ÒÑÖªX¡¢Y¡¢Z¡¢W¡¢RÎåÖÖÔªËØ,Ô­×ÓÐòÊýÒÀ´ÎÔö´ó,ÇÒÔ­×ÓÐòÊý¶¼Ð¡ÓÚ20,XÔªËصÄÔ­×ÓÊÇËùÓÐÔªËصÄÔ­×ÓÖа뾶×îСµÄ,Y¡¢WͬÖ÷×å,Z¡¢WͬÖÜÆÚ,YÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµÄ3±¶,Z¡¢R·Ö±ðÊÇͬÖÜÆÚÖнðÊôÐÔ×îÇ¿µÄÔªËØ¡£ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨  £©
A£®·Ðµã:X2Y£¾X2W
B£®ÓÉX¡¢Y¡¢Z¡¢WËÄÖÖÔªËØ×é³ÉµÄ»¯ºÏÎï¼Èº¬Óй²¼Û¼üÓÖº¬Àë×Ó¼ü
C£®Ô­×Ӱ뾶:X£¼Y£¼Z£¼W£¼R
D£®YÓëWÐγɵĻ¯ºÏÎïWY2ÊÇÐγÉËáÓêµÄÖ÷ÒªÎïÖÊÖ®Ò»
C
ÓÉÓÚXÔªËصÄÔ­×ÓÊÇËùÓÐÔªËصÄÔ­×ÓÖа뾶×îСµÄ,¹ÊXΪH;ÓÖÒòYÔªËØÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇ´ÎÍâ²ãµÄ3±¶,ÔòYΪO,Y¡¢WͬÖ÷×å,ÔòWΪS,Z¡¢WͬÖÜÆÚÇÒZ¡¢R·Ö±ðÊÇͬÖÜÆÚÖнðÊôÐÔ×îÇ¿µÄÔªËØ,ÔòZΪNa,RΪK¡£X2Y¼´ÎªH2O,X2W¼´ÎªH2S,·ÐµãÇ°Õ߸ß,AÏîÕýÈ·;X¡¢Y¡¢Z¡¢W×é³ÉµÄ»¯ºÏÎïΪNaHSO4(»òNaHSO3),¼Èº¬ÓÐÀë×Ó¼üÓÖº¬¹²¼Û¼ü,BÏîÕýÈ·;Ô­×Ӱ뾶:X(H)£¼Y(O)£¼W(S)£¼Z(Na)£¼R(K),CÏî´íÎó;WY2¼´SO2,ÊÇÐγÉËáÓêµÄÖ÷ÒªÎïÖÊÖ®Ò»,DÏîÕýÈ·¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑ֪ijÀë×ӵĽṹʾÒâͼΪ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨     £©
A£®¸ÃÔªËØλÓÚµÚ¶þÖÜÆÚµÚ¢òA×å
B£®¸ÃÔªËØλÓÚµÚ¶þÖÜÆÚµÚ¢ø×å
C£®¸ÃÔªËØλÓÚµÚÈýÖÜÆÚµÚ¢òA×å
D£®¸ÃÔªËØλÓÚµÚÈýÖÜÆÚ0×å

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

X¡¢Y¡¢ZÊÇÈýÖÖ¶ÌÖÜÆÚµÄÖ÷×åÔªËØ,ÔÚÖÜÆÚ±íµÄλÖÃÈçͼ,XÔ­×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇÆä´ÎÍâ²ãµç×ÓÊýµÄ3±¶,ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)
A£®Ô­×Ӱ뾶:Y>Z>X
B£®Æø̬Ç⻯ÎïµÄÈÈÎȶ¨ÐÔ:X<Z
C£®YºÍZµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔӦˮ»¯Îï¾ùΪǿËá
D£®ÈôZµÄ×î¸ßÕý¼ÛΪ+m,ÔòXµÄ×î¸ßÕý¼ÛÒ²Ò»¶¨Îª+m

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

2009Ä꡶×ÔÈ»¡·ÔÓÖ¾±¨µÀÁËÎÒ¹ú¿Æѧ¼Òͨ¹ý²âÁ¿SiO2ÖÐ26AlºÍ10BeÁ½ÖÖÔªËصıÈÀýÈ·¶¨¡°±±¾©ÈË¡±ÄêÁäµÄÑо¿½á¹û£¬ÕâÖÖ²âÁ¿·½·¨½Ð¡°ÂÁîë²âÄê·¨¡±¡£Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©10BeºÍ9Be________¡£
a£®ÊÇͬһÖÖÔ­×Ó
b£®¾ßÓÐÏàͬµÄÖÐ×ÓÊý
c£®¾ßÓÐÏàͬµÄ»¯Ñ§ÐÔÖÊ
£¨2£©AlºÍBe¾ßÓÐÏàËƵĻ¯Ñ§ÐÔÖÊ£¬Ð´³öBeCl2Ë®½â·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________¡£
£¨3£©Ñо¿±íÃ÷26Al¿ÉÒÔË¥±äΪ26Mg£¬¿ÉÒԱȽÏÕâÁ½ÖÖÔªËؽðÊôÐÔÇ¿ÈõµÄ·½·¨ÊÇ_________
_______________________________________________________________¡£
a£®±È½ÏÕâÁ½ÖÖÔªËصĵ¥ÖʵÄÓ²¶ÈºÍÈÛµã
b£®ÔÚÂÈ»¯ÂÁºÍÂÈ»¯Ã¾µÄÈÜÒºÖзֱðµÎ¼Ó¹ýÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº
c£®½«´òÄ¥¹ýµÄþ´øºÍÂÁƬ·Ö±ðºÍÈÈË®×÷Ó㬲¢µÎÈë·Ó̪ÈÜÒº
d£®½«¿ÕÆøÖзÅÖÃÒѾõÄÕâÁ½ÖÖÔªËصĵ¥ÖÊ·Ö±ðºÍÈÈË®×÷ÓÃ
£¨4£©Ä¿Ç°»¹ÓÐÒ»ÖÖ²âÁ¿·½·¨½Ð¡°¼Øë²²âÄê·¨¡±¡£Ð´³öºÍArºËÍâµç×ÓÅŲ¼ÏàͬµÄÒõÀë×ӵİ뾶ÓÉ´óµ½Ð¡µÄ˳Ðò________(Óû¯Ñ§·ûºÅ±íʾ)£»ÆäÖÐÒ»ÖÖÀë×ÓÓë¼ØÏàÁÚÔªËصÄÀë×ÓËùÐγɵĻ¯ºÏÎï¿ÉÓÃ×ö¸ÉÔï¼Á£¬´Ë»¯ºÏÎïµÄµç×ÓʽÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔªËØXÐγɵÄÀë×ÓÓë¸ÆÀë×ӵĺËÍâµç×ÓÅŲ¼Ïàͬ£¬ÇÒXµÄÀë×Ӱ뾶СÓÚ¸º¶þ¼ÛÁòÀë×ӵİ뾶£¬XÔªËØΪ£¨  £©
A£®AlB£®PC£®ArD£®K

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

X¡¢Y¡¢Z¡¢M¡¢N¡¢QΪԪËØÖÜÆÚ±íÇ°ËÄÖÜÆÚµÄÁùÖÖÔªËØ¡£ÆäÖУ¬XÔ­×ÓºËÍâµÄM²ãÖÐÖ»ÓÐÁ½¶Ô³É¶Ôµç×Ó£¬YÔ­×ÓºËÍâµÄL²ãµç×ÓÊýÊÇK²ãµÄÁ½±¶£¬ZÊǵؿÇÄÚº¬Á¿£¨ÖÊÁ¿·ÖÊý£©×î¸ßµÄÔªËØ£¬MµÄÄÚ²ãµç×ÓÊýÊÇ×îÍâ²ãµç×ÓÊýµÄ9±¶£¬NµÄÔ­×ÓÐòÊý±ÈMС1, QÔÚÔªËØÖÜÆÚ±íµÄ¸÷ÔªËØÖе縺ÐÔ×î´ó¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©XÔªËØÔÚÖÜÆÚ±íÖеÄλÖÃÊÇ              £¬ÊôÓÚ      ÇøÔªËØ£¬ËüµÄÍâΧµç×ӵĵç×ÓÅŲ¼Í¼Îª                                ¡£

£¨2£©XZ2·Ö×ÓµÄÁ¢Ìå½á¹¹ÊÇ              £¬YZ2·Ö×ÓÖÐYµÄÔÓ»¯¹ìµÀÀàÐÍΪ         £¬ÏàͬÌõ¼þÏÂÁ½ÕßÔÚË®ÖеÄÈܽâ¶È½Ï´óµÄÊÇ                          £¨Ð´·Ö×Óʽ£©£¬ÀíÓÉÊÇ                                                                     ¡£
£¨3£©º¬ÓÐÔªËØNµÄÑεÄÑæÉ«·´Ó¦Îª    É«£¬Ðí¶à½ðÊôÑζ¼¿ÉÒÔ·¢ÉúÑæÉ«·´Ó¦£¬ÆäÔ­ÒòÊÇ                                    ¡£
£¨4£©ÔªËØMÓëÔªËØQÐγɾ§Ìå½á¹¹ÈçͼËùʾ£¬ÉèÆ侧°û±ß³¤Îªa pm£¬ÔòaλÖÃÓëbλÖÃÖ®¼äµÄ¾àÀëΪ_______pm£¨Ö»ÒªÇóÁÐËãʽ£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂÃæÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬²ÎÕÕÔªËآ٣­¢àÔÚ±íÖеÄλÖã¬ÇëÓû¯Ñ§ÓÃÓï»Ø´ðÏÂÁÐÎÊÌ⣺
×å
ÖÜÆÚ
IA
 
0
1
¢Ù
¢òA
¢óA
¢ôA
¢õA
¢öA
¢÷A
 
2
 
 
 
¢Ú
¢Û
¢Ü
 
 
3
¢Ý
 
¢Þ
¢ß
 
 
¢à
 
 
£¨1£©¢Ü¡¢¢Ý¡¢¢ÞµÄÔ­×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪ(ÔªËØ·ûºÅ)________________________¡£
£¨2£©¢Ú¡¢¢Û¡¢¢ßµÄ×î¸ß¼Ûº¬ÑõËáµÄËáÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ(Ìѧʽ)________     ¡£
£¨3£©¢Ù¡¢¢Ü¡¢¢Ý¡¢¢àÖеÄijЩԪËØ¿ÉÐγɼȺ¬Àë×Ó¼üÓÖº¬¼«ÐÔ¹²¼Û¼üµÄ»¯ºÏÎд³öÆäÖÐÒ»ÖÖ»¯ºÏÎïµÄ»¯Ñ§Ê½£º_______________¡£
£¨4£©Óɢں͢Ü×é³ÉµÄ»¯ºÏÎïÓë¢ÝµÄͬÖÜÆÚÏàÁÚÖ÷×åÔªËصĵ¥ÖÊ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ:_______¡£
£¨5£©¢Þµ¥ÖÊÓë¢ÝµÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯Îï·´Ó¦µÄÀë×Ó·½³ÌʽΪ                    ¡£
£¨6£©ÈôÓâ٢Ú×é³É×î¼òµ¥µÄÓлúÎï×÷ΪȼÁϵç³ØµÄÔ­ÁÏ£¬Çëд³öÔÚ¼îÐÔ½éÖÊÖÐȼÁϵç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½:                                                         ¡£
£¨7£©È¼Ãº·ÏÆøÖеĺ¬ÓеªÑõ»¯ÎNOx£©¡¢¶þÑõ»¯Ì¼µÈÆøÌ壬³£ÓÃÏÂÁз½·¨¶Ôȼú·ÏÆø½øÐÐÍÑÏõ´¦Àíʱ£¬³£ÀûÓü×Íé´ß»¯»¹Ô­µªÑõ»¯Îï¡£
È磺CH4(g)£«4NO2(g)=4NO(g)£«CO2(g)£«2H2O(g) £¬ ¡÷H=£­574 kJ¡¤mol£­1
CH4(g)£«4NO(g)=2N2(g)£«CO2(g)£«2H2O(g) £¬ ¡÷H=£­1160 kJ¡¤mol£­1
ÔòCH4(g)½«NO2(g)»¹Ô­ÎªN2(g)µÈµÄÈÈ»¯Ñ§·½³ÌʽΪ                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

A¡¢B¡¢C¡¢D¡¢E¡¢FÔªËØÊÇÔ­×ÓÐòÊýÒÀ´ÎÔö´óµÄ6ÖÖ¶ÌÖÜÆÚÔªËØ¡£ÒÑÖªAÊǶÌÖÜÆÚÔªËØÖÐÔ­×Ӱ뾶×îСµÄÔªËØ£¬AºÍBÐγɵÄ18µç×ӵĻ¯ºÏÎïX³£ÓÃ×÷»ð¼ýµÄȼÁÏ£¬CÔ­×Ó×îÍâ²ãµç×ÓÊýÓëºËÍâµç×Ó×ÜÊýÖ®±ÈΪ3¡Ã4£¬EÓëCͬÖ÷×壬DºÍC¿ÉÒÔÐγÉÔ­×Ó¸öÊý±ÈΪ1¡Ã1ºÍ2¡Ã1µÄÁ½ÖÖÀë×Ó»¯ºÏÎï¡£
¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)DºÍCÐγÉ1¡Ã1µÄ»¯ºÏÎïÖÐÒõÑôÀë¸öÊý±ÈΪ____________¡£
(2)»¯ºÏÎïXµÄ½á¹¹Ê½Îª____________¡£1 molÆø̬XÔÚÊÊÁ¿C2ÖÐȼÉÕ£¬Éú³ÉB2ºÍÆø̬A2C, ·Å³ö534 kJµÄÈÈÁ¿£¬1 molҺ̬A2CÍêÈ«Æû»¯ÐèÎüÊÕ44 kJÈÈÁ¿¡£Çëд³öÆø̬XÔÚC2ÖÐȼÉÕÉú³ÉB2ºÍҺ̬A2CʱµÄÈÈ»¯Ñ§·½³Ìʽ_____________________________________
(3)ij»¯ºÏÎïÓÉÉÏÊö6ÖÖÔªËØÖеÄ3ÖÖÔªËØ×é³É£¬Îª³£¼û¼ÒÓÃÏû¶¾¼ÁµÄÖ÷Òª³É·Ö£¬ÆäÖл¯Ñ§¼üÀàÐÍΪ__________________£»¸Ã»¯ºÏÎïË®ÈÜÒº²»³ÊÖÐÐÔµÄÔ­ÒòÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)__________________________________£¬¸Ã»¯ºÏÎï¿ÉÒÔͨ¹ýµç½âDºÍFÐγɻ¯ºÏÎïµÄË®ÈÜÒº»ñµÃ£¬µç½âʱ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________________
(4)д³öÒ»¸ö¿ÉÒÔÖ¤Ã÷CµÄ·Ç½ðÊôÐÔ´óÓÚEµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º_____________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÅÅÁÐ˳ÐòÕýÈ·µÄÊÇ£¨¡¡¡¡£©¡£
¢ÙÈÈÎȶ¨ÐÔ£ºH2O£¾HF£¾H2S¡¡¢ÚÔ­×Ӱ뾶£ºNa£¾Mg£¾O
¢ÛËáÐÔ£ºH3PO4£¾H2SO4£¾HClO4¡¡¢Üʧµç×ÓÄÜÁ¦£ºCs£¾Na
A£®¢Ù¢ÛB£®¢Ú¢ÜC£®¢Ù¢ÜD£®¢Ú¢Û

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸