ÌâÄ¿ÄÚÈÝ
×ÛºÏÀûÓú£Ë®¿ÉÒÔÖƱ¸Ê³ÑΡ¢´¿¼î¡¢½ðÊôþµÈÎïÖÊ£¬ÆäÁ÷³ÌÈçÏÂͼËùʾ£º
£¨1£©·´Ó¦¢Ù¡«¢ÝÖУ¬ÊôÓÚÑõ»¯»¹Ô·´Ó¦µÄÊÇ £¨Ìî±àºÅ£©¡£
£¨2£©Ð´³ö·´Ó¦¢ÚµÄÀë×Ó·½³Ìʽ ¡£
£¨3£©XÈÜÒºÖеÄÖ÷ÒªÑôÀë×ÓÊÇNa+ºÍ ¡£
£¨4£©´ÖÑÎÖк¬ÓÐNa2SO4¡¢MgCl2¡¢CaCl2µÈ¿ÉÈÜÐÔÔÓÖÊ£¬ÎªÖƵô¿¾»µÄNaCl¾§Ì壬²Ù×÷ÈçÏ£º
¢ÙÈܽ⣻¢ÚÒÀ´Î¼ÓÈë¹ýÁ¿µÄBaCl2ÈÜÒº¡¢NaOHÈÜÒº¡¢Na2CO3ÈÜÒº£»¢Û £»¢Ü¼ÓÊÊÁ¿ÑÎË᣻¢Ý ¡££¨Ç벹ȫȱÉÙµÄʵÑé²½Ö裩
£¨5£©¼ìÑé´¿¼îÑùÆ·ÖÐÊÇ·ñº¬NaClӦѡÓõÄÊÔ¼ÁÊÇ ¡£
£¨1£©·´Ó¦¢Ù¡«¢ÝÖУ¬ÊôÓÚÑõ»¯»¹Ô·´Ó¦µÄÊÇ £¨Ìî±àºÅ£©¡£
£¨2£©Ð´³ö·´Ó¦¢ÚµÄÀë×Ó·½³Ìʽ ¡£
£¨3£©XÈÜÒºÖеÄÖ÷ÒªÑôÀë×ÓÊÇNa+ºÍ ¡£
£¨4£©´ÖÑÎÖк¬ÓÐNa2SO4¡¢MgCl2¡¢CaCl2µÈ¿ÉÈÜÐÔÔÓÖÊ£¬ÎªÖƵô¿¾»µÄNaCl¾§Ì壬²Ù×÷ÈçÏ£º
¢ÙÈܽ⣻¢ÚÒÀ´Î¼ÓÈë¹ýÁ¿µÄBaCl2ÈÜÒº¡¢NaOHÈÜÒº¡¢Na2CO3ÈÜÒº£»¢Û £»¢Ü¼ÓÊÊÁ¿ÑÎË᣻¢Ý ¡££¨Ç벹ȫȱÉÙµÄʵÑé²½Ö裩
£¨5£©¼ìÑé´¿¼îÑùÆ·ÖÐÊÇ·ñº¬NaClӦѡÓõÄÊÔ¼ÁÊÇ ¡£
£¨1£©¢Ý £¨2·Ö£©
£¨2£©Mg (OH)2+2H+ = Mg2+ + H2O £¨2·Ö£©
£¨3£©NH4+£¨2·Ö£©
£¨4£©¹ýÂË £¨2·Ö£© ¡¢Õô·¢½á¾§£¨ÆäËü´ð°¸ºÏÀíÒ²¸ø·Ö£©£¨2·Ö£©
£¨5£©Ï¡ÏõËá¡¢AgNO3ÈÜÒº£¨2·Ö£©
£¨2£©Mg (OH)2+2H+ = Mg2+ + H2O £¨2·Ö£©
£¨3£©NH4+£¨2·Ö£©
£¨4£©¹ýÂË £¨2·Ö£© ¡¢Õô·¢½á¾§£¨ÆäËü´ð°¸ºÏÀíÒ²¸ø·Ö£©£¨2·Ö£©
£¨5£©Ï¡ÏõËá¡¢AgNO3ÈÜÒº£¨2·Ö£©
ÊÔÌâ·ÖÎö£º£¨1£©·´Ó¦¢Ù¢ÚÊôÓÚ¸´·Ö½â·´Ó¦£¬¢ÛÊôÓÚ»¯ºÏ·´Ó¦£¬¢ÜÊôÓڷֽⷴӦ£¬¢ÝΪÑõ»¯»¹Ô·´Ó¦£»£¨2£©Ëá¼îÖкͷ´Ó¦£¬Mg (OH)2²»ÈÜÓÚˮд³É»¯Ñ§Ê½£¬Mg (OH)2+2H+ = Mg2+ + H2O£»£¨3£©¸Ã·´Ó¦Îª°±Ë®¡¢¶þÑõ»¯Ì¼¡¢ÂÈ»¯ÄÆ·´Ó¦Éú³É̼ËáÇâÄƳÁµíºÍÂÈ»¯ï§£¬XÈÜÒºÖеÄÖ÷ÒªÑôÀë×ÓÊÇNH4+ºÍNa+£»£¨4£©¼ÓÈëÑÎËáÇ°Éú³ÉµÄ³ÁµíÄÜÈܽâÓÚÑÎËᣬ¹ÊÓ¦¹ýÂ˳ýÈ¥³ÁµíºóÔÙ¼ÓÑÎËᣬȻºóͨ¹ýÕô·¢½á¾§µÃµ½ÂÈ»¯Äƾ§Ì壻£¨5£©¼ìÑéÑùÆ·ÖÐÊÇ·ñº¬ÓÐÂÈÀë×Ó£¬ÓÃÏõËáËữµÄÏõËáÒø¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿