»¯Ñ§¿ÎÍâÐËȤС×éѧÉúÔÚʵÑéÊÒÀïÖÆÈ¡µÄÒÒÏ©Öг£»ìÓÐÉÙÁ¿µÄ¶þÑõ»¯Áò£¬ÀÏʦÆô·¢ËûÃDz¢ÓÉËûÃÇ×Ô¼ºÉè¼ÆÁËÏÂÁÐʵÑéͼÒÔÈ·ÈÏÉÏÊö»ìºÏÆøÌåÖÐÓÐC2H4ºÍSO2£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¢ñ¡¢¢ò¡¢¢ó¡¢¢ô×°ÖÿÉÊ¢·ÅµÄÊÔ¼ÁÊÇ£º¢ñ
 
£»¢ò
 
£»¢ó
 
£»¢ô
 
£¨½«ÏÂÁÐÓйØÊÔ¼ÁµÄÐòºÅÌîÈë¿Õ¸ñÄÚ£©£®
A£®Æ·ºìÈÜÒº   B£®NaOHÈÜÒº   C£®Å¨ÁòËá   D£®ËáÐÔKMnO4ÈÜÒº
£¨2£©ÄÜ˵Ã÷SO2ÆøÌå´æÔÚµÄÏÖÏóÊÇ
 
£®
£¨3£©Ê¹ÓÃ×°ÖÃIIµÄÄ¿µÄÊÇ
 
£®
£¨4£©Ê¹ÓÃ×°ÖÃIIIµÄÄ¿µÄÊÇ
 
£®
£¨5£©È·¶¨º¬ÓÐÒÒÏ©µÄÏÖÏóÊÇ
 
£®
£¨6£©¢ôºÅÊÔ¼Á¿ÉÓÃ
 
´úÌæ£¬³öÏÖµÄÏÖÏóΪ
 
£®
¿¼µã£ºÐÔÖÊʵÑé·½°¸µÄÉè¼Æ
רÌ⣺ʵÑéÉè¼ÆÌâ
·ÖÎö£º¶þÑõ»¯ÁòµÄ¼ìÑéÓÃÆ·ºìÈÜÒº£¬ÒÒÏ©µÄ¼ìÑéÓøßÃÌËá¼ØËáÐÔÈÜÒº£¬ÒÒÏ©ºÍ¶þÑõ»¯Áò¶¼ÄÜʹ¸ßÃÌËá¼ØËáÐÔÈÜÒºÍÊÉ«£¬ËùÒÔÏȼìÑé¶þÑõ»¯Áò£¬È»ºó¼ìÑéÒÒÏ©£¬Í¬ÔÚ¼ìÑéÒÒϩ֮ǰÓÃNaOHÈÜÒº³ýÈ¥SO2£¬ÔÙͨ¹ýÆ·ºìÈÜÒº²»ÍÊɫȷÈÏSO2Òѳý¸É¾»£¬×îºóÓøßÃÌËá¼ØËáÐÔÈÜÒºÍÊÉ«¼ìÑéÒÒÏ©£¬È»ºó»Ø´ðÎÊÌ⣮
£¨1£©¶þÑõ»¯ÁòÊÇ·ñ´æÔÚ¿ÉÓÃÆ·ºìÈÜÒº¼ìÑ飻¼ìÑéÒÒÏ©¿ÉÒÔÓÃäåË®»ò¸ßÃÌËá¼ØËáÐÔÈÜÒº£»ÒÒÏ©ºÍ¶þÑõ»¯Áò¶¼ÄÜʹäåË®»ò¸ßÃÌËá¼ØËáÐÔÈÜÒºÍÊÉ«£»ÒÒÏ©ÓëäåË®·¢Éú¼Ó³É·´Ó¦Ê¹äåË®ÍÊÉ«£ºCH2=CH2+Br2¡úCH2Br-CH2Br£¬ÒÒÏ©±»ËáÐÔ¸ßÃÌËá¼ØÑõ»¯Ê¹ÆäÍÊÉ«£®¶þÑõ»¯ÁòÓëäåË®¡¢ËáÐÔ¸ßÃÌËá¼Ø·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬5SO2+2KMnO4+2H2O¨TK2SO4+2MnSO4+2H2SO4£¬SO2+Br2+H2O¨TH2SO4+2HBr£»ÒÒÏ©²»ÓëNaOHÈÜÒº·´Ó¦£¬µ«¶þÑõ»¯ÁòÄÜÓë¼î·´Ó¦£¨SO2+2NaOH=Na2SO3+H2O£©£¬¹ÊÒÒÏ©µÄ¼ìÑéÓ¦·ÅÔÚÅųýSO2µÄ¸ÉÈźó½øÐУ¬Ñ¡Í¨¹ýÆ·ºìÈÜÒºÍÊÉ«¼ìÑéSO2µÄ´æÔÚ£»ÔÙͨ¹ýNaOHÈÜÒº³ýÈ¥SO2£¬ÔÙͨ¹ýÆ·ºìÈÜÒº²»ÍÊɫȷÈÏSO2Òѳý¸É¾»£¬×îºóÓøßÃÌËá¼ØËáÐÔÈÜÒºÍÊɫʵÑé¼ìÑéÒÒÏ©£»
£¨2£©Í¨¹ýÆ·ºìÈÜÒºÍÊÉ«¼ìÑéSO2µÄ´æÔÚ£»
£¨3£©SO2+Br2+H2O¨TH2SO4+2HBr£¬³ýÈ¥¶þÑõ»¯ÁòÆøÌ壬ÒÔÃâ¸ÉÈÅÒÒÏ©µÄʵÑ飻
£¨4£©ÔÙͨ¹ýÆ·ºìÈÜÒº²»ÍÊɫȷÈÏSO2Òѳý¸É¾»£»
£¨5£©ÓøßÃÌËá¼ØËáÐÔÈÜÒºÍÊɫʵÑé¼ìÑéÒÒÏ©£»
£¨6£©¿ÉÒÔÀûÓÃäåË®¼ìÑéÒÒÏ©µÄ´æÔÚ£»
½â´ð£º ½â£º£¨1£©ÒÒÏ©²»ÓëNaOHÈÜÒº·´Ó¦£¬µ«¶þÑõ»¯ÁòÄÜÓë¼î·´Ó¦£¨SO2+2NaOH=Na2SO3+H2O£©£¬¶þÑõ»¯ÁòÊÇ·ñ´æÔÚ¿ÉÓÃÆ·ºìÈÜÒº¼ìÑ飮ÒÒÏ©µÄ¼ìÑéÓ¦·ÅÔÚÅųýSO2µÄ¸ÉÈźó½øÐУ¬Ñ¡Í¨¹ýÆ·ºìÈÜÒºÍÊÉ«¼ìÑéSO2µÄ´æÔÚ£»ÔÙͨ¹ýNaOHÈÜÒº³ýÈ¥SO2£¬ÔÙͨ¹ýÆ·ºìÈÜÒº²»ÍÊɫȷÈÏSO2Òѳý¸É¾»£®×îºóÓøßÃÌËá¼ØËáÐÔÈÜÒºÍÊɫʵÑé¼ìÑéÒÒÏ©£¬
Òò×°ÖâñÓÃÀ´¼ìÑéSO2£¬ÊÔ¹ÜÖÐÆ·ºìÈÜÒºÍÊÉ«£¬ËµÃ÷º¬ÓÐSO2£¬×°ÖâòÊÔ¹Ü×°ÓÐNaOHÈÜÒº³ýÈ¥SO2£¬×°ÖâóÊÔ¹Üͨ¹ýÆ·ºìÈÜÒº²»ÍÊɫȷÈÏSO2Òѳý¸É¾»£¬×°Öâôͨ¹ý¸ßÃÌËá¼ØËáÐÔÈÜÒºÍÊÉ«¼ìÑéÒÒÏ©£¬
¹Ê´ð°¸Îª£ºA£¬B£¬A£¬D£»
£¨2£©¶þÑõ»¯ÁòÊÇ·ñ´æÔÚ¿ÉÓÃÆ·ºìÈÜÒº¼ìÑ飬ƷºìÈÜÒºÍÊɫ˵Ã÷º¬ÓжþÑõ»¯Áò£¬
¹Ê´ð°¸Îª£º×°ÖâñÖÐÆ·ºìÈÜÒºÍÊÉ«£»
£¨3£©ÒÒÏ©ºÍ¶þÑõ»¯Áò¶¼ÄÜʹäåË®»ò¸ßÃÌËá¼ØËáÐÔÈÜÒºÍÊÉ«£¬¶þÑõ»¯ÁòµÄ´æÔÚÓ°ÏìÒÒÏ©µÄ¼ìÑ飬¹Ê¼ìÑéÒÒϩʱӦÏȳýÈ¥¶þÑõ»¯Áò£¬
¹Ê´ð°¸Îª£º³ýÈ¥¶þÑõ»¯ÁòÆøÌ壬ÒÔÃâ¸ÉÈÅÒÒÏ©µÄʵÑ飻
£¨4£©Í¨¹ýNaOHÈÜÒº³ýÈ¥SO2£¬ÔÙͨ¹ýÆ·ºìÈÜÒº²»ÍÊɫȷÈÏSO2Òѳý¸É¾»£¬
¹Ê´ð°¸Îª£º¼ìÑé¶þÑõ»¯ÁòÊÇ·ñ³ý¾¡£»
£¨5£©×îºóÓøßÃÌËá¼ØËáÐÔÈÜÒºÍÊɫʵÑé¼ìÑéÒÒÏ©£¬×°ÖâôÖеÄËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬ËµÃ÷º¬ÓÐÒÒÏ©£¬
¹Ê´ð°¸Îª£º×°ÖâóÖÐµÄÆ·ºìÈÜÒº²»ÍÊÉ«£¬×°ÖâôÖеÄËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£»
£¨6£©¿ÉÒÔÀûÓÃäåË®¼ìÑéÒÒÏ©µÄ´æÔÚ£¬´æÔÚµÄÏÖÏóÊÇäåË®ÍÊÉ«£»
¹Ê´ð°¸Îª£ºäåË®£¬äåË®ÍÊÉ«£»
µãÆÀ£º±¾Ì⿼²éÒÒÏ©µÄ»¯Ñ§ÐÔÖÊ¡¢ÖƱ¸ÒÔ¼°³£¼ûÆøÌåµÄ¼ìÑ飬עÒâʵÑéµÄÏȺó˳Ðò£¬ÌâÄ¿ÄѶȲ»´ó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

±µÔÚÑõÆøÖÐȼÉÕʱµÄµÃµ½Ò»ÖÖ±µµÄÑõ»¯Îï¾§Ì壬Æð½á¹¹ÈçͼËùʾ£¬ÓйØËµ·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢¸Ã¾§ÌåÊôÓÚÀë×Ó¾§Ìå
B¡¢¾§ÌåµÄ»¯Ñ§Ê½ÎªBa2O2
C¡¢¸Ã¾§Ìå¾§°û½á¹¹ÓëCsClÏàËÆ
D¡¢Óëÿ¸öBa2+¾àÀëÏàµÈÇÒ×î½üµÄBa2+¹²ÓÐ12¸ö

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¼îʽ̼ËáÍ­µÄ³É·ÖÓжàÖÖ£¬Æä»¯Ñ§Ê½Ò»°ã¿É±íʾΪxCu£¨OH£©2?yCuCO3£®¿×ȸʯ³ÊÂÌÉ«£¬ÊÇÒ»ÖÖÃû¹óµÄ±¦Ê¯£¬ÆäÖ÷Òª³É·ÖÊÇCu£¨OH£©2?CuCO3£®Ä³ÐËȤС×éΪ̽¾¿ÖÆÈ¡¿×ȸʯµÄ×î¼Ñ·´Ó¦Ìõ¼þ£¬Éè¼ÆÁËÈçÏÂʵÑ飺
ʵÑé1£º½«2.0mL 0.50mol/LCu£¨NO3£©2ÈÜÒº¡¢2.0mL 0.50mol/LNaOHÈÜÒººÍ0.25mol/L
Na2CO3ÈÜÒº°´±í¢ñËùʾÌå»ý»ìºÏ£®
ʵÑé2£º½«ºÏÊʱÈÀýµÄ»ìºÏÎïÔÚ±í¢òËùʾζÈÏ·´Ó¦£®
ʵÑé¼Ç¼ÈçÏ£º
                     ±í¢ñ±í¢ò
±àºÅV £¨Na2CO3£©/mL³ÁµíÇé¿ö ±àºÅ·´Ó¦Î¶È/¡æ³ÁµíÇé¿ö
12.8¶à¡¢À¶É« 140¶à¡¢À¶É«
22.4¶à¡¢À¶É« 260ÉÙ¡¢Ç³ÂÌÉ«
32.0½Ï¶à¡¢ÂÌÉ« 375½Ï¶à¡¢ÂÌÉ«
41.6½ÏÉÙ¡¢ÂÌÉ« 480½Ï¶à¡¢ÂÌÉ«£¨ÉÙÁ¿ºÖÉ«£©
£¨1£©ÊµÑéÊÒÖÆÈ¡ÉÙÐí¿×ȸʯ£¬Ó¦¸Ã²ÉÓõÄ×î¼Ñ·´Ó¦Ìõ¼þ£º¢ÙCu£¨NO3£©2ÈÜÒºÓëNa2CO3ÈÜÒºµÄÌå»ý±ÈΪ
 
£»¢Ú·´Ó¦Î¶ÈÊÇ
 
£®
£¨2£©80¡æÊ±£¬ËùÖÆµÃµÄ¿×ȸʯÓÐÉÙÁ¿ºÖÉ«ÎïÖʵÄÔ­ÒòÊÇ
 
£®
ʵÑéС×éΪ²â¶¨ÉÏÊöijÌõ¼þÏÂËùÖÆµÃµÄ¼îʽ̼ËáÍ­ÑùÆ·×é³É£¬ÀûÓÃÏÂͼËùʾµÄ×°Ö㨼гÖÒÇÆ÷Ê¡ÂÔ£©½øÐÐʵÑ飺

²½Öè1£º¼ì²é×°ÖÃµÄÆøÃÜÐÔ£¬½«¹ýÂË¡¢Ï´µÓ²¢¸ÉÔï¹ýµÄÑùÆ·ÖÃÓÚÆ½Ö±²£Á§¹ÜÖУ®
²½Öè2£º´ò¿ª»îÈûK1¡¢K2£¬¹Ø±ÕK3£¬¹ÄÈë¿ÕÆø£¬Ò»¶Îʱ¼äºó¹Ø±Õ»îÈûK1¡¢K2£¬´ò¿ªK3£¬³ÆÁ¿Ïà¹Ø×°ÖõÄÖÊÁ¿£®
²½Öè3£º¼ÓÈÈ×°ÖÃBÖ±ÖÁ×°ÖÃCÖÐÎÞÆøÅݲúÉú£®
£¨3£©²½Öè4£º
 
£®
²½Öè5£º³ÆÁ¿Ïà¹Ø×°ÖõÄÖÊÁ¿£®
£¨4£©×°ÖÃAµÄ×÷ÓÃÊÇ
 
£»ÈôÎÞ×°ÖÃE£¬ÔòʵÑé²â¶¨µÄx/yµÄ±ÈÖµ½«
 
£¨Ñ¡Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©£®
£¨5£©Ä³Í¬Ñ§ÔÚʵÑé¹ý³ÌÖвɼ¯ÁËÈçÏÂÊý¾Ý£º
a£®·´Ó¦Ç°²£Á§¹ÜÓëÑùÆ·µÄÖÊÁ¿163.8g
b£®·´Ó¦ºó²£Á§¹ÜÖвÐÁô¹ÌÌåÖÊÁ¿56.0g
c£®×°ÖÃCʵÑéºóÔöÖØ9.0g
d£®×°ÖÃDʵÑéºóÔöÖØ8.8g
Ϊ²â¶¨
x
y
µÄ±ÈÖµ£¬ÄãÈÏΪ¿ÉÒÔÑ¡ÓÃÉÏÊöËù²É¼¯Êý¾ÝÖеÄ
 
£¨Ð´³öËùÓÐ×éºÏµÄ×Öĸ´úºÅ£©Ò»×é¼´¿É½øÐмÆË㣮
£¨6£©¸ù¾ÝÄãµÄ¼ÆËã½á¹û£¬Ð´³ö¸ÃÑùÆ·×é³ÉµÄ»¯Ñ§Ê½
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÎÞËØÔ­×ӵĺËÍâÓÐËĸöÄܲ㣬×îÍâÄܲãÓÐ1¸öµç×Ó£¬¸ÃÔ­×ÓºËÄÚµÄÖÊ×ÓÊý²»ÄÜΪ£¨¡¡¡¡£©
A¡¢24B¡¢18C¡¢19D¡¢29

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÑéÊÒÓÃÈçͼËùʾµÄ×°ÖÃÖÆÈ¡ÒÒËáÒÒõ¥£®
£¨1£©Ð´³öÖÆÈ¡ÒÒËáÒÒõ¥µÄ»¯Ñ§·½³Ìʽ£º
 

£¨2£©×°ÖÃÖÐͨÕôÆøµÄµ¼¹ÜÒª²åµ½±¥ºÍ̼ËáÄÆÈÜÒºÒºÃæÉÏ£¬¶ø²»ÄܲåÈëÒºÃæÒÔÏ£¬Ä¿µÄÊÇ·ÀÖ¹
 
ʹʵķ¢Éú£® ÊÕ¼¯ÁËÒÒËáÒÒõ¥µÄÊÔ¹ÜÕñµ´Ê±ÓÐÆøÅݲúÉú£¬Çë½âÎö²úÉúÆøÅݵÄÔ­Òò£¨Óû¯Ñ§·½³Ìʽ±íʾ£©
 
£®
£¨3£©ÊµÑéÖÆÈ¡µÄÒÒËáÒÒõ¥£¬¿ÉÓÃ
 
·½·¨·ÖÀëÒÒËáÒÒõ¥Óë±¥ºÍµÄ̼ËáÄÆÈÜÒº£¬ÇëÓÃÓÃÎÄ×Ö˵Ã÷ʹÓÃÕâÖÖ·½·¨µÄÀíÓÉ£º
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÎïÖÊE¿É×öÏãÁÏ£¬Æä½á¹¹¼òʽΪ£¬ºÏ³É·ÏßÈçÏ£º

ÒÑÖª£º
R-CH=CH2+CBr¡ú
R-CH=CH2+HBr
H2O2
£¨R-´ú±íÌþ»ù£©
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖªAΪ·¼ÏãÌþ£¬ÏÂÁйØÓÚAµÄ˵·¨ÖУ¬ÕýÈ·µÄÊÇ
 
£¨ÌîÐòºÅ£©
a£®ÃܶȱÈË®´ó    b£®ËùÓÐÔ­×Ó¾ùÔÚÍ¬Ò»Æ½ÃæÉÏ   c£®Ò»ÂÈ´úÎïÖ»ÓÐÒ»ÖÖ
£¨2£©BµÄ½á¹¹¼òʽ¿ÉÄÜÊÇ
 
£»
£¨3£©Ð´³ö·´Ó¦¢ÜµÄ»¯Ñ§·½³ÌʽÊÇ
 

£¨4£©Ð´³ö·´Ó¦¢ÞµÄ»¯Ñ§·½³ÌʽÊÇ
 

£¨5£©EÓжàÖÖͬ·ÖÒì¹¹Ì壬д³ö·ûºÏÒÔÏÂÌõ¼þµÄ½á¹¹¼òʽ
 

£¨Ö»Ð´·´Ê½½á¹¹£©£®
¢Ù¾ßÓÐ˳·´½á¹¹
¢ÚÄÜÓëNaOHÈÜÒº·´Ó¦
¢Û·Ö×ÓÖб½»·ÉϵÄÒ»äå´úÎïÓÐÁ½ÖÖ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

a g Mg¡¢AlºÏ½ðÍêÈ«ÈܽâÔÚC1 mol?L-1¡¢V1L HClÈÜÒºÖУ¬²úÉúb g H2£®ÔÙÏò·´Ó¦ºóµÄÈÜÒºÖмÓÈëC2mol?L-1¡¢V2L NaOHÈÜÒº£¬Ç¡ºÃʹ³Áµí´ïµ½×î´óÖµ£¬ÇÒ³ÁµíÖÊÁ¿Îªd g£®ÏÂÁйØÏµ´íÎóµÄÊÇ£¨¡¡¡¡£©
A¡¢ÂÁΪ 
24b-a
9
mol
B¡¢C1=
C2V2
V1
C¡¢d=a+17b
D¡¢Óë½ðÊô·´Ó¦ºóÊ£ÓàÑÎËáΪ£¨C1V1-b£©mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÓлúÎï µÄÃû³ÆÊÇ£¨¡¡¡¡£©
A¡¢2£¬5-¶þ¼×»ù-4-ÒÒ»ù-3£¬6-¸ý¶þÏ©
B¡¢1£¬1£¬4-Èý¼×»ù-3-ÒÒ»ù-2£¬5-¼º¶þÏ©
C¡¢3£¬6-¶þ¼×»ù-4-ÒÒ»ù-1£¬4-¸ý¶þÏ©
D¡¢2£¬5-¶þ¼×»ù-4-ÒÒ»ù-3£¬6-¶þ¸ýÏ©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÐðÊö»ò²Ù×÷ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢Å¨ÁòËá¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Ï¡ÁòËáÎÞÑõ»¯ÐÔ
B¡¢Å¨ÁòËáµÎµ½À¶·¯ÉÏ£¬À¶·¯±ä³É°×É«·ÛÄ©£¬ÌåÏÖÁËŨÁòËáµÄÍÑË®ÐÔ
C¡¢Ï¡ÊÍŨÁòËáʱӦ½«Å¨ÁòËáÑØ×ÅÉÕ±­±ÚÂýÂýµØ×¢ÈëÊ¢ÓÐË®µÄÉÕ±­ÖУ¬²¢²»¶Ï½Á°è
D¡¢Å¨ÁòËáÓëÍ­µÄ·´Ó¦ÖУ¬Å¨ÁòËá½ö±íÏÖÇ¿Ñõ»¯ÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸