ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿(1)°´ÏµÍ³ÃüÃû·¨ÃüÃû£º

__________________.

(2)°´ÒªÇóд³öÓлúÎïµÄ½á¹¹¼òʽ£ºÖ§Á´Ö»ÓÐÒ»¸öÒÒ»ùÇÒÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄÍéÌþ________.

(3)ôÇ»ùµÄµç×ÓʽΪ________.

(4)ÓëBr2·¢Éú1£¬4¼Ó³É·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________.

(5)ÏÂÁÐ˵·¨´íÎóµÄΪ________(ÌîÑ¡Ïî×Öĸ)

a.ʵÑéÊÒÖÆÈ¡ÒÒϩʱ£¬½«Î¶ȼƵÄË®ÒøÇò²åÈëÒºÃæÒÔÏÂ

b.½«ÂÈÒÒÍéÈÜÓÚµÎÓÐHNO3µÄAgNO3ÈÜÒºÖУ¬¿É¼ìÑéÂÈÒÒÍé·Ö×ÓÖк¬ÓеÄÂÈÔ­×Ó

c.äåÒÒÍéºÍNaOHµÄ´¼ÈÜÒº»ìºÏÎï¼ÓÈÈ£¬½«²úÉúµÄÆøÌåÖ±½ÓͨÈëËáÐÔKMnO4ÈÜÒºÀ´¼ìÑé·´Ó¦ÊÇ·ñÉú³ÉÁËÒÒÏ©

d.½«Í­Ë¿ÈƳÉÂÝÐý×´£¬Ôھƾ«µÆÉϼÓÈȱäºÚºó£¬Á¢¼´ÉìÈëÊ¢ÓÐÎÞË®ÒÒ´¼µÄÊÔ¹ÜÖУ¬ÎÞË®ÒÒ´¼¿É±»Ñõ»¯ÎªÒÒÈ©

(6)AºÍBµÄ·Ö×Óʽ¾ùΪC2H4Br2£¬AµÄºË´Å¹²ÕñÇâÆ×Ö»ÓÐÒ»¸öÎüÊշ壬ÔòAµÄ½á¹¹¼òʽΪ___£¬ÇëÔ¤²âBµÄºË´Å¹²ÕñÇâÆ×ÓÐ___¸öÎüÊÕ·å¡£

¡¾´ð°¸¡¿3£¬3£¬5£¬5£­Ëļ׻ù¸ýÍé +Br2¡ú bc BrCH2CH2Br 2

¡¾½âÎö¡¿

£¨1£©¸ù¾ÝÍéÌþÃüÃûÔ­Ôò£¬¸ÃÓлúÎïΪ3£¬3£¬5£¬5£­Ëļ׻ù¸ýÍ飻

£¨2£©Ö§Á´Ö»ÓÐÒ»¸öÒÒ»ù£¬ÇÒÏà¶Ô·Ö×ÓÖÊÁ¿×îСµÄÍéÌþΪ3£­ÒÒ»ùÎìÍ飬¼´½á¹¹¼òʽΪ£»

£¨3£©ôÇ»ùµÄµç×ÓʽΪ£»

£¨4£©·¢Éú1£¬4¼Ó³É·´Ó¦µÄ·½³ÌʽΪ+Br2¡ú£»

£¨5£©a¡¢ÊµÑéÊÒÖƱ¸ÒÒÏ©£¬Æä·´Ó¦·½³ÌʽΪCH3CH2OHCH2=CH2¡ü£«H2O£¬Î¶ȼƵÄË®ÒøÇò²åÈëÒºÃæÒÔÏ£¬¹Êa˵·¨ÕýÈ·£»

b¡¢ÂÈÒÒÍéÖÐClΪԭ×Ó£¬¼ìÑéÂÈÒÒÍéÐèÒª½«ClÔ­×Óת»¯³ÉCl£­£¬¼´ÂÈÒÒÍé·¢ÉúË®½â·´Ó¦»òÏûÈ¥·´Ó¦£¬ÐèÒªÔÚNaOHË®ÈÜÒº»òNaOH´¼ÈÜÒºÖнøÐУ¬È»ºó¼ÓÈëÏõËáËữÈÜÒº£¬×îºóµÎ¼ÓAgNO3ÈÜÒº£¬¹Û²ì³ÁµíÑÕÉ«£¬¹Êb´íÎó£»

c¡¢²úÉúµÄÆøÌåÖпÉÄܺ¬Óд¼£¬´¼Ò²ÄÜʹËáÐÔ¸ßÃÌËá¼ØÈÜÒºÍÊÉ«£¬¶ÔʵÑé²úÉú¸ÉÈÅ£¬¹Êc˵·¨´íÎó£»

d¡¢Ôھƾ«µÆÉÏ×ÆÉÕ£¬·¢Éú2Cu£«O2 2CuO£¬È»ºóCuO£«CH3CH2OH CH3CHO£«Cu£«H2O£¬¹Êd˵·¨ÕýÈ·£»

£¨6£©AµÄºË´Å¹²ÕñÇâÆ×Ö»ÓÐÒ»¸öÎüÊշ壬˵Ã÷Ö»º¬Ò»ÖÖÇ⣬¼´AµÄ½á¹¹¼òʽΪBrCH2CH2Br£¬ÔòBµÄ½á¹¹¼òʽΪCH3CHBr2£¬ÓÐ2ÖÖ²»Í¬µÄÇâÔ­×Ó£¬¼´ºË´Å¹²ÕñÇâÆ×ÓÐ2ÖÖ·å¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿×ÊÔ´»¯ÀûÓÃCO2£¬¿ÉÒÔ¼õÉÙÎÂÊÒÆøÌåÅÅ·Å£¬»¹¿ÉÒÔ»ñµÃȼÁÏ»òÖØÒªµÄ»¯¹¤²úÆ·¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)CO2µÄ²¶¼¯

¢ÙÓñ¥ºÍNa2CO3ÈÜÒº×öÎüÊÕ¼Á¿É¡°²¶¼¯¡±CO2¡£Ð´³ö¡°²¶¼¯¡±CO2·´Ó¦µÄÀë×Ó·½Ê½_____________¡£

¢Ú¾ÛºÏÀë×ÓÒºÌåÊÇÄ¿Ç°¹ã·ºÑо¿µÄCO2Îü¸½¼Á¡£½áºÏͼÏñ·ÖÎö¾ÛºÏÀë×ÓÒºÌåÎü¸½CO2µÄÓÐÀûÌõ¼þÊÇ_________________________¡£

(2)Éú²úÄòËØ£º

¹¤ÒµÉÏÒÔCO2¡¢NH3ΪԭÁÏÉú²úÄòËØ[CO(NH2)2]£¬¸Ã·´Ó¦·ÖΪ¶þ²½½øÐУº

µÚÒ»²½£º2NH3(g)+CO2(g)H2NCOONH4(s) ¡÷H = - 159.5 kJ¡¤mol-1

µÚ¶þ²½£ºH2NCOONH4(s)CO(NH2)2(s)+ H2O(g) ¡÷H = +116.5 kJ¡¤mol-1

¢Ùд³öÉÏÊöºÏ³ÉÄòËصÄÈÈ»¯Ñ§·½³Ìʽ___________________________¡£¸Ã·´Ó¦»¯Ñ§Æ½ºâ³£ÊýKµÄ±í´ïʽ£º_________________________¡£

¢ÚijʵÑéС×éÄ£Ä⹤ҵÉϺϳÉÄòËØ£¬ÔÚÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖÐͶÈë4mol NH3ºÍ1mol CO2£¬ÊµÑé²âµÃ·´Ó¦Öи÷×é·ÖÎïÖʵÄÁ¿Ëæʱ¼äµÄ±ä»¯ÈçͼËùʾ£º

ÒÑÖª×Ü·´Ó¦µÄ¿ìÂýÓÉÂýµÄÒ»²½·´Ó¦¾ö¶¨£¬ÔòºÏ³ÉÄòËØ×Ü·´Ó¦µÄ¿ìÂýÓɵÚ__________²½·´Ó¦¾ö¶¨£¬×Ü·´Ó¦½øÐе½___________minʱµ½´ïƽºâ

(3)ºÏ³ÉÒÒË᣺Öйú¿Æѧ¼ÒÊ×´ÎÒÔCH3OH¡¢CO2ºÍH2ΪԭÁϸßЧºÏ³ÉÒÒËᣬÆ䷴Ӧ·¾¶ÈçͼËùʾ£º

¢ÙÔ­ÁÏÖеÄCH3OH¿Éͨ¹ýµç½â·¨ÓÉCO2ÖÆÈ¡£¬ÓÃÏ¡ÁòËá×÷µç½âÖÊÈÜÒº£¬Ð´³öÉú³ÉCH3OHµÄµç¼«·´Ó¦Ê½_______________________¡£

¢Ú¸ù¾Ýͼʾ£¬Ð´³ö×Ü·´Ó¦µÄ»¯Ñ§·½³Ì___________¡£

¡¾ÌâÄ¿¡¿ÒÒËáÒÒõ¥ÊÇÖØÒªµÄÓлúºÏ³ÉÖмäÌ壬¹ã·ºÓ¦ÓÃÓÚ»¯Ñ§¹¤Òµ¡£ÊµÑéÊÒÀûÓÃÏÂͼA×°ÖÃÖƱ¸ÒÒËáÒÒõ¥¡£

(1)ÈôʵÑéÖÐÓÃÒÒËáºÍº¬18OµÄÒÒ´¼×÷Ó㬸÷´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ£º______£®Óë½Ì²Ä²ÉÓõÄʵÑé×°Öò»Í¬£¬´Ë×°ÖÃÖвÉÓÃÁËÇòÐθÉÔï¹Ü£¬Æä×÷ÓÃÊÇ______£®

(2)ΪÁËÖ¤Ã÷ŨÁòËáÔڸ÷´Ó¦ÖÐÆðµ½ÁË´ß»¯¼ÁºÍÎüË®¼ÁµÄ×÷Óã¬Ä³Í¬Ñ§ÀûÓÃÉÏͼËùʾװÖýøÐÐÁËÒÔÏÂ4¸öʵÑ飮ʵÑ鿪ʼÏÈÓþƾ«µÆ΢ÈÈ3min£¬ÔÙ¼ÓÈÈʹ֮΢΢·ÐÌÚ3min£®ÊµÑé½áÊøºó³ä·ÖÕñµ´Ð¡ÊԹܢòÔÙ²âÓлú²ãµÄºñ¶È£¬ÊµÑé¼Ç¼ÈçÏ£º

ʵÑé±àºÅ

ÊÔ¹ÜIÖеÄÊÔ¼Á

ÊÔ¹ÜIIÖеÄÊÔ¼Á

Óлú²ãµÄºñ¶È/cm

A

2mLÒÒ´¼¡¢1mLÒÒËá

1mL18molL-1ŨÁòËá

±¥ºÍNa2CO3

3.0

B

2mLÒÒ´¼¡¢1mLÒÒËá

0.1

C

2mLÒÒ´¼¡¢1mLÒÒËá

3mL2molL-1H2SO4

0.6

D

2mLÒÒ´¼¡¢1mLÒÒËá¡¢ÑÎËá

0.6

¢ÙʵÑéDµÄÄ¿µÄÊÇÓëʵÑéCÏà¶ÔÕÕ£¬Ö¤Ã÷H+¶Ôõ¥»¯·´Ó¦¾ßÓд߻¯×÷Óã®ÊµÑéDÖÐÓ¦¼ÓÈëÑÎËáµÄÌå»ýºÍŨ¶È·Ö±ðÊÇ______mLºÍ______molL-1¡£

¢Ú·ÖÎöʵÑé______(ÌîʵÑé±àºÅ)µÄÊý¾Ý£¬¿ÉÒÔÍƲâ³öŨH2SO4µÄÎüË®ÐÔÌá¸ßÁËÒÒËáÒÒõ¥µÄ²úÂÊ¡£

(3)ÈôÏÖÓÐÒÒËá90g£¬ÒÒ´¼138g·¢Éúõ¥»¯·´Ó¦µÃµ½88gÒÒËáÒÒõ¥£¬ÊÔ¼ÆËã¸Ã·´Ó¦µÄ²úÂÊΪ______¡£

(4)Ϊ³ä·ÖÀûÓ÷´Ó¦Î¼×¡¢ÒÒÁ½Î»Í¬Ñ§·Ö±ðÉè¼ÆÁËÈçͼ¼×¡¢ÒÒÁ½¸ö×°ÖÃ(ÒÒͬѧ´ý·´Ó¦Íê±ÏÀäÈ´ºó£¬ÔÙÓñ¥ºÍ̼ËáÄÆÈÜÒºÌáÈ¡ÉÕÆ¿ÖеIJúÎï)£®ÄãÈÏΪ×îºÏÀíµÄÊÇ______¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø