ijÆø̬ÌþAÔÚ±ê×¼×´¿öϵÄÃܶÈΪ1.25g/L£¬Æä²úÁ¿¿ÉÒÔÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½¡£BºÍD¶¼ÊÇÉú»îÖг£¼ûµÄÓлúÎDÄܸú̼ËáÇâÄÆ·´Ó¦£¬FÓÐÏãζ¡£ËüÃÇÖ®¼äµÄת»¯¹ØϵÈçÏÂͼËùʾ£º

£¨1£©AµÄ½á¹¹Ê½Îª          £¬BÖйÙÄÜÍŵĵç×ÓʽΪ                      £¬

DÖйÙÄÜÍŵÄÃû³ÆΪ                 ¡£

£¨2£©·´Ó¦¢ÙµÄ·´Ó¦ÀàÐÍÊÇ            £¬

·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽΪ                                              ¡£

£¨3£©·´Ó¦¢ÚÔÚCu×ö´ß»¯¼ÁµÄÌõ¼þϽøÐУ¬¸ÃʵÑéµÄ²½ÖèÊǽ«ºìÁÁµÄÍ­Ë¿ÖÃÓھƾ«µÆÉϼÓÈÈ£¬´ýÍ­Ë¿±äΪºÚɫʱ£¬Ñ¸ËÙ½«Æä²åÈëµ½×°ÓÐBµÄÊÔ¹ÜÖУ¨ÈçͼËùʾ£©¡£Öظ´²Ù×÷2£­3´Î¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                              ¡£

£¨4£©DÓë̼ËáÇâÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ                          ¡£

£¨5£©B¡¢DÔÚŨÁòËáµÄ×÷ÓÃÏÂʵÏÖ·´Ó¦¢Ü£¬ÊµÑé×°ÖÃÈçÏÂͼËùʾ£º

ÊÔ¹Ü1ÖÐ×°ÈëÒ©Æ·ºó¼ÓÈÈ¡£Í¼ÖÐXµÄ»¯Ñ§Ê½Îª                     ¡£

Æä×÷ÓÃÊÇ                   ¡£

ÊÔ¹Ü1·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                        ¡£

£¨1£© ôÈ»ù¡£

£¨2£©¼Ó³É£»CH2=CH2 + Br2¡úCH2BrCH2Br

£¨3£©

£¨4£©CH3COOH+HCO3-¡úCH3COO-+H2O+CO2¡ü

£¨5£©Na2CO3£»ÈܽâÒÒ´¼£¬ÎüÊÕÒÒËᣬÓÐÀûÓÚÒÒËáÒÒõ¥·Ö²ã¡£

  


½âÎö:

ÂÔ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ijÆø̬ÌþAÔÚ±ê×¼×´¿öϵÄÃܶÈΪ1.25g/L£¬Æä²úÁ¿¿ÉÒÔÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½£®BºÍD¶¼ÊÇÉú»îÖг£¼ûµÄÓлúÎDÄܸú̼ËáÇâÄÆ·´Ó¦£¬FÓÐÏãζ£®ËüÃÇÖ®¼äµÄת»¯¹ØϵÈçͼ1Ëùʾ£º

£¨1£©AµÄ½á¹¹Ê½Îª
CH2=CH2
CH2=CH2
£¬BÖйÙÄÜÍŵĵç×ÓʽΪ
£¬DÖйÙÄÜÍŵÄÃû³ÆΪ
ôÈ»ù
ôÈ»ù
£®
£¨2£©·´Ó¦¢ÙµÄ·´Ó¦ÀàÐÍÊÇ
¼Ó³É·´Ó¦
¼Ó³É·´Ó¦
£¬·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽΪ
CH2=CH2+Br2¡úCH2BrCH2Br
CH2=CH2+Br2¡úCH2BrCH2Br
£®
£¨3£©·´Ó¦¢ÚÔÚCu×ö´ß»¯¼ÁµÄÌõ¼þϽøÐУ¬¸ÃʵÑéµÄ²½ÖèÊǽ«ºìÁÁµÄÍ­Ë¿ÖÃÓھƾ«µÆÉϼÓÈÈ£¬´ýÍ­Ë¿±äΪºÚɫʱ£¬Ñ¸ËÙ½«Æä²åÈëµ½×°ÓÐBµÄÊÔ¹ÜÖУ¨Èçͼ2Ëùʾ£©£®Öظ´²Ù×÷2-3´Î£¬¹Û²ìµ½µÄÏÖÏóÊÇ
Í­Ë¿ÓɺÚÉ«±äºìÉ«£¬²úÉú´Ì¼¤ÐÔÆøζ
Í­Ë¿ÓɺÚÉ«±äºìÉ«£¬²úÉú´Ì¼¤ÐÔÆøζ
£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
2CH3CH2OH+O2
Cu
¡÷
2CH3CHO+2H2O
2CH3CH2OH+O2
Cu
¡÷
2CH3CHO+2H2O
£®
£¨4£©DÓë̼ËáÇâÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ
CH3COOH+HCO3-¡úCH3COO-+H2O+CO2¡ü
CH3COOH+HCO3-¡úCH3COO-+H2O+CO2¡ü
£®
£¨5£©B¡¢DÔÚŨÁòËáµÄ×÷ÓÃÏÂʵÏÖ·´Ó¦¢Ü£¬ÊµÑé×°ÖÃÈçͼ3Ëùʾ£º
ÊÔ¹Ü1ÖÐ×°Ò©Æ·µÄ˳ÐòΪÏÈÈ¡
B
B
£¨Ìî¡°B¡±¡¢¡°D¡±¡¢¡°Å¨ÁòËᡱ£¬ÏÂͬ£©£¬ÔÙ¼ÓÈë
ŨÁòËá
ŨÁòËá
£¬×îºó¼ÓÈë
D
D
£®Í¼ÖÐXµÄ»¯Ñ§Ê½Îª
Na2CO3
Na2CO3
£®Å¨ÁòËáµÄ×÷ÓÃÊÇ
´ß»¯¼Á¡¢ÎüË®¼Á
´ß»¯¼Á¡¢ÎüË®¼Á
£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
CH3CH2OH+CH3COOH
ŨÁòËá
¡÷
CH3COOCH2CH3+H2O
CH3CH2OH+CH3COOH
ŨÁòËá
¡÷
CH3COOCH2CH3+H2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê±±¾©ÎåÖи߶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

ijÆø̬ÌþAÔÚ±ê×¼×´¿öϵÄÃܶÈΪ1.25g/L£¬Æä²úÁ¿¿ÉÒÔÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½¡£BºÍD¶¼ÊÇÉú»îÖг£¼ûµÄÓлúÎDÄܸú̼ËáÇâÄÆ·´Ó¦£¬FÓÐÏãζ¡£ËüÃÇÖ®¼äµÄת»¯¹ØϵÈçÏÂͼËùʾ£º

£¨1£©AµÄ½á¹¹Ê½Îª         £¬BÖйÙÄÜÍŵĵç×ÓʽΪ                     £¬
DÖйÙÄÜÍŵÄÃû³ÆΪ                ¡£
£¨2£©·´Ó¦¢ÙµÄ·´Ó¦ÀàÐÍÊÇ           £¬
·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽΪ                                             ¡£
£¨3£©·´Ó¦¢ÚÔÚCu×ö´ß»¯¼ÁµÄÌõ¼þϽøÐУ¬¸ÃʵÑéµÄ²½ÖèÊǽ«ºìÁÁµÄÍ­Ë¿ÖÃÓھƾ«µÆÉϼÓÈÈ£¬´ýÍ­Ë¿±äΪºÚɫʱ£¬Ñ¸ËÙ½«Æä²åÈëµ½×°ÓÐBµÄÊÔ¹ÜÖУ¨ÈçͼËùʾ£©¡£Öظ´²Ù×÷2£­3´Î¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                             ¡£
£¨4£©DÓë̼ËáÇâÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ                         ¡£
£¨5£©B¡¢DÔÚŨÁòËáµÄ×÷ÓÃÏÂʵÏÖ·´Ó¦¢Ü£¬ÊµÑé×°ÖÃÈçÏÂͼËùʾ£º

ÊÔ¹Ü1ÖÐ×°ÈëÒ©Æ·ºó¼ÓÈÈ¡£Í¼ÖÐXµÄ»¯Ñ§Ê½Îª                    ¡£

Æä×÷ÓÃÊÇ                  ¡£
ÊÔ¹Ü1·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄê¸ß¿¼»¯Ñ§ÈýÂÖ¸´Ï°ÅŲé רÌâ15Óлú»¯Ñ§»ù´¡Ñ¡ÐÞ5Á·Ï°¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

ijÆø̬ÌþAÔÚ±ê×¼×´¿öϵÄÃܶÈΪ1.25 g/L£¬Æä²úÁ¿¿ÉÒÔÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½¡£BºÍD¶¼ÊÇÉú»îÖг£¼ûµÄÓлúÎDÄܸú̼ËáÇâÄÆ·´Ó¦£¬FÓÐÏãζ¡£ËüÃÇÖ®¼äµÄת»¯¹ØϵÈçÏÂËùʾ£º

£¨1£©AµÄ½á¹¹Ê½Îª¡¡¡¡¡¡¡¡¡£BÖйÙÄÜÍŵĵç×ÓʽΪ¡¡¡¡¡¡¡¡¡£

£¨2£©·´Ó¦¢ÙµÄ·´Ó¦ÀàÐÍÊÇ¡¡¡¡¡¡¡¡£¬·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

£¨3£©·´Ó¦¢ÚÔÚCu×÷´ß»¯¼ÁµÄÌõ¼þϽøÐУ¬¸ÃʵÑéµÄ²½ÖèÊǽ«ºìÁÁµÄÍ­Ë¿ÖÃÓھƾ«µÆÉϼÓÈÈ£¬´ýÍ­Ë¿±äΪºÚɫʱ£¬Ñ¸ËÙ½«Æä²åÈëµ½×°ÓÐBµÄÊÔ¹ÜÖУ¨ÈçͼËùʾ£©¡£Öظ´²Ù×÷2¡«3´Î¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________________¡£

£¨4£©DÓë̼ËáÇâÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

£¨5£©B¡¢DÔÚŨÁòËáµÄ×÷ÓÃÏ·¢Éú·´Ó¦¢Ü£¬ÊµÑé×°ÖÃÈçͼËùʾ£º

ÊÔ¹ÜÖÐ×°ÈëÒ©Æ·ºó¼ÓÈÈ¡£Í¼ÖÐXµÄ»¯Ñ§Ê½Îª¡¡¡¡¡¡£¬Æä×÷ÓÃÊÇ____________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê±±¾©ÎåÖи߶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

ijÆø̬ÌþAÔÚ±ê×¼×´¿öϵÄÃܶÈΪ1.25g/L£¬Æä²úÁ¿¿ÉÒÔÓÃÀ´ºâÁ¿Ò»¸ö¹ú¼ÒµÄʯÓÍ»¯¹¤·¢Õ¹Ë®Æ½¡£BºÍD¶¼ÊÇÉú»îÖг£¼ûµÄÓлúÎDÄܸú̼ËáÇâÄÆ·´Ó¦£¬FÓÐÏãζ¡£ËüÃÇÖ®¼äµÄת»¯¹ØϵÈçÏÂͼËùʾ£º

£¨1£©AµÄ½á¹¹Ê½Îª          £¬BÖйÙÄÜÍŵĵç×ÓʽΪ                      £¬

DÖйÙÄÜÍŵÄÃû³ÆΪ                 ¡£

£¨2£©·´Ó¦¢ÙµÄ·´Ó¦ÀàÐÍÊÇ            £¬

·´Ó¦¢ÛµÄ»¯Ñ§·½³ÌʽΪ                                              ¡£

£¨3£©·´Ó¦¢ÚÔÚCu×ö´ß»¯¼ÁµÄÌõ¼þϽøÐУ¬¸ÃʵÑéµÄ²½ÖèÊǽ«ºìÁÁµÄÍ­Ë¿ÖÃÓھƾ«µÆÉϼÓÈÈ£¬´ýÍ­Ë¿±äΪºÚɫʱ£¬Ñ¸ËÙ½«Æä²åÈëµ½×°ÓÐBµÄÊÔ¹ÜÖУ¨ÈçͼËùʾ£©¡£Öظ´²Ù×÷2£­3´Î¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                              ¡£

£¨4£©DÓë̼ËáÇâÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ                          ¡£

£¨5£©B¡¢DÔÚŨÁòËáµÄ×÷ÓÃÏÂʵÏÖ·´Ó¦¢Ü£¬ÊµÑé×°ÖÃÈçÏÂͼËùʾ£º

ÊÔ¹Ü1ÖÐ×°ÈëÒ©Æ·ºó¼ÓÈÈ¡£Í¼ÖÐXµÄ»¯Ñ§Ê½Îª                     ¡£

Æä×÷ÓÃÊÇ                   ¡£

ÊÔ¹Ü1·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                        ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸