º£ÑóÊÇ×ÊÔ´µÄ±¦¿â£¬Ô̲Ø×ŷḻµÄ»¯Ñ§ÔªËØ£¬ÈçÂÈ¡¢äå¡¢µâ¡¢Ã¾µÈ¡£°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¿ÉÒÔÖ¤Ã÷äå±ÈµâµÄÑõ»¯ÐÔÇ¿µÄÀë×Ó·´Ó¦·½³ÌʽΪ____________________________¡£
£¨2£©´Óº£´ø»ÒÖÐÌáÈ¡µÄµâµ¥ÖÊ£¬ÍùÍùº¬ÓÐÒ»¶¨µÄÔÓÖÊ£¬Í¨³£ÀûÓüÓÈȵķ½·¨³ýÈ¥ÔÓÖÊ£¬Õâ¸ö¹ý³ÌÖÐÖ÷ÒªÀûÓõâÐÔÖÊÖеÄ___________£¬Ò²¿ÉÒÔͨ¹ýÝÍÈ¡·ÖÒºµÄ·½·¨ÌáÈ¡µâ£¬ÊµÑéÊÒÖзÖҺʱ£¬·ÖҺ©¶·ÖеÄÉϲãÒºÌåÓ¦´Ó·ÖҺ©¶·µÄ_______£¨Ìî¡°ÉÏ¡±»ò¡°Ï¡±£©¿Úµ¹³ö¡£
£¨3£©µãȼµÄþÌõ¿ÉÒÔÔÚ¶þÑõ»¯Ì¼ÆøÌåÖмÌÐøȼÉÕ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________¡£

£¨1£©2I£­£«Br2£½I2£«2Br£­ £¨2£©µ¥ÖʵâÒ×Éý»ª£» ÉÏ £¨3£©2Mg£«CO22MgO£«C

½âÎöÊÔÌâ·ÖÎö£º£¨1£©µ¥ÖÊäåÄܰѵâÀë×ÓÑõ»¯Éú³Éµ¥Öʵ⣬¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦ÖÐÑõ»¯¼ÁµÄÑõ»¯ÐÔÇ¿ÓÚÑõ»¯²úÎïµÄÑõ»¯ÐÔ¿ÉÖª£¬äå±ÈµâÑõ»¯ÐÔÇ¿£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ2I£­£«Br2£½I2£«2Br£­¡£
£¨2£©ÓÉÓڵⵥÖÊÒ×Éý»ª£¬ËùÒÔͨ³£ÀûÓüÓÈȵķ½·¨³ýÈ¥µ¥ÖʵâÖеÄÔÓÖÊ¡£·ÖҺʱ·ÖҺ©¶·ÖеÄÉϲãÒºÌåÓ¦´Ó·ÖҺ©¶·µÄÉÏ¿Úµ¹³ö£¬Ï²ãÒºÌå´ÓС¿ÚÁ÷³ö¡£
£¨3£©Ã¾¿ÉÒÔÔÚCO2ÖÐȼÉÕÉú³ÉÑõ»¯Ã¾ºÍ̼£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Mg£«CO22MgO£«C¡£
¿¼µã£º¿¼²éÑõ»¯»¹Ô­·´Ó¦µÄÓ¦Óᢵ¥ÖʵâµÄÐÔÖÊ¡¢·ÖÒº²Ù×÷ÒÔ¼°Ã¾ÔÚCO2ÖÐȼÉÕ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ñõ»¯»¹Ô­·´Ó¦ÖÐʵ¼ÊÉÏ°üº¬Ñõ»¯ºÍ»¹Ô­Á½¸ö¹ý³Ì¡£ÏÂÃæÊÇÒ»¸ö»¹Ô­¹ý³ÌµÄ·´Ó¦Ê½£ºNO£«4H£«£«3e£­=NO¡ü£«2H2O£»KMnO4¡¢Na2CO3¡¢Cu2O¡¢Fe2(SO4)3ËÄÖÖÎïÖÊÖеÄÒ»ÖÖÎïÖÊ(¼×)ÄÜʹÉÏÊö»¹Ô­¹ý³Ì·¢Éú¡£
£¨1£©Ð´³ö²¢Åäƽ¸ÃÑõ»¯»¹Ô­·´Ó¦µÄ·½³Ìʽ£º__________________________________¡£
£¨2£©·´Ó¦ÖÐÏõËáÌåÏÖÁË______________¡¢______________µÄÐÔÖÊ¡£
£¨3£©·´Ó¦ÖÐÈô²úÉú0.2 molÆøÌ壬ÔòתÒƵĵç×ÓµÄÎïÖʵÄÁ¿ÊÇ________ mol¡£
£¨4£©Èô1 mol¼×ÓëijŨ¶ÈÏõËᷴӦʱ£¬±»»¹Ô­ÏõËáµÄÎïÖʵÄÁ¿Ôö¼Ó£¬Ô­ÒòÊÇ________________________________________________________________________________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©ÏõËáÊÇÖØÒªµÄ¹¤ÒµÔ­ÁÏ¡£
¢Ù¿ÉÓÃÂÁÖÆÈÝÆ÷´æ·ÅŨÏõËáµÄÀíÓÉÊÇ                      £»
¢ÚijÏõË᳧´¦ÀíβÆøNO2µÄ·½·¨ÊÇ£º´ß»¯¼Á´æÔÚʱÓÃH2½«NO2»¹Ô­ÎªN2¡£
ÒÑÖª£º2H2(g) + O2(g) £½ 2H2O(g)  ¡÷H£½£­483 kJ¡¤moL£­1
N2(g) + 2O2(g) £½ 2NO2(g)  ¡÷H£½£«68 kJ¡¤moL£­1
ÔòH2»¹Ô­NO2Éú³ÉË®ÕôÆøµÄÈÈ»¯Ñ§·½³ÌʽÊÇ£º
                                                           ¡£
£¨2£©Ä³Ñо¿Ð¡×éÒÔCaCl2ºÍH2ΪԭÁÏÖƱ¸+1¼ÛCaµÄ»¯ºÏÎ²úÎïÖÐÖ»Óм׺ÍÒÒÁ½ÖÖ»¯ºÏÎï¡£Ñо¿·¢ÏÖ£º»¯ºÏÎï¼×µÄ×é³ÉÖиơ¢ÂÈÔªËصÄÖÊÁ¿·ÖÊý·Ö±ðΪ52.29%¡¢46.41%£»»¯ºÏÎïÒÒµÄË®ÈÜÒºÏÔËáÐÔ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÒҵĻ¯Ñ§Ê½Îª              £¬¼×ÓëË®·´Ó¦¿ÉµÃH2£¬Æ仯ѧ·½³ÌʽÊÇ£º                            £»
¢Úд³öÓÉCaCl2ͨ¹ý»¯ºÏ·´Ó¦ÖƱ¸CaClµÄ»¯Ñ§·½³Ìʽ£º                                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Æ«¶þ¼×ëÂÓëN2O4Êdz£ÓõĻð¼ýÍƽø¼Á£¬¶þÕß·¢ÉúÈçÏ»¯Ñ§·´Ó¦£º
(CH3)2NNH2 (l)£«2N2O4 (l)£½2CO2 (g)£«3N2(g)£«4H2O(g)      £¨I£©
(1)·´Ó¦£¨I£©ÖÐÑõ»¯¼ÁÊÇ_______¡£
(2)»ð¼ý²Ðº¡Öг£ÏÖºì×ØÉ«ÆøÌ壬ԭÒòΪ£ºN2O4 (g)  2NO2 (g) £¨¢ò£© Ò»¶¨Î¶ÈÏ£¬·´Ó¦£¨¢ò£©µÄìʱäΪ¦¤H¡£ÏÖ½«1 mol N2O4 ³äÈëÒ»ºãѹÃܱÕÈÝÆ÷ÖУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ________¡£

ÈôÔÚÏàͬζÈÏ£¬ÉÏÊö·´Ó¦¸ÄÔÚÌå»ýΪ1LµÄºãÈÝÃܱÕÈÝÆ÷ÖнøÐУ¬Æ½ºâ³£Êý________£¨Ìî¡°Ôö´ó¡±¡°²»±ä¡±»ò¡°¼õС¡±£©£¬·´Ó¦3sºóNO2µÄÎïÖʵÄÁ¿Îª0.6mol£¬Ôò0¡«3sÄÚµÄƽ¾ù·´Ó¦ËÙÂÊv£¨N2O4£©£½________mol¡¤L-1¡¤s-1¡£
(3)NO2¿ÉÓð±Ë®ÎüÊÕÉú³ÉNH4NO3¡£25¡æʱ£¬½«amol NH4NO3ÈÜÓÚË®£¬ÈÜÒºÏÔËáÐÔ£¬Ô­ÒòÊÇ_____
_______________________________________________£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£Ïò¸ÃÈÜÒºµÎ¼ÓbL°±Ë®ºóÈÜÒº³ÊÖÐÐÔ£¬ÔòµÎ¼Ó°±Ë®µÄ¹ý³ÌÖеÄË®µÄµçÀëƽºâ½«______£¨Ìî¡°ÕýÏò¡±¡°²»¡±»ò¡°ÄæÏò¡±£©Òƶ¯£¬ËùµÎ¼Ó°±Ë®µÄŨ¶ÈΪ_______mol¡¤L-1¡££¨NH3¡¤H2OµÄµçÀëƽºâ³£ÊýÈ¡Kb£½2¡Á10-5 mol¡¤L-1£¬¼ÙÉèÈÜÒºµÄÌå»ýÊÇbL£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÀûÓû¯ºÏ¼ÛºÍÎïÖÊÀà±ðÍƲâÎïÖʵÄÐÔÖÊÊÇ»¯Ñ§Ñо¿µÄÖØÒªÊֶΡ£
£¨1£©´Ó»¯ºÏ¼ÛµÄ½Ç¶È¿ÉÒÔÔ¤²âÎïÖʵÄÐÔÖÊ¡£
¢ÙSO2µÄÐÔÖÊ___________£¨ÌîÐòºÅ£¬ÏÂͬ£©£»
A£®Ö»ÓÐÑõ»¯ÐÔ   B£®Ö»Óл¹Ô­ÐÔ   C£®¼ÈÓÐÑõ»¯ÐÔÓÖÓл¹Ô­ÐÔ
¢Ú½«SO2ͨÈëËáÐÔKMnO4ÈÜÒºÖУ¬ÈÜÒºÓÉ×ÏÉ«ÍÊÖÁÎÞÉ«¡£·´Ó¦½áÊøºó£¬ÁòÔªËØ´æÔÚÐÎʽºÏÀíµÄÊÇ__________¡£
A£®S2¡ª      B£®S    C£®SO32¡ª    D£®SO42¡ª
£¨2£©´ÓÎïÖÊ·ÖÀàµÄ½Ç¶È¿ÉÒÔÍƲâÎïÖʵÄÐÔÖÊ¡£ÏÖÓÐÓÉMgO¡¢Al2O3¡¢Fe2O3¡¢SiO2×é³ÉµÄij»ìºÏÎïÊÔÑù¡£
¢ÙÆäÖÐAl2O3ÊôÓÚ_______Ñõ»¯ÎMgOºÍFe2O3ÊôÓÚ_________Ñõ»¯ÎÌî¡°ËáÐÔ¡±¡¢¡°¼îÐÔ¡±»ò¡°Á½ÐÔ¡±£©£»
¢Ú½«ÊÔÑùÈÜÓÚ¹ýÁ¿µÄÑÎËáÖУ¬¹ýÂË£¬ÂËÔüµÄÖ÷Òª³É·ÖÊÇ_________£»ÔÙÏòÂËÒºÖмÓÈëNaOHÈÜÒºÖÁ¹ýÁ¿£¬¹ýÂË£¬ÂËÔüÖеÄÖ÷Òª³É·ÖÊÇ_________£»
¢ÛÈô½«¸ÃÊÔÑùÖ±½ÓÈÜÓÚ¹ýÁ¿µÄNaOHÈÜÒºÖУ¬Ëù·¢ÉúµÄ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ______________________£¨¿ÉÈÎÒâдÆäÖÐÒ»¸ö£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(4·Ö) ʵÑéÊÒ¿ÉÒÔÓøßÃÌËá¼ØºÍŨÑÎËá·´Ó¦ÖÆÈ¡ÂÈÆø£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÈçÏÂ
2KMnO4+16HCl(Ũ) = 2KCl+2MnCl2+5Cl2¡ü+8H2O
£¨1£©Ôڸ÷´Ó¦ÖУ¬»¹Ô­¼ÁÊÇ     ¡£
£¨2£©ÈôÔÚ·´Ó¦ÖÐÉú³ÉÁ˱ê¿öÏÂ2.24LÂÈÆø£¬Ôòµç×ÓתÒƵĸöÊýÊÇ    NA¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

µªÆø¼°º¬µªµÄ»¯ºÏÎïÔÚ¹úÃñ¾­¼ÃÖÐÕ¼ÓÐÖØÒªµØλ¡£ºÏ³É°±¹¤ÒµÖУ¬ºÏ³ÉËþÖÐÿ²úÉú2 mol NH3£¬·Å³ö92.4 kJÈÈÁ¿¡£
£¨1£©ÈôÆðʼʱÏòÈÝÆ÷ÄÚ·ÅÈë2 mol N2ºÍ6 mol H2£¬´ïƽºâºó·Å³öµÄÈÈÁ¿ÎªQ£¬ÔòQ_____184.8kJ£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£© ¡£ Ò»¶¨Ìõ¼þÏ£¬ÔÚÃܱպãÈݵÄÈÝÆ÷ÖУ¬Äܱíʾ·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ____________¡£
a£®3vÄæ(N2)=vÕý(H2)  ¡¡¡¡¡¡   b£®2vÕý(H2)= vÕý(NH3)
c£®»ìºÏÆøÌåÃܶȱ£³Ö²»±ä ¡¡   d£®c(N2)£ºc(H2)£ºc(NH3)=1£º3£º2
¹¤ÒµÉú²úÄòËصÄÔ­ÀíÊÇÒÔNH3ºÍCO2ΪԭÁϺϳÉÄòËØ[CO(NH2)2]£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2NH3 (g)+ CO2 (g) CO(NH2)2 (l) + H2O (l)¡£
£¨2£©ÔÚÒ»¶¨Î¶ȺÍѹǿÏ£¬ÈôÔ­ÁÏÆøÖеÄNH3ºÍCO2µÄÎïÖʵÄÁ¿Ö®±È£¨°±Ì¼±È£©£¬ÓÒͼÊÇ°±Ì¼±È£¨x£©ÓëCO2ƽºâת»¯ÂÊ£¨¦Á£©µÄ¹Øϵ¡£¦ÁËæ×ÅxÔö´ó¶øÔö´óµÄÔ­ÒòÊÇ___________¡£
£¨3£©Í¼ÖеÄBµã´¦£¬NH3µÄƽºâת»¯ÂÊΪ_______¡£

ÒÑÖª£º3Cl2+2NH3¡úN2+6HCl     ¨D¨D¢Ù   3Cl2+8NH3¡úN2+6NH4Cl   ¨D¨D¢Ú
£¨4£©Íê³É²¢ÅäƽÏÂÁÐÑõ»¯»¹Ô­·´Ó¦·½³Ìʽ£¬ÔÙ±ê³öµç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º
12Cl2+15NH3¡ú                      ¨D¨D¢Û
£¨5£©·´Ó¦¢ÛÖеĻ¹Ô­¼ÁÊÇ                £¬»¹Ô­²úÎïÊÇ                      ¡£
£¨6£©Èô°´¢Û·´Ó¦ºó²úÉúÆøÌå9.408L£¨±ê×¼×´¿ö£©£¬Ôò±»Ñõ»¯µÄÆøÌåµÄÎïÖʵÄÁ¿ÊÇ        mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

I£¨1£©ÔÚµí·Ûµâ»¯¼ØÈÜÒºÖÐͨÈëÉÙÁ¿ÂÈÆø£¬Á¢¼´»á¿´µ½ÈÜÒº±äÀ¶É«£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ                                     ¡£
£¨2£©ÔÚµâºÍµí·ÛÐγɵÄÀ¶É«ÈÜÒºÖÐͨÈëSO2ÆøÌ壬·¢ÏÖÀ¶É«Öð½¥Ïûʧ£¬·´Ó¦µÄÀë×Ó·½³ÌÊÇ             ¡£
£¨3£©¶Ô±È£¨1£©ºÍ£¨2£©ÊµÑéËùµÃµÄ½á¹û£¬½«Cl¡¢ISO2°´»¹Ô­ÐÔÓÉÇ¿µ½Èõ˳ÐòÅÅÁÐΪ             ¡£
II £¨4£© ³ýÈ¥Ìú·ÛÖлìÓÐÂÁ·ÛµÄÊÔ¼ÁÊÇ                    £¬Àë×Ó·½³ÌʽΪ                                         
£¨5£© 1mol¹ýÑõ»¯ÄÆÓë2mol̼ËáÇâÄƹÌÌå»ìºÏºó£¬ÔÚÃܱÕÈÝÆ÷ÖмÓÈȳä·Ö·´Ó¦£¬ÅųöÆøÌåÎïÖʺóÀäÈ´£¬²ÐÁôµÄ¹ÌÌåÎïÖÊÊÇ                   

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖªÖÆÈ¡ÂÈÆøÒ²¿ÉÓÃŨÑÎËáÓë¸ßÃÌËá¼ØΪԭÁÏ£¬Æ仯ѧ·½³ÌʽΪ
2KMnO4+16HCl(Ũ)£½2MnCl2+2KCl+5Cl2¡ü+8H2O¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃË«ÏßÇÅ·¨±ê³ö¸Ã»¯Ñ§Ê½µç×ÓתÒƵķ½ÏòÓëÊýÄ¿£º
2KMnO4+16HCl(Ũ)£½2MnCl2+2KCl+5Cl2¡ü+8H2O¡£
£¨2£©·´Ó¦Öб»»¹Ô­µÄÔªËØΪ           £¨Ð´Ãû³Æ£©£»±ê×¼×´¿öϵ±Éú³É112 LÂÈÆøʱ£¬·´Ó¦ÖÐתÒƵĵç×ÓÊýĿΪ           ¡£
£¨3£©ÈôÓÐ4 molHCl±»Ñõ»¯£¬Ôò¿ÉÉú³É±ê×¼×´¿öϵÄÂÈÆø                 L¡£
£¨4£©ÈôÓÐ1.58g¸ßÃÌËá¼ØºÍ100mL10moL/LŨÑÎËá³ä·Ö·´Ó¦£¨²»¿¼ÂÇÑÎËá»Ó·¢£¬ºöÂÔÈÜÒºÌå»ýµÄÇ°ºó±ä»¯£©£¬·´Ó¦ÖÁÖÕµãʱ¸ßÃÌËá¼Ø×ÏÉ«ÍêÈ«ÍÊÈ¥£¬Ôò±»Ñõ»¯µÄHClµÄÎïÖʵÄÁ¿
Ϊ            mol¡£½«·´Ó¦ºóµÄÈÜҺȡ³ö10mL£¬¼ÓÈë×ãÁ¿µÄÏõËáÒøÈÜÒº£¬¿ÉµÃµ½³ÁµíµÄÎïÖʵÄÁ¿Îª           mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸