ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿²ÝËáÊÇÒ»ÖÖËᣬ²ÝËᾧÌ壨H2C2O42H2O£©Ò×ÈÜÓÚË®£¬ÈÛµã½ÏµÍ£¬¼ÓÈÈ»áÈÛ»¯¡¢Æø»¯ºÍ·Ö½â¡£²ÝËᣨH2C2O4£©ÓëÇâÑõ»¯¸ÆµÄ·´Ó¦£ºH2C2O4+Ca(OH)2=CaC2O4¡ý(°×É«)+2H2O¡£

£¨·ÖÎöÌÖÂÛ£©

ʵÑéÊÒ¿ÉÓüÓÈȲÝËᾧÌå·Ö½âµÄ·½·¨»ñµÃCO

£¨1£©ÏȼÓÈȲÝËᾧÌåÉú³ÉCO¡¢CO2ºÍH2O£¬Æ仯ѧ·½³ÌʽÊÇ_____£»

£¨2£©×îºóÓÃÈçͼװÖÃÊÕ¼¯CO£¬ÆøÌåÓ¦´Ó_____¶Ë½øÈ루ѡÌî¡°a¡±»ò¡°b¡±£©¡£

£¨ÊµÑ鷴˼£©

£¨3£©¼×ÈÏΪÈçͼÖÐÊԹܿÚÓ¦ÂÔÏòÏÂÇãб£¬ÀÏʦ¡¢Í¬Ñ§ÌÖÂÛºóÒ»ÖÂÈÏΪװÖÃÊÇÕýÈ·µÄ£¬ÀíÓÉÊÇ_____¡£

£¨4£©ÊµÑéÖй۲쵽³ÎÇåʯ»ÒË®±ä»ë×Ç£¬ÒÒÈÏΪһ¶¨ÊÇÓɲÝËᾧÌåÊÜÈÈ·Ö½â²úÉúµÄCO2Ëùµ¼Ö£¬±ûÈÏΪÒҵĽáÂÛ²»ÑÏÃÜ£¬ÀíÓÉÊÇ_____¡£

£¨Éú»îÁ´½Ó£©²¤²ËÓªÑø·á¸»£¬µ«²¤²Ë¶¹¸¯Í¬Ê³£¬Ò׵ýáʯ£¨Ö÷Òª³É·Ö²ÝËá¸Æ¾§Ì壩£¬Ð¡×éͬѧ¶Ô²ÝËá¸Æ¾§ÌåµÄÐÔÖʼ°×é³É²úÉúÁËÐËȤ¡£

£¨ÐÔÖÊ̽¾¿£©°´ÏÂͼװÖý«²ÝËá¸Æ¾§Ì壨CaC2O4¡¤xH2O£©ÑùÆ·¸ßμÓÈÈ£¬Ê¹ÆäÍêÈ«·Ö½â²¢¼ìÑé²úÉúµÄÆøÌå¡£

£¨ÎÊÌâÌÖÂÛ£©

£¨5£©BÖй۲쵽_____ÏÖÏó£¬ËµÃ÷·´Ó¦Éú³ÉÁËË®£»

£¨6£©C¡¢GÖгÎÇåʯ»ÒË®¾ù±ä»ë×Ç£¬ËµÃ÷·´Ó¦»¹Éú³É_____ºÍ_____ÆøÌå¡£

£¨7£©ÓÐͬѧÈÏΪÉÏÊö½áÂÛ²»ÑϽ÷£¬²¢²»ÄܵóöÓÐÒ»Ñõ»¯Ì¼Éú³ÉµÄ½áÂÛ¡£Çë˵Ã÷ÀíÓÉ_____¡£

£¨×é³ÉÈ·¶¨£©

£¨8£©ÀûÓÃÈÈ·ÖÎöÒǶԲÝËá¸Æ¾§Ì壨CaC2O4¡¤xH2O£©½øÐÐÈȷֽ⣬»ñµÃÏà¹ØÊý¾Ý£¬»æÖƳɹÌÌåÖÊÁ¿¡ª·Ö½âζȵĹØϵÈçͼ¡£

¢ÙζÈΪ200¡æ×óÓÒʱ£¬¾§ÌåÈ«²¿Ê§È¥½á¾§Ë®£¬¾§ÌåÖнᾧˮµÄÖÊÁ¿Îª_____g¡£

¢Ú¼ÆËãCaC2O4¡¤xH2OÖеÄx£¨CaC2O4µÄÏà¶Ô·Ö×ÓÖÊÁ¿£º128£©£¬x=_____¡£

¢Û800¡æÒÔÉÏʱ£¬¾­¼ìÑé²ÐÁô¹ÌÌåΪÑõ»¯ÎͼÖÐmµÄÖµÊÇ_____¡£

¢Üд³ö¹ÌÌåÖÊÁ¿ÓÉ12.8g±äΪ10.0gʱµÄ»¯Ñ§·½³Ìʽ_____¡£

¡¾´ð°¸¡¿H2C2O42H2O CO¡ü+CO2¡ü+3H2O¡ü b ²ÝËáÈÛµã½ÏµÍ£¬¼ÓÈÈ·Ö½âʱ£¬ÏÈÈÛ»¯ºó·Ö½âËùÒÔÊԹܿÚÂÔÏòÉÏÇãб ¿ÉÄÜÊDzÝËáÆøÌå½øÈë³ÎÇåʯ»ÒË®£¬Éú³É²ÝËá¸Æ°×É«³Áµí ÎÞË®ÁòËáÍ­±äÀ¶ CO CO2 ¿ÉÄÜÇâÑõ»¯ÄÆûÓн«¶þÑõ»¯Ì¼ÍêÈ«ÎüÊÕ 1.8 1 5.6 CaC2O4CaCO3+CO¡ü

¡¾½âÎö¡¿

·ÖÎöÌÖÂÛ£º

£¨1£©¸ù¾Ý¡°¼ÓÈȲÝËᾧÌåÉú³ÉCO¡¢CO2ºÍH2O¡±£¬¹ÊÆ仯ѧ·½³ÌʽÊÇH2C2O42H2O CO¡ü+CO2¡ü+3H2O¡ü£»

£¨2£©CO¼«ÄÑÈÜÓÚË®£¬ÃܶȱÈˮС£¬ÓÃͼ1×°ÖÃÅÅË®·¨ÊÕ¼¯CO£¬ÆøÌåÓ¦´Ób¶Ë½øÈ룻

ʵÑ鷴˼£º

£¨3£©¼×ÈÏΪͼ2ÖÐÊԹܿÚÓ¦ÂÔÏòÏÂÇãб£¬Ô­ÒòÊDzÝËáÈÛµã½ÏµÍ£¬¼ÓÈÈ·Ö½âʱ£¬ÏÈÈÛ»¯ºó·Ö½â£¬ËùÒÔÊԹܿÚÂÔÏòÉÏÇãб£»

£¨4£©ÊµÑéÖй۲쵽³ÎÇåʯ»ÒË®±ä»ë×Ç£¬ÒÒÈÏΪһ¶¨ÊÇÓɲÝËᾧÌåÊÜÈÈ·Ö½â²úÉúµÄCO2Ëùµ¼Ö£¬±ûÈÏΪÒҵĽáÂÛ²»ÑÏÃÜ£¬ÀíÓÉÊDzÝËá¼ÓÈÈÆø»¯²úÉú²ÝËᣨH2C2O4£©ÆøÌåÓëÇâÑõ»¯¸Æ·´Ó¦Éú³ÉCaC2O4³Áµí£»

£¨5£©ÁòËáÍ­·ÛÄ©ÓöË®±äÀ¶É«£¬½«²ÝËá¸Æ¾§Ì壨CaC2O4xH2O£©ÑùÆ·¸ßμÓÈÈ£¬BÖй۲쵽ÁòËáÍ­·ÛÄ©±äÀ¶ÏÖÏó£¬ËµÃ÷·´Ó¦Éú³ÉÁËË®£»

£¨6£©CÖгÎÇåʯ»ÒË®¾ù±ä»ë×Ç£¬ËµÃ÷·´Ó¦»¹Éú³ÉCO2ÆøÌ壻GÖгÎÇåʯ»ÒË®¾ù±ä»ë×Ç£¬ËµÃ÷·´Ó¦»¹Éú³ÉCOÆøÌ壬ÒòΪһÑõ»¯Ì¼ÓëÑõ»¯Í­·´Ó¦»áÉú³É¶þÑõ»¯Ì¼£»

£¨7£©¸ù¾ÝÉÏÊöʵÑé·ÖÎö¿ÉÖª²ÝËá¼ÓÈÈÓжþÑõ»¯Ì¼Éú³É£¬Èô¾­¹ýÇâÑõ»¯ÄÆÈÜÒº£¬²»ÄÜʹ¶þÑõ»¯Ì¼ÍêÈ«ÎüÊÕ£¬GÖгÎÇåʯ»ÒˮҲ»á±ä»ë×Ç£¬¿ÉÔÚD ºÍ E ×°Ö®¼äÔÙÔö¼ÓÒ»¸öÊ¢³ÎÇåʯ»ÒË®µÄÏ´ÆøÆ¿£¬Èô¸ÃÏ´ÆøÆ¿ÖгÎÇåʯ»ÒË®²»±ä»ë×Ç£¬ÈôGÖгÎÇåʯ»ÒË®±ä»ë×ÇÔò˵Ã÷²ÝËá·Ö½â²úÉúÁ˶þÑõ»¯Ì¼¡£ÓÐͬѧÈÏΪÉÏÊö½áÂÛ²»ÑϽ÷£¬²¢²»ÄܵóöÓÐÒ»Ñõ»¯Ì¼Éú³ÉµÄ½áÂÛ¡£ÀíÓÉ¿ÉÄÜÇâÑõ»¯ÄÆûÓн«¶þÑõ»¯Ì¼ÍêÈ«ÎüÊÕ¡£

£¨8£©¢ÙζÈΪ200¡æ×óÓÒʱ£¬¾§ÌåÈ«²¿Ê§È¥½á¾§Ë®£¬¾§ÌåÖнᾧˮµÄÖÊÁ¿Îª14.6g-12.8g=1.8g£»

¢Ú¸ù¾ÝͼÏñ¿ÉÖª0¡«200¡æÊǾ§Ìåʧȥ½á¾§Ë®µÄ¹ý³Ì£¬14.6¿ËCaC2O4xH2OʧȥˮºóÉú³É12.8¿ËCaC2O4£¬

£¬½âµÃx¨T1£»

¢Û800¡æÒÔÉÏʱ£¬¾­¼ìÑé²ÐÁô¹ÌÌåΪÑõ»¯Î¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ¿ÉÖª£¬¸Ã¹ÌÌåÑõ»¯ÎïÊÇÑõ»¯¸Æ£¬ÉèÉú³ÉÑõ»¯¸ÆµÄÖÊÁ¿Îªy¡£

£¬½âµÃx=5.6g£¬¼´Í¼ÖÐmµÄÖµÊÇ5.6g£»

¢Ü¸ù¾ÝͼÏñ¿ÉÖª800¡æÒÔÉÏʱ£¬·´Ó¦Éú³ÉÁËÑõ»¯¸Æ£¬Ñõ»¯¸Æ¿ÉÄÜÊÇÓÉ̼Ëá¸Æ·Ö½âËùµÃ£¬400¡æʱ²ÝËá¸Æ·Ö½â£¬¿ÉÄÜÉú³ÉÁË̼Ëá¸Æ£¬ÈôÉú³É̼Ëá¸Æ£¬Ôò10gµÄ¹ÌÌåΪ̼Ëá¸Æ£¬10g̼Ëá¸Æ¹ÌÌåÖУ¬¸ÆÔªËصÄÖÊÁ¿= £¬Ì¼ÔªËصÄÖÊÁ¿=£¬ÑõÔªËصÄÖÊÁ¿=10g-4g-1.2g=4.8g£»12.8g²ÝËá¸ÆÖиÆÔªËصÄÖÊÁ¿=£»Ì¼ÔªËصÄÖÊÁ¿=£¬ÑõÔªËصÄÖÊÁ¿=12.8g-2.4g-4g=6.4g£¬¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ¿ÉÖª£¬²ÝËá¸Æ·Ö½âºóµÄ²úÎïÖгýÁË̼Ëá¸ÆÍ⣬ÆäËûÎïÖÊÖÐ̼ԪËصÄÖÊÁ¿=2.4g-1.2g=1.2g£»ÑõÔªËصÄÖÊÁ¿=6.4g-4.8g=1.6g£¬ÆäËûÎïÖʵÄÖÊÁ¿=1.2g+1.6g=2.8g£¬ÓëÌâÉèÖÐÉú³ÉÆäËûÎïÖʵÄÖÊÁ¿=12.8g-10g=2.8gÏàͬ¡£¼ÙÉè³ÉÁ¢£¬Ì¼Ô­×ÓÓëÑõÔ­×ӵĸöÊý±È=£¬ËùÒÔ²ÝËá¸Æ·Ö½âºóµÄ²úÎïÖгýÁË̼Ëá¸ÆÍ⣬»¹ÓÐCO£»·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£ºCaC2O4CaCO3+CO¡ü¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿°±ÓÃÓÚÖÆÔ쵪·Ê£¨ÄòËØ¡¢Ì¼ï§µÈ£©¡¢¸´ºÏ·ÊÁÏ¡¢ÏõËá¡¢´¿¼îµÈ£¬¹ã·ºÓ¦ÓÃÓÚ»¯¹¤¡¢ÇṤ¡¢»¯·ÊÖÆÒ©¡¢ºÏ³ÉÏËάµÈÁìÓò¡£

¢ñ£®¹¤ÒµÖÆ°±ÊÇÒÔ¹þ²®·¨Í¨¹ýµªÆøºÍÇâÆøÔÚ¸ßθßѹºÍ´ß»¯¼ÁµÄ×÷ÓÃÏ»¯ºÏÉú³ÉµÄ£¬ÇëËã³öµªÆøÓëÇâÆøÇ¡ºÃÍêÈ«·´Ó¦Ê±µÄÖÊÁ¿±È________£»

¢ò£®Ä¿Ç°ÓÐÒ»ÖÖ¡°È˹¤¹Ìµª¡±µÄз½·¨·´Ó¦¹ý³ÌʾÒâͼÈçÏ£º

£¨1£©¸ù¾Ý·´Ó¦µÄ΢¹ÛʾÒâͼд³ö»¯Ñ§·½³Ìʽ£º_____________£»

£¨2£©°±Æø½øÈë´óÆøºóÓëÓêË®×÷Ó㬿ÉÄÜÐγɡ°¼îÓꡱ¡£Ð´³ö¼îÓêÖмîµÄ»¯Ñ§Ê½£¬²¢±ê³öÆäÖеªÔªËصĻ¯ºÏ¼Û£º_________£»

¢ó£®°±Êǵª·ÊµÄÖØÒªÔ­ÁÏ¡£Ä³»¯·Ê³§Éú²úï§Ì¬µª·Ê(NH4)2SO4µÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©²Ù×÷¢ñÖеIJ£Á§°ô×÷ÓÃÊÇ_________£»

£¨2£©¼ÓÈëµÄXӦΪ_____£¨Ìî×Öĸ£©

A H2SO4 B CaSO4 C SO2

¢ô£®°±Ò²Êǹ¤ÒµË÷ά¶û·¨ÖÆ´¿¼îµÄÖØÒªÔ­ÁÏ¡£

ijÐËȤС×é²ÉÓÃÏÂÁÐ×°ÖÃÄ£ÄâË÷ά¶û·¨ÖƱ¸Ì¼ËáÇâÄÆ£¬ÊµÑé²Ù×÷ÈçÏ£º

a£®¹Ø±ÕK1£¬´ò¿ªK2ͨÈëNH3£¬µ÷½ÚÆøÁ÷ËÙÂÊ£¬´ýÆäÎȶ¨ºó£¬´ò¿ªK1ͨÈëCO2£»

b£®´ýÈý¾±ÉÕÆ¿ÄÚ³öÏֽ϶àNaHCO3¹ÌÌåʱ£¬¹Ø±ÕK2ֹͣͨNH3£¬Ò»¶Îʱ¼äºó£¬¹Ø±ÕK1ֹͣͨCO2£»

c£®½«Èý¾±ÉÕÆ¿Äڵķ´Ó¦»ìºÏÎï¹ýÂË¡¢Ï´µÓ¡¢µÍθÉÔ²¢½«ËùµÃ¹ÌÌåÖÃÓÚ³¨¿ÚÈÝÆ÷ÖмÓÈÈ£¬¼Ç¼ʣÓà¹ÌÌåÖÊÁ¿¡£

¼ÓÈÈʱ¼ä/min

t0

t1

t2

t3

t4

t5

Ê£Óà¹ÌÌåÖÊÁ¿/g

δ¼Ç¼

30.6

27.4

23.8

21.2

21.2

£¨×ÊÁÏ£©³£Î³£Ñ¹Ï£¬1Ìå»ýˮԼÄÜÈܽâ700Ìå»ý°±Æø£¬1Ìå»ýˮԼÄÜÈܽâ1Ìå»ý¶þÑõ»¯Ì¼¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Èý¾±ÉÕÆ¿ÄÚÉú³ÉµÄÁíÒ»²úÎïÊÇÒ»ÖÖï§Ì¬µª·Ê£¬Ð´³öÆäÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º__________£»¼ÓÈȹýÂ˵õ½µÄNaHCO3£¬·¢Éú·´Ó¦£º2NaHCO3 Na2CO3 + CO2¡ü+ H2O¡£

£¨2£©Èý¾±ÉÕÆ¿ÉÏÁ¬½ÓµÄ³¤¾±Â©¶·µÄÖ÷Òª×÷ÓÃÊÇ__________£»

£¨3£©´ò¿ªK2ͨÈëNH3Ò»¶Îʱ¼äºó£¬²Å´ò¿ªK1ͨÈëCO2£¬Ô­ÒòÊÇ__________£»

£¨4£©±¥ºÍ̼ËáÇâÄÆÈÜÒºµÄ×÷ÓÃÊÇ______________£»

£¨5£©¸ù¾ÝʵÑé¼Ç¼£¬¼ÆËãt2 minʱNaHCO3¹ÌÌåµÄ·Ö½âÂÊ£¨ÒÑ·Ö½âµÄNaHCO3ÖÊÁ¿Óë¼ÓÈÈÇ°Ô­NaHCO3ÖÊÁ¿µÄ±ÈÖµ£©£¬Çëд³ö¼ÆËã¹ý³Ì__________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø