ÌâÄ¿ÄÚÈÝ

ʵÑéÊÒÒªÅäÖÆ100mL 0.1mol?L-1µÄNaClÈÜÒº£¬ÊԻشðÏÂÁи÷Ì⣺

(1)ÏÂÁÐÒÇÆ÷ÖУ¬²»»áÓõ½µÄÊÇ 

A£®×¶ÐÎÆ¿    B£®200mLÈÝÁ¿Æ¿  C£®Á¿Í²   D£®½ºÍ·µÎ¹Ü  E£®100mLÈÝÁ¿Æ¿  F£®Ììƽ

(2)ÈôҪʵʩÅäÖÆ£¬³ýÉÏÊöÒÇÆ÷Í⣬ÉÐȱµÄÒÇÆ÷»òÓÃÆ·ÊÇ_____________________¡£

(3)ÈÝÁ¿Æ¿ÉϳýÓп̶ÈÏßÍ⻹Ӧ±êÓÐ___________£¬ÔÚʹÓÃÇ°±ØÐë¼ì²éÈÝÁ¿Æ¿ÊÇ·ñÍêºÃÒÔ¼°________________´¦ÊÇ·ñ©ˮ¡£(ÌîÈÝÁ¿Æ¿µÄÊܼ첿λ)

(4)ÈËÃdz£½«ÅäÖƹý³Ì¼òÊöΪÒÔϸ÷²½Ö裺

A£®ÀäÈ´    B£®³ÆÁ¿    C£®Ï´µÓ    D£®¶¨ÈÝ      E£®Èܽ⠠  F£®Ò¡ÔÈ    G£®ÒÆÒº

ÆäÕýÈ·µÄ²Ù×÷˳ÐòÓ¦ÊÇ_____________________(Ìî¸÷²½ÖèÐòºÅ)¡£

(5)ÅäÖÆÍê±Ïºó£¬½Ìʦָ³öÓÐËÄλͬѧ¸÷½øÐÐÁËÏÂÁÐijһÏî´íÎó²Ù×÷£¬ÄãÈÏΪÕâËÄÏî´íÎó²Ù×÷»áµ¼ÖÂËùµÃÈÜҺŨ¶ÈÆ«¸ßµÄÊÇ      

A£®¶¨ÈÝʱÑöÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß

B£®¶¨ÈÝʱ¸©ÊÓÈÝÁ¿Æ¿¿Ì¶ÈÏß

C£®½«ÈܽâÀäÈ´µÄÈÜҺתÈëÈÝÁ¿Æ¿ºó¾ÍÖ±½ÓתÈ붨ÈݲÙ×÷

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

£¨6£©Í¨¹ý¼ÆËã¿ÉµÃ³ö¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡NaCl¹ÌÌå______________¿Ë¡£ÈôÓÃ4mol/LµÄNaClÈÜÒºÅäÖÆÓ¦ÓÃÁ¿Í²Á¿È¡________________mL¸ÃÈÜÒº¡£

£¨1£©A B C                                    (2)ÉÕ±­¡¢²£Á§°ô¡¢Ò©³×  

£¨3£©Î¶ȡ¢ÈÝÁ¿   Æ¿Èû               £¨4£©B E A G C D F           

£¨5£©B                                        £¨6£©0.6         2.5

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¢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¡±£©£®
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£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø