ʵÑéÊÒÒªÅäÖÆ100mL¡¢10mol¡¤L£­1µÄNaClÈÜÒº£¬ÊԻشðÏÂÁи÷Ì⣺

£¨1£©£¨2·Ö£©¾­¼ÆË㣬Ӧ¸ÃÓÃÍÐÅÌÌìƽ³ÆÈ¡NaCl¹ÌÌå      g¡£

£¨2£©£¨2·Ö£©ÅäÖÆÈÜҺʱһ°ã¿É·ÖΪÒÔϼ¸¸ö²½Ö裺 ¢Ù³ÆÁ¿   ¢Ú¼ÆËã   ¢Û¶¨ÈÝ

¢ÜÒÆÒº ¢ÝÏ´µÓ   ¢ÞÈܽ⠠ ÆäÕýÈ·µÄ²Ù×÷˳ÐòΪ              ¡£

£¨3£©£¨2·Ö£¬Ã¿¿Õ1·Ö£©¸ÃʵÑéÁ½´ÎÓõ½²£Á§°ô£¬Æä×÷Ó÷ֱðÊÇ£º

¢Ù Èܽâʱ²£Á§°ôµÄ×÷Óà          ¢Ú ÒÆҺʱ²£Á§°ôµÄ×÷Óà                                   

£¨4£©£¨2·Ö£©ÈÝÁ¿Æ¿ÉÏÐè±êÓÐÒÔÏÂ5ÏîÖÐµÄ ¢ÙÎÂ¶È ¢ÚŨ¶È ¢ÛÈÝÁ¿ ¢Üѹǿ ¢Ý¿Ì¶ÈÏߣ¨   £©        

A£®¢Ù¢Û¢Ý     B£®¢Û¢Ý¢Þ   C£®¢Ù¢Ú¢Ü    D£®¢Ú¢Ü¢Þ

£¨5£©(2·Ö)ÈôÓÃNaCl¹ÌÌåÅäÖÆÈÜÒº£¬ÏÂÁÐÒÇÆ÷ÖУ¬²»ÐèÒªÓõ½µÄÊÇ_____ ¡££¨ÌîÐòºÅ£©

A.Õô·¢Ãó  B.100mLÈÝÁ¿Æ¿  C.ÉÕ±­  D.½ºÍ·µÎ¹Ü  E.Ò©³×  F.ÍÐÅÌÌìƽ  G.²£Á§°ô

£¨6£©(2·Ö)ÏÂÁдíÎó²Ù×÷»áµ¼ÖÂËùµÃÈÜҺŨ¶ÈÆ«µÍµÄÊÇ       £¨¶àÏÌîÐòºÅ£©¡£

A. ¶¨ÈÝʱÑöÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß    

 B. ÈÝÁ¿Æ¿ÖÐÔ­ÓÐÉÙÁ¿ÕôÁóË®

C. ¶¨Èݺ󣬰ÑÈÝÁ¿Æ¿µ¹ÖÃÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬±ã²¹³ä¼¸µÎË®ÖÁ¿Ì¶È´¦

D. ÅäÖƺÃÈÜÒººó£¬ÈÝÁ¿Æ¿Î´ÈûºÃ£¬È÷³öһЩÈÜÒº

 

£¨1£©    58.5      g    £¨2£©ÆäÕýÈ·µÄ²Ù×÷˳ÐòΪ   ¢Ú¢Ù¢Þ¢Ü¢Ý¢Û      ¡£

£¨3£©¢ÙÈܽâʱ²£Á§°ôµÄ×÷Óà   ½Á°è£¬¼Ó¿ìÈܽâµÄËٶȠ                      

¢ÚÒÆҺʱ²£Á§°ôµÄ×÷Óà    ÒýÁ÷µÄ×÷Óà                                

£¨4£©  A        ¡££¨5£©     A      ¡££¨6£© ACD         ¡£

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ʵÑéÊÒÒªÅäÖÆ100mL 0.5mol?L-1µÄNaClÈÜÒº£¬ÊԻشðÏÂÁи÷Ì⣮
£¨1£©¾­¼ÆË㣬Ӧ¸ÃÓÃÍÐÅÌÌìƽ³ÆÈ¡NaCl¹ÌÌå
2.9
2.9
g£®
£¨2£©ÈôÓÃNaCl¹ÌÌåÅäÖÆÈÜÒº£¬ÐèʹÓõIJ£Á§ÒÇÆ÷ÓÐ
ÉÕ±­¡¢²£Á§°ô¡¢100mLÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü
ÉÕ±­¡¢²£Á§°ô¡¢100mLÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü

£¨3£©ÅäÖƹý³ÌÓÐÒÔϲÙ×÷£ºA£®ÒÆÒº£¬B£®³ÆÁ¿£¬C£®Ï´µÓ£¬D£®¶¨ÈÝ£¬E£®Èܽ⣬F£®Ò¡ÔÈ£®ÆäÕýÈ·µÄ²Ù×÷˳ÐòÓ¦ÊÇ
B¡úE¡úA¡úC¡úD¡úF
B¡úE¡úA¡úC¡úD¡úF
 £¨ÌîÐòºÅ£©£®
£¨4£©ÏÂÁвÙ×÷»áµ¼ÖÂËùµÃÈÜҺŨ¶ÈÆ«µÍµÄÊÇ
ABD
ABD
£¨ÌîÐòºÅ£©£®
A£®³ÆÁ¿Ê±£¬×ó±ßµÄÍÐÅÌ·ÅíÀÂ룬ÓұߵÄÍÐÅÌ·ÅNaCl¹ÌÌå
B£®¶¨ÈÝʱÑöÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß
C£®½«ÈܽâδÀäÈ´µÄÈÜҺתÈëÈÝÁ¿Æ¿ºó¾ÍÖ±½Ó¶¨ÈÝ
D£®¶¨Èݺ󣬰ÑÈÝÁ¿Æ¿µ¹×ªÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬±ã²¹³ä¼¸µÎË®ÖÁ¿Ì¶È´¦£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ʵÑéÊÒÒªÅäÖÆ100mL 0.5mol?L-1µÄNaClÈÜÒº£¬ÊԻشðÏÂÁи÷Ì⣺
£¨1£©ÏÂÁÐÒÇÆ÷ÖУ¬¿Ï¶¨²»»áÓõ½µÄÊÇ
AB
AB
£ºA£®×¶ÐÎÆ¿ B£®200mLÈÝÁ¿Æ¿ C£®Á¿Í² D£®½ºÍ·µÎ¹Ü  E£®100mLÈÝÁ¿Æ¿  F£®Ììƽ
£¨2£©ÈôҪʵʩÅäÖÆ£¬³ýÉÏÊöÒÇÆ÷Í⣬»¹È±ÉÙµÄÒÇÆ÷»òÓÃÆ·ÊÇ
ÉÕ±­¡¢²£Á§°ô¡¢Ò©³×
ÉÕ±­¡¢²£Á§°ô¡¢Ò©³×
£®
£¨3£©ÈÝÁ¿Æ¿ÉϳýÓп̶ÈÏßÍ⻹Ӧ±êÓÐ
ÈÝ»ý¡¢Ê¹ÓÃζÈ
ÈÝ»ý¡¢Ê¹ÓÃζÈ
£¬ÈÝÁ¿Æ¿ÔÚʹÓÃÇ°±ØÐë½øÐеÄÒ»²½²Ù×÷ÊÇ
²é©
²é©
£®
£¨4£©ÅäÖÆÍê±Ïºó£¬ÀÏʦָ³öÓÐËÄλͬѧ¸÷½øÐÐÁËÏÂÁÐijһÏî´íÎó²Ù×÷£¬ÄãÈÏΪÕâËÄÏî´íÎó²Ù×÷»áµ¼ÖÂËùµÃÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ
B
B
£º
A£®¶¨ÈÝʱÑöÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß     B£®¶¨ÈÝʱ¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß
C£®½«ÈܽâÀäÈ´ºóµÄÈÜÒºÖ±½ÓתÈëÈÝÁ¿Æ¿ºó¾Í½øÐж¨ÈݲÙ×÷
D£®¶¨Èݺ󣬰ÑÈÝÁ¿Æ¿µ¹ÖÃÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬±ã²¹³ä¼¸µÎË®ÖÁ¿Ì¶È´¦
£¨5£©Í¨¹ý¼ÆËã¿ÉµÃ³ö¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡NaCl¹ÌÌå
2.9
2.9
¿Ë£®ÈôÓÃ4mol/LµÄNaClŨÈÜÒºÅäÖÆ100mL 0.5mol?L-1µÄÏ¡ÈÜÒº£¬Ó¦ÓÃÁ¿Í²Á¿È¡
12.5
12.5
mL¸ÃŨÈÜÒº£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñ£®ÈçͼËùʾÊǼ¸ÖÖʵÑéÖг£ÓõÄÒÇÆ÷£º

д³öÐòºÅËù´ú±íµÄÒÇÆ÷µÄÃû³Æ£ºA
©¶·
©¶·
£»B
ÈÝÁ¿Æ¿
ÈÝÁ¿Æ¿
£»C
ÀäÄý¹Ü
ÀäÄý¹Ü
£»D
·ÖҺ©¶·
·ÖҺ©¶·
£®
¢ò£®ÊµÑéÊÒÒªÅäÖÆ100mL 2mol/L NaOHÈÜÒº£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÅäÖƹý³ÌÖÐʹÓõĻ¯Ñ§ÒÇÆ÷ÓÐ
C
C
£¨ÌîÑ¡ÏîµÄ×Öĸ£©£®
A£®ÉÕ±­¡¡¡¡ B£®100mLÈÝÁ¿Æ¿¡¡¡¡ C£®Â©¶·¡¡¡¡ D£®½ºÍ·µÎ¹Ü ¡¡¡¡¡¡E£®²£Á§°ô
£¨2£©ÓÃÍÐÅÌÌìƽ³ÆÈ¡ÇâÑõ»¯ÄÆ£¬ÆäÖÊÁ¿Îª
8.0
8.0
g£®
£¨3£©ÏÂÁÐÖ÷Òª²Ù×÷²½ÖèµÄÕýȷ˳ÐòÊÇ
¢Ù¢Û¢Ý¢Ú¢Ü
¢Ù¢Û¢Ý¢Ú¢Ü
£¨ÌîÐòºÅ£©£®
¢Ù³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÇâÑõ»¯ÄÆ£¬·ÅÈëÉÕ±­ÖУ¬ÓÃÊÊÁ¿ÕôÁóË®Èܽ⣻
¢Ú¼ÓË®ÖÁÒºÃæÀëÈÝÁ¿Æ¿¾±¿Ì¶ÈÏßÏÂ1-2ÀåÃ×ʱ£¬¸ÄÓýºÍ·µÎ¹ÜµÎ¼ÓÕôÁóË®ÖÁ°¼ÒºÃæÓë¿Ì¶ÈÏßÏàÇУ»
¢Û´ýÀäÈ´ÖÁÊÒκ󣬽«ÈÜҺתÒƵ½100mL ÈÝÁ¿Æ¿ÖУ»
¢Ü¸ÇºÃÆ¿Èû£¬·´¸´ÉÏϵߵ¹£¬Ò¡ÔÈ£»
¢ÝÓÃÉÙÁ¿µÄÕôÁóˮϴµÓÉÕ±­ÄڱںͲ£Á§°ô2¡«3´Î£¬Ï´µÓҺתÒƵ½ÈÝÁ¿Æ¿ÖУ®
£¨4£©Èç¹ûʵÑé¹ý³ÌÖÐȱÉÙ²½Öè¢Ý£¬»áʹÅäÖƳöµÄNaOHÈÜҺŨ¶È
Æ«µÍ
Æ«µÍ
£¨Ìî¡°Æ«¸ß¡±»ò¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ʵÑéÊÒÒªÅäÖÆ100mL 0.1mol?L-1µÄNaClÈÜÒº£¬ÊԻشðÏÂÁи÷Ì⣺
£¨1£©ÏÂÁÐÒÇÆ÷ÖУ¬Ò»¶¨²»»áÓõ½µÄÊÇ
AB
AB

A£®×¶ÐÎÆ¿      B£®200mLÈÝÁ¿Æ¿      C£®Á¿Í²      D£®½ºÍ·µÎ¹Ü     E£®100mLÈÝÁ¿Æ¿        F£®ÍÐÅÌÌìƽ     G¡¢ÉÕ±­
£¨2£©ÈôҪʵʩÅäÖÆ£¬³ýÉÏÊöÒÇÆ÷Í⣬ÉÐȱµÄÒÇÆ÷»òÓÃÆ·ÊÇ
Ô¿³×ºÍ²£Á§°ô
Ô¿³×ºÍ²£Á§°ô
£®
£¨3£©ÈÝÁ¿Æ¿ÉϳýÓп̶ÈÏß¡¢ÈÝ»ýÍ⻹Ӧ±êÓÐ
ζÈ
ζÈ
£¬ÔÚʹÓÃÇ°±ØÐë¼ì²éÈÝÁ¿Æ¿ÊÇ·ñÍêºÃÒÔ¼°
Æ¿Èû
Æ¿Èû
´¦ÊÇ·ñ©ˮ£®£¨ÌîÈÝÁ¿Æ¿µÄÊܼ첿룩
£¨4£©ÈÜÒºÅäÖƹý³ÌÖ÷Òª°üÀ¨Èçͼ¸÷²½Ö裺ÆäÕýÈ·µÄ²Ù×÷˳ÐòÓ¦ÊÇ
DCBFAE
DCBFAE
£¨Ìî¸÷²½ÖèÐòºÅ£©£®
£¨5£©ÅäÖÆÍê±Ïºó£¬½Ìʦָ³öÓÐËÄλͬѧ¸÷½øÐÐÁËÏÂÁÐijһÏî´íÎó²Ù×÷£¬ÄãÈÏΪÕâËÄÏî´íÎó²Ù×÷»áµ¼ÖÂËùµÃÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ
B
B

A£®¶¨ÈÝʱÑöÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß        B£®¶¨ÈÝʱ¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß
C£®½«ÈܽâÀäÈ´µÄÈÜҺתÈëÈÝÁ¿Æ¿ºóδϴµÓ¾ÍÖ±½ÓתÈ붨ÈݲÙ×÷
D£®¶¨Èݺ󣬰ÑÈÝÁ¿Æ¿µ¹ÖÃÒ¡ÔȺó·¢ÏÖÒºÃæµÍÓڿ̶ÈÏߣ¬±ã²¹³ä¼¸µÎË®ÖÁ¿Ì¶È´¦
£¨6£©Í¨¹ý¼ÆËã¿ÉµÃ³ö¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡NaCl¹ÌÌå
0.6
0.6
¿Ë£®ÈôÓÃ4mol/LµÄNaClÈÜÒºÅäÖÆÓ¦ÓÃÁ¿Í²Á¿È¡
2.5
2.5
mL¸ÃÈÜÒº£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

NaClÔÚÉú»î¡¢Éú²úºÍ¿ÆÑÐÖж¼Óй㷺µÄÓÃ;£®
I´Óº£Ë®ÖÐÌáÈ¡µÄ´ÖÑκ¬ÓÐCa2+¡¢Mg2+¡¢SO4-µÈÀë×Ó£¬ÎªÖƱ¸¾«ÑοÉʹÓÃÒÔÏÂËÄÖÖÊÔ¼Á£º¢ÙNa2CO3ÈÜÒº  ¢ÚBaCl2ÈÜÒº  ¢ÛNaOHÈÜÒº  ¢ÜÑÎËᣨÓÃÓÚ³ÁµíµÄÊÔ¼Á¾ùÉÔ¹ýÁ¿£©£®
£¨1£©¼ÓÈëÊÔ¼ÁµÄºÏÀí˳ÐòÊÇ
 
£¨ÌîÑ¡Ï£®
a£®¢Ù¢Ú¢Û¢Üb£®¢Û¢Ù¢Ú¢Üc£®¢Ú¢Ù¢Û¢Üd£®¢Û¢Ú¢Ù¢Ü
£¨2£©¼ÓÈë¹ýÁ¿Na2CO3ÈÜÒºµÄ×÷ÓÃÊÇ
 
£®
II£®ÊµÑéÊÒÒªÅäÖÆ100mL 0.2mol?L-1 NaClÈÜÒº£®
£¨1£©ÓÃÍÐÅÌÌìƽ³ÆÂÈ»¯ÄƹÌÌåµÄÖÊÁ¿ÊÇ
 
g£®
£¨2£©ÅäÖÆÈÜÒºÐèҪʹÓõÄÖ÷Òª²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜºÍ
 
£®
£¨3£©ÏÂÁвÙ×÷ÖУ¬¿ÉÄÜÔì³ÉËùÅäÖÆÈÜÒºµÄŨ¶ÈÆ«µÍµÄÊÇ
 
£¨ÌîÑ¡Ï£®
a£®Ã»ÓÐÏ´µÓÉÕ±­ºÍ²£Á§°ô    b£®¶¨ÈÝʱ£¬¸©Êӿ̶ÈÏß
c£®Ï´µÓºóµÄÈÝÁ¿Æ¿ÖвÐÁôÉÙÁ¿ÕôÁóË®£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸