¡¾ÌâÄ¿¡¿Ä³Ñ§Ï°Ð¡×éÓÃÈçͼËùʾװÖòⶨÂÁþºÏ½ðÖÐÂÁµÄÖÊÁ¿·ÖÊýºÍÂÁµÄÏà¶ÔÔ­×ÓÖÊÁ¿¡£

(1)AÖÐÊÔ¼ÁΪ_________________¡£

(2)ʵÑéÇ°£¬ÏȽ«ÂÁþºÏ½ðÔÚÏ¡ËáÖнþÅÝƬ¿Ì£¬ÆäÄ¿µÄÊÇ__________________________¡£

(3)¼ì²éÆøÃÜÐÔ£¬½«Ò©Æ·ºÍË®×°Èë¸÷ÒÇÆ÷ÖУ¬Á¬½ÓºÃ×°Öúó£¬Ðè½øÐеIJÙ×÷»¹ÓУº¢Ù¼Ç¼CµÄÒºÃæλÖ㻢ڽ«BÖÐÊ£Óà¹ÌÌå¹ýÂË£¬Ï´µÓ£¬¸ÉÔ³ÆÖØ£»¢Û´ýBÖв»ÔÙÓÐÆøÌå²úÉú²¢»Ö¸´ÖÁÊÒκ󣬼ǼCµÄÒºÃæλÖ㻢ÜÓÉAÏòBÖеμÓ×ãÁ¿ÊÔ¼Á¡£ÉÏÊö²Ù×÷µÄ˳ÐòÊÇ__________________ (ÌîÐòºÅ)£»¼Ç¼CµÄÒºÃæλÖÃʱ£¬³ýƽÊÓÍ⣬»¹Ó¦_________________¡£

(4)BÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________________________________¡£

(5)ÈôʵÑéÓÃÂÁþºÏ½ðµÄÖÊÁ¿Îªa g£¬²âµÃÇâÆøÌå»ýΪb mL(ÒÑ»»ËãΪ±ê×¼×´¿ö)£¬BÖÐÊ£Óà¹ÌÌåµÄÖÊÁ¿Îªc g£¬ÔòÂÁµÄÏà¶ÔÔ­×ÓÖÊÁ¿Îª_____________________¡£

¡¾´ð°¸¡¿NaOHÈÜÒº ³ýÈ¥ÂÁþºÏ½ð±íÃæµÄÑõ»¯Ä¤ ¢Ù¢Ü¢Û¢Ú ʹDºÍCµÄÒºÃæÏàƽ 2Al+2NaOH+2H2O£½2NaAlO2+3H2¡ü

¡¾½âÎö¡¿

ÂÁÄÜÓëÇ¿¼î·´Ó¦¶øþ²»ÄÜ£¬ÀûÓÃÕâÒ»ÌØÊâÐÔÖʿɴﵽʵÑéÄ¿µÄ¡£²â¶¨ºÏ½ðÖÐÂÁµÄÖÊÁ¿·ÖÊý¿É³ÆÈ¡Ò»¶¨Á¿ºÏ½ðÑùÆ·£¬²â³öÆäÖÐÂÁµÄÖÊÁ¿£»²âÁ¿ÂÁµÄÏà¶ÔÔ­×ÓÖÊÁ¿£¬¿É²âÆäĦ¶ûÖÊÁ¿=ÂÁµÄÖÊÁ¿/ÂÁµÄÎïÖʵÄÁ¿£¬²âÁ¿Éú³ÉÇâÆøÌå»ý¿ÉÇóÂÁµÄÎïÖʵÄÁ¿¡£

(1)AÖÐÊÔ¼ÁΪǿ¼îÈÜÒº£¬¿ÉÒÔÊÇNaOHÈÜÒºµÈ¡£

(2)ÂÁ¡¢Ã¾¶¼ÊÇ»îÆýðÊô£¬ÔÚ¿ÕÆøÖбíÃæÐγÉÖÂÃÜÑõ»¯Ä¤¡£ÊµÑéÇ°£¬½«ÂÁþºÏ½ðÔÚÏ¡ËáÖнþÅÝƬ¿Ì£¬¾ÍÊÇÒª³ýÈ¥±íÃæµÄÑõ»¯Ä¤¡£

(3)ʵÑéÖÐÐèÒª²â³öÑùÆ·ÖÐÂÁµÄÖÊÁ¿¡¢·´Ó¦Éú³ÉÇâÆøµÄÌå»ý¡£ÔÚ¼ì²éÆøÃÜÐÔ¡¢Ò©Æ·ºÍË®×°ÈëÒÇÆ÷¡¢Á¬½ÓºÃ×°Öú󣬢ټǼCµÄÒºÃæλÖ㻢ÜÓÉAÏòBÖеμÓ×ãÁ¿ÊÔ¼Á£»¢Û´ýBÖв»ÔÙÓÐÆøÌå²úÉú²¢»Ö¸´ÖÁÊÒκ󣬼ǼCµÄÒºÃæλÖ㻢ڽ«BÖÐÊ£Óà¹ÌÌå¹ýÂË£¬Ï´µÓ£¬¸ÉÔ³ÆÖØ£¬¼´²Ù×÷˳ÐòÊǢ٢ܢۢڡ£¼Ç¼CµÄÒºÃæλÖÃʱ£¬Ó¦µ÷ÕûD¹Ü¸ß¶ÈʹDºÍCÖÐÒºÃæÏàƽ£¬ÔÙƽÊÓ¶ÁÊý¡£

(4)ÊÔ¹ÜBÄÚ£¬ºÏ½ðÖеÄÂÁÓëNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Al+2NaOH+2H2O£½2NaAlO2+3H2¡ü¡£

(5)ÉèAlµÄÏà¶ÔÔ­×ÓÖÊÁ¿Îªx£¬Ôò

2Al ~ 3H2

2x g 3¡Á22.4 L

(a-c) g b¡Á10-3 L

µÃx=¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÊµÑéÊÒ´Óº£´øÖÐÌáÈ¡µâµÄÁ÷³ÌÈçͼ£º

ÏÂÁвÙ×÷×°Öò»·ûºÏÒªÇóµÄÊÇ

A. B. C. D.

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£ºN2(g)+3H2(g)2NH3(g) ¡÷H£¼0£¬Æ仯ѧƽºâ³£ÊýKÓëζÈtµÄ¹ØϵÈçÏÂ±í£¬Íê³ÉÏÂÁÐÎÊÌâ¡£

t/¡æ

25

125

225

¡­

K

4.1¡Á105

K1

K2

¡­

£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪK=________£»K1______K2£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±£©£»ÈôÔö´óѹǿʹƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬Ôòƽºâ³£Êý_________£¨Ìî¡°±ä¡±»ò¡°²»±ä¡±£©¡£

£¨2£©Åжϸ÷´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÒÀ¾ÝÊÇ____________£¨ÌîÐòºÅ£©£º

A£®2¦ÔH2£¨Õý£©= 3¦ÔNH3£¨Ä棩 B£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä

C£®ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä D£®N2µÄÏûºÄËÙÂʵÈÓÚH2µÄÏûºÄËÙÂÊ

E£®ÈÝÆ÷ÖÐÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»Ëæʱ¼ä¶ø±ä»¯ F£®»ìºÏÆøÌåµÄÑÕÉ«±£³Ö²»±ä

£¨3£©½«²»Í¬Á¿µÄN2ºÍH2·Ö±ðͨÈëµ½Ìå»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬½øÐÐÉÏÊö·´Ó¦µÃµ½ÈçÏÂÁ½×éÊý¾Ý£º

ʵÑé×é

ζȣ¨¡æ£©

ÆðʼÁ¿£¨mol£©

ƽºâÁ¿£¨mol£©

´ïµ½Æ½ºâËùÐèʱ¼ä£¨min£©

N2

H2

NH3

1

650

2

4

0.9

9

2

900

1

2

0.3

0.01

ʵÑé1ÖÐÒÔ¦Ô(NH3)±íʾµÄ·´Ó¦ËÙÂÊΪ______£¬ÊµÑé2µÄËÙÂʱÈʵÑé1¿ìµÄÔ­ÒòÊÇ____________£»

£¨4£©ÓÐÈËÉè¼Æ²ÉÓøßÐÔÄÜÖÊ×Óµ¼µç²ÄÁÏSCYÌÕ´É£¨ÄÜ´«µÝH+£©£¬ÊµÏÖÁ˳£Ñ¹Ï¼ÈÄܺϳɰ±ÓÖÄÜ·¢µçµÄʵÑé×°Öã¨Èçͼ£©¡£ÔòÆäÕý¼«µÄµç¼«·´Ó¦Îª_____________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿AÔÚÒ»¶¨Î¶ÈÏÂÓзֽⷴӦA (s) = B(s) + C(g) + 4D(g)£¬Èô²âµÃÉú³ÉµÄÆøÌåµÄÖÊÁ¿ÊÇͬÎÂѹÏ£¬ÏàͬÌå»ýÇâÆøµÄ10±¶£¬ÇÒµ±ËùÉú³ÉµÄÆøÌåÔÚ±ê¿öϵÄÌå»ýΪ22.4Lʱ£¬ËùµÃBµÄÖÊÁ¿Îª30.4g£¬AµÄĦ¶ûÖÊÁ¿Îª( )

A.120.4g/molB.50.4g/molC.182.4g/molD.252g/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ïØÊÇÖÆ×÷¸ßÐÔÄÜ°ëµ¼ÌåµÄÖØÒªÔ­ÁÏ¡£¹¤ÒµÉϳ£´Óп¿óÒ±Á¶µÄ·ÏÔüÖлØÊÕïØ¡£ÒÑ֪ijп¿óÔüÖ÷Òªº¬Zn¡¢Si¡¢Pb¡¢Fe¡¢GaµÄÑõ»¯ÎÀûÓøÿóÔüÖÆïصŤÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢ÙïØÔÚÔªËØÖÜÆÚ±íÖÐλÓÚµÚËÄÖÜÆÚµÚ¢óA×壬»¯Ñ§ÐÔÖÊÓëÂÁÏàËÆ¡£

¢Úlg2=0.3 lg3=0.48¡£

¢Û²¿·ÖÎïÖʵÄKspÈçϱíËùʾ¡£

ÎïÖÊ

Zn(OH)2

Ga(OH)3

Fe(OH)2

Fe(OH)3

Ksp

1.6¡Á10£­17

2.7¡Á10£­31

8¡Á10£­16

2.8¡Á10£­39

(1)ΪÁËÌá¸ßËá½þËÙÂÊ£¬³ýÊʵ±Ôö¼ÓÁòËáŨ¶ÈÍ⣬Ӧ²ÉÈ¡µÄ´ëÊ©ÓÐ__________________£¨Ð´³öÁ½Ìõ£©¡£ÂËÔü1µÄÖ÷Òª³É·ÖÊÇÁòËáǦ¼°_________________(д»¯Ñ§Ê½)¡£

(2)¼ÓÈëH2O2µÄÄ¿µÄÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)______________________ ¡£

(3)ÊÒÎÂÌõ¼þÏ£¬Èô½þ³öÒºÖи÷ÑôÀë×ÓµÄŨ¶È¾ùΪ0.01mo/L£¬µ±ÈÜÒºÖÐijÖÖÀë×ÓŨ¶ÈСÓÚ1¡Á10-5mol/Lʱ¼´ÈÏΪ¸ÃÀë×ÓÒÑÍêÈ«³ýÈ¥£¬ÔòpHÓ¦µ÷½ÚµÄ·¶Î§Îª___________________¡£

(4)²Ù×÷D°üÀ¨£º_______________¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔï¡£

(5)µç½â·¨ÖƱ¸½ðÊôïØ¡£ÓöèÐԵ缫µç½âNaGaO2ÈÜÒº¼´¿ÉÖƵýðÊôïØ£¬Ð´³öÒõ¼«µç¼«·´Ó¦Ê½________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓÐA¡¢B¡¢C¡¢D¡¢E¡¢F¡¢G¡¢H¡¢IµÈÈ«²¿ÓɶÌÖÜÆÚÔªËØ×é³ÉµÄ¾ÅÖÖ³£¼ûÎïÖÊ¡£ÒÑÖª£º¢ÙÆäÖÐÖ»ÓÐB¡¢C¡¢FΪµ¥ÖÊ£¬ÇÒ³£Î³£Ñ¹Ï£¬BΪ¹Ì̬£¬C¡¢FΪÆø̬£»¢Ú»¯ºÏÎïÖнöA¡¢EΪÆø̬£»¢ÛÔÚ³£ÎÂÏ£¬DΪµ­»ÆÉ«¹ÌÌ壻¢ÜHµÄÑæÉ«·´Ó¦³Ê»ÆÉ«¡£ËüÃǵÄת»¯¹Øϵ·´Ó¦Ìõ¼þδעÃ÷ÈçͼËùʾ£º

ÊԻشð£º(1)д³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºE________£¬F________£¬H________£¬I________¡£

(2)д³öAÓëD·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_______________¡£

(3)д³öBÓëG·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚ100 mL FeBr2ÈÜÒºÖÐͨÈë±ê×¼×´¿öÏÂ2.24 L Cl2£¬ÈÜÒºÖÐÓеÄBr£­±»Ñõ»¯³Éµ¥ÖÊBr2£¬ÔòÔ­FeBr2ÈÜÒºÖÐFeBr2µÄÎïÖʵÄÁ¿Å¨¶ÈΪ(¡¡¡¡)

A.4 mol L-1B. mol L-1C. mol L-1D.mol L-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿µª»¯Åð£¨BN£©¾§ÌåÓжàÖÖÏà½á¹¹¡£Áù·½Ï൪»¯ÅðÊÇͨ³£´æÔÚµÄÎȶ¨Ï࣬ÓëʯīÏàËÆ£¬¾ßÓвã×´½á¹¹£¬¿É×÷¸ßÎÂÈ󻬼Á£¬µ«²»Äܵ¼µç¡£Á¢·½Ï൪»¯ÅðÊdz¬Ó²²ÄÁÏ£¬ÓÐÓÅÒìµÄÄÍÄ¥ÐÔ¡£ËüÃǵľ§Ìå½á¹¹ÈçͼËùʾ¡£¹ØÓÚÕâÁ½ÖÖ¾§ÌåµÄ˵·¨£¬ÕýÈ·µÄÊÇ

A. Á¢·½Ï൪»¯Åðº¬Åäλ¼üB¡úN

B. Áù·½Ï൪»¯Åð²ã¼ä×÷ÓÃÁ¦Ð¡£¬ËùÒÔÖʵØÈí£¬ÈÛµãµÍ

C. Á½ÖÖµª»¯ÅðÖеÄÅðÔ­×Ó¶¼ÊDzÉÓÃsp2ÔÓ»¯

D. Áù·½Ï൪»¯Åð¾§ÌåÆä½á¹¹ÓëʯīÏàËÆÈ´²»µ¼µç£¬Ô­ÒòÊÇûÓпÉÒÔ×ÔÓÉÒƶ¯µÄµç×Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿1.52 gͭþºÏ½ðÍêÈ«ÈܽâÓÚ50 mLÃܶÈΪ1.40 g/mL¡¢ÖÊÁ¿·ÖÊýΪ63%µÄŨÏõËáÖУ¬µÃµ½NO2ºÍN2O4µÄ»ìºÏÆøÌå1120 mL(±ê×¼×´¿ö)£¬Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈë1.0 mol/L NaOHÈÜÒº£¬µ±½ðÊôÀë×ÓÈ«²¿³Áµíʱ£¬µÃµ½2.54 g³Áµí¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©

A.¸ÃºÏ½ðÖÐÍ­ÓëþµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ2:1

B.¸ÃŨÏõËáÖÐHNO3µÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ14.0 mol/L

C.µÃµ½2.54 g³Áµíʱ£¬¼ÓÈëNaOHÈÜÒºµÄÌå»ýÊÇ600 mL

D.NO2ºÍN2O4µÄ»ìºÏÆøÌåÖУ¬NO2µÄÌå»ý·ÖÊýÊÇ80%

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸