ÌâÄ¿ÄÚÈÝ

¿ÆѧÉϰѺ¬Óн϶àCa2+¡¢Mg2+µÄË®³ÆΪӲˮ¡£ÎÒ¹ú¹æ¶¨ÒûÓÃË®µÄÓ²¶È²»Äܳ¬¹ý25¶È£¬Ó²¶ÈµÄ±íʾ·½·¨ÊÇ£º½«Ë®ÖеÄCa2+ºÍMg2+¶¼¿´³ÉCa2+£¬²¢½«ÆäÕÛËã³ÉCaOµÄÖÊÁ¿¡£Í¨³£°Ñ1LË®Öк¬ÓÐ10 mgCaO³ÆΪ1¶È¡£
¢ÅʹӲˮÖÐËùº¬Ca2+¡¢Mg2+¡¢HCO3£­µÈ³ýÈ¥µÄ·½·¨ÊÇ£º¼ÓÈëʯ»ÒºóÉú³ÉCa(OH)2£¬½ø¶ø·¢ÉúÈô¸É¸´·Ö½â·´Ó¦£¬ÔÚ³Á½µ³ØÖÐËùµÃ³ÁµíµÄÖ÷Òª³É·ÖΪ_____________ºÍ___________£¨»¯Ñ§Ê½£©¡£
¢ÆFeSO4¡¤7H2OÊdz£ÓõÄÄý¾Û¼Á£¬ËüÔÚË®ÖÐ×îÖÕÉú³É½º×´Fe(OH)3³Áµí¡£Äý¾Û¼Á³ýÈ¥Ðü¸¡¹ÌÌå¿ÅÁ£Æðµ½¾»Ë®µÄ×÷ÓõĹý³Ì______________£¨Ìîд±àºÅ£©¡£
¢Ù·¢ÉúË®½â   ¢Ú·¢ÉúÑõ»¯»¹Ô­·´Ó¦   ¢Û½ºÌåµÄ¾Û³Á
¢Ç¾­¹ý¾»»¯ºóµÄˮҪ½øÒ»²½Ïû¶¾£¬³£ÓõÄÏû¶¾¼ÁÊÇCl2£¬Ó¦ÓÃCl2Ïû¶¾µÄʵÖÊÊÇCl2ºÍË®·´Ó¦Éú³ÉÁË__________£¨Ìѧʽ£©£¬Cl2ÓëË®·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º_______________¡£ÓÉÓÚCl2ÓëË®ÖеÄÓлúÎï×÷ÓòúÉú¶ÔÈËÌå²»ÀûµÄÎïÖÊ£¬ËùÒÔÊÀ½çÎÀÉú×éÖ¯ÒªÇóÍ£Ö¹ÓÃCl2¶Ô×ÔÀ´Ë®½øÐÐÏû¶¾£¬¶øÓÃÆäËûµÄ´úÓÃÆ·£¬ÎÒ¹úÒѾ­¿ªÊ¼Ê¹ÓÃClO2´úÌæCl2¶Ô×ÔÀ´Ë®½øÐÐÏû¶¾£¬ClO2ºÍCl2µÄÏû¶¾Ð§ÂʱÈΪ____________£¨µ¥Î»ÖÊÁ¿×ªÒƵç×ÓÊýÖ®±È£©¡£´ÓÑõ»¯»¹Ô­·´Ó¦½Ç¶È¿´£¬ÏÂÁÐÎïÖÊÖУ¬___________¿ÉÒÔ×÷ΪÆøÌåCl2µÄ´úÓÃÆ·£¨Ìîд±àºÅ£©¡£
¢ÙCa(ClO)2  ¢ÚNH2Cl(ÂÈ°±)  ¢ÛK2FeO4  ¢ÜSO2  ¢ÝAl2(SO4)3
¢ÈȡijˮÑù10mL£¬¾­ÊµÑé²â¶¨£¬ÆäÖк¬Ca2+0.001g£¬º¬Mg2+0.00048g£¬ÊÔͨ¹ý¼ÆËã»Ø´ð£¬¸ÃË®Ñù´ÓÓ²¶È½Ç¶È¿´£¬ÊÇ·ñ·ûºÏÒûÓÃË®±ê×¼£¿£¨Ð´³ö¼ÆËã¹ý³Ì£©
¢ÅMg(OH)2£»CaCO3
¢Æ¢Ù¢Ú¢Û
¢ÇHClO£»Cl2+H2O==H++Cl£­+HClO£»2.63£»¢Ù¢Ú¢Û
¢È10 mLË®ÖУ¬º¬Ca2+ 0.001 g£¬º¬Mg2+ 0.00048 g£¬Ôò1LË®Öк¬Ca2+ 100 mg£¬Mg2+ 48 mg¡£½«Mg2+ 48 mg ÕۺϳÉCa2+µÄÖÊÁ¿Îª80 mg£¬¿É¿´³É1LË®Öй²º¬Ca2+ 180 mg£¬ÓɸÆÀë×Ó¿ÉÇó³öÑõ»¯¸ÆµÄÖÊÁ¿m(CaO)== 252 mg£¬ÔòË®ÑùµÄÓ²¶ÈΪ252 mg/10 mL=25.2 ¶È£¾25 ¶È£¬²»·ûºÏÒûÓÃË®±ê×¼¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¡¾ÈýÑ¡Ò»¡ª»¯Ñ§Óë¼¼Êõ¡¿
¿ÆѧÉϰѺ¬Óн϶àCa2+¡¢Mg2+µÄË®³ÆΪӲˮ¡£ÎÒ¹ú¹æ¶¨ÒûÓÃË®µÄÓ²¶È²»Äܳ¬¹ý25¶È£¬Ó²¶ÈµÄ±íʾ·½·¨ÊÇ£º½«Ë®ÖеÄCa2+ºÍMg2+¶¼¿´³ÉCa2+£¬²¢½«ÆäÕÛËã³ÉCaOµÄÖÊÁ¿¡£Í¨³£°Ñ1LË®Öк¬ÓÐ10 mgCaO³ÆΪ1¶È¡£
¢ÅʹӲˮÖÐËùº¬Ca2+¡¢Mg2+¡¢HCO3£­µÈ³ýÈ¥µÄ·½·¨ÊÇ£º¼ÓÈëʯ»ÒºóÉú³ÉCa(OH)2£¬½ø¶ø·¢ÉúÈô¸É¸´·Ö½â·´Ó¦£¬ÔÚ³Á½µ³ØÖÐËùµÃ³ÁµíµÄÖ÷Òª³É·ÖΪ_____________ºÍ___________£¨»¯Ñ§Ê½£©¡£
¢ÆFeSO4¡¤7H2OÊdz£ÓõÄÄý¾Û¼Á£¬ËüÔÚË®ÖÐ×îÖÕÉú³É½º×´Fe(OH)3³Áµí¡£Äý¾Û¼Á³ýÈ¥Ðü¸¡¹ÌÌå¿ÅÁ£Æðµ½¾»Ë®µÄ×÷ÓõĹý³Ì______________£¨Ìîд±àºÅ£©¡£
¢Ù·¢ÉúË®½â   ¢Ú·¢ÉúÑõ»¯»¹Ô­·´Ó¦   ¢Û½ºÌåµÄ¾Û³Á
¢Ç¾­¹ý¾»»¯ºóµÄˮҪ½øÒ»²½Ïû¶¾£¬³£ÓõÄÏû¶¾¼ÁÊÇCl2£¬Ó¦ÓÃCl2Ïû¶¾µÄʵÖÊÊÇCl2ºÍË®·´Ó¦Éú³ÉÁË__________£¨Ìѧʽ£©£¬Cl2ÓëË®·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º_______________¡£ÓÉÓÚCl2ÓëË®ÖеÄÓлúÎï×÷ÓòúÉú¶ÔÈËÌå²»ÀûµÄÎïÖÊ£¬ËùÒÔÊÀ½çÎÀÉú×éÖ¯ÒªÇóÍ£Ö¹ÓÃCl2¶Ô×ÔÀ´Ë®½øÐÐÏû¶¾£¬¶øÓÃÆäËûµÄ´úÓÃÆ·£¬ÎÒ¹úÒѾ­¿ªÊ¼Ê¹ÓÃClO2´úÌæCl2¶Ô×ÔÀ´Ë®½øÐÐÏû¶¾£¬ClO2ºÍCl2µÄÏû¶¾Ð§ÂʱÈΪ____________£¨µ¥Î»ÖÊÁ¿×ªÒƵç×ÓÊýÖ®±È£©¡£´ÓÑõ»¯»¹Ô­·´Ó¦½Ç¶È¿´£¬ÏÂÁÐÎïÖÊÖУ¬___________¿ÉÒÔ×÷ΪÆøÌåCl2µÄ´úÓÃÆ·£¨Ìîд±àºÅ£©¡£
¢ÙCa(ClO)2  ¢ÚNH2Cl(ÂÈ°±)  ¢ÛK2FeO4  ¢ÜSO2  ¢ÝAl2(SO4)3
¢ÈȡijˮÑù10mL£¬¾­ÊµÑé²â¶¨£¬ÆäÖк¬Ca2+0.001g£¬º¬Mg2+0.00048g£¬ÊÔͨ¹ý¼ÆËã»Ø´ð£¬¸ÃË®Ñù´ÓÓ²¶È½Ç¶È¿´£¬ÊÇ·ñ·ûºÏÒûÓÃË®±ê×¼£¿£¨Ð´³ö¼ÆËã¹ý³Ì£©

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

¾«Ó¢¼Ò½ÌÍø