£¨12·Ö£©ÁòËáÍ­ÔÚÉú²ú¡¢Éú»îÖÐÓ¦Óù㷺¡£Ä³»¯¹¤³§Óú¬ÉÙÁ¿ÌúµÄ·ÏÍ­ÔüΪԭÁÏÉú²úµ¨·¯µÄÁ÷³ÌÈçÏ£º
  
£¨1£©Ð´³ö½þ³öʱͭÓëÏ¡ÁòËᡢϡÏõËá·´Ó¦Éú³ÉÁòËáÍ­µÄ»¯Ñ§·½³Ìʽ£º     ¡£
£¨2£©È¡Ñù¼ìÑéÊÇΪÁËÈ·ÈÏFe3+ÊÇ·ñ³ý¾»£¬ÄãµÄ¼ìÑé·½·¨ÊÇ     ¡£
£¨3£©ÂËÔücÊÇ     ¡£
£¨4£©ÆøÌåa¿ÉÒÔ±»Ñ­»·ÀûÓã¬Óû¯Ñ§·½³Ìʽ±íʾÆøÌåa±»Ñ­»·ÀûÓõÄÔ­ÀíΪ£º
2NO+O2 =2NO2¡¢     ¡£
£¨5£©Ò»¶¨Î¶ÈÏ£¬ÁòËáÍ­ÊÜÈÈ·Ö½âÉú³ÉCuO¡¢SO2ÆøÌå¡¢SO3ÆøÌåºÍO2ÆøÌ壬д³öÁòËáÍ­ÊÜÈÈ·Ö½âµÄ»¯Ñ§·½³Ìʽ£º     ¡£
ijͬѧÉè¼ÆÁËÈçÏÂͼËùʾµÄʵÑé×°Ö÷ֱð²â¶¨Éú³ÉµÄSO2ÆøÌå¡¢SO3ÆøÌåµÄÖÊÁ¿ºÍO2ÆøÌåµÄÌå»ý¡£´ËÉè¼ÆÓв»ºÏÀíÖ®´¦£¬Çë˵Ã÷ÀíÓÉ£º     ¡£

¹²12·Ö¡£
£¨1£©3Cu + 2HNO3 + 3H2SO4 =3CuSO4 + 2NO¡ü+ 4H2O£¨2·Ö£©
£¨2£©ÏòÊÔÑùÖеμÓKSCNÈÜÒº£¬ÈôÈÜÒºÏÔºìÉ«£¬ÔòFe3+δ³ý¾»£¬·ñÔòFe3+³ý¾»£¨2·Ö£©
£¨3£©Fe(OH)3£¨2·Ö£©
£¨4£©3NO2+H2O=2HNO3+NO£¨2·Ö£©
£¨5£©3CuSO43CuO + SO3¡ü+ 2SO2¡ü+ O2¡ü£¨ºÏÀí´ð°¸¾ù¿É£©£¨2·Ö£©
NaHSO3³ýÁËÎüÊÕSO3ÍâºÍ»¹ÎüÊÕ²¿·ÖO2£¨ºÏÀí´ð°¸¾ù¿É£©£¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦ÀíÂÛ£¬Í­ÓëÏ¡ÁòËᡢϡÏõËáµÄ»ìºÏÒº·´Ó¦Éú³ÉÁòËáÍ­¡¢Ò»Ñõ»¯µª¡¢Ë®£¬»¯Ñ§·½³ÌʽΪ3Cu + 2HNO3 + 3H2SO4 =3CuSO4 + 2NO¡ü+ 4H2O£»
£¨2£©Fe3+µÄ¼ìÑé·½·¨ÊÇ£ºÏòÊÔÑùÖеμÓKSCNÈÜÒº£¬ÈôÈÜÒºÏÔºìÉ«£¬ÔòFe3+δ³ý¾»£¬·ñÔòFe3+³ý¾»£»
£¨3£©·ÏÍ­ÔüÖк¬ÓÐÔÓÖÊÌú£¬ËùÒÔµ÷½ÚpHÄ¿µÄÊÇʹÌúÀë×Ó³Áµí£¬ÔòÂËÔücÊÇFe(OH)3£»
£¨4£©ÆøÌåaÊÇNO£¬NOÓëÑõÆø·´Ó¦Éú³É¶þÑõ»¯µª£¬¶þÑõ»¯µªÈÜÓÚË®ÓÖµÃÏõËáºÍNO£¬»¯Ñ§·½³ÌʽΪ3NO2+H2O=2HNO3+NO£»
£¨5£©ÁòËáÍ­ÊÜÈÈ·Ö½âÉú³ÉCuO¡¢SO2ÆøÌå¡¢SO3ÆøÌåºÍO2ÆøÌ壬Åäƽ·½³ÌʽµÃ3CuSO43CuO + SO3¡ü+ 2SO2¡ü+ O2¡ü£»±¥ºÍNaHSO3ÈÜÒº²»ÎüÊÕ¶þÑõ»¯Áò£¬µ«¿ÉÒÔÓëÈýÑõ»¯Áò·´Ó¦ÓÖÉú³É¶þÑõ»¯Áò£¬µ¼Ö¶þÑõ»¯ÁòÖÊÁ¿Ôö´ó¡£
¿¼µã£º¿¼²é»¯Ñ§·½³ÌʽµÄÅжÏÓëÊéд£¬Àë×Ó¼ìÑ飬ÎïÖÊÅжϣ¬×°ÖõÄÅжÏ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂͼÊÇÖÐѧ»¯Ñ§Öг£ÓÃÓÚ»ìºÏÎïµÄ·ÖÀëºÍÌá´¿µÄ×°Öã¬Çë¸ù¾Ý×°ÖûشðÎÊÌ⣺

£¨1£©´ÓÂÈ»¯¼ØÈÜÒºÖеõ½ÂÈ»¯¼Ø¹ÌÌ壬ѡÔñ×°Öà      (Ìî´ú±í×°ÖÃͼµÄ×Öĸ£¬ÏÂͬ)£»³ýÈ¥×ÔÀ´Ë®ÖеÄCl£­µÈÔÓÖÊ£¬Ñ¡Ôñ×°Öà       ¡£
£¨2£©´ÓµâË®ÖзÖÀë³öI2£¬Ñ¡Ôñ×°Öà       £¬¸Ã·ÖÀë·½·¨µÄÃû³ÆΪ             ¡£
£¨3£©×°ÖÃAÖТٵÄÃû³ÆÊÇ       £¬½øË®µÄ·½ÏòÊÇ´Ó     ¿Ú½øË®¡£×°ÖÃBÔÚ·ÖҺʱΪʹҺÌå˳ÀûÏµΣ¬Ó¦½øÐеľßÌå²Ù×÷ÊÇ                                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ijͭ¿óʯÖ÷Òªº¬Cu2(OH)2CO3£¬»¹º¬ÉÙÁ¿Fe¡¢SiµÄ»¯ºÏÎʵÑéÊÒÒÔ´ËÍ­¿óʯΪԭÁÏÖƱ¸CuSO4¡¤5H2O ¼°CaCO3£¬²¿·Ö²½ÖèÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÈÜÒºA³ýº¬ÓÐCu2+Í⣬»¹¿ÉÄܺ¬ÓеĽðÊôÀë×ÓÓÐ________(ÌîÀë×Ó·ûºÅ)£¬ÑéÖ¤Ëùº¬Àë×ÓËùÓõÄÊÔ¼ÁÊÇ______________¡£
£¨2£©¿ÉÓÃÉú³ÉµÄCO2ÖÆÈ¡ÓÅÖÊ̼Ëá¸Æ¡£ÖƱ¸Ê±£¬ÏÈÏòÂÈ»¯¸ÆÈÜÒºÖÐͨÈë°±Æø£¬ÔÙͨÈëCO2¡£
¢ÙʵÑéÊÒͨ³£²ÉÓüÓÈÈÂÈ»¯ï§ºÍÇâÑõ»¯¸Æ»ìºÏÎïµÄ·½·¨ÖÆÈ¡°±Æø¡£Ä³Ñ§Ï°Ð¡×éÑ¡È¡ÏÂͼËù¸ø²¿·Ö×°ÖÃÖÆÈ¡²¢ÊÕ¼¯´¿¾»µÄ°±Æø¡£

Èç¹û°´ÆøÁ÷·½ÏòÁ¬½Ó¸÷ÒÇÆ÷½Ó¿Ú£¬ÄãÈÏΪÕýÈ·µÄ˳ÐòΪa¡ú______¡¢______¡ú______¡¢______¡ú i¡£ÆäÖÐÓëiÏàÁ¬Â©¶·µÄ×÷ÓÃÊÇ______________¡£
¢ÚʵÑéÊÒÖл¹¿ÉÓùÌÌåÇâÑõ»¯ÄƺÍŨ°±Ë®ÖÆÈ¡ÉÙÁ¿°±Æø£¬ÏÂÁÐ×îÊʺÏÍê³É¸ÃʵÑéµÄ¼òÒ××°ÖÃÊÇ_________(Ìî±àºÅ)

£¨3£©²â¶¨Í­¿óʯÖÐCu2(OH)2CO3ÖÊÁ¿°Ù·Öº¬Á¿µÄ·½·¨ÊÇ£ºa£®½«1.25gÍ­¿óʯÖÆÈ¡µÄCuSO4¡¤5H2OÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈëÊÊÁ¿Ë®ÍêÈ«Èܽ⣻b£®ÏòÈÜÒºÖмÓÈë100mL0.25mol/LµÄÇâÑõ»¯ÄÆÈÜҺʹCu2+ÍêÈ«³Áµí£»c£®¹ýÂË£»d£®ÂËÒºÖеÄÇâÑõ»¯ÄÆÈÜÒºÓÃ0.5mol/LÑÎËáµÎ¶¨ÖÁÖյ㣬ºÄÓÃ10mLÑÎËá¡£ÔòÍ­¿óʯÖÐCu2(OH)2CO3ÖÊÁ¿·ÖÊýΪ_____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ʵÑéÊÒÖƱ¸1£¬2-¶þäåÒÒÍéµÄ·´Ó¦Ô­ÀíÈçÏ£º
CH3CH2OHCH2=CH2+H2O
CH2=CH2+Br2¡úBrCH2CH2Br
¿ÉÄÜ´æÔÚµÄÖ÷Òª¸±·´Ó¦ÓУºÒÒ´¼ÔÚŨÁòËáµÄ´æÔÚÏÂÔÚ140¡æÍÑË®Éú³ÉÒÒÃÑ£»Å¨ÁòËá°ÑÒÒ´¼Ñõ»¯ÎªCO2µÈ¡£
ÓÃÉÙÁ¿µÄäåºÍ×ãÁ¿µÄÒÒ´¼ÖƱ¸1£¬2-¶þäåÒÒÍéµÄ×°ÖÃÈçÏÂͼËùʾ£º

ÓйØÊý¾ÝÁбíÈçÏÂ:

 
ÒÒ´¼
1,2-¶þäåÒÒÍé
ÒÒÃÑ
״̬
ÎÞÉ«ÒºÌå
ÎÞÉ«ÒºÌå
ÎÞÉ«ÒºÌå
ÃܶÈ/g¡¤cm3
0£®79
2£®2
0£®71
·Ðµã/oC
78£®5
132
34£®6
ÈÛµã/oC
-130
9
-116
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©A×°ÖÃÉÏ·½Ê¹ÓõÎҺ©¶·µÄÓŵãÊÇ£º_________________________£»Èç¹û¼ÓÈÈÒ»¶Îʱ¼äºó·¢ÏÖÍü¼Ç¼Ó´ÉƬ£¬Ó¦¸Ã²ÉÈ¡µÄÕýÈ·²Ù×÷ÊÇ_____________________(ÌîÕýÈ·´ð°¸±êºÅ)¡£
A£®Á¢¼´²¹¼Ó    B£®ÀäÈ´ºó²¹¼Ó    C£®²»Ðè²¹¼Ó    D£®ÖØÐÂÅäÁÏ
£¨2£©B×°ÖõÄ×÷ÓÃÊÇ_____________________________________¡£
£¨3£©ÔÚ×°ÖÃCÖÐÓ¦¼ÓÈë________(ÌîÕýÈ·Ñ¡ÏîÇ°µÄ×Öĸ)£¬ÆäÄ¿µÄÊÇ______________¡£
a£®Ë®    b£®Å¨ÁòËá    c£®ÇâÑõ»¯ÄÆÈÜÒº    d£®±¥ºÍ̼ËáÇâÄÆÈÜÒº
£¨4£©ÅжϸÃÖƱ¸·´Ó¦ÒѾ­½áÊøµÄ×î¼òµ¥·½·¨ÊÇ____________________________£®
£¨5£©D×°ÖþßÖ§ÊÔ¹ÜÄÚÈôÓÃäåË®´úÌæÒºäå(¼Ù¶¨²úÎïÏàͬ)£¬·ÖÎöÆäÓŵã________________¡£
£¨6£©·´Ó¦¹ý³ÌÖÐÓ¦ÓÃÀäË®ÀäÈ´×°ÖÃD£¬ÆäÖ÷ҪĿµÄÊÇ_____________£»µ«ÓÖ²»Äܹý¶ÈÀäÈ´(ÈçÓñùË®)£¬ÆäÔ­ÒòÊÇ_________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

Îíö²ÑÏÖØÓ°ÏìÈËÃǵÄÉú»îÓ뽡¿µ¡£Ä³µØÇøµÄÎíö²ÖпÉÄܺ¬ÓÐÈçÏ¿ÉÈÜÐÔÎÞ»úÀë×Ó£ºNa+¡¢NH4+¡¢Mg2+¡¢Al3+¡¢SO42£­¡¢NO3£­¡¢Cl£­ ¡£Ä³Í¬Ñ§ÊÕ¼¯Á˸õØÇøµÄÎíö²£¬¾­±ØÒªµÄÔ¤´¦ÀíºóÊÔÑùÈÜÒº£¬Éè¼Æ²¢Íê³ÉÁËÈçϵÄʵÑ飺

ÒÑÖª£º3NO3£­+ 8Al + 5OH£­ + 2H2O3NH3 + 8AlO2£­
¸ù¾ÝÒÔÉϵÄʵÑé²Ù×÷ÓëÏÖÏ󣬸ÃͬѧµÃ³öµÄ½áÂÛ²»ÕýÈ·µÄÊÇ
ÊÔÑùÖп϶¨´æÔÚNH4+¡¢Mg2+¡¢SO42£­ºÍNO3£­
ÊÔÑùÖÐÒ»¶¨²»º¬Al3+
ÊÔÑùÖпÉÄÜ´æÔÚNa+¡¢Cl£­
¸ÃÎíö²ÖпÉÄÜ´æÔÚNaNO3¡¢NH4ClºÍMgSO4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

£¨16·Ö£©°±»ù¼×Ëá泥¨NH2COONH4£©ÊÇÒ»ÖÖ°×É«¹ÌÌ壬ÊÜÈÈÒ׷ֽ⡣ijС×éÄ£ÄâÖƱ¸°±»ù¼×Ëá泥¬·´Ó¦ÈçÏ£¨ÇÒζȶԷ´Ó¦µÄÓ°Ïì±È½ÏÁéÃô£©£º2NH3(g)+CO2(g)NH2COONH4(s)  ¦¤H£¼0
£¨1£©ÈçÓÃÏÂͼI×°ÖÃÖÆÈ¡°±Æø£¬¿ÉÑ¡ÔñµÄÊÔ¼ÁÊÇ                        ¡£
£¨2£©ÖƱ¸°±»ù¼×Ëá淋Ä×°ÖÃÈçÏÂͼ¢òËùʾ£¬°ÑNH3ºÍCO2ͨÈëËÄÂÈ»¯Ì¼ÖУ¬²»¶Ï½Á°è»ìºÏ£¬Éú³ÉµÄ°±»ù¼×Ëá淋ÄС¾§ÌåÐü¸¡ÔÚCCl4ÖС£ µ±Ðü¸¡Îï½Ï¶àʱ£¬Í£Ö¹ÖƱ¸¡£

×¢£ºCCl4ÓëÒºÌåʯÀ¯¾ùΪ¶èÐÔ½éÖÊ¡£
¢Ù·¢ÉúÆ÷ÓñùË®ÀäÈ´µÄÔ­ÒòÊÇ_________________________________________£¬ÒºÌåʯÀ¯¹ÄÅÝÆ¿µÄ×÷ÓÃÊÇ                                                               ¡£
¢Ú´Ó·´Ó¦ºóµÄ»ìºÏÎïÖзÖÀë³ö²úÆ·µÄʵÑé·½·¨ÊÇ            £¨Ìîд²Ù×÷Ãû³Æ£©¡£ÎªÁ˵õ½¸ÉÔï²úÆ·£¬Ó¦²ÉÈ¡µÄ·½·¨ÊÇ________£¨ÌîдѡÏîÐòºÅ£©¡£
a£®³£Ñ¹¼ÓÈȺæ¸É              b£®¸ßѹ¼ÓÈȺæ¸É           c£®¼õѹ40¡æÒÔϺæ¸É
£¨3£©ÖƵõݱ»ù¼×Ëá刺ÉÄܺ¬ÓÐ̼ËáÇâ李¢Ì¼Ëáï§ÖеÄÒ»ÖÖ»òÁ½ÖÖ¡£
¢ÙÉè¼Æ·½°¸£¬½øÐгɷÖ̽¾¿£¬ÇëÌîд±íÖпոñ¡£
ÏÞÑ¡ÊÔ¼Á£ºÕôÁóË®¡¢Ï¡HNO3¡¢BaCl2ÈÜÒº¡¢³ÎÇåʯ»ÒË®¡¢AgNO3ÈÜÒº¡¢Ï¡ÑÎËá¡£

ʵÑé²½Öè
Ô¤ÆÚÏÖÏóºÍ½áÂÛ
²½Öè1£ºÈ¡ÉÙÁ¿¹ÌÌåÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓÈëÕôÁóË®ÖÁ¹ÌÌåÈܽ⡣
µÃµ½ÎÞÉ«ÈÜÒº
²½Öè2£ºÏòÊÔ¹ÜÖмÓÈë¹ýÁ¿µÄBaCl2ÈÜÒº£¬¾²ÖÃ
ÈôÈÜÒº²»±ä»ë×Ç£¬Ö¤Ã÷¹ÌÌåÖв»º¬Ì¼Ëá李£
²½Öè3£ºÏòÊÔ¹ÜÖмÌÐø¼ÓÈë                         
                                        ¡£
                                £¬Ö¤Ã÷¹ÌÌåÖк¬ÓÐ̼ËáÇâ李£
¢Ú¸ù¾Ý¢ÙµÄ½áÂÛ£ºÈ¡°±»ù¼×Ëáï§ÑùÆ·3.95 g£¬ÓÃ×ãÁ¿ÇâÑõ»¯±µÈÜÒº³ä·Ö´¦Àíºó£¬¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ²âµÃ³ÁµíÖÊÁ¿Îª1.97 g¡£ÔòÑùÆ·Öа±»ù¼×Ëá淋ÄÎïÖʵÄÁ¿·ÖÊýΪ_______________¡£[Mr(NH2COONH4)=78¡¢Mr(NH4HCO3) =79¡¢Mr(BaCO3)=197]

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ʵÑéÊÒÓÃÉÙÁ¿µÄäåË®ºÍ×ãÁ¿µÄÒÒ´¼ÖƱ¸1£¬2¡ª¶þäåÒÒÍéµÄ×°ÖÃÈçÏÂͼËùʾ£º

ÓйØÊý¾ÝÁбíÈçÏ£º

 
ÒÒ´¼
1,2-¶þäåÒÒÍé
   ÒÒÃÑ
״̬
ÎÞÉ«ÒºÌå
  ÎÞÉ«ÒºÌå
 ÎÞÉ«ÒºÌå
Ãܶȣ¯g ¡¤ cm-3
    0.79
    2.2
    0.71
·Ðµã£¯¡æ
    78.5
    132
    34.6
È۵㣯¡æ
    -l30
    9
    -1l6
 

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÉÕÆ¿DÖз¢ÉúµÄÖ÷ÒªµÄ·´Ó¦·½³Ìʽ                        ¡£
£¨2£©°²È«Æ¿B¿ÉÒÔ·Àµ¹Îü,²¢¿ÉÒÔ¼ì²éʵÑé½øÐÐʱÊÔ¹ÜDÊÇ·ñ·¢Éú¶ÂÈû¡£Çëд³ö·¢Éú¶ÂÈûʱƿBÖеÄÏÖÏó                               ¡£
£¨3£©ÔÚ×°ÖÃCÖÐÓ¦¼ÓÈë      (Ìî×Öĸ) £¬ÆäÄ¿µÄÊÇ_______________¡£
a£®Ë®         b£®Å¨ÁòËá       c£®ÇâÑõ»¯ÄÆÈÜÒº
£¨4£©Èô²úÎïÖÐÓÐÉÙÁ¿Î´·´Ó¦µÄBr2£¬×îºÃÓà    Ï´µÓ³ýÈ¥¡££¨Ìî×Öĸ£©
a£®Ë®      b£®ÇâÑõ»¯ÄÆÈÜÒº       c£®µâ»¯ÄÆÈÜÒº          d£®ÒÒ´¼
²ÉÈ¡·ÖÀëµÄÖ÷Òª²£Á§ÒÇÆ÷ÊÇ           £»
£¨5£©Èô²úÎïÖÐÓÐÉÙÁ¿¸±²úÎïÒÒÃÑ£¬¿ÉÓà       µÄ·½·¨³ýÈ¥¡£
£¨6£©·´Ó¦¹ý³ÌÖÐÓ¦ÓÃÀäË®ÀäÈ´×°ÖÃD£¬µ«ÓÖ²»Äܹý¶ÈÀäÈ´£¨ÈçÓñùË®£©£¬ÆäÔ­ÒòÊÇ         ¡£
£¨7£©ÅжϸÃÖƸ÷·´Ó¦ÒѾ­½áÊøµÄ×î¼òµ¥·½·¨ÊÇ               £»

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

ij»¯Ñ§ÐËȤС×éµÄͬѧÀûÓÃÏÂͼËùʾʵÑé×°ÖýøÐÐijЩÆøÌåµÄÖƱ¸¡¢ÐÔÖʵÈʵÑé(ͼÖмгÖ×°ÖÃÓÐÊ¡ÂÔ)¡£Çë°´ÒªÇóÌî¿Õ£º

¢ñ.̽¾¿ÂÈÆøÓë°±ÆøµÄ·´Ó¦
(1)ΪÖÆÈ¡¸ÉÔï°±Æø£¬¿É½«×°ÖÃCÓë________(Ìî×°ÖñàºÅ)Á¬½Ó£»×°ÖÃCÖеÄÉÕÆ¿ÄÚ¹ÌÌåÒËÑ¡ÓÃ________¡£
a£®¼îʯ»Ò   b£®ÂÈ»¯¸Æ   c£®ÎåÑõ»¯¶þÁ×   d£®Éúʯ»Ò
(2)×°ÖÃA¡¢E¡¢EÁ¬½Ó¿ÉÖÆÈ¡´¿¾»¡¢¸ÉÔïµÄÂÈÆø£¬ÔòÁ½¸öE×°ÖÃÄÚµÄÒ©Æ·ÒÀ´ÎÊÇ________¡£
(3)×°ÖÃF¿ÉÓÃÓÚ̽¾¿ÂÈÆøÓë°±Æø(ÒÑÖªÂÈÆøÓë°±Æø¿É·¢Éú·´Ó¦£º3Cl2£«2NH3=N2£«6HCl)µÄ·´Ó¦¡£ÊµÑéʱ´ò¿ª¿ª¹Ø1¡¢3£¬¹Ø±Õ2£¬ÏÈÏòÉÕÆ¿ÖÐͨÈë________£¬È»ºó¹Ø±Õ1¡¢3£¬´ò¿ª2£¬ÏòÉÕÆ¿ÖлºÂýͨÈëÒ»¶¨Á¿µÄÁíÒ»ÖÖÆøÌ塣ʵÑéÒ»¶Îʱ¼äºóÉÕÆ¿ÄÚ³öÏÖŨºñµÄ°×Ñ̲¢ÔÚÈÝÆ÷ÄÚ±ÚÄý½á£¬ÇëÉè¼ÆÒ»¸öʵÑé·½°¸¼ø¶¨¸Ã¹ÌÌåÖеÄÑôÀë×Ó________¡£
¢ò.̽¾¿Ä³Ð©ÎïÖʵÄÐÔÖÊ
(1)ÀûÓÃ×°ÖÃA¡¢E£¬¿ÉÉè¼ÆʵÑé±È½ÏCl£­ºÍBr£­µÄ»¹Ô­ÐÔÇ¿Èõ£¬ÄÜÖ¤Ã÷½áÂÛµÄʵÑéÏÖÏóÊÇ________¡£
(2)ÈôÀûÓÃ×°ÖÃA¡¢E½øÐÐÒÒÏ©ÓëäåË®·´Ó¦µÄʵÑ飬±ØÐë¶Ô×°ÖÃA½øÐеĸĶ¯ÊÇ________¡£
(3)½«×°ÖÃB¡¢C·Ö±ðÓëFÏàÁ¬ºó£¬½øÐÐH2SÓëSO2·´Ó¦µÄʵÑé¡£FµÄÉÕÆ¿Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________£»FµÄÉÕ±­ËùÆðµÄ×÷ÓÃÊÇ_________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ

¼×ÍéÖлìÓÐÒÒÏ©£¬Óû³ýÈ¥ÒÒÏ©µÃµ½´¿¾»µÄ¼×Í飬×îºÃÒÀ´Îͨ¹ýÊ¢ÓÐÏÂÁÐÄÄЩÊÔ¼ÁµÄÏ´ÆøÆ¿£¨   £©

A£®³ÎÇåʯ»ÒË®£¬Å¨H2SO4B£®ËáÐÔKMnO4£¬Å¨H2SO4
C£®äåË®£¬Å¨H2SO4D£®Å¨H2SO4£¬äåË®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸