15£®Ò»Ïî¿ÆѧÑо¿³É¹û±íÃ÷£¬Í­ÃÌÑõ»¯ÎCuMn2O4£©ÄÜÔÚ³£ÎÂÏ´߻¯Ñõ»¯¿ÕÆøÖеÄÒ»Ñõ»¯Ì¼ºÍ¼×È©£¨HCHO£©£®
£¨1£©ÏòÒ»¶¨ÎïÖʵÄÁ¿Å¨¶ÈµÄCu£¨NO3£©2 ºÍMn£¨NO3£©2 ÈÜÒºÖмÓÈëNa2CO3 ÈÜÒº£¬ËùµÃ³Áµí¾­¸ßÎÂ×ÆÉÕ£¬¿ÉÖƵÃCuMn2O4£®
¢Ùд³ö»ù̬MnÔ­×ӵļ۵ç×ÓÅŲ¼Ê½Îª3d54S2£®
¢ÚCO32-µÄ¿Õ¼ä¹¹ÐÍÊÇƽÃæÈý½ÇÐΣ¨ÓÃÎÄ×ÖÃèÊö£©£®
£¨2£©ÔÚÍ­ÃÌÑõ»¯ÎïµÄ´ß»¯Ï£¬CO ±»Ñõ»¯ÎªCO2£¬HCHO ±»Ñõ»¯ÎªCO2 ºÍH2O£®
¢Ù¸ù¾ÝµÈµç×ÓÌåÔ­Àí£¬CO ·Ö×ӵĽṹʽΪC¡ÔO£®
¢ÚCO2 ·Ö×ÓÖÐC Ô­×Ó¹ìµÀµÄÔÓ»¯ÀàÐÍΪspÔÓ»¯£®
¢Û1mol¼×È©£¨HCHO£©·Ö×ÓÖк¬ÓеĦҼüÊýĿΪ3¡Á6.02¡Á1023¸ö£¨»ò3NA£©£®
£¨3£©ÏòCuSO4ÈÜÒºÖмÓÈë¹ýÁ¿NH3©qH2OÈÜÒº¿ÉÉú³É[Cu £¨NH3£©4]2+£®²»¿¼Âǿռ乹ÐÍ£¬[Cu £¨NH3£©4]2+µÄ½á¹¹¿ÉÓÃʾÒâͼ±íʾΪ2+£®

·ÖÎö £¨1£©¢ÙMnÔ­×ÓºËÍâµç×ÓÊýΪ25£¬´¦ÓÚµÚËÄÖÜÆÚ¢÷B×壻CuÔ­×ÓºËÍâµç×ÓÊýΪ29£¬¸ù¾ÝÄÜÁ¿×îµÍÔ­ÀíÊéдºËÍâµç×ÓÅŲ¼Ê½£»
¢Ú¼ÆËãCÔ­×Ó¼Û²ãµç×Ó¶ÔÊý¡¢¹Âµç×Ó¶ÔÊýÈ·¶¨¿Õ¼ä¹¹ÐÍ£»
£¨2£©¢ÙCOÓëN2»¥ÎªµÈµç×ÓÌ壬¶þÕ߽ṹÏàËÆ£¬CO·Ö×ÓÖÐCÔ­×ÓÓëÑõÔ­×ÓÖ®¼äÐγÉ3¶Ô¹²Óõç×Ó¶Ô£»
¢ÚCO2 ·Ö×ÓÖÐCÔ­×ÓÐòÊý2¸ö¦Ò¼ü£¬Ã»Óйµç×Ó¶Ô£¬ÔÓ»¯¹ìµÀÊýĿΪ2£»
¢ÛHCHO·Ö×ÓÖк¬ÓÐ2¸öC-H¼ü¡¢1¸öC=OË«¼ü£¬·Ö×ÓÖк¬ÓÐ3¸ö¦Ò¼ü£»
£¨3£©[Cu £¨NH3£©4]2+ÖÐÓëCu2+Óë4¸öNH3ÐγÉÅäλ¼ü£®

½â´ð ½â£º£¨1£©¢ÙMnÔ­×ÓºËÍâµç×ÓÊýΪ25£¬´¦ÓÚµÚËÄÖÜÆÚ¢÷B×壬¼Ûµç×ÓÅŲ¼Ê½Îª3d54s2£¬¹Ê´ð°¸Îª£º3d54s2£»
¢ÚCO32-ÖÐCÔ­×ӹµç×Ó¶ÔÊý=$\frac{1}{2}$£¨4+2-2¡Á3£©=0£¬¼Û²ãµç×Ó¶ÔÊý=3+0=3£¬¿Õ¼ä¹¹ÐÍΪƽÃæÈý½ÇÐΣ¬¹Ê´ð°¸Îª£ºÆ½ÃæÈý½ÇÐΣ»
£¨2£©¢ÙCOÓëN2»¥ÎªµÈµç×ÓÌ壬¶þÕ߽ṹÏàËÆ£¬CO·Ö×ÓÖÐCÔ­×ÓÓëÑõÔ­×ÓÖ®¼äÐγÉ3¶Ô¹²Óõç×Ó¶Ô£¬CO½á¹¹Ê½ÎªC¡ÔO£¬¹Ê´ð°¸Îª£ºC¡ÔO£»
¢ÚCO2 ·Ö×ÓÖÐCÔ­×ÓÐòÊý2¸ö¦Ò¼ü£¬Ã»Óйµç×Ó¶Ô£¬ÔÓ»¯¹ìµÀÊýĿΪ2£¬CÔ­×Ó²ÉÈ¡spÔÓ»¯£¬¹Ê´ð°¸Îª£ºsp£»
¢ÛHCHO·Ö×ÓÖк¬ÓÐ2¸öC-H¼ü¡¢1¸öC=OË«¼ü£¬·Ö×ÓÖк¬ÓÐ3¸ö¦Ò¼ü£¬1mol¼×È©£¨HCHO£©·Ö×ÓÖк¬ÓеĦҼüÊýĿΪ3¡Á6.02¡Á1023£¨»ò3 NA£©£¬
¹Ê´ð°¸Îª£º3¡Á6.02¡Á1023£¨»ò3 NA£©£»
£¨3£©[Cu£¨NH3£©4]2+ÖÐÓëCu2+Óë4¸öNH3ÐγÉÅäλ¼ü£¬Ôò[Cu£¨NH3£©4]2+µÄ½á¹¹¿ÉÓÃʾÒâͼ±íʾΪ£º2+£¬
¹Ê´ð°¸Îª2+£®

µãÆÀ ±¾ÌâÊǶÔÎïÖʽṹÓëÐÔÖʵĿ¼²é£¬Éæ¼°ºËÍâµç×ÓÅŲ¼¡¢ÔÓ»¯·½Ê½Óë¿Õ¼ä¹¹ÐÍÅжϡ¢µÈµç×ÓÌå¡¢»¯Ñ§¼ü¡¢ÅäºÏÎïµÈ£¬×¢Òâ¶Ô»ù´¡ÖªÊ¶µÄÀí½âÕÆÎÕ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

5£®ÏÂÁйØÓÚ»¯Ñ§¼üµÄ˵·¨Öв»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®»¯Ñ§¼ü¿ÉÒÔʹÀë×ÓÏà½áºÏ£¬Ò²¿ÉÒÔʹԭ×ÓÏà½áºÏ
B£®·Ç½ðÊôÔªËصÄÔ­×ÓÖ®¼äÖ»ÄÜÐγɹ²¼Û¼ü
C£®»¯Ñ§·´Ó¦¹ý³ÌÖУ¬·´Ó¦Îï·Ö×ÓÄڵĻ¯Ñ§¼ü¶ÏÁÑ£¬²úÎï·Ö×ÓÖл¯Ñ§¼üÐγÉ
D£®»¯Ñ§¼üÊÇÒ»ÖÖ×÷ÓÃÁ¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®X¡¢Y¡¢ZÈýÖÖ¶ÌÖÜÆÚÔªËØ£¬XÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµç×ÓÊýµÄ2±¶£¬YÔ­×ӵĴÎÍâ²ãµç×ÓÊýÊÇ×îÍâ²ãµç×ÓÊýµÄ2±¶£¬ZÔ­×ӵĴÎÍâ²ãµç×ÓÊýÊÇ×îÍâ²ãµç×ÓÊýµÄ4±¶£®ÔòX¡¢Y¡¢ZÈýÖÖÔªËØ£¬¿ÉÄܵÄ×éºÏÊÇ£¨¡¡¡¡£©
A£®C¡¢Mg¡¢LiB£®Li¡¢C¡¢MgC£®C¡¢Si¡¢MgD£®C¡¢O¡¢Mg

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

3£®ÏÂÁÐʵÑéÉè¼ÆÄÜ´ïµ½ÏàӦʵÑéÄ¿µÄÊÇ£¨¡¡¡¡£©
Ñ¡ÏîʵÑéÉè¼ÆʵÑéÄ¿µÄ
AÔÚÈȵÄNaOHÈÜÒºÖеÎÈë±¥ºÍFeCl3ÈÜÒºÖƱ¸Fe£¨OH£©3½ºÌå
B½«SO2ͨÈëKMnO4ÈÜÒºÑéÖ¤SO2µÄƯ°×ÐÔ
CÏòº¬ÉÙÁ¿Fe3+µÄMgCl2ÈÜÒºÖмÓÈëÊÊÁ¿MgCO3·ÛÄ©£¬¼ÓÈÈ¡¢½Á°è²¢¹ýÂ˳ýÈ¥MgCl2ÈÜÒºÖÐÉÙÁ¿µÄFe3+
D½«0.1mol•L-1µÄNa2SO4ÈÜÒºµÎÈëBaCl2ÈÜÒºÖÁ²»ÔÙÓгÁµí²úÉú£¬ÔٵμÓ0.1mol•L-1µÄNa2CO3ÈÜÒº±È½ÏBaCO3ÓëBaSO4ÈܶȻýµÄ´óС
A£®AB£®BC£®CD£®D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

10£®ÀûÓÃÈçͼװÖòⶨÖкÍÈȵÄʵÑé²½ÖèÈçÏ£º
¢ÙÁ¿È¡50mL 0.25mol/L H2SO4ÈÜÒºµ¹ÈëСÉÕ±­ÖУ¬²âÁ¿Î¶ȣ»
¢ÚÁ¿È¡50mL 0.55mol/L NaOHÈÜÒº£¬²âÁ¿Î¶ȣ»
¢Û½«NaOHÈÜÒºµ¹ÈëСÉÕ±­ÖУ¬»ìºÏ¾ùÔȺó²âÁ¿»ìºÏҺζȣ®Çë»Ø´ð£º
£¨1£©ÈçͼËùʾ£¬ÒÇÆ÷AµÄÃû³ÆÊÇ»·Ðβ£Á§½Á°è°ô£»
£¨2£©NaOHÈÜÒºÉÔ¹ýÁ¿µÄÔ­ÒòÈ·±£ÁòËá±»ÍêÈ«Öкͣ®
£¨3£©¼ÓÈëNaOHÈÜÒºµÄÕýÈ·²Ù×÷ÊÇB£¨Ìî×Öĸ£©£®
A£®Ñز£Á§°ô»ºÂý¼ÓÈë   B£®Ò»´ÎѸËÙ¼ÓÈë   C£®·ÖÈý´Î¼ÓÈë
£¨4£©Ê¹ÁòËáÓëNaOHÈÜÒº»ìºÏ¾ùÔȵÄÕýÈ·²Ù×÷ÊÇÓû·Ðβ£Á§°ôÇáÇá½Á¶¯£®
£¨5£©ÉèÈÜÒºµÄÃܶȾùΪ1g•cm-3£¬ÖкͺóÈÜÒºµÄ±ÈÈÈÈÝc=4.18J•£¨g•¡æ£©-1£¬Çë¸ù¾ÝʵÑéÊý¾Ýд³ö¸ÃÖкÍÈȵÄÈÈ»¯Ñ§·½³ÌʽH2SO4£¨aq£©+2NaOH£¨aq£©Na2SO4£¨aq£©+2H2O£¨l£©¡÷H=-113.7kJ•mol-1
ζÈ
ʵÑé´ÎÊý
ÆðʼζÈt1/¡æÖÕֹζÈ
t2/¡æ
ζȲîƽ¾ùÖµ
£¨t2-t1£©/¡æ
H2SO4NaOHƽ¾ùÖµ
125.025.2 28.5 
224.925.1 28.3
325.526.5 31.8
425.625.4 29.0
£¨6£©ÉÏÊöʵÑéÊýÖµ½á¹ûÓë57.3kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔ­Òò¿ÉÄÜÊÇ£¨Ìî×Öĸ£©abc
a£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î
b£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÁòËáµÄСÉÕ±­ÖÐ
c£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζÈ
£¨7£©ÔõÑù²ÅÄÜÈ·±£¶ÁÈ¡»ìºÏÒºµÄ×î¸ßζȣ¿²»¶Ï¶ÁȡζÈÊý¾Ý£¬²¢¼Ç¼£¬Ö±µ½³öÏÖϽµ£¬È¡×î´óÖµ£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®ÔÚ»ù̬¶àµç×ÓÔ­×ÓÖУ¬¹ØÓÚºËÍâµç×ÓÄÜÁ¿µÄÐðÊö´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®×îÒ×ʧȥµÄµç×ÓÄÜÁ¿×î¸ß
B£®µçÀëÄÜ×îСµÄµç×ÓÄÜÁ¿×î¸ß
C£®2p¹ìµÀµç×ÓÄÜÁ¿¸ßÓÚ2s¹ìµÀµç×ÓÄÜÁ¿
D£®ÔÚÀëºË×î½üÇøÓòÄÚÔ˶¯µÄµç×ÓÄÜÁ¿×î¸ß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

7£®ÏÂÁи÷×é·Ö×ÓÖУ¬¶¼ÊôÓÚº¬¼«ÐÔ¼üµÄ·Ç¼«ÐÔ·Ö×ÓµÄÊÇ£¨¡¡¡¡£©
A£®CO2¡¢H2O2B£®C2H4¡¢CH4C£®C60¡¢C2H4D£®NH3¡¢HCl

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®¶ÌÖÜÆÚµÄÁ½ÖÖÔªËØAºÍB£¬ËüÃǵÄÀë×ÓA-ºÍB2+¾ßÓÐÏàͬµÄµç×Ó²ã½á¹¹£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Ô­×ÓÐòÊýA£¾BB£®µç×Ó×ÜÊýA-£¾B2+C£®Àë×Ӱ뾶A-£¾B2+D£®Ô­×Ӱ뾶A£¾B

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

5£®ÏÂÁÐÐðÊö²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®µí·Û¡¢ÏËάËØ¡¢ÓÍÖ¬¶¼ÊôÓڸ߷Ö×Ó»¯ºÏÎï
B£®ÓÐÌðζµÄÎïÖʲ»Ò»¶¨ÊôÓÚÌÇÀà
C£®ÌìÈ»µ°°×ÖÊË®½âµÄ×îÖÕ²úÎï¾ùΪ°±»ùËá
D£®ÓÍ֬ˮ½âµÃµ½µÄ´¼ÊDZûÈý´¼

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸