½âÎö¡¿(1)²½Öè¢Ù·´Ó¦£ºSrSO4+2CSrS+2CO2¡ü£¬ÈôÔÚ±ê×¼×´¿öϲúÉú3.36 LCO2£¬ÎïÖʵÄÁ¿Îª0.15mol£¬·´Ó¦ÖÐCµÄ»¯ºÏ¼Û¹²Éý¸ß8£¬Ôò²úÉú0.15mol CO2£¬×ªÒƵç×ÓÊý0.6mol£¬¹Ê´ð°¸Îª£º0.6NA£»
(2) ÌìÇàʯ¿ó¸ßÎÂÏÂÓë̼·´Ó¦Éú³ÉSrS£¬ïÈÔÚ¡°ÂËÒº1¡±ÖÐÖ÷ÒªÒÔSr(HS)2¡¢Sr(OH)2ÐÎʽ´æÔÚ£¬Òò´Ë²½Öè¢ÚÖС°Ë®½þ¡±µÄÖ÷Òª·´Ó¦»¯Ñ§·½³ÌʽΪ2SrS+2H2O= Sr(HS)2+Sr(OH)2¡£²½Öè¢Ú¡°¹ýÂË¡±²Ù×÷ÐèÒªµÄÖ÷Òª²£Á§ÒÇÆ÷ÓÐÉÕ±¡¢Â©¶·¡¢²£Á§°ô£¬¹Ê´ð°¸Îª£º2SrS+2H2O= Sr(HS)2+Sr(OH)2£»ÉÕ±¡¢Â©¶·¡¢²£Á§°ô£»
(3) Sr(HS)2ÈÜÒºÖÐͨÈë¶þÑõ»¯Ì¼»á·Å³öÁò»¯ÇâµÈÎÛȾ¿ÕÆøµÄÎïÖÊ£¬¹Ê´ð°¸Îª£º²úÉúÎÛȾÎïH2SÆøÌ壻
(4)ÈôÏòº¬ÓÐSr2+¡¢Ba2+µÄ¡°ÂËÒº1¡±ÖеμÓÏ¡ÁòËᣬµ±Á½ÖÖ³Áµí¹²´æʱ£¬c(Sr2+)£ºc(Ba2+)====3200£¬¹Ê´ð°¸Îª£º3200£»
(5)¡°Ì¼»¯¡±¹ý³ÌÖÐSr(HS)2ºÍSr(OH)2Óë¶þÑõ»¯Ì¼·´Ó¦Éú³ÉSrCO3ºÍH2S£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪSr(HS)2+Sr(OH)2+2CO2=2SrCO3¡ý+2H2S¡ü£¬¹Ê´ð°¸Îª£ºSr(HS)2+Sr(OH)2+2CO2 =2SrCO3¡ý+2H2S¡ü£»
(6)¸ù¾ÝSrÔªËØÊغ㣬SrSO4~ SrCO3£¬Ì¼ËáïȵÄÎïÖʵÄÁ¿=£¬Òò´Ë̼ËáïȵIJúÂÊ=¡Á100%=¡Á100%£¬¹Ê´ð°¸Îª£º¡Á100%¡£
¡¾ÌâÄ¿¡¿¸ù¾ÝÒªÇó»Ø´ðÏÂÁÐÓйØÎÊÌâ¡£
(1)¸ß¯Á¶ÌúÊÇÒ±Á¶ÌúµÄÖ÷Òª·½·¨£¬·¢ÉúµÄÖ÷Òª·´Ó¦Îª£ºFe2O3(s)+3CO(g)2Fe(s)+3CO2(g) H=-28.5kJ/mol£¬Ò±Á¶Ìú·´Ó¦µÄƽºâ³£Êý±í´ïʽK=____________£¬Î¶ÈÉý¸ßºó£¬K Öµ________£¨Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±£©¡£
(2)ÒÑÖª£º¢ÙFe2O3(s)+3C(ʯī) = 2Fe(s)+3CO(g) H1= +489.0kJ/mol
¢ÚFe2O3(s)+3CO(g)2Fe(s)+3CO2(g) H2= -28.5kJ/mol
¢ÛC£¨Ê¯Ä«£©+ CO2(g) = 2CO(g£© H3= akJ/mol ,
Ôòa=__________kJ/mol¡£
(3)ÔÚT¡æʱ£¬·´Ó¦Fe2O3(s)+3CO(g)2Fe(s)+3CO2(g)µÄƽºâ³£ÊýK=64£¬ÔÚ2LºãÈÝÃܱÕÈÝÆ÷ÖУ¬°´Ï±íËùʾ¼ÓÈëÎïÖÊ£¬¾¹ýÒ»¶Îʱ¼äºó´ïµ½Æ½ºâ¡£
Fe2O3 | CO | Fe | CO2 | |
ʼ̬mol | 1.0 | 1.0 | 1.0 | 1.0 |
¢ÙƽºâʱCO µÄת»¯ÂÊΪ_______¡£
¢ÚÏÂÁÐÇé¿ö±êÖ¾·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ________£¨Ìî×Öĸ£©¡£
a£®ÈÝÆ÷ÄÚÆøÌåÃܶȱ£³Ö²»±ä
b£®ÈÝÆ÷ÄÚÆøÌåѹǿ±£³Ö²»±ä
c£®COµÄÏûºÄËÙÂʺÍCO2µÄÉú³ÉËÙÂÊÏàµÈ
¡¾ÌâÄ¿¡¿CrSi¡¢Ge-GaAs¡¢ZnGeAs2¡¢¾ÛßÁ¿©¡¢Ì¼»¯¹èºÍÑõ»¯ÑÇͶ¼ÊÇÖØÒªµÄ°ëµ¼Ì廯ºÏÎï¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)»ù̬¸õÔ×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª___________£¬ÆäÖÐδ³É¶Ôµç×ÓÊýΪ____________¡£
(2) Ge-GaAsÖÐÔªËØGe¡¢Ga¡¢AsµÄµÚÒ»µçÀëÄÜ´Ó´óµ½Ð¡µÄ˳ÐòΪ_______________¡£ZnGeAs2ÖÐZn¡¢Ge¡¢AsµÄµç¸ºÐÔ´Ó´óµ½Ð¡µÄ˳ÐòΪ________________¡£
(3)¾ÛßÁ¿©µÄµ¥ÌåΪßÁ¿©()£¬¸Ã·Ö×ÓÖеªÔ×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ__________£»·Ö×ÓÖЦҼüÓë¦Ð¼üµÄÊýÄ¿Ö®±ÈΪ________________¡£
(4)̼»¯¹è¡¢¾§Ìå¹è¼°½ð¸ÕʯµÄÈÛµãÈçÏÂ±í£º
Á¢·½Ì¼»¯¹è | ¾§Ìå¹è | ½ð¸Õʯ | |
ÈÛµã/¡æ | 2973 | 1410 | 3550¡«4000 |
·ÖÎöÈÛµã±ä»¯¹æÂɼ°Æä²îÒìµÄÔÒò£º__________________________________________________¡£
(5)Ñõ»¯ÑÇ͵ÄÈÛµãΪ1235¡æ£¬Æä¹Ì̬ʱµÄµ¥¾§°ûÈçÏÂͼËùʾ¡£
¢ÙÑõ»¯ÑÇÍÊôÓÚ__________¾§Ìå¡£
¢ÚÒÑÖªCu2OµÄ¾§°û²ÎÊýa=425.8pm,ÔòÆäÃܶÈΪ__________ g¡¤cm-3(Áгö¼ÆËãʽ¼´¿É)¡£