£¨10·Ö£©Ï±íÊÇA¡¢B¶þÖÖÓлúÎïµÄÓйØÐÅÏ¢£»                                     

A
B
¢ÙÄÜʹº¬äåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ«£»
¢Ú±ÈÀýÄ£ÐÍΪ£º

¢ÛÄÜÓëË®ÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦
¢ÙÓÉC¡¢HÁ½ÖÖÔªËØ×é³É£»
¢ÚÇò¹÷Ä£ÐÍΪ£º

¸ù¾Ý±íÖÐÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AÓ뺬äåµÄËÄÂÈ»¯Ì¼ÈÜÒº·´Ó¦µÄÉú³ÉÎïµÄÃû³Æ½Ð×ö                £»
д³öÔÚÒ»¶¨Ìõ¼þÏ£¬AÉú³É¸ß·Ö×Ó»¯ºÏÎïµÄ»¯Ñ§·´Ó¦·½³Ìʽ__                       _¡£
£¨2£© AÓëÇâÆø·¢Éú¼Ó³É·´Ó¦£¬ÆäÉú³ÉÎïµÄͬϵÎïµÄ×é³ÉͨʽΪ________¡£
£¨3£©B¾ßÓеÄÐÔÖÊÊÇ      £¨ÌîÐòºÅ£©£º
¢ÙÎÞÉ«ÎÞζҺÌå¡¢  ¢ÚÓж¾¡¢   ¢Û²»ÈÜÓÚË®¡¢   ¢ÜÃܶȱÈË®´ó¡¢
¢ÝÓëËáÐÔKMnO4ÈÜÒººÍäåË®·´Ó¦ÍÊÉ«¡¢  ¢ÞÈκÎÌõ¼þϲ»ÓëÇâÆø·´Ó¦£»
д³öŨÁòËáºÍŨÏõËáµÄ»ìºÍÎïÓëBÓë·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                       ¡£

(10·Ö,ÿ¿Õ2·Ö)

(1) 1,2-¶þäåÒÒÍé                                   
(2) CnH2nÊ®2  (3)¢Ú¢Û      

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2009?ÄϾ©¶þÄ££©Ñ¡×öÌ⣬±¾ÌâÓÐA¡¢BÁ½Ì⣬·Ö±ð¶ÔÓ¦ÓÚ¡°ÎïÖʽṹÓëÐÔÖÊ¡±ºÍ¡°ÊµÑ黯ѧ¡±Á½¸öÑ¡ÐÞÄ£¿éµÄÄÚÈÝ£¬ÇëÑ¡ÔñÆäÖÐÒ»Ìâ×÷´ð£¬²¢°ÑËùÑ¡ÌâÄ¿¶ÔÓ¦×ÖĸºóµÄ·½¿òÍ¿ºÚ£®ÈôÁ½Ìâ¶¼×÷´ð£¬½«°´AÌâÆÀ·Ö£®
A£®¿ÉÒÔÓÉÏÂÁз´Ó¦ºÏ³ÉÈý¾ÛÇè°·£º
CaO+3C
 ¸ßΠ
.
 
CaC2+CO¡ü       CaC2+N2
 ¸ßΠ
.
 
CaCN2+C¡ü    CaCN2+2H2O=NH2CN+Ca£¨OH£©2£¬NH2CNÓëË®·´Ó¦Éú³ÉÄòËØ[CO£¨NH2£©2]£¬ÄòËØºÏ³ÉÈý¾ÛÇè°·£®
£¨1£©Ð´³öÓëCaÔÚͬһÖÜÆÚÇÒ×îÍâ²ãµç×ÓÊýÏàͬ¡¢ÄÚ²ãÅÅÂúµç×ӵĻù̬ԭ×ӵĵç×ÓÅŲ¼Ê½£º
1s22s22p63s23p63d104s2»ò[Ar]3d104s2
1s22s22p63s23p63d104s2»ò[Ar]3d104s2
£®CaCN2ÖÐÒõÀë×ÓΪCN22-£¬ÓëCN22-»¥ÎªµÈµç×ÓÌåµÄ·Ö×ÓÓÐN2OºÍ
CO2
CO2
£¨Ìѧʽ£©£¬ÓÉ´Ë¿ÉÒÔÍÆÖªCN22-Àë×ӵĿռ乹ÐÍΪ
Ö±ÏßÐÎ
Ö±ÏßÐÎ
£®
£¨2£©ÄòËØ·Ö×ÓÖÐCÔ­×Ó²ÉÈ¡
sp2
sp2
ÔÓ»¯£®ÄòËØ·Ö×ӵĽṹ¼òʽÊÇ
£¬ÆäÖÐ̼ÑõÔ­×ÓÖ®¼äµÄ¹²¼Û¼üÊÇ
C
C
£¨Ìî×Öĸ£©
A£®2¸ö¦Ò¼ü          B£®2¸ö¦Ð¼ü          C£®1¸ö¦Ò¼ü¡¢1¸ö¦Ð¼ü
£¨3£©Èý¾ÛÇè°·£¨£©Ë׳ơ°µ°°×¾«¡±£®¶¯ÎïÉãÈëÈý¾ÛÇè°·ºÍÈý¾ÛÇèËᣨ    £©ºó£¬Èý¾ÛÇèËáÓëÈý¾ÛÇè°··Ö×ÓÏ໥֮¼äͨ¹ý
·Ö×Ó¼äÇâ¼ü
·Ö×Ó¼äÇâ¼ü
½áºÏ£¬ÔÚÉöÔàÄÚÒ×Ðγɽáʯ£®
£¨4£©CaO¾§°ûÈçͼAËùʾ£¬CaO¾§ÌåÖÐCa2+µÄÅäλÊýΪ
6
6
£®CaO¾§ÌåºÍNaCl¾§ÌåµÄ¾§¸ñÄÜ·Ö±ðΪ£ºCaO 3401kJ/mol¡¢NaCl 786kJ/mol£®µ¼ÖÂÁ½Õß¾§¸ñÄܲîÒìµÄÖ÷ÒªÔ­ÒòÊÇ
CaO¾§ÌåÖÐCa 2+¡¢O 2-µÄ´øµçÁ¿´óÓÚNaCl¾§ÌåÖÐNa+¡¢Cl-µÄ´øµçÁ¿
CaO¾§ÌåÖÐCa 2+¡¢O 2-µÄ´øµçÁ¿´óÓÚNaCl¾§ÌåÖÐNa+¡¢Cl-µÄ´øµçÁ¿
£®
B£®ÊµÑéÊÒÓÃÒÒËáºÍÕý¶¡´¼ÖƱ¸ÒÒËáÕý¶¡õ¥£®ÓйØÎïÖʵÄÎïÀíÐÔÖÊÈçÏÂ±í£®Çë»Ø´ðÓйØÎÊÌ⣮
»¯ºÏÎï ÃܶÈ/g?cm-3 ·Ðµã/¡æ Èܽâ¶È/100gË®
Õý¶¡´¼ 0.810 118.0 9
±ù´×Ëá 1.049 118.1 ¡Þ
ÒÒËáÕý¶¡õ¥ 0.882 126.1 0.7
¢ñ£®ÒÒËáÕý¶¡õ¥´Ö²úÆ·µÄÖÆ±¸
ÔÚ¸ÉÔïµÄ50mLÔ²µ×ÉÕÆ¿ÖУ¬×°Èë·Ðʯ£¬¼ÓÈë11.5mLÕý¶¡´¼ºÍ9.4mL±ù´×ËᣬÔÙ¼Ó3¡«4µÎŨÁòËᣮȻºó°²×°·ÖË®Æ÷£¨×÷ÓãºÊµÑé¹ý³ÌÖв»¶Ï·ÖÀë³ýÈ¥·´Ó¦Éú³ÉµÄË®£©¡¢Î¶ȼƼ°»ØÁ÷ÀäÄý¹Ü£¬¼ÓÈÈÀäÄý»ØÁ÷·´Ó¦£®
£¨1£©±¾ÊµÑé¹ý³ÌÖпÉÄܲúÉú¶àÖÖÓлú¸±²úÎд³öÆäÖÐÁ½ÖֵĽṹ¼òʽ£º
CH3CH2CH2CH2OCH2CH2CH2CH3
CH3CH2CH2CH2OCH2CH2CH2CH3
¡¢
CH2=CHCH2CH3
CH2=CHCH2CH3
£®
£¨2£©ÊµÑéÖÐΪÁËÌá¸ßÒÒËáÕý¶¡õ¥µÄ²úÂÊ£¬²ÉÈ¡µÄ´ëÊ©ÊÇ£º
Ó÷ÖË®Æ÷¼°Ê±ÒÆ×ß·´Ó¦Éú³ÉµÄË®£¬¼õÉÙÉú³ÉÎïµÄŨ¶È£»
Ó÷ÖË®Æ÷¼°Ê±ÒÆ×ß·´Ó¦Éú³ÉµÄË®£¬¼õÉÙÉú³ÉÎïµÄŨ¶È£»
¡¢
ʹÓùýÁ¿´×ËᣬÌá¸ßÕý¶¡´¼µÄת»¯ÂÊ
ʹÓùýÁ¿´×ËᣬÌá¸ßÕý¶¡´¼µÄת»¯ÂÊ
£®
¢ò£®ÒÒËáÕý¶¡õ¥´Ö²úÆ·µÄÖÆ±¸
£¨1£©½«ÒÒËáÕý¶¡õ¥´Ö²úÆ·ÓÃÈçϵIJÙ×÷½øÐо«ÖÆ£º¢Ùˮϴ  ¢ÚÕôÁó  ¢ÛÓÃÎÞË®MgSO4¸ÉÔï ¢ÜÓÃ10%̼ËáÄÆÏ´µÓ£¬ÕýÈ·µÄ²Ù×÷²½ÖèÊÇ
C
C
£¨Ìî×Öĸ£©£®
A£®¢Ù¢Ú¢Û¢ÜB£®¢Û¢Ù¢Ü¢ÚC£®¢Ù¢Ü¢Ù¢Û¢ÚD£®¢Ü¢Ù¢Û¢Ú¢Û
£¨2£©½«õ¥²ã²ÉÓÃÈçͼBËùʾװÖÃÕôÁó£®
1£®³öͼÖÐÒÇÆ÷AµÄÃû³Æ
ÀäÄý¹Ü
ÀäÄý¹Ü
£®ÀäÈ´Ë®´Ó
ÏÂ
ÏÂ
¿Ú½øÈ루Ìî×Öĸ£©£®
2£®¢ÚÕôÁóÊÕ¼¯ÒÒËáÕý¶¡õ¥²úƷʱ£¬Ó¦½«Î¶ȿØÖÆÔÚ
126.1¡æ
126.1¡æ
×óÓÒ£®
¢ó£®¼ÆËã²úÂÊ
²âÁ¿·ÖË®Æ÷ÄÚÓÉÒÒËáÓëÕý¶¡´¼·´Ó¦Éú³ÉµÄË®Ìå»ýΪ1.8mL£¬¼ÙÉèÔÚÖÆÈ¡ÒÒËáÕý¶¡õ¥¹ý³ÌÖз´Ó¦ÎïºÍÉú³ÉÎïûÓÐËðʧ£¬ÇÒºöÂÔ¸±·´Ó¦£¬¼ÆËãÒÒËáÕý¶¡õ¥µÄ²úÂÊ
79.4%
79.4%
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

A£®£¨1£©Çëд³öÒÔÒÒϩΪÓлúÔ­ÁϺϳÉÒÒËáÒÒõ¥µÄ¸÷²½·´Ó¦·½³Ìʽ£¬²¢×¢Ã÷·´Ó¦ÀàÐÍ£®
CH2=CH2+H2O
´ß»¯¼Á
CH3CH2OH
CH2=CH2+H2O
´ß»¯¼Á
CH3CH2OH
£º¼Ó³É·´Ó¦
CH3COOH+CH3CH2OH
ŨÁòËá
¡÷
CH3COOCH2CH3+H2O
CH3COOH+CH3CH2OH
ŨÁòËá
¡÷
CH3COOCH2CH3+H2O
£º
õ¥»¯·´Ó¦
õ¥»¯·´Ó¦
   
2CH3CHO+O2
´ß»¯¼Á
2CH3COOH
2CH3CHO+O2
´ß»¯¼Á
2CH3COOH
£ºÑõ»¯·´Ó¦
£¨2£©Ï±íÊÇA¡¢B¶þÖÖÓлúÎïµÄÓйØÐÅÏ¢£»
A B
 ¢Ù·Ö×ÓʽΪC3H8O£»
 
 ¢ÚÔÚCuµÄ×÷ÓÃÏ´߻¯Ñõ»¯µÄ²úÎï
 
²»ÄÜ·¢ÉúÒø¾µ·´Ó¦
 
 
 
  ¢ÙÓÉC¡¢HÁ½ÖÖÔªËØ×é³É£»
 
  ¢ÚÇò¹÷Ä£ÐÍΪ£º
¸ù¾Ý±íÖÐÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙAµÄ½á¹¹¼òʽΪ
CH3CH£¨OH£©CH3
CH3CH£¨OH£©CH3
£®
¢Úд³öÔÚŨÁòËá×÷ÓÃÏÂA·¢ÉúÏûÈ¥·´Ó¦µÄ»¯Ñ§·½³Ìʽ
CH3CH£¨OH£©CH3
ŨH2SO4
¡÷
CH2=CHCH3+H2O
CH3CH£¨OH£©CH3
ŨH2SO4
¡÷
CH2=CHCH3+H2O
£®
¢ÛB¾ßÓеÄÐÔÖÊÊÇ
bc
bc
£¨ÌîÐòºÅ£©£º
a£®ÎÞÉ«ÎÞζҺÌ壻   b£®Óж¾£»   c£®²»ÈÜÓÚË®£»  d£®ÃܶȱÈË®´ó£»  e£®ÓëËáÐÔKMnO4 ÈÜÒººÍäåË®·´Ó¦ÍÊÉ«£»  f£®ÈκÎÌõ¼þϲ»ÓëÇâÆø·´Ó¦£»
д³öÔÚŨÁòËá×÷ÓÃÏ£¬BÓëŨÏõËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
+HNO3£¨Å¨£©
ŨH2SO4
¡÷
+HNO3£¨Å¨£©
ŨH2SO4
¡÷
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2012?ºÓ¶«ÇøÒ»Ä££©ÊÒÄÚ¼×È©ÎÛȾÒѱ»ÁÐÈë¶Ô¹«ÖÚ½¡¿µÓ°Ïì×î´óµÄ»·¾³ÒòËØÖ®Ò»£¬ÊÐÃæÉÏÓÐÒ»ÖÖ¹â´ß»¯Í¿ÁÏ£¬Äܹ»Æðµ½½µµÍÊÒÄÚ¼×È©º¬Á¿µÄ×÷Óã®Ä³»¯Ñ§ÐËȤС×éÏëҪͨ¹ý´ËÍ¿ÁÏÀ´½øÐм×È©µÄ¹â´ß»¯Ñõ»¯·´Ó¦£¬²¢²â¶¨±»Ñõ»¯µÄ¼×È©µÄÁ¿£®Éè¼ÆÊµÑé×°ÖÃÈçÏ£º

ʵÑéʱÏÈ¿ªÆô×ÏÍâµÆ£¬»º»º¹ÄÈë¿ÕÆø£»Ò»¶Îʱ¼äºóÍ£Ö¹·´Ó¦£¬¹Ø±Õ×ÏÍâµÆ£¬ÔÙ¼ÌÐøÍ¨ÈëÒ»»á¶ù¿ÕÆø£¬ÏòBÖмÓÈë×ãÁ¿BaCl2ÈÜÒº£¬µÃµ½°×É«³Áµí5.91g£®
£¨1£©¸£¶ûÂíÁÖÊÇ´ß»¯Ñõ»¯¼×´¼ÖƵõĺ¬¼×È©37%µÄË®ÈÜÒº£¬Òò´Ëº¬ÓжþÖÖÓлúÔÓÖÊ£¬ËüÃÇÊÇ
¼×´¼
¼×´¼
£¬
¼×Ëá
¼×Ëá
£¬¸£¶ûÂíÁÖµÄÒ»¸öÓÃ;ÊÇ
½þÖÆÉúÎï±ê±¾
½þÖÆÉúÎï±ê±¾
£®
£¨2£©ÉÏͼװÖÃAÖУ¬¼îʯ»ÒµÄ×÷ÓÃÊÇ
ÎüÊÕ¿ÕÆøÖеÄCO2
ÎüÊÕ¿ÕÆøÖеÄCO2
£¬ºãÎÂˮԡµÄÄ¿µÄÊÇ
ʹ»ìºÏÆøÖм×È©º¬Á¿Îȶ¨£¨Ð´Ê¹¼×È©»Ó·¢Òà¿É£©
ʹ»ìºÏÆøÖм×È©º¬Á¿Îȶ¨£¨Ð´Ê¹¼×È©»Ó·¢Òà¿É£©
£®
£¨3£©Ó²Öʲ£Á§¹ÜÖУ¬¼×È©ÔÚ¹â´ß»¯Ñõ»¯Ìõ¼þϵķ´Ó¦·½³Ìʽ£º
HCHO+O2
´ß»¯¼Á
×ÏÍâ¹â
CO2+H2O
HCHO+O2
´ß»¯¼Á
×ÏÍâ¹â
CO2+H2O
£»
£¨4£©ÏòBÖмÓÈë×ãÁ¿BaCl2ÈÜÒººóÖÁµÃµ½5.91g°×É«³ÁµíµÄһϵÁвÙ×÷ÒÀ´ÎÊÇ£º¹ýÂË¡¢
Ï´µÓ
Ï´µÓ
¡¢¸ÉÔï¡¢
³ÆÁ¿
³ÆÁ¿
£»¹ýÂËʱÐèÓõIJ£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ôÍ⻹ÓÐ
ÆÕͨ©¶·
ÆÕͨ©¶·
£®
£¨5£©¼ÆËã±»Ñõ»¯µÄ¼×È©µÄÎïÖʵÄÁ¿Îª
0.03
0.03
mol
£¨6£©·´Ó¦½áÊøºó£¬¼ÌÐøÔâÈëÒ»»á¶ù¿ÕÆøµÄÄ¿µÄÊÇ
½«²ÐÁôÔÚ×°ÖÃÖеġ¢CO2Åųö£¬±»NaOHÎüÊÕ£¬¼õСÎó²î
½«²ÐÁôÔÚ×°ÖÃÖеġ¢CO2Åųö£¬±»NaOHÎüÊÕ£¬¼õСÎó²î
£®
£¨7£©¼×ͬѧÈÏΪ£¬·´Ó¦¹ý³ÌÖв¿·Ö¼×È©¿ÉÄܱ»Ñõ»¯Îª¼×ËᣮΪÑéÖ¤Æä´æÔÚ£¬¼×ͬѧȡBÖÐÈÜÒº£¬·Ö±ðÑ¡ÓÃϱíÊÔ¼Á½øÐмìÑ飮µ«ÒÒͬѧÈÏΪÊÔ¼ÁÑ¡Ôñ¾ù²»ºÏÀí£¬ËûµÄÀíÓÉÊÇ£º
¼×ͬѧѡÔñµÄÊÔ¼Á ÒÒͬѧÈÏΪ²»ºÏÀíµÄÀíÓÉ
×ÏɫʯÈïÊÔÒº
×ÏɫʯÈïÊÔÒºBÖÐÈÜÒº³Ê¼îÐÔ£¨NaOH×ãÁ¿£©£¬ÎÞ·¨Ö¤Ã÷ÊǼ×ËᣨËáÐÔ£©
×ÏɫʯÈïÊÔÒºBÖÐÈÜÒº³Ê¼îÐÔ£¨NaOH×ãÁ¿£©£¬ÎÞ·¨Ö¤Ã÷ÊǼ×ËᣨËáÐÔ£©
ÐÂÖÆÇâÑõ»¯Í­
BÖпÉÄÜÈÜÓйýÁ¿µÄ¼×È©£¬ÎÞ·¨Ö¤Ã÷ÊǼ×ËáÖÐÈ©»ù
BÖпÉÄÜÈÜÓйýÁ¿µÄ¼×È©£¬ÎÞ·¨Ö¤Ã÷ÊǼ×ËáÖÐÈ©»ù

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ͼÖРA¡¢B¡¢C¡¢D¡¢E¾ùΪÓлú»¯ºÏÎÒÑÖª£ºCÄܸúNaHCO3·¢Éú·´Ó¦£¬CºÍDµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÏàµÈ£¬ÇÒEΪÎÞÖ§Á´µÄ»¯ºÏÎ

¸ù¾Ýͼ»Ø´ðÎÊÌ⣺
£¨1£©C·Ö×ÓÖеĹÙÄÜÍÅÃû³ÆÊÇ£º
ôÈ»ù
ôÈ»ù
£»
ÏÂÁз´Ó¦ÖУ¬»¯ºÏÎïB²»ÄÜ·¢ÉúµÄ·´Ó¦ÊÇ
e
e
£¨Ìî×ÖĸÐòºÅ£©£º
a¡¢¼Ó³É·´Ó¦  b¡¢È¡´ú·´Ó¦  c¡¢ÏûÈ¥·´Ó¦    d¡¢õ¥»¯·´Ó¦  e¡¢Ë®½â·´Ó¦  f¡¢Öû»·´Ó¦
£¨2£©·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ
CH3COOH+CH3CH2CH2OH
ŨÁòËá
¼ÓÈÈ
CH3COOCH2CH2CH3+H2O
CH3COOH+CH3CH2CH2OH
ŨÁòËá
¼ÓÈÈ
CH3COOCH2CH2CH3+H2O
£®
£¨3£©AµÄ½á¹¹¼òʽÊÇ
£®
£¨4£©Í¬Ê±·ûºÏÏÂÁÐÈý¸öÌõ¼þµÄBµÄͬ·ÖÒì¹¹ÌåµÄÊýÄ¿ÓÐ
4
4
¸ö£®
¢ñ£®º¬Óмä¶þÈ¡´ú±½»·½á¹¹£»¢ò£®ÊôÓÚ·Ç·¼ÏãËáõ¥£»¢ó£®Óë FeCl3 ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£®
д³öÆäÖÐÈÎÒâÒ»¸öͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ
д³öËÄÕßÖ®Ò»¼´¿É
д³öËÄÕßÖ®Ò»¼´¿É

£¨5£©³£ÎÂÏ£¬½«CÈÜÒººÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºpHÈçÏÂ±í£º
ʵÑé±àºÅ CÎïÖʵÄÁ¿Å¨¶È£¨mol?L-1£© NaOHÎïÖʵÄÁ¿Å¨¶È£¨mol?L-1£© »ìºÏÈÜÒºµÄpH
m 0.1 0.1 pH=9
n 0.2 0.1 pH£¼7
´Óm×éÇé¿ö·ÖÎö£¬ËùµÃ»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨OH-£©=
10-5
10-5
mol?L-1£®
n×é»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©
c£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£ººÓ¶«ÇøÒ»Ä£ ÌâÐÍ£ºÎÊ´ðÌâ

ÊÒÄÚ¼×È©ÎÛȾÒѱ»ÁÐÈë¶Ô¹«ÖÚ½¡¿µÓ°Ïì×î´óµÄ»·¾³ÒòËØÖ®Ò»£¬ÊÐÃæÉÏÓÐÒ»ÖÖ¹â´ß»¯Í¿ÁÏ£¬Äܹ»Æðµ½½µµÍÊÒÄÚ¼×È©º¬Á¿µÄ×÷Óã®Ä³»¯Ñ§ÐËȤС×éÏëҪͨ¹ý´ËÍ¿ÁÏÀ´½øÐм×È©µÄ¹â´ß»¯Ñõ»¯·´Ó¦£¬²¢²â¶¨±»Ñõ»¯µÄ¼×È©µÄÁ¿£®Éè¼ÆÊµÑé×°ÖÃÈçÏ£º

¾«Ó¢¼Ò½ÌÍø

ʵÑéʱÏÈ¿ªÆô×ÏÍâµÆ£¬»º»º¹ÄÈë¿ÕÆø£»Ò»¶Îʱ¼äºóÍ£Ö¹·´Ó¦£¬¹Ø±Õ×ÏÍâµÆ£¬ÔÙ¼ÌÐøÍ¨ÈëÒ»»á¶ù¿ÕÆø£¬ÏòBÖмÓÈë×ãÁ¿BaCl2ÈÜÒº£¬µÃµ½°×É«³Áµí5.91g£®
£¨1£©¸£¶ûÂíÁÖÊÇ´ß»¯Ñõ»¯¼×´¼ÖƵõĺ¬¼×È©37%µÄË®ÈÜÒº£¬Òò´Ëº¬ÓжþÖÖÓлúÔÓÖÊ£¬ËüÃÇÊÇ______£¬______£¬¸£¶ûÂíÁÖµÄÒ»¸öÓÃ;ÊÇ______£®
£¨2£©ÉÏͼװÖÃAÖУ¬¼îʯ»ÒµÄ×÷ÓÃÊÇ______£¬ºãÎÂˮԡµÄÄ¿µÄÊÇ______£®
£¨3£©Ó²Öʲ£Á§¹ÜÖУ¬¼×È©ÔÚ¹â´ß»¯Ñõ»¯Ìõ¼þϵķ´Ó¦·½³Ìʽ£º______£»
£¨4£©ÏòBÖмÓÈë×ãÁ¿BaCl2ÈÜÒººóÖÁµÃµ½5.91g°×É«³ÁµíµÄһϵÁвÙ×÷ÒÀ´ÎÊÇ£º¹ýÂË¡¢______¡¢¸ÉÔï¡¢______£»¹ýÂËʱÐèÓõIJ£Á§ÒÇÆ÷³ýÉÕ±­¡¢²£Á§°ôÍ⻹ÓÐ______£®
£¨5£©¼ÆËã±»Ñõ»¯µÄ¼×È©µÄÎïÖʵÄÁ¿Îª______mol
£¨6£©·´Ó¦½áÊøºó£¬¼ÌÐøÔâÈëÒ»»á¶ù¿ÕÆøµÄÄ¿µÄÊÇ______£®
£¨7£©¼×ͬѧÈÏΪ£¬·´Ó¦¹ý³ÌÖв¿·Ö¼×È©¿ÉÄܱ»Ñõ»¯Îª¼×ËᣮΪÑéÖ¤Æä´æÔÚ£¬¼×ͬѧȡBÖÐÈÜÒº£¬·Ö±ðÑ¡ÓÃϱíÊÔ¼Á½øÐмìÑ飮µ«ÒÒͬѧÈÏΪÊÔ¼ÁÑ¡Ôñ¾ù²»ºÏÀí£¬ËûµÄÀíÓÉÊÇ£º
¼×ͬѧѡÔñµÄÊÔ¼Á ÒÒͬѧÈÏΪ²»ºÏÀíµÄÀíÓÉ
×ÏɫʯÈïÊÔÒº ______
ÐÂÖÆÇâÑõ»¯Í­ ______

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸