£¨8·Ö£©¼ÆËãÒÔÏÂÁ½Ð¡Ìâʱ£¬³ý±ØÐëÓ¦ÓÃËùÓиø³öµÄÊý¾ÝÍ⣬»¹¸÷ȱÉÙÒ»¸öÊý¾Ý£¬Ö¸³ö¸ÃÊý¾ÝµÄÃû³Æ£¨·Ö±ðÒÔaºÍb±íʾ£©£¬²¢Áгö¼ÆËãʽ¡£
£¨1£©ÔÚζÈΪt¡æºÍѹǿpPaµÄÇé¿öÏ£¬19.5g AÓë11.0g BÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³É¹ÌÌåCºÍ3.00LµÄDÆøÌ壬¼ÆËãÉú³ÉµÄCµÄÖÊÁ¿(m)¡£È±ÉÙµÄÊý¾ÝÊÇ                            £¬
¼ÆËãʽΪm=                                      ¡£
£¨2£©0.48g½ðÊôþÓë10mLÑÎËá·´Ó¦£¬¼ÆËãÉú³ÉµÄH2ÔÚ±ê×¼×´¿öϵÄÌå»ý[V£¨H2£©]¡£È±ÉÙµÄÊý¾ÝÊÇ                    £¬¼ÆËãʽΪ                                   
                                                                            
                                                                           

£¨1£©t¡æʱpPaÌõ¼þÏÂÆøÌåDµÄÃܶȣ¨ ag/L£©£¬£¨1·Ö£©¼ÆËãʽΪm= mA+mB-VD?a¡££¨2·Ö£©£¨2£©È±ÉÙµÄÊý¾ÝÊÇÑÎËáµÄÎïÖʵÄÁ¿Å¨¶È£¨ b mol/L£© £¬£¨1·Ö£©¼ÆËãʽΪ   ÈôÑÎËá¹ýÁ¿£¬V(H2)="0.48g/(24g/mol)¡Á(22.4L/mol)" £¬ Èôþ¹ýÁ¿V(H2)= (0.010Lbmol/L)/2¡Á(22.4L/mol)¡££¨4·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¼ÆËãÒÔÏÂÁ½Ð¡Ìâʱ£¬³ý±ØÐëÓ¦ÓÃËùÓиø³öµÄÊý¾ÝÍ⣬»¹¸÷ȱÉÙÒ»¸öÊý¾Ý£¬Ö¸³ö¸ÃÊý¾ÝµÄÃû³Æ(·Ö±ðÒÔaºÍb±íʾ)£¬²¢Áгö¼ÆËãʽ¡£

(1)ÔÚζÈΪt ¡æºÍѹǿΪp PaµÄÇé¿öÏ£¬19.5 g AÓë11.0 g BÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³É¹ÌÌåCºÍ3.00 LµÄDÆøÌ壬¼ÆËãÉú³ÉµÄCµÄÖÊÁ¿m£¬È±ÉÙµÄÊý¾ÝÊÇ£º___________£»¼ÆËãʽΪm=_______________________________________________¡£

(2)0.48 g½ðÊôþÓë10 mLÑÎËá·´Ó¦£¬¼ÆËãÉú³ÉµÄH2ÔÚ±ê×¼×´¿öϵÄÌå»ýV(H2)£¬È±ÉÙµÄÊý¾ÝÊÇ£º_____________________£»¼ÆËãʽΪV(H2)=_______________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨8·Ö£©¼ÆËãÒÔÏÂÁ½Ð¡Ìâʱ£¬³ý±ØÐëÓ¦ÓÃËùÓиø³öµÄÊý¾ÝÍ⣬»¹¸÷ȱÉÙÒ»¸öÊý¾Ý£¬Ö¸³ö¸ÃÊý¾ÝµÄÃû³Æ£¨·Ö±ðÒÔaºÍb±íʾ£©£¬²¢Áгö¼ÆËãʽ¡£

£¨1£©ÔÚζÈΪt¡æºÍѹǿpPaµÄÇé¿öÏ£¬19.5g AÓë11.0g BÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³É¹ÌÌåCºÍ3.00LµÄDÆøÌ壬¼ÆËãÉú³ÉµÄCµÄÖÊÁ¿(m)¡£È±ÉÙµÄÊý¾ÝÊÇ                            £¬

¼ÆËãʽΪm=                                      ¡£

£¨2£©0.48g½ðÊôþÓë10mLÑÎËá·´Ó¦£¬¼ÆËãÉú³ÉµÄH2ÔÚ±ê×¼×´¿öϵÄÌå»ý[V£¨H2£©]¡£È±ÉÙµÄÊý¾ÝÊÇ                    £¬¼ÆËãʽΪ                                   

                                                                            

                                                                           

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ººÓÄÏÊ¡10-11ѧÄê¸ßÒ»µÚÒ»´ÎÔ¿¼£¨»¯Ñ§£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©¼ÆËãÒÔÏÂÁ½Ð¡Ìâʱ£¬³ý±ØÐëÓ¦ÓÃËùÓиø³öµÄÊý¾ÝÍ⣬»¹¸÷ȱÉÙÒ»¸öÊý¾Ý£¬Ö¸³ö¸ÃÊý¾ÝµÄÃû³Æ£¨·Ö±ðÒÔaºÍb±íʾ£©£¬²¢Áгö¼ÆËãʽ¡£

£¨1£©ÔÚζÈΪt¡æºÍѹǿpPaµÄÇé¿öÏ£¬19.5g AÓë11.0g BÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³É¹ÌÌåCºÍ3.00LµÄDÆøÌ壬¼ÆËãÉú³ÉµÄCµÄÖÊÁ¿(m)¡£È±ÉÙµÄÊý¾ÝÊÇ                             £¬

¼ÆËãʽΪm=                                       ¡£

£¨2£©0.48g½ðÊôþÓë10mLÑÎËá·´Ó¦£¬¼ÆËãÉú³ÉµÄH2ÔÚ±ê×¼×´¿öϵÄÌå»ý[V£¨H2£©]¡£È±ÉÙµÄÊý¾ÝÊÇ                     £¬¼ÆËãʽΪ                                   

                                                                            

                                                                           

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(8·Ö)¼ÆËãÒÔÏÂÁ½Ð¡Ìâʱ£¬³ý±ØÐëÓ¦ÓÃËùÓиø³öµÄÊý¾ÝÍ⣬»¹¸÷ȱÉÙÒ»¸öÊý¾Ý£¬Ö¸³ö¸ÃÊý¾ÝµÄÃû³Æ£¨·Ö±ðÒÔaºÍb±íʾ£©£¬²¢Áгö¼ÆËãʽ¡£

¢ÅÔÚζÈΪt¡æºÍѹÁ¦ÎªpPaµÄÇé¿öÏ£¬16.5g A Óë8.0g BÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³É¹ÌÌåCºÍ2.00LµÄDÆøÌ壬¼ÆËãÉú³ÉµÄCµÄÖÊÁ¿£¨m£©¡£            

ȱÉÙµÄÊý¾ÝÊÇ£º

¼ÆËãʽΪm=

¢Æ0.24g½ðÊôþÓë16 gÑÎËá·´Ó¦£¬¼ÆËãÉú³ÉH2µÄÖÊÁ¿¡£

ȱÉÙµÄÊý¾ÝÊÇ£º

¼ÆËãʽΪ£º 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸