£¨1£©´×Ëá¿ÉÓÃÓÚ³ýȥˮºøÖеÄË®¹¸£¬³É·ÝÖ÷ÒªÊÇCaCO3ºÍMg£¨OH£©2£®Ð´³ö´×ËáÓëMg£¨OH£©2·´Ó¦µÄÀë×Ó·½³Ìʽ
Mg£¨OH£©2+2CHCOOH=2CH3COO-+Mg2++2H2O
Mg£¨OH£©2+2CHCOOH=2CH3COO-+Mg2++2H2O
£®
£¨2£©½«11.2L£¨±ê×¼×´¿ö£©ÒÒÏ©ºÍÒÒÍéµÄ»ìºÏÆøÌåͨÈë×ãÁ¿µÄäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÖУ¬³ä·Ö·´Ó¦ºó£¬äåµÄËÄÂÈ»¯Ì¼ÈÜÒºÖÊÁ¿Ôö¼ÓÁË8.4¿Ë£¬ÔòÔ­»ìºÏÆøÌåÖÐÏ©ºÍÒÒÍéµÄÖÊÁ¿Ö®±ÈΪ£º
7£º5
7£º5

£¨3£©ÔÚ2LÃܱÕÈÝÆ÷ÖУ¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦A£¨g£©+3B £¨g£©   2C£¨g£©£¬ÔÚ10ÃëÄÚ·´Ó¦ÎïAµÄŨ¶ÈÓÉ1mol/L½µµ½0.6mol/L£¬ÔòÓÃCŨ¶ÈµÄ±ä»¯±íʾµÄ¸Ã·´Ó¦ÔÚÕâ¶Îʱ¼äÄÚµÄƽ¾ù·´Ó¦ËÙÂÊΪ
0.08mol/£¨L?s£©
0.08mol/£¨L?s£©
£®
·ÖÎö£º£¨1£©¸ù¾Ý´×Ëá¾ßÓÐËáµÄͨÐÔ£¬ÄܺͼӦÉú³ÉÑκÍË®£»
£¨2£©ÒÒÏ©º¬ÓÐË«¼ü£¬ÄÜÓëäåË®·¢Éú¼Ó³É·´Ó¦£¬ÒÒÏ©ºÍÒÒÍéµÄ»ìºÏÆøÌåͨÈë×ãÁ¿äåË®ÖУ¬³ä·Ö·´Ó¦ºó£¬äåË®µÄÖÊÁ¿Ôö¼ÓÁË8.4g£¬¼´ÎªÒÒÏ©µÄÖÊÁ¿£®¸ù¾Ýn=
m
M
=
V
Vm 
¼ÆËãÎïÖʵÄÁ¿¡¢ÖÊÁ¿¹Øϵ£»
£¨3£©ÏÈÇó³öAµÄƽ¾ù·´Ó¦ËÙÂÊ£¬ÔÙ¸ù¾Ýͬһ·´Ó¦£¬²»Í¬ÎïÖÊ·´Ó¦ËÙÂÊËÙÂÊÖ®±ÈµÈÓÚ¼ÆÁ¿ÊýÖ®±È£¬Çó³öCµÄƽ¾ù·´Ó¦ËÙÂÊ£®
½â´ð£º½â£º£¨1£©ÒòCHCOOHÄܺÍMg£¨OH£©2£¬Éú³É£¨CH3COO£©2MgºÍË®£¬»¯Ñ§·½³ÌʽΪ£º2CHCOOH+Mg£¨OH£©2¨T£¨CH3COO£©2Mg+2H2O£¬¹Ê´ð°¸Îª£ºMg£¨OH£©2+2CHCOOH=2CH3COO-+Mg2++2H2O£¬
£¨2£©11.2L»ìºÏÆøÌåµÄÎïÖʵÄÁ¿Îªn=
V
Vm 
=
11.2L
22.4L/mol
=0.5mol£¬ÒÒÏ©º¬ÓÐË«¼ü£¬ÄÜÓëäåË®·¢Éú¼Ó³É·´Ó¦£¬ÒÒÏ©ºÍÒÒÍéµÄ»ìºÏÆøÌåͨÈë×ãÁ¿äåË®ÖУ¬³ä·Ö·´Ó¦ºó£¬äåË®µÄÖÊÁ¿Ôö¼ÓÁË8.4g£¬¼´ÎªÒÒÏ©µÄÖÊÁ¿£¬ËùÒÔÒÒÏ©µÄÎïÖʵÄÁ¿Îªn=
m
M
=
8.4g
28g/mol
=0.3mol£¬ÔòÒÒÍéµÄÎïÖʵÄÁ¿Îª£º0.5mol-0.3mol=0.2mol£¬ÖÊÁ¿Îª£º0.2mol¡Á30g/mol=6g£¬ËùÒÔ£¬ÒÒÏ©ÓëÒÒÍéµÄÖÊÁ¿Ö®±ÈΪ8.4g£º6g=7£º5£¬¹Ê´ð°¸Îª£º7£º5£¬
£¨3£©ÔÚ10ÃëÄÚv£¨A£©=
¡÷C
¡÷t
=
1.0mol/L-0.6mol/L
10s
=0.04mol?L-1?s-1£¬¶øͬһ·´Ó¦£¬²»Í¬ÎïÖÊ·´Ó¦ËÙÂÊËÙÂÊÖ®±ÈµÈÓÚ¼ÆÁ¿ÊýÖ®±È£¬ËùÒÔv£¨C£©=2 v£¨A£©=2¡Á0.04mol?L-1?s-1=0.08mol?L-1?s-1£¬¹Ê´ð°¸Îª£º0.08mol/£¨L?s£©£®
µãÆÀ£º±¾ÌâÎÊÌ⣨2£©Ö÷Òª¿¼²éÎïÖʵÄÁ¿µÄÏà¹Ø¼ÆË㣬ÌâÄ¿ÄѶȲ»´ó£¬×¢ÒâÒÒÏ©ÓëÒÒÍéµÄÐÔÖʲî±ðÊǽâÌâµÄ¹Ø¼ü£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Äð¾ÆºÍÔì´×ÊǹŴúÀͶ¯ÈËÃñµÄÖǻ۽ᾧ£¬°×¾ÆºÍ´×Ò²ÊÇÈÕ³£Éú»îÖг£¼ûµÄÓлúÎ
£¨1£©´×Ëá¿ÉÓÃÓÚ³ýȥˮºøÖеÄË®¹¸£¬³É·ÝÖ÷ÒªÊÇCaCO3ºÍMg£¨OH£©2£®Ð´³ö´×ËáÓëMg£¨OH£©2·´Ó¦µÄÀë×Ó·½³Ìʽ
2CH3COOH+CaCO3¨T2CH3COO-+Ca2++CO2¡ü+H2O
2CH3COOH+CaCO3¨T2CH3COO-+Ca2++CO2¡ü+H2O
£®Ê³ÑÎÖг£¼ÓÈëKIO3À´²¹µâ£¬ÎªÑé֤ʳÑÎÖеÄKIO3£¬¿ÉÔÚʳÑÎÖмÓÈë´×ËᣬÔÙ¼ÓÈëKIºÍµí·ÛÈÜÒº£®·´Ó¦µÄÀë×Ó·½³ÌʽΪ
IO3-+5I-+6CH3COOH=6CH3COO-+3H2O+3I2
IO3-+5I-+6CH3COOH=6CH3COO-+3H2O+3I2
£®
£¨2£©ÄðÖÆÃ׾ƵĹý³ÌÖÐÊÇÃ×Öеĵí·ÛÔÚøµÄ×÷ÓÃÏÂÉú³ÉÆÏÌÑÌÇ£®Ð´³ö»¯Ñ§·½³Ìʽ
£¨C6H10O5£©n£¨µí·Û£©+nH2O
µí·Ûø
nC6H12O6£¨ÆÏÌÑÌÇ£©
£¨C6H10O5£©n£¨µí·Û£©+nH2O
µí·Ûø
nC6H12O6£¨ÆÏÌÑÌÇ£©
£¬È»ºóÆÏÌÑÌÇÔÚøµÄ×÷ÓÃÏÂÉú³ÉÒÒ´¼£®
£¨3£©ÔÚÄðÖÆÃ׾ƵĹý³ÌÖУ¬ÓпÉÄܱäËᣬÕâÀïÒòΪ²úÉúÁË´×ËᣮΪÁ˳ýÈ¥Ã×¾ÆÖеÄËáζÓÐÈ˽¨ÒéÏòÃ×¾ÆÖмÓÈëºìÈȵÄǦ£¬Ê¹ËüÓë´×ËáÉú³ÉÓÐÌðζµÄ´×ËáǦ£®ÄãÈÏΪ´Ë·¨ÊÇ·ñ¿ÉÐÐ
²»¿ÉÐÐ
²»¿ÉÐÐ
£¨ÌîÐУ¬²»¿ÉÐУ©£¬Ô­ÒòÊÇ
ǦÊÇÖؽðÊô£¬Pb2+¶ÔÈËÌåÓж¾
ǦÊÇÖؽðÊô£¬Pb2+¶ÔÈËÌåÓж¾
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨9·Ö£©Äð¾ÆºÍÔì´×ÊǹŴúÀͶ¯ÈËÃñµÄÖǻ۽ᾧ£¬°×¾ÆºÍ´×Ò²ÊÇÈÕ³£Éú»îÖг£¼ûµÄÓлúÎï¡£

£¨1£©´×Ëá¿ÉÓÃÓÚ³ýȥˮºøÖеÄË®¹¸,³É·ÝÖ÷ÒªÊÇCaCO3ºÍMg(OH) 2¡£Ð´³ö´×ËáÓëMg(OH) 2·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________¡£Ê³ÑÎÖг£¼ÓÈëKIO3À´²¹µâ£¬ÎªÑé֤ʳÑÎÖеÄKIO3£¬¿ÉÔÚʳÑÎÖмÓÈë´×ËᣬÔÙ¼ÓÈëKIºÍµí·ÛÈÜÒº¡£·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________¡£

£¨2£©ÄðÖÆÃ׾ƵĹý³ÌÖÐÊÇÃ×Öеĵí·ÛÔÚøµÄ×÷ÓÃÏÂÉú³ÉÆÏÌÑÌÇ¡£Ð´³ö»¯Ñ§·½³Ìʽ_____________________________£¬È»ºóÆÏÌÑÌÇÔÚøµÄ×÷ÓÃÏÂÉú³ÉÒÒ´¼¡£

£¨3£©ÔÚÄðÖÆÃ׾ƵĹý³ÌÖУ¬ÓпÉÄܱäËᣬÕâÀïÒòΪ²úÉúÁË´×ËᡣΪÁ˳ýÈ¥Ã×¾ÆÖеÄËáζÓÐÈ˽¨ÒéÏòÃ×¾ÆÖмÓÈëºìÈȵÄǦ£¬Ê¹ËüÓë´×ËáÉú³ÉÓÐÌðζµÄ´×ËáǦ¡£ÄãÈÏΪ´Ë·¨ÊÇ·ñ¿ÉÐÐ_____£¨ÌîÐУ¬²»¿ÉÐУ©£¬Ô­ÒòÊÇ___________________________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äêºþ±±Ê¡Ð¢¸Ð¸ßÖиßÒ»ÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

£¨9·Ö£©Äð¾ÆºÍÔì´×ÊǹŴúÀͶ¯ÈËÃñµÄÖǻ۽ᾧ£¬°×¾ÆºÍ´×Ò²ÊÇÈÕ³£Éú»îÖг£¼ûµÄÓлúÎï¡£
£¨1£©´×Ëá¿ÉÓÃÓÚ³ýȥˮºøÖеÄË®¹¸,³É·ÝÖ÷ÒªÊÇCaCO3ºÍMg(OH) 2¡£Ð´³ö´×ËáÓëMg(OH) 2·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________¡£Ê³ÑÎÖг£¼ÓÈëKIO3À´²¹µâ£¬ÎªÑé֤ʳÑÎÖеÄKIO3£¬¿ÉÔÚʳÑÎÖмÓÈë´×ËᣬÔÙ¼ÓÈëKIºÍµí·ÛÈÜÒº¡£·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________¡£
£¨2£©ÄðÖÆÃ׾ƵĹý³ÌÖÐÊÇÃ×Öеĵí·ÛÔÚøµÄ×÷ÓÃÏÂÉú³ÉÆÏÌÑÌÇ¡£Ð´³ö»¯Ñ§·½³Ìʽ_____________________________£¬È»ºóÆÏÌÑÌÇÔÚøµÄ×÷ÓÃÏÂÉú³ÉÒÒ´¼¡£
£¨3£©ÔÚÄðÖÆÃ׾ƵĹý³ÌÖУ¬ÓпÉÄܱäËᣬÕâÀïÒòΪ²úÉúÁË´×ËᡣΪÁ˳ýÈ¥Ã×¾ÆÖеÄËáζÓÐÈ˽¨ÒéÏòÃ×¾ÆÖмÓÈëºìÈȵÄǦ£¬Ê¹ËüÓë´×ËáÉú³ÉÓÐÌðζµÄ´×ËáǦ¡£ÄãÈÏΪ´Ë·¨ÊÇ·ñ¿ÉÐÐ_____£¨ÌîÐУ¬²»¿ÉÐУ©£¬Ô­ÒòÊÇ___________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äêºþ±±Ê¡¸ßÒ»ÏÂѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

£¨9·Ö£©Äð¾ÆºÍÔì´×ÊǹŴúÀͶ¯ÈËÃñµÄÖǻ۽ᾧ£¬°×¾ÆºÍ´×Ò²ÊÇÈÕ³£Éú»îÖг£¼ûµÄÓлúÎï¡£

£¨1£©´×Ëá¿ÉÓÃÓÚ³ýȥˮºøÖеÄË®¹¸,³É·ÝÖ÷ÒªÊÇCaCO3ºÍMg(OH) 2¡£Ð´³ö´×ËáÓëMg(OH) 2·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________¡£Ê³ÑÎÖг£¼ÓÈëKIO3À´²¹µâ£¬ÎªÑé֤ʳÑÎÖеÄKIO3£¬¿ÉÔÚʳÑÎÖмÓÈë´×ËᣬÔÙ¼ÓÈëKIºÍµí·ÛÈÜÒº¡£·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________¡£

£¨2£©ÄðÖÆÃ׾ƵĹý³ÌÖÐÊÇÃ×Öеĵí·ÛÔÚøµÄ×÷ÓÃÏÂÉú³ÉÆÏÌÑÌÇ¡£Ð´³ö»¯Ñ§·½³Ìʽ_____________________________£¬È»ºóÆÏÌÑÌÇÔÚøµÄ×÷ÓÃÏÂÉú³ÉÒÒ´¼¡£

£¨3£©ÔÚÄðÖÆÃ׾ƵĹý³ÌÖУ¬ÓпÉÄܱäËᣬÕâÀïÒòΪ²úÉúÁË´×ËᡣΪÁ˳ýÈ¥Ã×¾ÆÖеÄËáζÓÐÈ˽¨ÒéÏòÃ×¾ÆÖмÓÈëºìÈȵÄǦ£¬Ê¹ËüÓë´×ËáÉú³ÉÓÐÌðζµÄ´×ËáǦ¡£ÄãÈÏΪ´Ë·¨ÊÇ·ñ¿ÉÐÐ_____£¨ÌîÐУ¬²»¿ÉÐУ©£¬Ô­ÒòÊÇ___________________________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸