Ϊ²â¶¨Ä³Óлú»¯ºÏÎïAµÄ½á¹¹£¬½øÐÐÈçÏÂʵÑ飮
ʵ Ñé ²½ Öè½â ÊÍ »ò Êµ Ñé ½á ÂÛ
£¨1£©¸ÃÓлú»¯ºÏÎïAÓÃÖÊÆ×ÒDzⶨʱ£¬²âµÃ·ÖÀë³öµÄ×î´óµÄËéƬµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª90£»ÊÔͨ¹ý¼ÆËãÌî¿Õ£º
£¨1£©AµÄĦ¶ûÖÊÁ¿Îª£º
 
£¨2£©½«´Ë9.0gAÔÚ×ãÁ¿´¿O2³ä·ÖȼÉÕ£¬²¢Ê¹Æä²úÎïÒÀ´Î»º»ºÍ¨¹ýŨÁòËá¡¢¼îʯ»Ò£¬·¢ÏÖÁ½Õß·Ö±ðÔöÖØ5.4gºÍ13.2g£»£¨2£©AµÄ·Ö×ÓʽΪ£º
 
£¨3£©ÁíÈ¡A 9.0g£¬¸ú×ãÁ¿µÄNaHCO3·ÛÄ©·´Ó¦£¬Éú³É2.24LCO2£¨±ê×¼×´¿ö£©£¬ÈôÓë×ãÁ¿½ðÊôÄÆ·´Ó¦ÔòÉú³É2.24LH2£¨±ê×¼×´¿ö£©£®£¨3£©¿ÉÍƳöAÖк¬ÓеĹÙÄÜÍÅ£º£¨Ð´Ãû³Æ£©
 
¡¢
 
£¨4£©AµÄºË´Å¹²ÕñÇâÆ×Èçͼ£º£¨4£©AÖк¬ÓÐ
 
ÖÖÇâÔ­×Ó
£¨5£©×ÛÉÏËùÊö£¬AµÄ½á¹¹¼òʽ
 
¿¼µã£ºÓлúÎïʵÑéʽºÍ·Ö×ÓʽµÄÈ·¶¨,ÓлúÎïµÄ½á¹¹Ê½
רÌ⣺ÓлúÎïµÄ»¯Ñ§ÐÔÖʼ°ÍƶÏ,Ìþ¼°ÆäÑÜÉúÎïµÄȼÉÕ¹æÂÉ
·ÖÎö£º£¨1£©¸ÃÓлú»¯ºÏÎïAÓÃÖÊÆ×ÒDzⶨʱ£¬²âµÃ·ÖÀë³öµÄ×î´óµÄËéƬµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª90£¬¼´Îª¸ÃÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿£»
£¨2£©¼ÆËã³öʵÑéʽ£¬¸ù¾ÝÏà¶Ô·Ö×ÓÖÊÁ¿¼ÆËã·Ö×Óʽ£»
£¨3£©9.0g¸ÃÓлúÎïµÄÎïÖʵÄÁ¿Îª£º
9.0g
90g/mol
=0.1mol£¬2.24LCO2µÄÎïÖʵÄÁ¿Îª£º
2.24L
22.4L/mol
=0.1mol£¬ÓÉ´Ë¿ÉÖª¸ÃÓлúÎï·Ö×ÓÖк¬1¸öôÈ»ù£»ºÍ×ãÁ¿½ðÊôÄÆ·´Ó¦ÔòÉú³É2.24LH2£¨±ê×¼×´¿ö£©£¬ÎïÖʵÄÁ¿Îª£º
2.24L
22.4L/mol
=0.1mol£¬ËµÃ÷¸ÃÓлúÎï·Ö×ÓÖл¹º¬ÓÐ1¸öôÇ»ù£»
£¨4£©ºË´Å¹²ÕñÇâÆ×ÄܶÔÓлúÎï·Ö×ÓÖв»Í¬»¯Ñ§»·¾³µÄÇâÔ­×Ó¸ø³ö²»Í¬µÄ·åÖµ£¨Ðźţ©£¬¸ù¾Ý·åÖµ£¨Ðźţ©¿ÉÒÔÈ·¶¨·Ö×ÓÖÐÇâÔ­×ÓµÄÖÖÀࣻ
£¨5£©¸ù¾Ý·Ö×ÓʽºÍËùº¬¹ÙÄÜÍż°ÇâÔ­×ÓÖÖÀàд³ö½á¹¹¼òʽ£®
½â´ð£º ½â£º£¨1£©¸ÃÓлú»¯ºÏÎïAÓÃÖÊÆ×ÒDzⶨʱ£¬²âµÃ·ÖÀë³öµÄ×î´óµÄËéƬµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª90£¬¼´Îª¸ÃÓлúÎïµÄÏà¶Ô·Ö×ÓÖÊÁ¿£¬¹Ê´ð°¸Îª£º90g/mol£»
£¨2£©Í¨¹ýŨÁòËá¡¢¼îʯ»Ò£¬·¢ÏÖÁ½Õß·Ö±ðÔöÖØ5.4gºÍ13.2g£¬¼´ÎªÉú³ÉË®ºÍ¶þÑõ»¯Ì¼µÄÖÊÁ¿£¬n£¨CO2£©=
13.2g
44g/mol
=0.3mol£¬n£¨H2O£©=
5.4g
18g/mol
=0.3mol£¬
ÔòÓлúÎïÖк¬n£¨O£©=
9.0g-0.3mo¡Á12g/mol-0.3mol¡Á2¡Á1g/mol
16g/mol
=0.3mol£¬ÔòÓлúÎïÖÐN£¨C£©£ºN£¨H£©£ºN£¨O£©=0.3mol£º0.6mol£º0.3mol=1£º2£º1£¬¹ÊÓлúÎïAµÄʵÑéʽΪCH2O£¬Éè·Ö×ÓʽΪCnH2nOn£¬ÔòÓÐ12n+2n+16n=90£¬½âµÃn=3£¬¹Ê·Ö×ÓʽΪ£ºC3H6O3£¬
¹Ê´ð°¸Îª£ºC3H6O3£»
 £¨3£©9.0g¸ÃÓлúÎïµÄÎïÖʵÄÁ¿Îª£º
9.0g
90g/mol
=0.1mol£¬2.24LCO2µÄÎïÖʵÄÁ¿Îª£º
2.24L
22.4L/mol
=0.1mol£¬ÓÉ´Ë¿ÉÖª¸ÃÓлúÎï·Ö×ÓÖк¬1¸öôÈ»ù£»ºÍ×ãÁ¿½ðÊôÄÆ·´Ó¦ÔòÉú³É2.24LH2£¨±ê×¼×´¿ö£©£¬ÎïÖʵÄÁ¿Îª£º
2.24L
22.4L/mol
=0.1mol£¬ËµÃ÷¸ÃÓлúÎï·Ö×ÓÖл¹º¬ÓÐ1¸öôÇ»ù£¬¹Ê´ð°¸Îª£ºôÈ»ù£»ôÇ»ù£»
£¨4£©ºË´Å¹²ÕñÇâÆ×ÄܶÔÓлúÎï·Ö×ÓÖв»Í¬»¯Ñ§»·¾³µÄÇâÔ­×Ó¸ø³ö²»Í¬µÄ·åÖµ£¨Ðźţ©£¬¸ù¾Ý·åÖµ£¨Ðźţ©¿ÉÒÔÈ·¶¨·Ö×ÓÖÐÇâÔ­×ÓµÄÖÖÀ࣬¸ÃÓлúÎïºË´Å¹²ÕñÇâÆ×ÓÐ4Ìõ²»Í¬µÄ·åÖµ£¬¹ÊA·Ö×ÓÖÐÓÐ4ÖÖ²»Í¬»¯Ñ§»·¾³µÄÇâÔ­×Ó£¬¹Ê´ð°¸Îª£º4£»
£¨5£©¸ù¾Ý·Ö×ÓʽC3H6O3ºÍ¹ÙÄÜÍźÍÇâÔ­×ÓÖÖÀà¿Éд³ö½á¹¹¼òʽΪ£ºCH3-CH£¨OH£©-COOH£¬¹Ê´ð°¸Îª£ºCH3-CH£¨OH£©-COOH£®
µãÆÀ£º±¾Ì⿼²éÓлúÎïµÄÍƶϣ¬ÌâÄ¿ÄѶȲ»´ó£¬×¢ÒâÓлúÎïʵÑéʽ·¨¼ÆËã·Ö×ÓʽºÍºË´Å¹²ÕñÇâÆ×µÄÓ¦Óã®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÐðÊö´íÎóµÄÊÇ£¨¡¡¡¡£©
A¡¢×ÜÖÊÁ¿Ò»¶¨µÄÆÏÌÑÌÇÓëÒÒËáµÄ»ìºÏÎÎÞÂÛÒÔºÎÖÖ±ÈÀý»ìºÏ£¬ÍêȫȼÉյĺÄÑõÁ¿²»±ä
B¡¢×ÜÎïÖʵÄÁ¿Ò»¶¨µÄÒÒ´¼ÓëÒÒÏ©µÄ»ìºÏÎÎÞÂÛÒÔºÎÖÖÎïÖʵÄÁ¿±ÈÀý»ìºÏ£¬ÍêȫȼÉյĺÄÑõÁ¿²»±ä
C¡¢×ÜÖÊÁ¿Ò»¶¨µÄµí·ÛºÍÏËάËØ»ìºÏÎÎÞÂÛÒÔºÎÖÖ±ÈÀý»ìºÏ£¬ÍêÈ«Ë®½âºÄË®Á¿²»±ä
D¡¢×ÜÖÊÁ¿Ò»¶¨µÄÓÍÖ¬£¬ÎÞÂÛÒÔºÎÖÖ±ÈÀý»ìºÏ£¬ÍêÈ«Ë®½âºóÉú³É¸ÊÓ͵ÄÁ¿²»±ä

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ãð»ðʱÍù¿ÉȼÎïÉÏÅç´óÁ¿Ë®µÄÖ÷Òª×÷ÓÃÊÇ£¨¡¡¡¡£©
A¡¢¸ô¾ø¿ÕÆø
B¡¢½µµÍζÈ
C¡¢Ê¹Ë®·Ö½â
D¡¢Ê¹Ë®±ä³ÉË®ÕôÆøÒÔ»Ó·¢

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª¢Ù £¨2£¬4-¶þ¼×»ù-1-¸ýÏ©£© ¢Ú£¬¢ÛCH3£¨CH2£©2CH=CH-CH=CH£¨CH2£©8CH3£®»Ø´ðÏÂÁÐÎÊÌ⣺£¨1£©µÄ·Ö×Óʽ
 
¢ÚµÄÃû³Æ
 

»¥ÎªÍ¬ÏµÎïµÄÊÇ
 
£¨ÌîÐòºÅ£©¼Ó¾Û·´Ó¦²úÎïÖл¹Óв»±¥ºÍ̼ԭ×ÓµÄÊÇ
 

£¨2£©Ï©ÌþAÊÇ¢ÚµÄÒ»ÖÖͬ·ÖÒì¹¹Ì壬AÓëÇâÆø¼Ó³ÉµÃµ½ÍéÌþB£¬BµÄÒ»ÂÈ´úÎïÖ»ÓÐ2ÖÖ£¬ÔòBµÄ½á¹¹¼òʽΪ£º
 

£¨3£©Óлú·Ö×ÓM£»ÓлúÎïÓëM»¥ÎªÍ¬·ÖÒì¹¹Ì壬ÄÜÓë̼ËáÇâÄÆÈÜÒº·´Ó¦·Å³öCO2£¬ÓлúÎïÖÐX¿ÉÄܵĽṹÓÐ
 
ÖÖ²»Í¬µÄ½á¹¹£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾Ý±¨µÀ£¬Ò»Á¾Ê¢Å¨ÁòËáµÄ²Û³µ£¬¿ªÖÁijתÍä´¦£¬ÓÉÓÚ³µËÙÌ«¿ì£¬³µÌåÇã·­£¬´óÁ¿Å¨ÁòËáÈ÷Âú·Ã森ΪÁ˼õÉÙË𺦣¬ÏÖÓÐÈçÏ·½·¨£º¢ÙÁ¢¼´µ÷ÓôóÁ¿µÄË®³åÏ´µØÃ棻¢ÚÁ¢¼´´Ó·»ùÅÔ±ßÈ¡ÍÁ½«Å¨ÁòËḲ¸Ç£»¢ÛÁ¢¼´µ÷ÓôóÁ¿Ê¯»ÒÈ鸲¸Ç£®ÄãÔ޳ɵķ½·¨ÊÇ
 
£¬²»Ô޳ɵķ½·¨ÊÇ
 
£¬ÀíÓÉÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÖÊÆ×ͼ±íÃ÷ijÓлúÎïµÄÖʺɱȵÄ×î´óֵΪ70£¬ºìÍâ¹âÆ×±íÕ÷µ½C¨TCºÍC¨TOµÄ´æÔÚ£¬1HºË´Å¹²ÕñÆ×Èçͼ£¨·åÃæ»ýÖ®±ÈÒÀ´ÎΪ1£º1£º1£º3£©£®·ÖÎöºË´Å¹²ÕñÆ×ͼ£¬¸ÃÓлúÎïµÄ½á¹¹¼òʽΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

È¡50.0mL̼ËáÄƺÍÁòËáÄƵĻìºÏÈÜÒº£¬¼ÓÈë¹ýÁ¿µÄÂÈ»¯±µÈÜÒººóµÃµ½14.51¿Ë³Áµí£¬ÓùýÁ¿Ï¡ÏõËá´¦Àíºó³ÁµíÖÊÁ¿¼õÉÙµ½4.66¿Ë£¬²¢ÓÐÆøÌå·Å³ö£®ÊÔ¼ÆË㣺
£¨1£©Ô­»ìºÏÈÜÒºÖÐ̼ËáÄƺÍÁòËáÄƵÄÎïÖʵÄÁ¿µÄŨ¶È
£¨2£©²úÉúµÄÆøÌåµÄÖÊÁ¿£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ϱíÊÇÓлúÎïA¡¢BµÄÓйØÐÅÏ¢£®
 A B
 ¢ÙÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«
¢Ú±ÈÀýÄ£ÐÍΪ£º
¢ÛÄÜÓëË®ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³É´¼
 ¢ÙÓÉC£¬HÁ½ÖÖÔªËØ×é³É
¢ÚÇò¹÷Ä£ÐÍΪ£º
¸ù¾Ý±íÖÐÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÔÚÒ»¶¨Ìõ¼þÏ£¬AÉú³É¸ß·Ö×Ó»¯ºÏÎïµÄ»¯Ñ§·´Ó¦·½³Ìʽ
 

£¨2£©AÓëÇâÆø·¢Éú¼Ó³É·´Ó¦ºóÉú³É·Ö×ÓC£¬CÔÚ·Ö×Ó×é³ÉºÍ½á¹¹ÉÏÏàËƵÄÓлúÎïÓÐÒ»´óÀࣨË׳ơ°Í¬ÏµÎ£©£¬ËüÃǾù·ûºÏͨʽCnH2n+2£®µ±n=
 
ʱ£¬ÕâÀàÓлúÎ↑ʼ³öÏÖͬ·ÖÒì¹¹Ì壮
£¨3£©B¾ßÓеÄÐÔÖÊÊÇ
 
£¨ÌîÐòºÅ£©£®
¢ÙÎÞÉ«ÎÞζҺÌå  ¢ÚÓж¾  ¢Û²»ÈÜÓÚË®  ¢ÜÃܶȱÈË®´ó¢ÝÓëËáÐÔKMnO4ÈÜÒººÍäåË®·´Ó¦Ê¹Ö®ÍÊÉ«   ¢ÞÈκÎÌõ¼þϲ»ÓëÇâÆø·´Ó¦£®
£¨4£©Ð´³öÔÚŨÁòËá×÷ÓÃÏ£¬BÓëŨÏõËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯ºÏÎïAÊÇÒ»ÖÖ²»Îȶ¨µÄÎïÖÊ£¬ËüµÄ·Ö×Ó×é³É¿ÉÓÃOxFy±íʾ£¬10mL AÆøÌåÄÜ·Ö½âÉú³É15mL O2ºÍ10mL F2£¨Í¬ÎÂͬѹÏ£©£®Ôò£º
£¨1£©ÎÒÃÇ¿ÉÒÔͨ¹ý¼ÆËãÈ·¶¨AµÄ»¯Ñ§Ê½£¬ÆäÖÐx=
 
£¬y=
 
£»
£¨2£©ÒÑÖªA·Ö×ÓÖÐx¸öÑõÔ­×ӳʡ­-O-O-O-¡­Á´×´ÅÅÁУ¬ÔòA·Ö×ӵĵç×ÓʽÊÇ
 
£¬½á¹¹Ê½ÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸