ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿»¯Ñ§ÓëÉú²ú¡¢Éú»îµÈϢϢÏà¹Ø£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ( )

A.¡°»ªÎªP30 Pro¡±ÊÖ»úÖÐ÷è÷ëоƬµÄÖ÷Òª³É·ÖÊǶþÑõ»¯¹è

B.ÕÔÃÏî\Ê«¾ä¡°·×·×²ÓÀÃÈçÐÇÔÉ£¬»ô»ôÐúÖðËƻ𹥡±¡£¡°²ÓÀÃÃÀÀöµÄÑÌ»¨ÊÇijЩ½ðÊôµÄÑæÉ«·´Ó¦£¬ÑæÉ«·´Ó¦ÊôÓÚÎïÀí±ä»¯

C.ÎÒ¹ú×ÔÖ÷Ñз¢µÄ¶«·½³¬»·(ÈËÔìÌ«Ñô)ʹÓÃµÄ 2H¡¢3HÓë 1H»¥ÎªÍ¬ËØÒìÐÎÌå

D.ÈÛÅç²¼ÊÇÒ½ÓÿÚÕÖ×îºËÐĵIJÄÁÏ£¬ÈÛÅç²¼Ö÷ÒªÒÔ¾Û±ûϩΪÖ÷ÒªÔ­ÁÏ£¬Æä½á¹¹¼òʽΪ

¡¾´ð°¸¡¿BD

¡¾½âÎö¡¿

A£®÷è÷ëоƬµÄÖ÷Òª³É·ÖÊǹ裬¹ÊA´íÎó£»

B£®ÑæÉ«·´Ó¦µÄ¹ý³ÌÖÐÎÞÐÂÎïÖÊÉú³É£¬ÊôÓÚÎïÀí±ä»¯£¬¹ÊBÕýÈ·£»

C£®2H¡¢3HÓë 1HÊÇͬÖÖÔªËصIJ»Í¬ºËËØ£¬»¥ÎªÍ¬Î»ËØ£¬¹ÊC´íÎó£»

D£®±ûÏ©µÄ½á¹¹¼òʽΪCH3CH=CH2£¬Ôò·¢Éú¼Ó¾Û·´Ó¦Éú³ÉµÄ¾Û±ûÏ©½á¹¹¼òʽΪ£¬¹ÊDÕýÈ·£»

¹Ê´ð°¸ÎªBD¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Á×»¯ÂÁ(AlP)ÊÇÒ»ÖÖ³£ÓÃÓÚÁ¸Ê³²Ö´¢µÄ¹ãÆ×ÐÔѬÕôɱ³æ¼Á£¬ÓöË®Á¢¼´²úÉú¸ß¶¾ÐÔÆøÌå PH3(·Ðµã£­89.7¡æ£¬»¹Ô­ÐÔÇ¿)¡£¹ú¼ÒÎÀ¼Æί¹æ¶¨Á¸Ê³ÖÐÁ×»¯Îï(ÒÔPH3¼Æ)µÄ²ÐÁôÁ¿²»³¬¹ý0.0500 mg/kgʱΪÖÊÁ¿ºÏ¸ñ£¬·´Ö®²»ºÏ¸ñ¡£Ä³»¯Ñ§ÐËȤС×éµÄͬѧÓÃÏÂÊö·½·¨²â¶¨Ä³Á¸Ê³ÑùÆ· ÖвÐÁôÁ×»¯ÎïµÄÖÊÁ¿ÒÔÅжÏÊÇ·ñºÏ¸ñ¡£

ÔÚCÖмÓÈë100gÔ­Á¸£¬EÖмÓÈë20.00 mL 2.50¡Á10-4 mol/ L KMnO4ÈÜÒº(H2SO4Ëữ)£¬ÍùCÖмÓÈë×ãÁ¿Ë®£¬³ä·Ö·´Ó¦ºó£¬ÓÃÑÇÁòËáÄƱê×¼ÈÜÒºµÎ¶¨EÖйýÁ¿µÄKMnO4ÈÜÒº¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)PH3µÄµç×ÓʽΪ___________¡£ÒÇÆ÷CµÄÃû³ÆÊÇ___________¡£

(2)AÖÐËáÐÔ¸ßÃÌËá¼ØÈÜÒºµÄ×÷ÓÃÊÇ___________ͨÈë¿ÕÆøµÄ×÷ÓÃÊÇ______________________¡£

(3)PH3Ò²¿É±»NaClOÑõ»¯¿ÉÓÃÓÚÖƱ¸NaH2PO2£¬ÖƵõÄNaH2PO2ºÍNiCl2ÈÜÒº¿ÉÓÃÓÚ»¯Ñ§¶ÆÄø£¬ ͬʱÉú³ÉÁ×ËáºÍÂÈ»¯ÎÇëд³ö»¯Ñ§¶ÆÄøµÄ»¯Ñ§·½³ÌʽΪ______________________________¡£

(4)×°ÖÃEÖÐPH3±»Ñõ»¯³ÉÁ×Ëᣬ³ä·Ö·´Ó¦ºóµÄÎüÊÕÒº£¬¼ÓˮϡÊÍÖÁ250mL£¬È¡25.00mLÓÚ׶ÐÎÆ¿ÖУ¬ÓÃ4.0¡Á10£­5mol/LµÄNa2SO3±ê×¼ÈÜÒºµÎ¶¨Ê£ÓàµÄKMnO4ÈÜÒº£¬ÏûºÄNa2SO3±ê×¼ÈÜÒº 20.00mL£¬Na2SO3ÓëKMnO4ÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºSO32£­+MnO4£­+H+¡úSO42£­+Mn2++H2O(δÅäƽ)£¬ÔòµÎ¶¨ÖÕµãµÄÏÖÏóΪ______________________________________________¡£

¸ÃÔ­Á¸ÑùÆ·ÖÐÁ×»¯Îï(ÒÔPH3¼Æ)µÄ²ÐÁôÁ¿Îª__________________mg/kg¡££¨±£ÁôÈýλÓÐЧÊý×Ö£©¡£

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

¾«Ó¢¼Ò½ÌÍø