ϱíÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£®ËùÁÐ×Öĸ·Ö±ð´ú±íÒ»ÖÖ»¯Ñ§ÔªËØ£®
a
bcdef
ghijklm
no
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©Çëд³ö£ºÔªËØoµÄ»ù̬ԭ×Óµç×ÓÅŲ¼Ê½
 
£»
£¨2£©kÔÚ¿ÕÆøÖÐȼÉÕ²úÎïµÄ·Ö×Ó¹¹ÐÍΪ
 
£¬ÖÐÐÄÔ­×ÓµÄÔÓ»¯ÐÎʽΪ
 
£¬¸Ã·Ö×ÓÊÇ
 
£¨Ìî¡°¼«ÐÔ¡±»ò¡°·Ç¼«ÐÔ¡±£©·Ö×Ó£»
£¨3£©º¬10µç×ÓµÄdµÄÇ⻯Îï·Ö×ÓµÄVSEPRÄ£ÐÍΪ
 
£»
£¨4£©g¡¢h¡¢iÈýÖÖÔªËصĵÚÒ»µçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪ
 
£¨ÌîÔªËØ·ûºÅ£©
£¨5£©ËùÓÐÔªËØÆäÖе縺ÐÔ×î´óµÄÊÇ
 
£¨ÌîͼÖеÄÐòºÅ»òÔªËØ·ûºÅ£©£¬ÔªËØkÓÐÁ½ÖÖÑõ»¯ÎËüÃǶÔÓ¦µÄË®»¯ÎïµÄËáÐÔÇ¿Èõ˳ÐòΪ
 
£®£¨Ìѧʽ£©
¿¼µã£ºÔªËØÖÜÆÚÂɺÍÔªËØÖÜÆÚ±íµÄ×ÛºÏÓ¦ÓÃ
רÌ⣺ԪËØÖÜÆÚÂÉÓëÔªËØÖÜÆÚ±íרÌâ
·ÖÎö£ºÓÉÔªËØÔÚÖÜÆÚ±íÖÐλÖ㬿ÉÖªaΪH¡¢bΪLi¡¢cΪC¡¢dΪN¡¢eΪO¡¢fΪF¡¢gΪNa¡¢hΪMg¡¢iΪAl¡¢jΪSi¡¢KΪS¡¢lΪCl¡¢mΪAr¡¢nΪK¡¢oΪFe£®
£¨1£©oΪFe£¬Ô­×ÓºËÍâÓÐ26¸öµç×Ó£¬¸ù¾ÝÄÜÁ¿×îµÍÔ­ÀíÊéдÆäºËÍâµç×ÓÅŲ¼Ê½£»
£¨2£©kΪS£¬ÔÚ¿ÕÆøÖÐȼÉÕ²úÎïΪSO2£¬¼ÆËãSÔ­×ӹµç×Ó¶ÔÊý¡¢¼Û²ãµç×Ó¶ÔÊý£¬È·¶¨·Ö×Ó¹¹ÐÍ£¬SÔ­×ÓµÄÔÓ»¯ÐÎʽ£¬½áºÏ·Ö×ӽṹÅжϷÖ×ÓÖÐÕý¸ºµçºÉÖØÐÄÊÇ·ñÖغϣ¬ÅжϷÖ×Ó¼«ÐÔ£»
£¨3£©º¬10µç×ÓµÄdµÄÇ⻯Îï·Ö×ÓΪNH3£¬¼ÆËãNÔ­×Ó¼Û²ãµç×Ó¶ÔÊý£¬È·¶¨ÆäVSEPRÄ£ÐÍ£»
£¨4£©Í¬ÖÜÆÚ×Ô×ó¶øÓÒÔªËصÚÒ»µçÀëÄܳÊÔö´óÇ÷ÊÆ£¬µ«MgÔªËØÔ­×Ó3sÄܼ¶ÈÝÄÉ2µÄµç×Ó£¬ÎªÈ«ÂúÎȶ¨×´Ì¬£¬ÄÜÁ¿½ÏµÍ£¬µÚÒ»µçÀëÄܸßÓÚAl£»
£¨5£©ËùÓÐÔªËØÆäÖе縺ÐÔ×î´óµÄÊÇ·ú£¬ÔªËØkÓÐÁ½ÖÖÑõ»¯Î·Ö±ðΪ¶þÑõ»¯Áò¡¢ÈýÑõ»¯Áò£¬ËüÃǶÔÓ¦µÄË®»¯Îï·Ö±ðΪÑÇÁòËá¡¢ÁòËᣬÑÇÁòËáÊôÓÚÈõËᣬ¶øÁòËáÊôÓÚÇ¿Ëᣮ
½â´ð£º ½â£ºÓÉÔªËØÔÚÖÜÆÚ±íÖÐλÖ㬿ÉÖªaΪH¡¢bΪLi¡¢cΪC¡¢dΪN¡¢eΪO¡¢fΪF¡¢gΪNa¡¢hΪMg¡¢iΪAl¡¢jΪSi¡¢KΪS¡¢lΪCl¡¢mΪAr¡¢nΪK¡¢oΪFe£®
£¨1£©oΪFe£¬Ô­×ÓºËÍâÓÐ26¸öµç×Ó£¬¸ù¾ÝÄÜÁ¿×îµÍÔ­Àí£¬ÆäºËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d64s2£¬
¹Ê´ð°¸Îª£º1s22s22p63s23p63d64s2£»
£¨2£©kΪS£¬ÔÚ¿ÕÆøÖÐȼÉÕ²úÎïΪSO2£¬SÔ­×ӹµç×Ó¶ÔÊý=
6-1¡Á2
2
=1¡¢¼Û²ãµç×Ó¶ÔÊý=2+1=3£¬¹Ê·Ö×Ó¹¹ÐÍΪVÐΣ¬SÔ­×ÓµÄÔÓ»¯ÐÎʽΪsp2£¬·Ö×ÓÖÐÕý¸ºµçºÉÖØÐIJ»·ñÖغϣ¬ÊôÓÚ¼«ÐÔ·Ö×Ó£¬
¹Ê´ð°¸Îª£ºVÐΣ»sp2£»¼«ÐÔ£»
£¨3£©º¬10µç×ÓµÄdµÄÇ⻯Îï·Ö×ÓΪNH3£¬NÔ­×Ó¼Û²ãµç×Ó¶ÔÊý=3+
5-1¡Á3
2
=4£¬ÆäVSEPRÄ£ÐÍΪÕýËÄÃæÌ壬¹Ê´ð°¸Îª£ºÕýËÄÃæÌ壻
£¨4£©Í¬ÖÜÆÚ×Ô×ó¶øÓÒÔªËصÚÒ»µçÀëÄܳÊÔö´óÇ÷ÊÆ£¬µ«MgÔªËØÔ­×Ó3sÄܼ¶ÈÝÄÉ2µÄµç×Ó£¬ÎªÈ«ÂúÎȶ¨×´Ì¬£¬ÄÜÁ¿½ÏµÍ£¬µÚÒ»µçÀëÄܸßÓÚAl£¬¹ÊµÚÒ»µçÀëÄÜ£ºM g£¾Al£¾Na£¬¹Ê´ð°¸Îª£ºM g£¾Al£¾Na£»
£¨5£©ËùÓÐÔªËØÆäÖе縺ÐÔ×î´óµÄÊÇ·ú£¬ÔªËØkÓÐÁ½ÖÖÑõ»¯Î·Ö±ðΪ¶þÑõ»¯Áò¡¢ÈýÑõ»¯Áò£¬ËüÃǶÔÓ¦µÄË®»¯Îï·Ö±ðΪÑÇÁòËá¡¢ÁòËᣬÑÇÁòËáÊôÓÚÈõËᣬ¶øÁòËáÊôÓÚÇ¿Ëᣬ¹ÊËáÐÔ£ºH2SO4£¾H2SO3£¬
¹Ê´ð°¸Îª£ºF£»H2SO4£¾H2SO3£®
µãÆÀ£º±¾Ìâ¿ÉÖªÔªËØÖÜÆÚ±íÓëÔªËØÖÜÆÚÂÉ¡¢·Ö×ӽṹÓëÐÔÖÊ¡¢ºËÍâµç×ÓÅŲ¼µÈ£¬ÄѶÈÖеȣ¬×¢ÒâÀí½âÕÆÎÕ·Ö×ӽṹµÄÅжϣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ±ê×¼×´¿öÏ£¬Ä³Æø̬ÌþµÄÃܶÈÊÇ1.34g?L-1£¬Ò»¶¨Ìå»ýµÄ¸ÃÌþ³ä·ÖȼÉÕºóÉú³É CO213.2g£¬Í¬Ê±Éú³ÉH2O 8.1g£¬Çó´ËÌþµÄ·Ö×Óʽ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ij¹¤³§ÒªÓÃH2SO4ÓëHF£¨Çâ·úËᣩµÄ»ìºÏÈÜÒº×÷¿óÎïÖÐÏ¡ÓÐÔªËصÄÝÍÈ¡Òº£¬ÒªÇó¸ÃÝÍÈ¡ÒºÖÐH2SO4¡¢HFµÄÎïÖʵÄÁ¿Å¨¶È¸÷Ϊ3.16mol/L¡¢9.00mol/L£®ÏÖÓÐ500L»ØÊÕËáÒº£¬¾­²â¶¨ÆäÖУºH2SO4¡¢HFµÄÎïÖʵÄÁ¿Å¨¶È¸÷2.15mol/L¡¢12.6mol/L£®ÏÖÒªÓôË500L»ØÊÕËáÒºÅäÖÆÉÏÊöÝÍÈ¡Òº£¬Ôò£º¢Ù´Ë»ØÊÕËáÒº¾­Ï¡Ê͹²¿ÉµÃµ½
 
LºÏ·ûÉÏÊöÒªÇóµÄÝÍÈ¡Òº£¬¢ÚÔÚ´Ë500LµÄ»ØÊÕËáÒºÖÐÓ¦¼ÓÈë
 
LÃܶÈΪ1.84g/ml¡¢Å¨¶ÈΪ98%µÄŨÁòËᣬ²Å·ûºÏÝÍÈ¡ÒºµÄ»ù±¾ÒªÇó£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÊÂʵ֤Ã÷£¬ÄÜÉè¼Æ³ÉÔ­µç³ØµÄ·´Ó¦Í¨³£ÊÇ·ÅÈÈ·´Ó¦£¬ÏÂÁл¯Ñ§·´Ó¦ÔÚÀíÂÛÉÏ¿ÉÒÔÉè¼Æ³ÉÔ­µç³ØµÄÊÇ
 
£®£¨Ìî×Öĸ£©
A£®C£¨s£©+CO2£¨g£©¨T2CO£¨g£©¡÷H£¾0
B£®NaOH£¨aq£©+HC1£¨aq£©¨TNaC1£¨aq£©+H2O£¨1£©¡÷H£¼0
C£®2CO£¨g£©+O2£¨g£©¨T2CO2£¨1£©¡÷H£¼0
£¨2£©ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒº£¬ÒÀ¾Ý£¨1£©ÖÐËùÑ¡·´Ó¦Éè¼ÆÒ»¸öÔ­µç³Ø£¬ÆäÕý¼«µÄµç¼«·´Ó¦Ê½Îª£º
 
£®
£¨3£©ÀûÓÃÈçͼװÖ㬿ÉÒÔÄ£ÄâÌúµÄµç»¯Ñ§·À»¤£® ÈôXΪ̼°ô£¬Îª¼õ»ºÌúµÄ¸¯Ê´£¬¿ª¹ØKÓ¦¸ÃÖÃÓÚ
 
£¨ÌîM»òÕßN£©´¦£¬´ËʱÈÜÒºÖеÄÒõÀë×ÓÏò
 
¼«£¨ÌîX»òÌú£©Òƶ¯£®ÈôXΪп£¬¿ª¹ØKÖÃÓÚM´¦£¬¸Ãµç»¯Ñ§·À»¤·¨³ÆΪ£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓÐËÄÖÖÒ»ÔªËáHA¡¢HB¡¢HC¡¢HD£¬ÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄNaDºÍNaBÈÜÒºµÄpH£¬Ç°Õ߱ȺóÕß´ó£¬NaAÈÜÒº³ÊÖÐÐÔ£¬1mol/LµÄKCÈÜÒºÓö·Ó̪ÊÔÒº³ÊºìÉ«£»Í¬Ìå»ý¡¢Í¬ÎïÖʵÄÁ¿Å¨¶ÈµÄHB¡¢HCÓÃÑùµÄ×°Ö÷ֱð×÷µ¼µçÐÔÊÔÑ飬·¢ÏÖºóÕߵĵÆÅݱÈÇ°ÕßÁÁ£¬ÔòÕâËÄÖÖËáµÄËáÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòΪ
 
£¨Ó໯ѧʽ¼°¡°£¾¡±±íʾ£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÀ¾ÝÐðÊö£¬Íê³ÉÏÂÁÐÈý¸öСÌ⣮
£¨1£©ÔÚ25¡æ¡¢101kPaÏ£¬1g¼×´¼£¨CH3OH£©È¼ÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22.68kJ£®Ôò±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ
 
£®
£¨2£©ÈôÊÊÁ¿µÄN2ºÍO2ÍêÈ«·´Ó¦£¬Ã¿Éú³É23g NO2ÐèÒªÎüÊÕ16.95kJÈÈÁ¿£®ÆäÈÈ»¯Ñ§·½³ÌʽΪ
 
£®
£¨3£©ÒÑÖªÁ½¸öÈÈ»¯Ñ§·½³Ìʽ£º
C£¨s£©+O2£¨g£©¨TCO2£¨g£©¡÷H=-393.5kJ/mol£¬2H2£¨g£©+O2£¨g£©¨T2H2O£¨g£©¡÷H=-483.6kJ/mol
ÏÖÓÐ0.2molµÄÌ¿·ÛºÍÇâÆø×é³ÉµÄÐü¸¡Æø¹Ì»ìºÏÎÔÚÑõÆøÖÐÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍË®ÕôÆø£¬¹²·Å³öÈÈÁ¿63.53kJ£¬ÔòÌ¿·ÛÓëÇâÆøµÄÎïÖʵÄÁ¿Ö®±ÈΪ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÇâÄÜ×÷ΪһÖÖÐÂÐÍÄÜÔ´¾ßÓÐȼÉÕÈÈÖµ¸ß¡¢×ÊÔ´·á¸»¡¢È¼ÉÕ²úÎïÎÞÎÛȾµÈÓŵ㣮ÒÑ֪ȼÉÕ4g H2Éú³ÉҺ̬ˮʱ·ÅÈÈΪ571.6kJ£¬ÊÔд³ö±íʾH2ȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÈçͼÊÇÅäÖÆ50mLËáÐÔKMnO4±ê×¼ÈÜÒºµÄ¹ý³ÌʾÒâͼ£®

£¨1£©ÇëÄã¹Û²ìͼʾÅжÏÆäÖв»ÕýÈ·µÄ²Ù×÷ÓÐ
 
£¨ÌîÐòºÅ£©£®
£¨2£©ÆäÖÐÈ·¶¨50mLÈÜÒºÌå»ýµÄÈÝÆ÷ÊÇ
 
£¨ÌîÃû³Æ£©£®
£¨3£©Èç¹û°´ÕÕͼʾµÄ²Ù×÷ËùÅäÖƵÄÈÜÒº½øÐÐʵÑ飬ÔÚÆäËû²Ù×÷¾ùÕýÈ·µÄÇé¿öÏ£¬Ëù²âµÃµÄʵÑé½á¹û½«
 
£¨Ìî¡°Æ«´ó¡±»ò¡°Æ«Ð¡¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª£ºH+H¨TH2£»¡÷H=-436KJ/mol£¬ÔòÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢Ù2¸öHÔ­×ÓµÄÄÜÁ¿¸ßÓÚ1¸öH2µÄÄÜÁ¿
¢Ú2¸öHÔ­×ÓµÄÄÜÁ¿µÍÓÚ1¸öH2µÄÄÜÁ¿
¢ÛH2·Ö×Ó±ÈHÔ­×ÓÎȶ¨
¢ÜHÔ­×Ó±ÈH2·Ö×ÓÎȶ¨£®
A¡¢¢Ú¢ÛB¡¢¢Ù¢ÛC¡¢¢Ù¢ÜD¡¢¢Ú¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸