ÌâÄ¿ÄÚÈÝ
½ñÓТÙCH3COOH£»¢ÚHCl£»¢ÛH2SO4ÈýÖÖÈÜÒº£¬Ñ¡ÔñÌî¿Õ£º
A£®¢Ù£¾¢Ú£¾¢Û B£®¢Ù£¼¢Ú£¼¢Û C£®¢Ù=¢Ú=¢Û E£®¢Ù£¾¢Ú=¢Û
D£®¢Ù=¢Û£¾¢Ú F£®¢Ù£¼¢Ú=¢Û G£®¢Ù=¢Ú£¼¢Û H£®¢Ù=¢Ú£¾¢Û
(1)µ±ËüÃÇpHÏàͬʱ£¬ÆäÎïÖʵÄÁ¿Å¨¶È¹ØÏµÊÇ___________¡£
(2)Ìå»ýºÍÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄ¢Ù¡¢¢Ú¡¢¢ÛÈýÈÜÒº£¬·Ö±ðÓëͬŨ¶ÈµÄÉÕ¼îÈÜÒº·´Ó¦£¬ÒªÊ¹·´Ó¦ºóµÄÈÜÒº³ÊÖÐÐÔ£¬ËùÐèÉÕ¼îÈÜÒºµÄÌå»ý¹ØÏµÎª_________________¡£
(3)µ±ËüÃÇpHÏàͬ¡¢Ìå»ýÏàͬʱ£¬·Ö±ð¼ÓÈë×ãÁ¿Ð¿£¬Ôò¿ªÊ¼Ê±·´Ó¦ËÙÂÊ_________£¬Ïàͬ״¿öϲúÉúÆøÌåÌå»ý¹ØÏµÎª____________¡£
(4)ijѧÉúÓÃ0.2000mol/L NaOHÈÜÒºµÎ¶¨Î´ÖªÅ¨¶ÈµÄÑÎËáÈÜÒº£¬Æä²Ù×÷¿É·Ö½âΪÈçϼ¸²½£º
£¨A£©ÒÆÈ¡20.00mL´ý²âµÄÑÎËáÈÜҺעÈë½à¾»µÄ×¶ÐÎÆ¿£¬²¢¼ÓÈë2-3µÎ·Ó̪
£¨B£©Óñê×¼ÈÜÒºÈóÏ´µÎ¶¨¹Ü2-3´Î
£¨C£©°ÑÊ¢Óбê×¼ÈÜÒºµÄ¼îʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚÒºÃæÊ¹µÎ¶¨¹Ü¼â×ì³äÂúÈÜÒº
£¨D£©È¡±ê×¼NaOHÈÜҺעÈë¼îʽµÎ¶¨¹ÜÖÁ0¿Ì¶ÈÒÔÉÏ2-3cm
£¨E£©µ÷½ÚÒºÃæÖÁ0»ò0¿Ì¶ÈÒÔÏ£¬¼Ç϶ÁÊý
£¨F£©°Ñ×¶ÐÎÆ¿·ÅÔڵζ¨¹ÜµÄÏÂÃæ£¬Óñê×¼NaOHÈÜÒºµÎ¶¨ÖÁÖյ㣬¼ÇÏµζ¨¹ÜÒºÃæµÄ¿Ì¶È
Íê³ÉÒÔÏÂÌî¿Õ£º
¢ÙÕýÈ·²Ù×÷µÄ˳ÐòÊÇ£¨ÓÃÐòºÅ×ÖĸÌîд£©______________________.
¢ÚʵÑéÖУ¬ÑÛ¾¦Ó¦×¢ÊÓ_______£¬Ö±ÖÁµÎ¶¨Öյ㡣
¢Û¼¸´ÎµÎ¶¨ÏûºÄNaOHÈÜÒºµÄÌå»ýÈçÏÂ:
|
ʵÑéÐòºÅ |
1 |
2 |
3 |
4 |
|
ÏûºÄNaOHÈÜÒºµÄÌå»ý(mL) |
20.05 |
20.00 |
18.80 |
19.95 |
Ôò¸ÃÑÎËáÈÜÒºµÄ׼ȷŨ¶ÈΪ_____________¡£(±£ÁôСÊýµãºó4λ)
£¨5£©¡¢Óñê×¼µÄNaOHµÎ¶¨Î´ÖªÅ¨¶ÈµÄÑÎËᣬѡÓ÷Ó̪Ϊָʾ¼Á£¬Ôì³É²â¶¨½á¹ûÆ«¸ßµÄÔÒò¿ÉÄÜÊÇ:____________ (´íѡȫ¿Û)
A. ÅäÖÆ±ê×¼ÈÜÒºµÄÇâÑõ»¯ÄÆÖлìÓÐNa2CO3ÔÓÖÊ
B. µÎ¶¨ÖÕµã¶ÁÊýʱ£¬¸©Êӵζ¨¹ÜµÄ¿Ì¶È£¬ÆäËü²Ù×÷¾ùÕýÈ·
C. ʢװδ֪ҺµÄ×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¹ý£¬Î´Óôý²âÒºÈóÏ´
D. µÎ¶¨µ½ÖÕµã¶ÁÊýʱ·¢Ïֵζ¨¹Ü¼â×ì´¦Ðü¹ÒÒ»µÎÈÜÒº
E. δÓñê×¼ÒºÈóÏ´¼îʽµÎ¶¨¹Ü
![]()
¡¾½âÎö¡¿ÂÔ