¡¾ÌâÄ¿¡¿CH3CH=CHCH3´æÔÚÒÔÏÂת»¯¹Øϵ£º

£¨1£©Õý¶¡ÍéµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽΪ________¡£

£¨2£©Ð´³ö¿òͼÖз´Ó¦¢ÙÔÚ´ß»¯¼ÁºÍ¼ÓÈÈÌõ¼þϵĻ¯Ñ§·½³Ìʽ£º________¡£

£¨3£©CH3CH=CHCH3ÄÜʹäåË®ºÍËáÐÔKMnO4ÈÜÒºÍÊÉ«£¬¶þÕßÍÊÉ«µÄÔ­ÀíÏàͬÂð£¿ËµÃ÷Ô­Òò¡£_______¡£

£¨4£©Ï©ÌþAÊÇCH3CH=CHCH3µÄÒ»ÖÖͬ·ÖÒì¹¹Ì壬ËüÔÚ´ß»¯¼Á×÷ÓÃÏÂÓëÇâÆø·´Ó¦µÄ²úÎï²»ÊÇÕý¶¡Í飬ÔòAµÄ½á¹¹¼òʽΪ___________£»A·Ö×ÓÖй²Æ½ÃæµÄ̼ԭ×Ó¸öÊýΪ___________¡£

£¨5£©·´Ó¦¢ÚµÄ²úÎïµÄͬ·ÖÒì¹¹ÌåÓÐ______ÖÖ¡£

¡¾´ð°¸¡¿CH(CH3)3 CH3CH=CHCH3£«H2CH3CH2CH2CH3 ²»Í¬£¬Ê¹äåË®ÍÊÉ«ÊÇ·¢ÉúÁ˼ӳɷ´Ó¦£¬Ê¹ËáÐÔKMnO4ÈÜÒºÍÊÉ«ÊÇ·¢ÉúÁËÑõ»¯·´Ó¦ CH2=C(CH3)2 4 3

¡¾½âÎö¡¿

£¨1£©Õý¶¡ÍéCH3CH2CH2CH3´æÔÚ̼Á´Òì¹¹£¬Í¬·ÖÒì¹¹ÌåÊÇÒ춡ÍéCH(CH3)3¡£

£¨2£©CH3CH=CHCH3ºÍÇâÆøÔÚ´ß»¯¼ÁÌõ¼þÏ·¢Éú¼Ó³É·´Ó¦Éú³ÉCH3CH2CH2CH3£¬·´Ó¦·½³ÌʽÊÇCH3CH=CHCH3£«H2CH3CH2CH2CH3£»

£¨3£©CH3CH=CHCH3ʹäåË®ÍÊÉ«ÊÇ·¢ÉúÁ˼ӳɷ´Ó¦£¬Ê¹ËáÐÔKMnO4ÈÜÒºÍÊÉ«ÊÇ·¢ÉúÁËÑõ»¯·´Ó¦£¬¶þÕßÍÊÉ«µÄÔ­Àí²»Í¬¡£

£¨4£©CH3CH=CHCH3µÄͬ·ÖÒì¹¹ÌåÓÐCH2=CHCH2CH3¡¢CH2=C(CH3)2£¬Ç°ÕßÓëÇâÆø¼Ó³ÉµÄ²úÎïÊÇÕý¶¡Í飬Òò´ËAΪCH2=C(CH3)2£»A·Ö×ÓÖк¬ÓÐ̼̼˫¼ü£¬Óë̼̼˫¼üÖеÄ̼ԭ×ÓÖ±½ÓÏàÁ¬µÄÔ­×Ó¼°Ì¼Ì¼Ë«¼üÖеÄÁ½¸ö̼ԭ×Ó£¬¹²6¸öÔ­×ӿ϶¨¹²Æ½Ã棬¹ÊA·Ö×ÓÖй²Æ½ÃæµÄ̼ԭ×Ó¸öÊýΪ4¡£

£¨5£©·´Ó¦¢ÚµÄ²úÎïµÄͬ·ÖÒì¹¹ÌåÓÐCH3CH2CH2CH2Br¡¢(CH3)2CHCH2Br¡¢(CH3)3CBr£¬¹²3ÖÖ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³¹¤ÒµÁ÷³ÌÖУ¬½øÈë·´Ó¦ËþµÄ»ìºÏÆøÌåÖÐNOºÍO2µÄÎïÖʵÄÁ¿·ÖÊý·Ö±ðΪ0.10ºÍ0.06£¬·¢Éú»¯Ñ§·´Ó¦2NO(g)+O2(g)=2NO2(g)£¬ÔÚÆäËûÌõ¼þÏàͬʱ£¬²âµÃʵÑéÊý¾ÝÈçÏÂ±í£º

ѹǿ/(¡Á105Pa)

ζÈ/¡æ

NO´ïµ½ËùÁÐת»¯ÂÊÐèҪʱ¼ä/s

50%

90%

98%

1.0

30

12

250

2830

90

25

510

5760

8.0

30

0.2

3.9

36

90

0.6

7.9

74

¸ù¾Ý±íÖÐÊý¾Ý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. Éý¸ßζȣ¬·´Ó¦ËÙÂʼӿì

B. Ôö´óѹǿ£¬·´Ó¦ËÙÂʱäÂý

C. ÔÚ1.0¡Á105Pa¡¢90¡æÌõ¼þÏ£¬µ±×ª»¯ÂÊΪ98%ʱµÄ·´Ó¦ÒѴﵽƽºâ

D. Èô½øÈë·´Ó¦ËþµÄ»ìºÏÆøÌåΪamol£¬·´Ó¦ËÙÂÊÒÔv=¡÷n/¡÷t±íʾ£¬ÔòÔÚ8.0¡Á105Pa¡¢30¡æÌõ¼þÏÂת»¯ÂÊ´Ó50%ÔöÖÁ90%ʱ¶ÎNOµÄ·´Ó¦ËÙÂÊΪ4a/370mol/s

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿°ÑÒ»¿éþÂÁºÏ½ðͶÈëµ½1mol/LÑÎËáÀ´ýºÏ½ðÍêÈ«Èܽâºó£¬ÍùÈÜÒºÀï¼ÓÈë1mol/L NaOHÈÜÒº£¬Éú³É³ÁµíµÄÎïÖʵÄÁ¿Ëæ¼ÓÈëNaOHÈÜÒºÌå»ý±ä»¯µÄ¹ØϵÈçͼAËùʾ¡£ÏÂÁÐ˵·¨Öв»ÕýÈ·µÄÊÇ£¨ £©

A.aµÄÈ¡Öµ·¶Î§Îª0¡Üa£¼50

B.µÄ×î´óֵΪ2.5

C.Èô½«¹Øϵͼ¸ÄΪBͼʱ£¬ÔòaµÄÈ¡Öµ·¶Î§Îª80£¼a£¼90

D.Èô½«¹Øϵͼ¸ÄΪCͼʱ£¬ÔòaµÄÈ¡Öµ·¶Î§Îª75£¼a£¼90

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¾§ÌåÅðµÄ½á¹¹ÈçÓÒͼËùʾ¡£ÒÑÖª¾§ÌåÅð½á¹¹µ¥ÔªÊÇÓÉÅðÔ­×Ó×é³ÉµÄÕý¶þÊ®ÃæÌ壬ÆäÖÐÓÐ20¸öµÈ±ßÈý½ÇÐεÄÃæºÍÒ»¶¨ÊýÄ¿µÄ¶¥µã£¬Ã¿¸öÏîµãÉϸ÷ÓÐ1¸öBÔ­×Ó¡£ÏÂÁÐÓйØ˵·¨²»ÕýÈ·µÄÊÇ

A.ÿ¸öÅð·Ö×Óº¬ÓÐ12¸öÅðÔ­×Ó

B.¾§ÌåÅðÊÇ¿Õ¼äÍø×´½á¹¹

C.¾§ÌåÅðÖмü½ÇÊÇ60¡ã

D.ÿ¸öÅð·Ö×Óº¬ÓÐ30¸öÅðÅðµ¥¼ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Á¬¶þÑÇÁòËáÄÆ(Na2S2O4)Ë׳Ʊ£ÏÕ·Û£¬¹ã·ºÓÃÓÚ·ÄÖ¯¹¤ÒµµÄ»¹Ô­ÐÔȾɫ¡¢ÇåÏ´¡¢Ó¡»¨¡¢ÍÑÉ«ÒÔ¼°Ö¯ÎïµÄƯ°×µÈ¡£ÖÆÈ¡±£ÏÕ·Ûͨ³£ÐèÒª¶þÑõ»¯Áò¡£

£¨1£©ÖƱ¸¶þÑõ»¯Áò

ÈôʹÓÃÉÏͼËùʾװÖÃÖƱ¸¸ÉÔïµÄSO2ÆøÌ壬»Ø´ðÒÔÏÂÎÊÌ⣺

¢ÙAÖÐÊ¢ÒºÌåµÄ²£Á§ÒÇÆ÷Ãû³ÆÊÇ____________£¬ÊµÑ鿪ʼºóAÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________________________________________¡£

¢ÚB×°ÖõÄ×÷ÓÃÊÇ_______________________£»C×°ÖõÄ×÷ÓÃ____________________¡£

¢ÛEÖйÌÌåÊÔ¼ÁΪ________________¡£

£¨2£©ÖƱ¸±£ÏÕ·Û

ÈçÏÂͼ£¬Î¬³Ö35¡«45¡æͨSO2ÖÁп·Û-Ë®Ðü¸¡Òº·´Ó¦Éú³ÉZnS2O4£»È»ºó¼ÓÈë18%µÄÇâÑõ»¯ÄÆÈÜÒº£¬ÔÚ28¡«35¡æÏ·´Ó¦Éú³ÉNa2S2O4ºÍZn(OH)2Ðü¸¡Òº¡£

·´Ó¦ÎᆳѹÂ˳ýÈ¥ÇâÑõ»¯Ð¿³Áµíºó£¬ÍùÂËÒºÖмÓÈëÂÈ»¯ÄÆ£¬²¢ÀäÈ´ÖÁ20¡æ£¬Ê¹Na2S2O4½á¾§Îö³ö£¬Â˳ö¾§ÌåºóÓþƾ«ÍÑË®¸ÉÔï¼´µÃ²úÆ·¡£

¢ÙÔÚÖÆÈ¡Na2S2O4¹ý³ÌÖз¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦£¬Ñõ»¯¼ÁÊÇ___________£»Éú³É1mol Na2S2O4תÒƵç×Ó______mol¡£

¢ÚÂËÒºÖмÓÈëÂÈ»¯ÄÆʹ_______Àë×ÓŨ¶ÈÔö´ó£¬´Ù½øNa2S2O4½á¾§Îö³ö£»Â˳ö¾§ÌåºóÓþƾ«ÍÑË®¸ÉÔïÊÇÒòΪNa2S2O4Ôھƾ«ÖеÄÈܽâ¶È_______£¨Ìî¡°½Ï´ó¡±»ò¡°½ÏС¡±£©£¬ÇҾƾ«Ò×»Ó·¢¡£

¢ÛÖÆÈ¡Na2S2O4Ò²³£Óü×ËáÄÆ·¨£¬¿ØÖÆζÈ70¡«80¡æ£¬ÔÚ¼×´¼ÈÜÒº£¨ÈܼÁ£©ÖÐÈܽâ¼×ËáÄÆ£¨HCOONa£©£¬ÔٵμÓNa2CO3ÈÜҺͬʱͨSO2ά³ÖÈÜҺ΢ËáÐÔ,¼´¿ÉÉú³ÉNa2S2O4£¬·´Ó¦µÄÀë×Ó·½³Ìʽ£º_________________________________________¡£

£¨3£©²â¶¨±£ÏÕ·Û´¿¶È

Na2S2O4ÊôÓÚÇ¿»¹Ô­¼Á£¬±©Â¶ÓÚ¿ÕÆøÖÐÒ×±»ÑõÆøÑõ»¯¡£Na2S2O4ÓöKMnO4ËáÐÔÈÜÒº·¢Éú·´Ó¦£º5Na2S2O4 + 6KMnO4 + 4H2SO4 = 5Na2SO4 + 3K2SO4 + 6MnSO4 + 4H2O

³ÆÈ¡3.0g Na2S2O4ÑùÆ·ÈÜÓÚÀäË®ÖУ¬Åä³É100mLÈÜÒº£¬È¡³ö10mL¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÓÃ0.10mol¡¤L-1µÄKMnO4ÈÜÒºµÎ¶¨¡£

Öظ´ÉÏÊö²Ù×÷2´Î£¬Æ½¾ùÏûºÄKMnO4ÈÜÒº18.00 mLÔò¸ÃÑùÆ·ÖÐNa2S2O4µÄÖÊÁ¿·ÖÊýΪ_____________£¨ÔÓÖʲ»²ÎÓë·´Ó¦£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿îë¼°Æ仯ºÏÎïµÄÓ¦ÓÃÕýÈÕÒæ±»ÖØÊÓ¡£

(1)×îÖØÒªµÄº¬îë¿óÎïÊÇÂÌÖùʯ£¬º¬2%¸õ(Cr)µÄÂÌÖùʯ¼´Îª×æĸÂÌ¡£»ù̬CrÔ­×Ó¼Ûµç×ӵĹìµÀ±íʾʽΪ__________¡£

(2)îëÓëÏàÁÚÖ÷×åµÄÂÁÔªËØÐÔÖÊÏàËÆ¡£ÏÂÁÐÓйØîëºÍÂÁµÄÐðÊöÕýÈ·µÄÓÐ______(Ìî×Öĸ)¡£

A£®¶¼ÊôÓÚpÇøÖ÷×åÔªËØ B£®µç¸ºÐÔ¶¼±Èþ´ó

C£®µÚÒ»µçÀëÄܶ¼±Èþ´ó D£®ÂÈ»¯ÎïµÄË®ÈÜÒºpH¾ùСÓÚ7

(3)ÂÈ»¯îëÔÚÆø̬ʱ´æÔÚBeCl2·Ö×Ó(a)ºÍ¶þ¾Û·Ö×Ó(BeCl2)2(b)£¬¹Ì̬ʱÔò¾ßÓÐÈçÏÂͼËùʾµÄÁ´×´½á¹¹(c)¡£

¢ÙaÊôÓÚ________(Ìî¡°¼«ÐÔ¡±»ò¡°·Ç¼«ÐÔ¡±)·Ö×Ó¡£

¢Ú¶þ¾Û·Ö×Ó(BeCl2)2ÖÐBeÔ­×ÓµÄÔÓ»¯·½Ê½Ïàͬ£¬ÇÒËùÓÐÔ­×Ó¶¼ÔÚͬһƽÃæÉÏ¡£b µÄ½á¹¹Ê½Îª__________(±ê³öÅäλ¼ü)¡£

(4)BeOÁ¢·½¾§°ûÈçÏÂͼËùʾ£¬ÈôBeO¾§ÌåµÄÃܶÈΪd g¡¤cm-3£¬Ôò¾§°û²ÎÊýa=______ nm¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔËÓÃÓйظÅÄîÅжÏÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨ £©

A.1mol H2ȼÉշųöµÄÈÈÁ¿ÎªH2µÄȼÉÕÈÈ

B.Na2SO3ºÍH2O2µÄ·´Ó¦ÎªÑõ»¯»¹Ô­·´Ó¦

C.ºÍ»¥ÎªÍ¬ÏµÎï

D.BaSO4µÄË®ÈÜÒº²»µ¼µç£¬¹ÊBaSO4ÊÇÈõµç½âÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿£¨1£©ÏÂÃæÊÇʯÀ¯ÓÍÔÚ³ãÈÈËé´ÉƬµÄ×÷ÓÃϲúÉúC2H4²¢¼ìÑéC2H4ÐÔÖʵÄʵÑ飬Íê³ÉÏÂÁи÷ÎÊÌâ¡£

¢ÙBÖÐÈÜÒºÍÊÉ«£¬ÊÇÒòΪÒÒÏ©±»______________¡£

¢ÚCÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________¡£

¢ÛÔÚD´¦µãȼʱ±ØÐë½øÐеIJÙ×÷ÊÇ_____________¡£

£¨2£©ÊµÑéÊÒÖÆÈ¡µÄÒÒÏ©Öг£»ìÓÐÉÙÁ¿µÄSO2£¬ÓÐÈËÉè¼ÆÈçͼʵÑé×°ÖÃÒÔÖ¤Ã÷ÉÏÊö»ìºÏÆøÌåÖк¬ÓÐÒÒÏ©ºÍ¶þÑõ»¯Áò¡£ÊԻشðÏÂÁÐÎÊÌâ:

¢ÙͼÖÐa¡¢b¡¢c¡¢d×°ÖÃÊ¢·ÅµÄÊÔ¼ÁÒÀ´ÎÊÇ___(ÌîÐòºÅ)¡£

A£®Æ·ºìÈÜÒº

B£®NaOHÈÜÒº

C£®Å¨ÁòËá

D£®ËáÐÔ¸ßÃÌËá¼ØÈÜÒº

¢ÚÄÜ˵Ã÷SO2´æÔÚµÄʵÑéÏÖÏóÊÇ_____________¡£

¢ÛʹÓÃ×°ÖÃbµÄÄ¿µÄÊÇ_____________¡£

¢ÜʹÓÃ×°ÖÃcµÄÄ¿µÄÊÇ_____________¡£

¢ÝÄÜÖ¤Ã÷»ìºÏÆøÌåÖк¬ÓÐÒÒÏ©µÄÏÖÏóÊÇ________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¶þÑõ»¯î棨CeO2£©ÊÇÒ»ÖÖÖØÒªµÄÏ¡ÍÁÑõ»¯Îƽ°åµçÊÓÏÔʾÆÁÉú²ú¹ý³ÌÖвúÉú´óÁ¿µÄ·Ï²£Á§·ÛÄ©£¨º¬SiO2¡¢Fe2O3¡¢CeO2ÒÔ¼°ÆäËûÉÙÁ¿¿ÉÈÜÓÚÏ¡ËáµÄÎïÖÊ£©¡£Ä³¿ÎÌâ×éÒÔ´Ë·ÛĩΪԭÁÏ»ØÊÕî棬Éè¼ÆʵÑéÁ÷³ÌÈçÏ£º

£¨1£©Ï´µÓÂËÔüAµÄÄ¿µÄÊÇΪÁËÈ¥³ý_______£¨ÌîÀë×Ó·ûºÅ£©£¬¼ìÑé¸ÃÀë×ÓÊÇ·ñÏ´µÓµÄ·½·¨ÊÇ_________________________________________________________¡£

£¨2£©µÚ¢Ú²½·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________________________£¬ÂËÔüBµÄÖ÷Òª³É·ÖÊÇ_________¡£

£¨3£©ÝÍÈ¡ÊÇ·ÖÀëÏ¡ÍÁÔªËصij£Ó÷½·¨£¬ÒÑÖª»¯ºÏÎïTBP×÷ΪÝÍÈ¡¼ÁÄܽ«îæÀë×Ó´ÓË®ÈÜÒºÖÐÝÍÈ¡³öÀ´£¬TBP________£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©ÓëË®»¥ÈÜ¡£ÊµÑéÊÒ½øÐÐÝÍÈ¡²Ù×÷ÊÇÓõ½µÄÖ÷Òª²£Á§ÒÇÆ÷ÓÐ_________¡¢ÉÕ±­¡¢²£Á§°ô¡¢Á¿Í²µÈ¡£

£¨4£©È¡ÉÏÊöÁ÷³ÌÖеõ½µÄCe(OH)4²úÆ·0.536g£¬¼ÓÁòËáÈܽâºó£¬ÓÃ0.1000molL-1FeSO4±ê×¼ÈÜÒºµÎ¶¨ÖÕµãÊÇ£¨îæ±»»¹Ô­ÎªCe3+£©£¬ÏûºÄ25.00mL±ê×¼ÈÜÒº£¬¸Ã²úÆ·ÖÐCe(OH)4µÄÖÊÁ¿·ÖÊýΪ__________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸