A¡¢B¡¢C¡¢DÊÇÔªËØÖÜÆÚ±íÖÐÇ°36ºÅÔªËØ,ËüÃǵĺ˵çºÉÊýÒÀ´ÎÔö´ó¡£AÔ­×ÓL²ãµÄ³É¶Ôµç×ÓÊýºÍδ³É¶Ôµç×ÓÊýÏàµÈ,BÔ­×ÓµÄ×îÍâ²ãp¹ìµÀµÄµç×ÓΪ°ë³äÂú½á¹¹,CÊǵؿÇÖк¬Á¿×î¶àµÄÔªËØ¡£DÊǵÚËÄÖÜÆÚÔªËØ,ÆäÔ­×ÓºËÍâ×îÍâ²ãµç×ÓÊýÓëÇâÔ­×ÓÏàͬ,ÆäÓà¸÷²ãµç×Ó¾ù³äÂú¡£Çë»Ø´ðÏÂÁÐÎÊÌâ:

(1)A¡¢B¡¢CµÄµÚÒ»µçÀëÄÜÓÉСµ½´óµÄ˳ÐòÊÇ¡¡¡¡¡¡¡¡(ÓöÔÓ¦µÄÔªËØ·ûºÅ±íʾ);»ù̬DÔ­×ӵĵç×ÓÅŲ¼Ê½Îª¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ 

(2)AµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯Îï·Ö×ÓÖÐ,ÆäÖÐÐÄÔ­×Ó²ÉÈ¡¡¡¡¡¡¡¡¡ÔÓ»¯;BµÄ¿Õ¼ä¹¹ÐÍΪ¡¡¡¡¡¡¡¡(ÓÃÎÄ×ÖÃèÊö)¡£ 

(3)1 mol AB-Öк¬ÓеĦмü¸öÊýΪ¡¡¡¡¡¡¡¡¡£ 

(4)ÈçͼÊǽðÊôCaºÍDËùÐγɵÄijÖֺϽðµÄ¾§°û½á¹¹Ê¾Òâͼ,Ôò¸ÃºÏ½ðÖÐCaºÍDµÄÔ­×Ó¸öÊý±ÈÊÇ¡¡¡¡¡¡¡¡¡£ 

(5)ïçÄøºÏ½ðÓëÉÏÊöºÏ½ð¶¼¾ßÓÐÏàͬÀàÐ͵ľ§°û½á¹¹XYn,ËüÃÇÓкÜÇ¿µÄ´¢ÇâÄÜÁ¦¡£ÒÑÖªïçÄøºÏ½ðLaNin¾§°ûÌå»ýΪ9.0¡Á10-23 cm3,´¢ÇâºóÐγÉLaNinH4.5ºÏ½ð(Çâ½øÈ뾧°û¿Õ϶,Ìå»ý²»±ä),ÔòLaNinÖÐn=¡¡¡¡¡¡¡¡(ÌîÊýÖµ);ÇâÔںϽðÖеÄÃܶÈΪ¡¡¡¡¡¡¡¡¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÕýÈ·ÕÆÎÕ»¯Ñ§ÓÃÓïºÍ»¯Ñ§»ù±¾¸ÅÄîÊÇѧºÃ»¯Ñ§µÄ»ù´¡¡£ÏÂÁÐÓйرíÊöÖÐÕýÈ·µÄÒ»×éÊÇ

A£®16OÓë18O»¥ÎªÍ¬Î»ËØ£»H216O¡¢D216O¡¢H218O¡¢D218O»¥ÎªÍ¬ËØÒìÐÎÌå

B£®SiH4¡¢PH3¡¢HClµÄÎȶ¨ÐÔÖð½¥ÔöÇ¿

C£®¹ýÑõÒÒËá(CH3COOOH)ÓëôÇ»ùÒÒËá(HOCH2COOH)Ëùº¬¹ÙÄÜÍÅÏàͬ£»Á½Õß»¥ÎªÍ¬·ÖÒì ¹¹Ìå

D£®Ca2+µÄ½á¹¹Ê¾ÒâͼΪ£¬ NH4ClµÄµç×ÓʽΪ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½«º£Ë®µ­»¯ÓëŨº£Ë®×ÊÔ´»¯½áºÏÆðÀ´ÊÇ×ÛºÏÀûÓú£Ë®µÄÖØҪ;¾¶Ö®Ò»¡£Ò»°ãÊÇÏȽ«º£Ë®µ­»¯»ñµÃµ­Ë®£¬ÔÙ´ÓÊ£ÓàµÄŨº£Ë®ÖÐͨ¹ýһϵÁй¤ÒÕÌáÈ¡ÆäËû²úÆ·¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÏÂÁиĽøºÍÓÅ»¯º£Ë®×ÛºÏÀûÓù¤ÒÕµÄÉèÏëºÍ×ö·¨¿ÉÐеÄÊÇ             £¨ÌîÐòºÅ£©¡£

¢ÙÓûìÄý·¨»ñÈ¡µ­Ë®          ¢ÚÌá¸ß²¿·Ö²úÆ·µÄÖÊÁ¿ 

¢ÛÓÅ»¯ÌáÈ¡²úÆ·µÄÆ·ÖÖ        ¢Ü¸Ä½ø¼Ø¡¢ä塢þµÄÌáÈ¡¹¤ÒÕ

£¨2£©²ÉÓá°¿ÕÆø´µ³ö·¨¡±´ÓŨº£Ë®Öдµ³öBr2£¬²¢Óô¿¼îÎüÊÕ¡£¼îÎüÊÕäåµÄÖ÷Òª·´Ó¦ÊÇBr2+Na2CO3+H2NaBr + NaBrO3+NaHCO3£¬ÎüÊÕ1mol Br2ʱ£¬×ªÒƵĵç×ÓÊýΪ         mol¡£

£¨3£©º£Ë®ÌáþµÄÒ»¶Î¹¤ÒÕÁ÷³ÌÈçÏÂͼ£º

Ũº£Ë®µÄÖ÷Òª³É·ÖÈçÏ£º

Àë×Ó

Na+

Mg2+

Cl-

SO42-

Ũ¶È/g/L

63.7

28.8

144.6

46.4

¸Ã¹¤ÒÕ¹ý³ÌÖУ¬ÍÑÁò½×¶ÎÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ                              £¬²úÆ·2µÄ»¯Ñ§Ê½Îª        £¬1LŨº£Ë®×î¶à¿ÉµÃµ½²úÆ·2µÄÖÊÁ¿Îª      g¡£

£¨4£©²ÉÓÃʯīÑô¼«¡¢²»Ðâ¸ÖÒõ¼«µç½âÈÛÈÚµÄÂÈ»¯Ã¾£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                  £»µç½âʱ£¬ÈôÓÐÉÙÁ¿Ë®´æÔÚ»áÔì³É²úƷþµÄÏûºÄ£¬Ð´³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ                         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


²Î¿¼Ï±íÖÐÎïÖʵÄÈ۵㣬»Ø´ðÓйØÎÊÌ⣺

ÎïÖÊ

NaF

NaCl

NaBr

NaI

NaCl

KCl

RbCl

CsCl

ÈÛµã/¡æ

995

801

755

651

801

776

715

646

ÎïÖÊ

SiF4

SiCl4

SiBr4

SiI4

SiCl4

GeCl4

SnCl4

PbCl4

ÈÛµã/¡æ

-90.4

£­70.4

5.2

120

£­70.4

£­49.5

£­36.2

£­15

(1)ÄƵı»¯Îï¼°¼î½ðÊôµÄÂÈ»¯ÎïµÄÈÛµãÓë±ËØÀë×Ó¼°¼î½ðÊôÀë×ÓµÄ________Óйأ¬Ëæ×Å________µÄÔö´ó£¬ÈÛµãÒÀ´Î½µµÍ¡£

(2)¹èµÄ±»¯ÎïµÄÈ۵㼰¹è¡¢Õà¡¢Îý¡¢Ç¦µÄÂÈ»¯ÎïµÄÈÛµãÓë________Óйأ¬Ëæ×Å________Ôö´ó£¬________Ôö´ó£¬¹ÊÈÛµãÒÀ´ÎÉý¸ß¡£

(3)ÄƵı»¯ÎïµÄÈÛµã±ÈÏàÓ¦µÄ¹èµÄ±»¯ÎïµÄÈÛµã¸ßµÃ¶à£¬ÕâÓë________Óйأ¬ÒòΪ_______________£¬

¹ÊÇ°ÕßµÄÈÛµãÔ¶¸ßÓÚºóÕß¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


µª»¯Åð(BN)¾§ÌåÓжàÖÖÏà½á¹¹¡£Áù·½Ï൪»¯ÅðÊÇͨ³£´æÔÚµÄÎȶ¨Ï࣬ÓëʯīÏàËÆ£¬¾ßÓвã×´½á¹¹£¬¿É×÷¸ßÎÂÈ󻬼Á¡£Á¢·½Ï൪»¯ÅðÊdz¬Ó²²ÄÁÏ£¬ÓÐÓÅÒìµÄÄÍÄ¥ÐÔ¡£ËüÃǵľ§Ìå½á¹¹ÈçÓÒͼËùʾ¡£

¢Å»ù̬ÅðÔ­×ӵĵç×ÓÅŲ¼Ê½Îª                               ¡£

¢Æ ¹ØÓÚÕâÁ½ÖÖ¾§ÌåµÄ˵·¨£¬ÕýÈ·µÄÊÇ                  (ÌîÐòºÅ)¡£

a.Á¢·½Ï൪»¯Åðº¬ÓЦҼüºÍ¦Ð¼ü£¬ËùÒÔÓ²¶È´ó  b.Áù·½Ï൪»¯Åð²ã¼ä×÷ÓÃÁ¦Ð¡£¬ËùÒÔÖʵØÈí

c.Á½ÖÖ¾§ÌåÖеÄB£­N¼ü¾ùΪ¹²¼Û¼ü          d.Á½ÖÖ¾§Ìå¾ùΪ·Ö×Ó¾§Ìå

¢ÇÁù·½Ï൪»¯Åð¾§Ìå²ãÄÚÒ»¸öÅðÔ­×ÓÓëÏàÁÚµªÔ­×Ó¹¹³ÉµÄ¿Õ¼ä¹¹ÐÍΪ         £¬Æä½á¹¹ÓëʯīÏàËÆÈ´²»µ¼µç£¬Ô­ÒòÊÇ                               ¡£

¢ÈÁ¢·½Ï൪»¯Åð¾§ÌåÖУ¬ÅðÔ­×ÓµÄÔÓ»¯¹ìµÀÀàÐÍΪ                    ¡£¸Ã¾§ÌåµÄÌìÈ»¿óÎïÔÚÇà²Ø¸ßÔ­ÔÚÏÂÔ¼300KmµÄ¹ÅµØ¿ÇÖб»·¢ÏÖ¡£¸ù¾ÝÕâÒ»¿óÎïÐγÉÊÂʵ£¬ÍƶÏʵÑéÊÒÓÉÁù·½Ï൪»¯ÅðºÏ³ÉÁ¢·½Ï൪»¯ÅðÐèÒªµÄÌõ¼þÓ¦ÊÇ                             ¡£

¢ÉNH4BF4(·úÅðËáï§)ÊǺϳɵª»¯ÅðÄÉÃ׹ܵÄÔ­ÁÏÖ®Ò»¡£1mo NH4BF4º¬ÓР             molÅäλ¼ü¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ϱíÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬±íÖÐËùÁеÄ×Öĸ·Ö±ð´ú±íÒ»ÖÖ»¯Ñ§ÔªËØ¡£

(1)д³öÓÉÉÏÊöÁ½ÖÖÔªËØ×é³ÉX2Y2ÐÍ»¯ºÏÎïµÄ»¯Ñ§Ê½·Ö±ðΪ____________£¬Æ侧ÌåÀàÐÍ·Ö±ðΪ__________________¡£

(2)B¡¢C¡¢DµÄÇ⻯ÎïÖУ¬·Ðµã×î¸ßµÄÇ⻯ÎïµÄ½á¹¹Ê½ÊÇ____________£¬¸»º¬·Ðµã×îµÍµÄÇ⻯ÎïµÄ¿óÎïµÄÃû³ÆÊÇ____________¡£

(3)ÉÏÊöÔªËØÐγɵĵ¥ÖÊÖУ¬ÊôÓÚÔ­×Ó¾§ÌåµÄÊÇ________(ÌîÃû³Æ)£¬Æä¿Õ¼ä¹¹ÐÍΪ________________¡£

(4)д³öÓÉÉÏÊöÔªËØÖÐÏàͬµÄËÄÖÖÔªËØ×é³ÉµÄÁ½ÖÖ»¯ºÏÎïµÄË®ÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º_____________________¡£

(5)д³öÓÉÉÏÊöÔªËØÖеÄÈýÖÖÐγɵļȺ¬ÓÐÀë×Ó¼üÓÖº¬Óм«ÐÔ¹²¼Û¼üµÄ»¯ºÏÎïµÄµç×Óʽ£º____________________(ÈÎдÁ½¸ö)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ô­×ÓÐòÊýÒÀ´ÎÔö´óµÄËÄÖÖÔªËØA¡¢B¡¢C¡¢D·Ö±ð´¦ÓÚµÚÒ»ÖÁµÚËÄÖÜÆÚ£¬×ÔÈ»½çÖдæÔÚ¶àÖÖAµÄ»¯ºÏÎBÔ­×ÓºËÍâµç×ÓÓÐ6ÖÖ²»Í¬µÄÔ˶¯×´Ì¬£¬BÓëC¿ÉÐγÉÕýËÄÃæÌåÐηÖ×Ó£¬DµÄ»ù̬ԭ×ÓµÄ×îÍâÄܲãÖ»ÓÐÒ»¸öµç×Ó£¬ÆäËûÄܲã¾ùÒѳäÂúµç×Ó¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÕâËÄÖÖÔªËØÖе縺ÐÔ×î´óµÄÔªËØ£¬Æä»ù̬ԭ×ӵļ۵ç×ÓÅŲ¼Í¼Îª________£¬µÚÒ»µçÀëÄÜ×îСµÄÔªËØÊÇ________(ÌîÔªËØ·ûºÅ)¡£

(2)CËùÔÚÖ÷×åµÄÇ°ËÄÖÖÔªËØ·Ö±ðÓëAÐγɵĻ¯ºÏÎ·ÐµãÓɸߵ½µÍµÄ˳ÐòÊÇ________(Ìѧʽ)£¬³ÊÏÖÈç´ËµÝ±ä¹æÂɵÄÔ­ÒòÊÇ___________________¡£

(3)BÔªËØ¿ÉÐγɶàÖÖµ¥ÖÊ£¬Ò»ÖÖ¾§Ìå½á¹¹ÈçͼһËùʾ£¬ÆäÔ­×ÓµÄÔÓ»¯ÀàÐÍΪ________¡£ÁíÒ»Öֵľ§°ûÈçͼ¶þËùʾ£¬Èô´Ë¾§°ûÖеÄÀⳤΪ356.6 pm£¬Ôò´Ë¾§°ûµÄÃܶÈΪ________________________________________________g·cm£­3(±£ÁôÁ½Î»ÓÐЧÊý×Ö)¡£(£½1.732)

 

(4)DÔªËØÐγɵĵ¥ÖÊ£¬Æ侧ÌåµÄ¶Ñ»ýÄ£ÐÍΪ________£¬DµÄ´×ËáÑξ§Ìå¾Ö²¿½á¹¹ÈçͼÈý£¬¸Ã¾§ÌåÖк¬ÓеĻ¯Ñ§¼üÊÇ________(ÌîÑ¡ÏîÐòºÅ)¡£

¢Ù¼«ÐÔ¼ü¡¡¡¡¢Ú·Ç¼«ÐÔ¼ü¡¡¡¡¢ÛÅäλ¼ü¡¡¡¡¢Ü½ðÊô¼ü

(5)ÏòDµÄÁòËáÑÎÈÜÒºÖеμӹýÁ¿°±Ë®£¬¹Û²ìµ½µÄÏÖÏóÊÇ________¡£Çëд³öÉÏÊö¹ý³ÌµÄÀë×Ó·½³Ìʽ£º_________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔÚÈý¸öÃܱÕÈÝÆ÷Öзֱð³äÈëNe¡¢H2¡¢O2ÈýÖÖÆøÌ壬µ±ËüÃǵÄζȺÍÃܶȶ¼Ïàͬʱ£¬ÕâÈýÖÖÆøÌåµÄѹǿ´Ó´óµ½Ð¡µÄ˳ÐòÊÇ

A£®p(Ne) £¾p(H2) £¾p(O2)        B£®p(O2) £¾p(Ne) £¾p(H2

C£®p(H2) £¾p(O2) £¾p(Ne)        D£®p(H2) £¾p(Ne) £¾p(O2)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


NA±íʾ°¢·ü¼ÓµÂÂÞ³£Êý£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ

A£®lmol FeI2Óë×ãÁ¿ÂÈÆø·´Ó¦Ê±×ªÒƵĵç×ÓÊýΪ2NA

B£®2 L0.5 mol • L£­1ÁòËá¼ØÈÜÒºÖÐÒõÀë×ÓËù´øµçºÉÊýΪNA

C£®1 mol Na2O2¹ÌÌåÖк¬Àë×Ó×ÜÊýΪ4NA

D£®±ûÏ©ºÍ»·±ûÍé×é³ÉµÄ42 g»ìºÏÆøÌåÖÐÇâÔ­×ӵĸöÊýΪ6 NA

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸