¡¾ÌâÄ¿¡¿¹ý̼ËáÄÆ(Na2CO4)ÊÇÒ»ÖֺܺõĹ©Ñõ¼Á£¬ÆäÓëÏ¡ÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Na2CO4£«4HCl=4NaCl£«2CO2¡ü£«O2¡ü£«2H2O¡£ÊÐÊÛ¹ý̼ËáÄÆÒ»°ã¶¼º¬ÓÐ̼ËáÄÆ£¬Îª²â¶¨Ä³¹ý̼ËáÄÆÑùÆ·(Ö»º¬Na2CO4ºÍNa2CO3)µÄ´¿¶È£¬Ä³»¯Ñ§ÐËȤС×é²ÉÓÃÒÔÏÂÁ½ÖÖ·½°¸ÊµÊ©£º
·½°¸Ò»£º
(1)²Ù×÷¢ÙºÍ¢ÛµÄÃû³Æ·Ö±ðΪ______________________¡£
(2)ÉÏÊö²Ù×÷ÖУ¬Ê¹Óõ½²£Á§°ôµÄÓÐ _________________(Ìî²Ù×÷ÐòºÅ)¡£
(3)Çë¼òÊö²Ù×÷¢ÛµÄ²Ù×÷¹ý³Ì£º________________________________________________¡£
·½°¸¶þ£º°´ÏÂͼËùʾ°²×°ºÃʵÑé×°Öã¬QΪһËÜÁÏÆø´ü£¬ËæÒâÈ¡ÊÊÁ¿ÑùÆ·ÓÚÆäÖУ¬´ò¿ª·ÖҺ©¶·»îÈû£¬½«Ï¡ÑÎËáµÎÈëÆø´üÖÐÖÁ³ä·Ö·´Ó¦¡£
(4)Ϊ²â¶¨·´Ó¦Éú³ÉÆøÌåµÄ×ÜÌå»ý£¬µÎÏ¡ÑÎËáÇ°±ØÐë¹Ø±Õ______(Ìî¡°K1¡±¡°K2¡±»ò¡°K3¡±£¬ÏÂͬ)£¬´ò¿ª________¡£µ¼¹ÜAµÄ×÷ÓÃÊÇ_____________________________________________¡£
(5)µ±ÉÏÊö·´Ó¦Í£Ö¹ºó£¬Ê¹K1¡¢K3´¦ÓڹرÕ״̬£¬K2´¦ÓÚ´ò¿ª×´Ì¬£¬ÔÙ»º»º´ò¿ªKl¡£ BÖÐ×°µÄ¹ÌÌåÊÔ¼ÁÊÇ________________________¡£
(6)ʵÑé½áÊøʱ£¬Á¿Í²¢ñÖÐÓÐx mLË®£¬Á¿Í²¢òÖÐÊÕ¼¯µ½ÁËy mLÆøÌ壬ÔòÑùÆ·Öйý̼ËáÄƵÄÖÊÁ¿·ÖÊýÊÇ___________________¡£
¡¾´ð°¸¡¿³ÆÁ¿¡¢Õô·¢½á¾§ ¢Ú¢Û ¼ÓÈÈÕô·¢ÖÁÕô·¢ÃóÖгöÏÖ´óÁ¿¾§Ìåʱ£¬Í£Ö¹¼ÓÈÈ£¬ÀûÓÃÓàÈÈÕô¸ÉÕô·¢ÃóÖеÄÊ£ÓàË®·Ö K1£¬K2 K3 ƽºâ·ÖҺ©¶·Äںͷ´Ó¦ÌåϵÄÚѹǿ£¬Ê¹Ï¡ÑÎËá˳ÀûµÎÏ£¬Í¬Ê±Ïû³ýµÎÏÂÏ¡ÑÎËáµÄÌå»ý¶ÔÆøÌåÌå»ýµÄÓ°Ïì ¼îʯ»Ò 122y/(53x-37y)
¡¾½âÎö¡¿
²â¶¨Ä³¹ý̼ËáÄÆÑùÆ·(Ö»º¬Na2CO4ºÍNa2CO3)µÄ´¿¶È£¬ÑùÆ·³ÆÁ¿ÖÊÁ¿µÃµ½ÑùÆ·ÖÊÁ¿Îªm1g£¬¼ÓÈëÏ¡ÑÎËáÈܽâºó¹ýÂ˵õ½ÈÜҺΪÂÈ»¯ÄÆÈÜÒº£¬Õô·¢Å¨Ëõ½á¾§Îö³öÂÈ»¯ÄÆ£¬³ÆÁ¿µÃµ½¹ÌÌåÂÈ»¯ÄƵÄÖÊÁ¿Îªm2g£¬¸ù¾Ý·½°¸Ò»Á÷³Ì¿ÉÅжϳö²Ù×÷¢ÙΪ³ÆÁ¿£¬²Ù×÷¢ÚΪÈܽ⣬²Ù×÷¢ÛΪÕô·¢½á¾§¡£·½°¸¶þÖÐÁ¿Í²¢ñÖÐË®µÄÌå»ýΪ²úÉúµÄCO2¡¢O2µÄÌå»ý£¬Á¿Í²¢òÖÐÊÕ¼¯µ½µÄÆøÌåΪO2£¬¸ù¾Ý·½°¸¶þÒÀ¾Ý²â¶¨µÄÑõÆøÌå»ý¼ÆËãÎïÖʵÄÁ¿£¬½áºÏ»¯Ñ§·½³Ìʽ¼ÆËã¹ý̼ËáÄÆÎïÖʵÄÁ¿¼ÆËãµÃµ½ÖÊÁ¿·ÖÊý¡£ÒÔ´Ë·ÖÎö½â´ð¡£
¸ù¾Ý·½°¸Ò»Á÷³Ì¿ÉÅжϳö²Ù×÷¢ÙΪ³ÆÁ¿£¬²Ù×÷¢ÚΪÈܽ⣬²Ù×÷¢ÛΪÕô·¢½á¾§¡£
£¨1£©ÓÉÉÏÊö·ÖÎö¿ÉÖª²Ù×÷¢ÙΪ³ÆÁ¿£¬²Ù×÷¢ÛΪÕô·¢½á¾§£¬
¹Ê´ð°¸Îª£º³ÆÁ¿¡¢Õô·¢½á¾§£»
£¨2£©ÉÏÊö²Ù×÷ÖÐÓõ½²£Á§°ôµÄÓÐÈܽâºÍÕô·¢½á¾§£¬
¹Ê´ð°¸Îª£º¢Ú¢Û£»
£¨3£©²Ù×÷¢ÛΪÕô·¢½á¾§£¬Æä¾ßÌå¹ý³ÌΪ£º¼ÓÈÈÕô·¢ÖÁÕô·¢ÃóÖгöÏÖ´óÁ¿¾§Ìåʱ£¬Í£Ö¹¼ÓÈÈ£¬ÀûÓÃÓàÈÈÕô¸ÉÕô·¢ÃóÖеĹÌÌ壬
¹Ê´ð°¸Îª£º¼ÓÈÈÕô·¢ÖÁÕô·¢ÃóÖгöÏÖ´óÁ¿¾§Ìåʱ£¬Í£Ö¹¼ÓÈÈ£¬ÀûÓÃÓàÈÈÕô¸ÉÕô·¢ÃóÖеÄÊ£ÓàË®·Ö£»
£¨4£©·´Ó¦²úÉúµÄCO2¡¢O2ʹÆø´ü±ä´ó£¬½«¹ã¿ÚÆ¿ÖÐÆøÌåÅųö£¬Ë®½øÈëÁ¿Í²¢ñÖУ¬ËùÒÔÁ¿Í²¢ñÖÐË®µÄÌå»ý¼´Îª²úÉúµÄCO2¡¢O2µÄÌå»ý£¬ËùÒÔµÎÏ¡ÑÎËáÇ°±ØÐë¹Ø±ÕK1¡¢K2£¬´ò¿ªK3¡£µ¼¹ÜAµÄ×÷ÓÃÊÇƽºâ·ÖҺ©¶·Äںͷ´Ó¦ÌåϵÄÚѹǿ£¬Ê¹Ï¡ÑÎËá˳ÀûµÎÏ£¬Í¬Ê±Ïû³ýµÎÏÂÏ¡ÑÎËáµÄÌå»ý¶ÔÆøÌåÌå»ýµÄÓ°Ï죬
¹Ê´ð°¸Îª£ºK1£¬K2 £»K3 £»Æ½ºâ·ÖҺ©¶·Äںͷ´Ó¦ÌåϵÄÚѹǿ£¬Ê¹Ï¡ÑÎËá˳ÀûµÎÏ£¬Í¬Ê±Ïû³ýµÎÏÂÏ¡ÑÎËáµÄÌå»ý¶ÔÆøÌåÌå»ýµÄÓ°Ï죻
£¨5£©B×°ÖÃÓÃÀ´³ýÈ¥Éú³ÉÆøÌåÖеÄCO2£¬ÔòBÖÐËù×°¹ÌÌåÊÔ¼ÁΪ¼îʯ»Ò¡£»ºÂý´ò¿ªK1ÊÇΪÁËÈÃÉú³ÉµÄCO2Äܹ»±»B×°ÖÃÖмîʯ»Ò³ä·ÖÎüÊÕ£¬´Ó¶øʹÁ¿Í²ÖÐÊÕ¼¯µ½½Ï´¿µÄO2£¬
¹Ê´ð°¸Îª£º¼îʯ»Ò£»
£¨6£©ÒÀÌâÒâ¿ÉÖª£¬Á¿Í²¢òÖÐÊÕ¼¯µ½µÄÆøÌåΪO2£¬Á¿Í²¢ñÖÐË®µÄÌå»ýµÈÓÚ·´Ó¦²úÉúµÄCO2ÓëO2µÄ×ÜÌå»ý£¬Ôò¿É½øÐÐÈçϼÆË㣺
n(Na2CO4)=2 n(O2)=mmol£¬n(CO2)=
mmol£¬Na2CO4ÓëÑÎËá·´Ó¦Éú³ÉCO2µÄÎïÖʵÄÁ¿Îªn(CO2)1=
mmol£¬ÓÖNa2CO3ÓëÑÎËá·´Ó¦Éú³ÉCO2µÄ¼ÆÁ¿¹ØϵΪNa2CO3~CO2£¬Ôòn(Na2CO3)= n(CO2)2=
mmol£¬
ËùÒÔNa2CO4µÄÖÊÁ¿·ÖÊýw( Na2CO4)%=100%=122y/(53x-37y)¡£
¹Ê´ð°¸Îª£º122y/(53x-37y)¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¹¤ÒµÉÏÀûÓÃÁòÌú¿óÉÕÔü£¨Ö÷Òª³É·ÖΪFe2O3¡¢FeO¡¢SiO2µÈ£©ÎªÔÁÏÖƱ¸¸ßµµÑÕÁÏÌúºì(Fe2O3)£¬¾ßÌåÉú²úÁ÷³ÌÈçͼ£º
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÂËÒºXÖк¬ÓеĽðÊôÑôÀë×ÓÊÇ___£¨ÌîÀë×Ó·ûºÅ£©¡£
£¨2£©²½Öè¢òÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____¡£
£¨3£©²½Öè¢óÖÐ×îºÃ¿ÉÑ¡ÓÃ___£¨Ìî×Öĸ£©µ÷½ÚÈÜÒºµÄpH¡£
A£®Ï¡ÏõËá B£®°±Ë® C£®ÇâÑõ»¯ÄÆÈÜÒº D£®¸ßÃÌËá¼ØÈÜÒº
£¨4£©²½Öè¢ôÖУ¬FeCO3³ÁµíÍêÈ«ºó£¬ÈÜÒºÖк¬ÓÐÉÙÁ¿Fe2£«£¬¼ìÑéFe2£«µÄ·½·¨ÊÇ___¡£
£¨5£©²½Öè¢ôµÄ·´Ó¦Î¶ÈÒ»°ãÐè¿ØÖÆÔÚ35¡æÒÔÏ£¬ÆäÄ¿µÄÊÇ______¡£
£¨6£©ÔÚ¿ÕÆøÖÐìÑÉÕFeCO3Éú³É²úÆ·Ñõ»¯ÌúµÄ»¯Ñ§·½³ÌʽΪ_____¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÔÚÁ½·ÝÏàͬµÄBa(OH)2ÈÜÒºÖУ¬·Ö±ðµÎÈëÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄH2SO4¡¢NaHSO4ÈÜÒº£¬Æäµ¼µçÄÜÁ¦ËæµÎÈëÈÜÒºÌå»ý±ä»¯µÄÇúÏßÈçͼËùʾ¡£ÏÂÁзÖÎöÕýÈ·µÄÊÇ
A.b¡úd·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºH++OH-=H2O
B.o¡úa·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºBa2++OH-+H++SO42-=BaSO4¡ý+H2O
C.cµãµ¼µçÄÜÁ¦Ïàͬ£¬ËùÒÔÁ½ÈÜÒºÖк¬ÓÐÏàͬÁ¿µÄOH¨C
D.a¡¢bÁ½µãBa2+¾ù³ÁµíÍêÈ«£¬ËùÒÔ¶ÔÓ¦µÄÈÜÒº¾ùÏÔÖÐÐÔ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¶ÌÖÜÆÚÔªËØW¡¢X¡¢Y¡¢ZµÄÔ×ÓÐòÊýÒÀ´ÎÔö¼Ó¡£m¡¢p¡¢rÊÇÓÉÕâЩԪËØ×é³ÉµÄ¶þÔª»¯ºÏÎqÊÇYµÄµ¥ÖÊÇÒΪµ»ÆÉ«¹ÌÌ壬nÊÇÔªËØZµÄµ¥ÖÊ£¬³£ÎÂÏÂÊÇ»ÆÂÌÉ«µÄÆøÌ壬0.01mol¡¤L¨C1rÈÜÒºµÄpHΪ2£¬sͨ³£ÊÇÄÑÈÜÓÚË®µÄ»ìºÏÎï¡£ÉÏÊöÎïÖʵÄת»¯¹ØϵÈçͼËùʾ¡£ÏÂÁÐ˵·¨Ò»¶¨ÕýÈ·µÄÊÇ
A.Ô×Ӱ뾶µÄ´óС£ºW<X<Y<Z
B.ÔªËصķǽðÊôÐÔ£ºX>Y
C.¼òµ¥Ç⻯ÎïµÄÎȶ¨ÐÔ£ºZ>Y>X
D.Ñõ»¯ÎïµÄ¶ÔӦˮ»¯ÎïµÄËáÐÔ£ºZ>Y
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿´¿¾»¸ÉÔïµÄÂÈÆøÓëÈÛÈÚ½ðÊôÎý·´Ó¦¿ÉÖƱ¸ SnCl4£¬Ä³»¯Ñ§Ð¡×éµÄͬѧÉè¼ÆÁËÈçÏÂʵÑé×°ÖýøÐÐÖƱ¸¡£
ÒÑÖª£º¢Ù½ðÊôÎýÈÛµãΪ 231¡æ£¬»¯Ñ§»îÆÃÐÔÓëÌúÏàËÆ£»¢ÚSnCl4 µÄ·ÐµãΪ 114¡æ£»¢ÛSnCl4 Ò×ÓëË®·´Ó¦¡£
Çë¸ù¾ÝÉÏͼװÖûشð:
£¨1£©×°ÖâôÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________¡£
£¨2£©ÊÔ¹ÜIIÖеÄÊÔ¼ÁÊÇ___________________£¬×°Öà V µÄ×÷ÓÃÊÇ______________¡£
£¨3£©ÑbÖâö×îºÃÑ¡ÓÃÏÂÁÐ×°ÖÃÖеÄ______________________(Ìî±êºÅ)¡£
£¨4£©ÊµÑé½áÊøºó£¬Óû»ØÊÕÀûÓÃ×°ÖâñÖÐδ·´Ó¦ÍêµÄ MnO2£¬ÐèÒªµÄ·ÖÀë·½·¨___________¡£
£¨5£©ÎªÁË˳ÀûÍê³ÉʵÑ飬µãȼ¾Æ¾«µÆµÄÕýÈ·²Ù×÷ÊÇ___________________(Ìî×Öĸ)¡£
A£®Ïȵãȼ I ´¦¾Æ¾«µÆ£¬ºóµãȼ¢ô´¦¾Æ¾«µÆ
B£®Ïȵãȼ¢ô´¦¾Æ¾«µÆ£¬ºóµãȼ I ´¦¾Æ¾«µÆ
C£®Í¬Ê±µãȼ I¡¢¢ôÁ½´¦¾Æ¾«µÆ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÂÌ·¯ÊǺ¬ÓÐÒ»¶¨Á¿½á¾§Ë®µÄÁòËáÑÇÌú£¬ÔÚ¹¤Å©ÒµÉú²úÖоßÓÐÖØÒªµÄÓÃ;¡£Ä³»¯Ñ§ÐËȤС×é¶ÔÂÌ·¯µÄһЩÐÔÖʽøÐÐ̽¾¿¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÈçºÎÓÃʵÑéÖ¤Ã÷ÂÌ·¯ÖеÄÌúÊǶþ¼ÛÌú¶ø²»ÊÇÈý¼ÛÌú_______¡£
(2)Ϊ²â¶¨ÂÌ·¯Öнᾧˮº¬Á¿£¬½«Ê¯Ó¢²£Á§¹Ü£¨´øÁ½¶Ë¿ª¹ØK1ºÍK2£©£¨ÉèΪװÖÃA£©³ÆÖØ£¬¼ÇΪm1g¡£½«ÑùÆ·×°ÈëʯӢ²£Á§¹ÜÖУ¬Ôٴν«×°ÖÃA³ÆÖØ£¬¼ÇΪm2g¡£°´ÈçͼÁ¬½ÓºÃ×°ÖýøÐÐʵÑé¡£
¢ÙÒÇÆ÷BµÄÃû³ÆÊÇ________¡£
¢Ú½«ÏÂÁÐʵÑé²Ù×÷²½ÖèÕýÈ·ÅÅÐò_______£¨Ìî±êºÅ£©£»Öظ´ÉÏÊö²Ù×÷²½Ö裬ֱÖÁAºãÖØ£¬¼ÇΪm3g¡£
a.µãȼ¾Æ¾«µÆ£¬¼ÓÈÈ b.´ò¿ªK1ºÍK2£¬»º»ºÍ¨ÈëN2 c.¹Ø±ÕK1ºÍK2
d.ϨÃð¾Æ¾«µÆ e.³ÆÁ¿A f.ÀäÈ´µ½ÊÒÎÂ
¢Û¸ù¾ÝʵÑé¼Ç¼£¬¼ÆËãÂÌ·¯»¯Ñ§Ê½ÖнᾧˮÊýÄ¿x=_______(ÁÐʽ±íʾ)¡£ÈôʵÑéʱ°´a¡¢b´ÎÐò²Ù×÷£¬Ôòʹx_______£¨Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©¡£
(3)Ϊ̽¾¿ÁòËáÑÇÌúµÄ·Ö½â²úÎ½«(2)ÖÐÒѺãÖصÄ×°ÖÃA½ÓÈëÈçͼËùʾµÄ×°ÖÃÖУ¬´ò¿ªK1ºÍK2£¬»º»ºÍ¨ÈëN2£¬¼ÓÈÈ¡£ÊµÑéºó·´Ó¦¹ÜÖвÐÁô¹ÌÌåΪºìÉ«·ÛÄ©¡£
¢ÙCÖеÄÈÜҺΪBaCl2£¬Æä×÷ÓÃÊÇ______¡£DÈÜҺΪ______£¬DÖпɹ۲쵽µÄÏÖÏóΪ_______¡£
¢Úд³öÁòËáÑÇÌú¸ßηֽⷴӦµÄ»¯Ñ§·½³Ìʽ________¡£
¢ÛÓÐͬѧÈÏΪ¸ÃʵÑé×°ÖôæÔÚÒ»¶¨È±ÏÝ£¬ÇëÄã̸̸ÍêÉƸÃÊÔÑé×°ÖõĴëÊ©____¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÊµÑéÊÒÓÃÏÂÊö×°ÖÃÖÆÈ¡ÂÈÆø£¬²¢ÓÃÂÈÆø½øÐÐÏÂÁÐʵÑé¡£¿´Í¼»Ø´ðÏÂÁÐÎÊÌ⣺
(1)A¡¢BÁ½ÒÇÆ÷µÄÃû³Æ£ºA________£¬B________¡£
(2)Ï´Æø×°ÖÃCÊÇΪÁ˳ýÈ¥Cl2ÖеÄHClÆøÌ壬DÊÇΪÁ˸ÉÔïCl2£¬ÔòC¡¢DÖÐÓ¦·Ö±ð·ÅÈëÏÂÁÐÈÜÒºÖеÄC______________£»D______________¡£
¢ÙNaOHÈÜÒº¡¡¢Ú±¥ºÍʳÑÎË®¡¡¢ÛAgNO3ÈÜÒº¡¡¢ÜŨH2SO4
(3)EÖÐΪºìÉ«¸É²¼Ìõ£¬FÖÐΪºìɫʪ²¼Ìõ£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ______________________________¡£
(4)GÊǽþÓеí·ÛKIÈÜÒºµÄÃÞ»¨Çò£¬G´¦ÏÖÏóÊÇÃÞ»¨Çò±íÃæ±ä³É____________¡£·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________________£¬HÊǽþÓÐNaBrÈÜÒºµÄÃÞ»¨Çò£¬H´¦ÏÖÏóÊÇÃÞ»¨Çò±íÃæ±ä³É________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÔÚ2010Äêθ绪¶¬¼¾°ÂÔË»áÉÏ£¬Óиö±ðÔ˶¯Ô±Òò·þÓÃÐ˷ܼÁ±»È¡Ïû²ÎÈüµÄ×ʸñ¡£ÏÂͼÊǼì²â³öÐ˷ܼÁµÄijÖÖͬϵÎïXµÄ½á¹¹£¬¹ØÓÚXµÄ˵·¨ÕýÈ·µÄÊÇ( )
A. X·Ö×ÓÖв»¿ÉÄÜËùÓÐÔ×Ó¶¼ÔÚͬһƽÃæÉÏ
B. X Óöµ½FeCl3ÈÜҺʱÏÔ×ÏÉ«£¬¶øÇÒÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«
C. 1 mol X Óë×ãÁ¿µÄŨäåË®·´Ó¦£¬×î¶àÏûºÄ5 mol Br2
D. 1 mol XÔÚÒ»¶¨Ìõ¼þÏÂÓë×ãÁ¿µÄÇâÆø·´Ó¦£¬×î¶àÏûºÄ1 mol H2
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Ô×ÓÐòÊýСÓÚ36µÄX¡¢Y¡¢Z¡¢R¡¢WÎåÖÖÔªËØ£¬ÆäÖÐXÊÇÖÜÆÚ±íÖÐÔ×Ӱ뾶×îСµÄÔªËØ£¬YÊÇÐγɻ¯ºÏÎïÖÖÀà×î¶àµÄÔªËØ£¬ZÔ×Ó»ù̬ʱ2pÔ×Ó¹ìµÀÉÏÓÐ3¸öδ³É¶ÔµÄµç×Ó£¬Rµ¥ÖÊÕ¼¿ÕÆøÌå»ýµÄ £»WµÄÔ×ÓÐòÊýΪ29¡£»Ø´ðÏÂÁÐÎÊÌâ:
£¨1£©Y2X4·Ö×ÓÖÐYÔ×Ó¹ìµÀµÄÔÓ»¯ÀàÐÍΪ________£¬1mol Z2X4º¬ÓЦҼüµÄÊýĿΪ ________¡£
£¨2£©»¯ºÏÎïZX3Ó뻯ºÏÎïX2RµÄVSEPR¹¹ÐÍÏàͬ£¬µ«Á¢Ìå¹¹ÐͲ»Í¬£¬ZX3µÄÁ¢Ìå¹¹ÐÍΪ ________£¬Á½ÖÖ»¯ºÏÎï·Ö×ÓÖл¯Ñ§¼üµÄ¼ü½Ç½ÏСµÄÊÇ________£¨Ó÷Ö×Óʽ±íʾ£¬ÏÂͬ)¡£
£¨3£©ÓëRͬÖ÷×åµÄÈýÖַǽðÊôÔªËØÓëX¿ÉÐγɽṹÏàËƵÄÈýÖÖÎïÖÊ£¬ÊÔÍƲâÈýÕßµÄÎȶ¨ÐÔÓÉ´óµ½Ð¡µÄ˳Ðò________£¬ÀíÓÉÊÇ ________£»ÈýÕߵķеãÓɸߵ½µÍµÄ˳ÐòÊÇ ________£¬½âÊÍÔÒò________¡£
£¨4£©ÔªËØYµÄÒ»ÖÖÑõ»¯ÎïÓëÔªËØZµÄµ¥ÖÊ»¥ÎªµÈµç×ÓÌ壬ԪËØYµÄÕâÖÖÑõ»¯ÎïµÄ·Ö×ÓʽÊÇ________¡£
£¨5£©WÔªËØÓÐ________¸öÔ˶¯×´Ì¬²»Í¬µÄµç×Ó£¬Æä»ù̬Ô×ӵļ۵ç×ÓÅŲ¼Ê½Îª________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com