ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿¸ù¾ÝÏÂÁи÷ÌâËù¸ø³öµÄÊý¾Ý£¬¿É·Ö±ðÇó³öÆä¡°ÈÜÖʵÄÖÊÁ¿·ÖÊý¡±»ò¡°ÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶È¡±£¬ÊÔÅжϲ¢Çó½â¡£
£¨1£©ÉèNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÊýÖµ£¬ÈôijÇâÑõ»¯ÄÆÈÜÒºV LÖк¬ÓÐN¸öOH-£¬Ôò¿ÉÇó³ö´ËÈÜÒºÖÐ______Ϊ______¡£
£¨2£©ÒÑ֪ijÇâÑõ»¯ÄÆÈÜÒºÖÐNa£«ÓëH2OµÄ¸öÊýÖ®±ÈΪ1¡Ãa£¬Ôò¿ÉÇó³ö´ËÈÜÒºÖÐ______Ϊ______¡£
£¨3£©ÒÑÖª±ê×¼×´¿öÏÂ1Ìå»ýË®ÄÜÈܽâ500Ìå»ýµÄÂÈ»¯Ç⣬Ôò¿ÉÇó³ö±ê×¼×´¿öÏÂÂÈ»¯Çâ±¥ºÍÈÜÒºÖÐ______Ϊ______¡£
£¨4£©ÒÑÖª½«100 mLÂÈ»¯ÂÁµÄË®ÈÜÒº¼ÓÈÈÕô¸É×ÆÉÕ£¬¿ÉµÃµ½°×É«¹ÌÌåb g£¬Ôò¿ÉÇó³öÔÂÈ»¯ÂÁÈÜÒºÖÐ______Ϊ______¡£
¡¾´ð°¸¡¿£¨1£©NaOHµÄÎïÖʵÄÁ¿Å¨¶È
£¨2£©NaOHµÄÖÊÁ¿·ÖÊý
£¨3£©HClµÄÖÊÁ¿·ÖÊý 44.9%
£¨4£©AlCl3µÄÎïÖʵÄÁ¿Å¨¶Èmol/L
¡¾½âÎö¡¿£¨1£©²»ÖªÈÜÒºµÄÃܶȣ¬Ö»ÄÜÇó³öÎïÖʵÄÁ¿Å¨¶È£¬c£½mol/L¡£
£¨2£©NaOHÓëH2OµÄÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ãa£¬¿ÉÇóÈÜÖʵÄÖÊÁ¿·ÖÊý£¬w£½¡£
£¨3£©²»ÖªÈÜÒºµÄÃܶȣ¬²»ÄܼÆËãÎïÖʵÄÁ¿Å¨¶È£¬¿ÉÇóÆäÖÊÁ¿·ÖÊý£½ ¡£
£¨4£©°×É«¹ÌÌåΪAl2O3£¬n£¨Al2O3£©£½mol£¬n£¨AlCl3£©£½mol£¬c£¨AlCl3£©£½£½mol/L¡£
¡¾ÌâÄ¿¡¿Ä³¿ÎÍâС×éÉè¼ÆʵÑéÊÒÖÆÈ¡²¢Ìá´¿ÒÒËáÒÒõ¥µÄ·½°¸ÈçÏ£º
ÒÑÖª£º¢ÙÂÈ»¯¸Æ¿ÉÓëÒÒ´¼ÐγÉCaCl2¡¤6C2H5OH
¢ÚÓйØÓлúÎïµÄ·Ðµã£º
ÊÔ¼Á | ÒÒÃÑ | ÒÒ´¼ | ÒÒËá | ÒÒËáÒÒõ¥ |
·Ðµã/¡æ | 34.7 | 78.5 | 118 | 77.1 |
¢Û2CH3CH2OHCH3CH2OCH2CH3£«H2O
I£®ÖƱ¸¹ý³Ì
×°ÖÃÈçͼËùʾ£¬AÖзÅÓÐŨÁòËᣬBÖзÅÓÐ9.5mLÎÞË®ÒÒ´¼ºÍ6mL±ù´×ËᣬDÖзÅÓб¥ºÍ̼ËáÄÆÈÜÒº¡£
£¨1£©Ð´³öÒÒËáÓëÒÒ´¼·¢Éúõ¥»¯·´Ó¦µÄ»¯Ñ§·½³Ìʽ______________________________¡£
£¨2£©ÊµÑé¹ý³ÌÖеμӴóÔ¼3mLŨÁòËᣬBµÄÈÝ»ý×îºÏÊʵÄÊÇ________£¨ÌîÈëÕýÈ·Ñ¡ÏîÇ°µÄ×Öĸ£©
A£®25mL B£®50mL C£®250mL D£®500mL
£¨3£©ÇòÐθÉÔï¹ÜµÄÖ÷Òª×÷ÓÃÊÇ_________________________¡£
£¨4£©±¥ºÍNa2CO3ÈÜÒºµÄ×÷ÓÃÊÇ__________________________________________________________________________________________________________________________________¡£
II£®Ìá´¿·½·¨£º¢Ù½«DÖлìºÏҺתÈë·ÖҺ©¶·½øÐзÖÒº¡£
¢ÚÓлú²ãÓÃ5mL±¥ºÍʳÑÎˮϴµÓ£¬ÔÙÓÃ5mL±¥ºÍÂÈ»¯¸ÆÈÜҺϴµÓ£¬×îºóÓÃˮϴµÓ¡£Óлú²ãµ¹ÈëÒ»¸ÉÔïµÄÉÕÆ¿ÖУ¬ÓÃÎÞË®ÁòËáþ¸ÉÔµÃ´Ö²úÎï¡£
¢Û½«´Ö²úÎïÕôÁó£¬ÊÕ¼¯77.1¡æµÄÁó·Ö£¬µÃµ½´¿¾»¸ÉÔïµÄÒÒËáÒÒõ¥¡£
£¨5£©µÚ¢Ù²½·ÖҺʱ£¬Ñ¡ÓõÄÁ½ÖÖ²£Á§ÒÇÆ÷µÄÃû³Æ·Ö±ðÊÇ__________¡¢_______¡£
£¨6£©µÚ¢Ú²½ÖÐÓñ¥ºÍʳÑÎË®¡¢±¥ºÍÂÈ»¯¸ÆÈÜÒº¡¢×îºóÓÃˮϴµÓ£¬·Ö±ðÖ÷Ҫϴȥ´Ö²úÆ·ÖеÄ________________£¬__________________£¬______________¡£
¡¾ÌâÄ¿¡¿ÓÃÉúʯ»Ò£¨CaO£©¡¢´¿¼î£¨Na2CO3£©ºÍʳÑΣ¨NaCl£©°´Ò»¶¨±ÈÀý»ìºÏ¿ÉÖƵÃÒ»ÖÖëçÖÆÔÁÏ£¬Ä³Í¬Ñ§ÎªÁË̽¾¿¸ÃÔÁϳɷÝ×öÁËÒÔÏÂʵÑ飺
¸Ãͬѧȡһ¶¨Á¿µÄÑùÆ·ÈÜÓÚË®£¬Ö÷Òª·¢ÉúµÄ»¯Ñ§·½³ÌʽÓУº____________________________, _____________________________________¡£
£¨2£©¸ÃͬѧÈÏΪ£¨1£©¹ýÂ˺óËùµÃµÄÂËÒºÖп϶¨º¬ÓдóÁ¿µÄNaOH¡¢NaClÈÜÖÊ£¬»¹¿ÉÄܺ¬ÓÐCa(OH)2»òNa2CO3 £¬ÎªÁË̽¾¿ËùµÃÂËÒºÖпÉÄܺ¬ÓеÄÎïÖÊÊÇ·ñ´æÔÚ£¬ÇëÄã°ïËûÍêÉÆʵÑé·½°¸¡£
ʵÑé·½°¸Éè¼ÆÈçϱíËùʾ£º
ʵ Ñé ²½ Öè | ʵ Ñé ÏÖ Ïó | ʵ Ñé ½á ÂÛ | ÓÃÀë×Ó·½³Ìʽ½âÊÍ |
¢ÙÈ¡ÉÙÁ¿ÂËÒº£¬µÎ¼ÓÊÊÁ¿K2CO3ÈÜÒº | Èô³öÏÖ°×É«³Áµí | ÂËÒºÖк¬Ca(OH)2 | ¢ñ:________________ |
ÈôÎÞ°×É«³Áµí | ÂËÒºÖÐÎÞCa(OH)2 | ||
¢ÚÈ¡ÉÙÁ¿ÂËÒº£¬______________________________________________ | ¢¡:Èô³ö___________ | ÂËÒºÖк¬Na2CO3 | ¢ò:_______________ |
¢¢:ÈôÎÞ___________ | ÂËÒºÖÐÎÞNa2CO3 |