ÌâÄ¿ÄÚÈÝ

ÏÂÃæ×ó±ßÊý¾ÝÊÇÈ¡×ÔijÕý³£ÈËÔ­ÄòºÍÄòÒºµÄÑùÆ·£¬ÓÒͼÊÇÉöµ¥Î»½á¹¹Ä£Ê½Í¼£®Çë¸ù¾Ý±í¸ñºÍʾÒâͼ»Ø´ðÎÊÌ⣮
ÎïÖÊ ÑùÆ·A£¨g/L£© ÑùÆ·B£¨g/L£©
ÆÏÌÑÌÇ 0 1.0
ÎÞ»úÑÎ 16.0 7.5
µ°°×ÖÊ 0 0
ÄòËØ 2.0 0.3
Ѫϸ°û ÎÞ ÎÞ£¨1£©¾Ý×ó±í·ÖÎö£¬Ô­ÄòÊÇÑùÆ·
B
B
£¨Ìî¡°A¡±»ò¡°B¡±£©£¬´æÔÚÓÚÓÒͼÖеÄ
[4]ÉöСÄÒ
[4]ÉöСÄÒ
ÖУ®
£¨2£©ÄòÒºÊÇÔ­Äò¾­ÓÒͼÖÐ
[6]ÉöС¹Ü
[6]ÉöС¹Ü
µÄÖØÎüÊÕºóÐγɵģ¬ÖØÎüÊÕµÄÎïÖÊ°üÀ¨È«²¿
µÄ
ÆÏÌÑÌÇ
ÆÏÌÑÌÇ
£¬´ó²¿·ÖµÄË®ºÍ²¿·ÖÎÞ»úÑΣ®Í¼ÖÐ[4]¡¢[5]¡¢[6]ºÏ³ÆΪ
Éöµ¥Î»
Éöµ¥Î»
£®
£¨3£©ÈôÄòÒºÖлá³öÏÖѪϸ°ûºÍ´ó·Ö×Ó
µ°°×ÖÊ
µ°°×ÖÊ
£¬Ôò˵Ã÷
ÉöСÇò
ÉöСÇò
ͨ͸ÐÔ±ä´ó£®
£¨4£©ÓÒͼÖеĢÙÊÇ
ÈëÇòС¶¯Âö
ÈëÇòС¶¯Âö
£¬Á÷
¶¯Âö
¶¯Âö
Ѫ£®
·ÖÎö£º´ËÌ⿼²éµÄÊÇÓйØÄòÒºµÄͼÐΡ¢Í¼±íÌ⣬Ҫ½â´ð´ËÌâÊ×ÏÈÒªÁ˽âÄòµÄÐγɵĽṹÉúÀí֪ʶ£¬Í¬Ê±»¹ÒªÃ÷ȷѪ½¬¡¢Ô­Äò¡¢ÄòÒºÈýÕߵijɷÖÇø±ð£®
½â´ð£º½â£ºµ±ÑªÒºÁ÷¾­ÉöСÇòʱ£¬³ýÁËѪϸ°ûºÍ´ó·Ö×ӵĵ°°×ÖÊÍ⣬Ѫ½¬ÖеÄÒ»²¿·ÖË®¡¢ÎÞ»úÑΡ¢ÆÏÌÑÌǺÍÄòËصÈÎïÖÊ£¬¶¼¿ÉÒÔ¾­¹ýÉöСÇòÂ˹ýµ½ÉöСÄÒÄÚ£¬ÐγÉÔ­Äò£»µ±Ô­ÄòÁ÷¾­ÉöС¹Üʱ£¬ÆäÖжÔÈËÌåÓÐÓõÄÎïÖÊ£¬°üÀ¨´ó²¿·ÖË®¡¢È«²¿ÆÏÌÑÌǺͲ¿·ÖÎÞ»úÑΣ¬±»ÉöС¹ÜÖØÐÂÎüÊÕ£¬²¢ÇÒ½øÈë°üÈÆÔÚÉöС¹ÜÍâÃæµÄëϸѪ¹ÜÖУ¬ÖØлص½ÑªÒºÀԭÄòÖÐʣϵÄÆäËû·ÏÎÈçÄòËØ¡¢Ò»²¿·ÖË®ºÍÎÞ»úÑεÈÓÉÉöС¹ÜÁ÷³ö£¬ÐγÉÄòÒº£®Òò´ËÔ­ÄòÖв»º¬ÓÐѪϸ°ûºÍ´ó·Ö×ӵĵ°°×Öʶøº¬ÓÐÆÏÌÑÌÇ£¬ÄòÒºÖв»º¬ÓÐѪϸ°û¡¢´ó·Ö×ӵĵ°°×ÖʺÍÆÏÌÑÌÇ£®
£¨1£©´Ó±í¸ñÖзÖÎö£¬ÑùÆ·AûÓÐÆÏÌÑÌǶøÑùÆ·BÓÐÆÏÌÑÌÇ£¬¿ÉÒÔÅжϳöÑùÆ·AÊÇÄòÒº£¬¶øÑùÆ·BÓÖÎÞ´ó·Ö×ÓµÄѪϸ°ûÓëµ°°×ÖÊ£¬ËùÒÔÑùÆ·BÊÇÔ­Äò£®Ô­ÄòÊÇѪҺÁ÷¾­ÉöСÇòʱ£¬ÑªÒºÖеÄÎïÖʳý´ó·Ö×ӵĵ°°×ÖʺÍѪϸ°ûÍ⣬¹ýÂ˵½[4]ÉöСÄÒÇ»ÖÐÒºÌ壮ÔÚÓÒͼÖÐ[1]ÊÇÈëÇòС¶¯Âö£¬[2]ÊdzöÇòС¶¯Âö£¬[3]ÊÇÉöС¹ÜÖÜΧµÄëϸѪ¹Ü£¬[4]ÊÇÉöСÄÒÇ»£¬[5]ÊÇÉöСÇò£¬[6]ÊÇÉöС¹Ü£®
£¨2£©ÔÚÓÒͼµÄģʽͼÖУ¬[6]ÊÇÉöС¹Ü£¬Æä×÷ÓÃÊÇ°ÑÔ­ÄòÖеÄÈ«²¿ÆÏÌÑÌÇ£¬´ó²¿·ÖË®ºÍ²¿·ÖÎÞ»úÎïÖØÐÂÎüÊÕ»ØѪҺ£®Éöµ¥Î»ÊÇÉöÔà½á¹¹ºÍ¹¦ÄܵĻù±¾µ¥Î»£¬ÓÉ[6]ÉöС¹ÜºÍÉöСÌå×é³É£¬ÉöСÌåÓÉ[5]ÉöСÇòºÍ[4]ÉöСÄÒ×é³É£®Òò´Ë£¬Í¼ÖÐ[4]¡¢[5]¡¢[6]ºÏ³ÆΪÉöµ¥Î»£®
£¨3£©Èç¹ûÉöСÇòÓÐÑ×Ö¢£¬»áµ¼ÖÂÆäͨ͸ÐÔ¼Ó´ó£¬Ô­ÏȲ»¹ýÂ˹ýµÄѪϸ°ûÓë´ó·Ö×Óµ°°×ÖʾÍÓпÉÒÔ±»¹ýÂ˵½ÉöСÄÒÇ»ÖУ¬È»ºóËæÄòÒºÅųöÌåÍ⣬µ¼ÖÂÄòÒºÖгöÏÖ´ó·Ö×Óµ°°×ÖʺÍѪϸ°û£®
£¨4£©Éö¶¯ÂöÊôÓÚÌåÑ­»·£¬ÀïÃæÁ÷µÄÊǺ¬Ñõ¶àµÄ¶¯ÂöѪ£®Í¨¹ýÈëÇòС¶¯Âö½øÈëµ½ÉöСÇò£¬ÔÚÁ÷¾­ÉöСÇòʱ£¬ÓÉÓÚÉöСÇòµÄÂ˹ý×÷Ó㬳ý´ó·Ö×ӵĵ°°×ÖʺÍѪϸ°û£¬ÆäËûС·Ö×ÓµÄÈçË®¡¢ÎÞ»úÑΡ¢ÆÏÌÑÌÇ¡¢ÄòËصÈÎïÖʱ»¹ýÂ˵½ÉöСÄÒÇ»ÐγÉÔ­Äò£¬¶øʣϵÄÎïÖÊͨ¹ý³öÇòС¶¯ÂöÁ÷³ö£¬ÓÉÓÚÉöСÇòµÄÂ˹ý×÷Ó㬲¢Ã»Óз¢ÉúÆøÌå½»»»£¬¹Êºìϸ°ûÔËÊäµÄÑõ²¢Ã»ÓмõÉÙ£¬¹ÊѪҺÈÔÈ»ÊǶ¯ÂöѪ£®ËùÒÔÓÒͼÖеÄ[1]ÊÇÈëÇòС¶¯ÂöÁ÷¶¯µÄÊǶ¯ÂöѪ£®
¹Ê´ð°¸Îª£º
£¨1£©B£»[4]ÉöСÄÒ
£¨2£©[6]ÉöС¹Ü£»ÆÏÌÑÌÇ£»Éöµ¥Î»£»
£¨3£©µ°°×ÖÊ£»ÉöСÇò£»
£¨4£©ÈëÇòС¶¯Âö£»¶¯Âö£®
µãÆÀ£ºÕýÈ··ÖÎö±í¸ñÖеÄÊý¾ÝÕÒ³öÖ÷Òª²»Í¬µã£¬Àí½âÉöµ¥Î»½á¹¹Ä£Ê½Í¼¸÷²¿·ÖËù´ú±íµÄ½á¹¹¼°¹¦ÄÜÊǽâ´ð´ËÌâµÄ¹Ø¼ü
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2011?ȪÖÝÖʼ죩ÏÂÃæ×ó±ßÊý¾ÝÊÇÈ¡×ÔijÕý³£ÈËÔ­ÄòºÍÄòÒºµÄÑùÆ·£¬Í¼ÊÇÉöµ¥Î»½á¹¹Ä£Ê½Í¼£®Çë¸ù¾Ý±í¸ñºÍʾÒâͼ»Ø´ðÎÊÌ⣮£¨¡°[]¡±ÖÐÌîдÐòºÅ£¬¡°
4
4
¡±ÄÚÌîдÎÄ×Ö£®£©
ÎïÖÊ ÑùÆ·A£¨g/L£© ÑùÆ·B£¨g/L£©
ÆÏÌÑÌÇ
ÎÞ»úÑÎ
µ°°×ÖÊ
ÄòËØ
Ѫϸ°û
0
16.0
0
2.0
ÎÞ
1.0
7.5
0
0.3
ÎÞ£¨1£©¸ù¾Ý×ó±ß±íÄÚÊý¾Ý·ÖÎö£¬Ô­ÄòÊÇ
ÑùÆ·B
ÑùÆ·B
£¨Ìî¡°ÑùÆ·A¡±»ò¡°ÑùÆ·B¡±£©£¬´æÔÚÓÚÓÒͼÖеÄ[]ÉöСÄÒÇ»ÖУ®
£¨2£©ÄòÒºÊÇÔ­Äò¾­ÓÒͼÖÐÉöС¹ÜµÄÖØÎüÊÕºóÐγɵģ¬ÖØÎüÊÕµÄÎïÖÊ°üÀ¨È«²¿µÄ
ÆÏÌÑÌÇ
ÆÏÌÑÌÇ
£¬´ó²¿·ÖµÄË®ºÍ²¿·Ö
ÎÞ»úÑÎ
ÎÞ»úÑÎ
£®Í¼ÖÐ[4][5][6]ºÏ³ÆΪ
Éöµ¥Î»
Éöµ¥Î»
£®
£¨3£©Èô[5]ÉöСÇòÓÐÑ×Ö¢£¬ÄòÒºÖлá³öÏÖѪϸ°ûºÍ´ó·Ö×Ó
µ°°×ÖÊ
µ°°×ÖÊ
£»Èô
ÒȵºËØ
ÒȵºËØ
·ÖÃÚµÄÒȵºËز»×㣬ÄòÒºÖлá³öÏÖÆÏÌÑÌÇ£®
£¨4£©´úл·ÏÎïÖеÄÄòËسýÁËÒÔÄòÒºÐÎʽÅųöÍ⣬»¹¿ÉÒÔͨ¹ýƤ·ôÒÔ
º¹Òº
º¹Òº
ÐÎʽÅųö£®
ÏÂÃæ×ó±ßÊý¾ÝÊÇÈ¡×ÔijÕý³£ÈËÔ­ÄòºÍÄòÒºµÄÑùÆ·£¬Í¼ÊÇÉöµ¥Î»½á¹¹Ä£Ê½Í¼£®Çë¸ù¾Ý±í¸ñºÍʾÒâͼ»Ø´ðÎÊÌ⣮£¨¡°[]¡±ÖÐÌîдÐòºÅ£¬¡°
1
1
¡±ÄÚÌîдÎÄ×Ö£®£©
ÎïÖÊ ÑùÆ·A
£¨g/L£©
ÑùÆ·B
£¨g/L£©
ÆÏÌÑÌÇ
ÎÞ»úÑÎ
µ°°×ÖÊ
ÄòËØ
Ѫϸ°û
0
16.0
0
2.0
ÎÞ
1.0
7.5
0
0.3
ÎÞ£¨1£©¸ù¾Ý±íÄÚÊý¾Ý·ÖÎö£¬Ô­ÄòÊÇ
ÑùÆ·B
ÑùÆ·B
£¨Ìî¡°ÑùÆ·A¡±»ò¡°ÑùÆ·B¡±£©£®
£¨2£©ÄòÒºÊÇÔ­Äò¾­Í¼ÖÐ[
5
5
]ÉöС¹ÜµÄÖØÎüÊÕºóÐγɵģ¬ÖØÎüÊÕµÄÎïÖÊ°üÀ¨È«²¿µÄ
ÆÏÌÑÌÇ
ÆÏÌÑÌÇ
£¬´ó²¿·ÖµÄ
Ë®
Ë®
ºÍ²¿·ÖÎÞ»úÎ
£¨3£©Èô[3]ÉöСÇòÓÐÑ×Ö¢£¬ÄòÒºÖлá³öÏÖѪϸ°ûºÍ´ó·Ö×Ó
µ°°×ÖÊ
µ°°×ÖÊ
£»ÈôÒȵºËØ·ÖÃÚ²»×㣬ÄòÒºÖлá³öÏÖ
ÆÏÌÑÌÇ
ÆÏÌÑÌÇ
£®
£¨2012?ȪÖÝ£©ÏÂÃæ×ó±ßÊý¾ÝÊÇÈ¡×ÔijÕý³£ÈËÔ­ÄòºÍÄòÒºµÄÑùÆ·£¬ÓÒͼÊÇÉöµ¥Î»½á¹¹Ä£Ê½Í¼£®Çë¸ù¾Ý±í¸ñºÍʾÒâͼ»Ø´ðÎÊÌ⣮£¨¡°[]¡±ÖÐÌîдÐòºÅ£¬¡°___¡±ÄÚÌîдÎÄ×Ö£®£©
ÎïÖÊ ÑùÆ·A£¨g/L£© ÑùÆ·B£¨g/L£©
ÆÏÌÑÌÇ 0 1.0
ÎÞ»úÑÎ 16.0 7.5
µ°°×ÖÊ 0 0
ÄòËØ 2.0 0.3
Ѫϸ°û ÎÞ ÎÞ£¨1£©¸ù¾Ý×ó±ß±íÄÚÊý¾Ý·ÖÎö£¬Ô­ÄòÊÇ
ÑùÆ·B
ÑùÆ·B
£¨Ìî¡°ÑùÆ·A¡±»ò¡°ÑùÆ·B¡±£©£¬´æÔÚÓÚÓÒͼÖеÄ[
4
4
]ÉöСÄÒÇ»ÖУ®
£¨2£©ÄòÒºÊÇÔ­Äò¾­ÓÒͼÖÐ[
6
6
]ÉöС¹ÜµÄÖØÎüÊÕºóÐγɵģ¬ÖØÎüÊÕµÄÎïÖÊ°üÀ¨È«²¿µÄ
ÆÏÌÑÌÇ
ÆÏÌÑÌÇ
£¬´ó²¿·ÖµÄ
Ë®
Ë®
ºÍ²¿·ÖÎÞ»úÎ
£¨3£©Èô[5]ÉöСÇòÓÐÑ×Ö¢£¬ÄòÒºÖлá³öÏÖѪϸ°ûºÍ´ó·Ö×Ó
µ°°×ÖÊ
µ°°×ÖÊ
£»Èô
Òȵº
Òȵº
·ÖÃÚµÄÒȵºËز»×㣬ÄòÒºÖлá³öÏÖ
ÆÏÌÑÌÇ£¨¡°ÌÇÄò¡±Ò²¸ø·Ö£©
ÆÏÌÑÌÇ£¨¡°ÌÇÄò¡±Ò²¸ø·Ö£©
£®
ÏÂÃæ×ó±ßÊý¾ÝÊÇÈ¡×ÔijÕý³£ÈËÔ­ÄòºÍÄòÒºµÄÑùÆ·£¬ÓÒͼÊÇÉöµ¥Î»½á¹¹Ä£Ê½Í¼£®Çë»Ø´ð£®
ÎïÖÊ ÆÏÌÑÌÇ ÎÞ»úÑÎ µ°°×ÖÊ ÄòËØ ÑªÏ¸°û
ÑùÆ·A£¨g/L£© 0 16.0 0 2.0 ÎÞ
ÑùÆ·B£¨g/L£© 1.0 7.5 0 0.3 ÎÞ£¨1£©¸ù¾Ý×ó±ß±íÄÚÊý¾Ý·ÖÎö£¬Ô­ÄòÊÇ
ÑùÆ·B
ÑùÆ·B
£¨Ìî¡°ÑùÆ·A¡±»ò¡°ÑùÆ·B¡±£©£¬´æÔÚÓÚͼÖеÄ
4
4
ÉöСÄÒÇ»ÖУ®
£¨2£©µ±ÑªÒºÁ÷¾­5ʱ£¬Ñª½¬ÖеĴ󲿷ÖË®¡¢ÎÞ»úÑΡ¢ÆÏÌÑÌǼ°ÄòËصÈÎïÖÊͨ¹ý
ÉöСÇòÂ˹ý
ÉöСÇòÂ˹ý
µ½ÉöСÄÒÖÐÐγÉÔ­Äò£®Í¼ÖÐ6 ´¦µÄ¼¸¸öÏòÉϵļýÍ·µÄº¬ÒåÊÇ
ÉöС¹ÜµÄÖØÎüÊÕ×÷ÓÃ
ÉöС¹ÜµÄÖØÎüÊÕ×÷ÓÃ
£®Í¼ÖТܢݢ޺ϳÆ
Éöµ¥Î»
Éöµ¥Î»
£®
£¨3£©Ô­ÄòºÍÄòÒºÏà±È£¬ÄòÒºÖв»º¬
ÆÏÌÑÌÇ
ÆÏÌÑÌÇ
£¬Èô[5]ÉöСÇòÓÐÑ×Ö¢£¬ÄòÒºÖлá³öÏÖѪϸ°ûºÍ´ó·Ö×Ó
µ°°×ÖÊ
µ°°×ÖÊ
£®

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

¾«Ó¢¼Ò½ÌÍø