ÌâÄ¿ÄÚÈÝ

ÓÐËÄ×éͬ×åÔªËصÄÎïÖÊ£¬ÔÚ101.3 kPaʱ²â¶¨ËüÃǵķе㣨¡æ£©ÈçϱíËùʾ£º

µÚÒ»×é

He £­268.8

(a) £­249.5

Ar £­185.8

Kr £­151.7

µÚ¶þ×é

F2 £­187.0

Cl2 £­33.6

(b) 58.7

I2 184.0

µÚÈý×é

(c) 19.4

HCl £­84.0

HBr £­67.0

HI £­35.3

µÚËÄ×é

H2O 100.0

H2S £­60.2

(d) £­42.0

H2Te £­1.8

ÏÂÁи÷ÏîÖÐÕýÈ·µÄÊÇ

A. a¡¢b¡¢cµÄ»¯Ñ§Ê½·Ö±ðΪNe2¡¢Br2¡¢HF

B. µÚ¶þ×éÎïÖÊÖ»±íÏÖÑõ»¯ÐÔ£¬²»±íÏÖ»¹Ô­ÐÔ

C. µÚÈý×éÎïÖÊÖÐCµÄ·Ðµã×î¸ß£¬ÊÇÒòΪC·Ö×ÓÄÚ´æÔÚÇâ¼ü

D. µÚËÄ×éÖи÷»¯ºÏÎïµÄÎȶ¨ÐÔ˳ÐòΪ£ºH2O>H2S>H2Se>H2Te

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÑÇÂÈËáÄÆ( NaCl02)ÊÇÒ»ÖÖ¸ßЧÑõ»¯¼ÁºÍƯ°×¼Á£¬Ö÷ÒªÓÃÓÚÃÞ·Ä¡¢Ö½ÕÅƯ°×¡¢Ê³Æ·Ïû¶¾¡¢Ë®´¦ÀíµÈ¡£ÒÑÖª£ºNaClO2±¥ºÍÈÜÒºÔÚζȵÍÓÚ38 ¡æʱÎö³öµÄ¾§ÌåÊÇNaClO23H2O£¬¸ßÓÚ38 ¡æʱÎö³ö¾§ÌåÊÇNaClO2£¬¸ßÓÚ60¡æʱNaClO2·Ö½â³ÉNaClO3ºÍNaCl¡£´¿ClO2Ò׷ֽⱬը¡£Ò»ÖÖÖƱ¸ÑÇÂÈËáÄÆ´Ö²úÆ·µÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£© ClO2·¢ÉúÆ÷ÖеÄÀë×Ó·½³ÌʽΪ £¬·¢ÉúÆ÷ÖйÄÈË¿ÕÆøµÄ×÷ÓÿÉÄÜÊÇ__________£¨Ñ¡ÌîÐòºÅ£©¡£

a£®½«SO2Ñõ»¯³ÉSO3£¬ÔöÇ¿ËáÐÔ

b£®Ï¡ÊÍClO2ÒÔ·ÀÖ¹±¬Õ¨

c£®½«NaClO3»¹Ô­ÎªClO2

£¨2£©ÎüÊÕËþÄÚ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ £¬ÎüÊÕËþµÄζȲ»Äܳ¬¹ý20¡æ£¬ÆäÔ­ÒòÊÇ_____________¡£

£¨3£©´Ó¡°Ä¸Òº¡±ÖпɻØÊÕµÄÖ÷ÒªÎïÖÊÊÇ ¡£

£¨4£©´ÓÎüÊÕËþÖпɻñµÃNaCl02ÈÜÒº£¬´ÓNaCl02ÈÜÒºµ½´Ö²úÆ·(NaClO2)¾­¹ýµÄ²Ù×÷²½ÖèÒÀ´ÎΪ£º¢Ù¼õѹ£¬55¡æÕô·¢½á¾§£»¢Ú £»¢Û £»¢ÜµÍÓÚ60¡æ¸ÉÔµÃµ½³ÉÆ·¡£

£¨5£©Îª²â¶¨´ÖÆ·ÖÐNaCl02µÄÖÊÁ¿·ÖÊý£¬×öÈçÏÂʵÑ飺

׼ȷ³ÆÈ¡ËùµÃÑÇÂÈËáÄÆÑùÆ·10.00 gÓÚÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®ºÍ¹ýÁ¿µÄµâ»¯¼Ø¾§Ì壬ÔÙµÎÈËÊÊÁ¿µÄÏ¡ÁòËᣬ³ä·Ö·´Ó¦£¨ClO2-+ 4I-+ 4H+= 2H2O+ 2I2+ Cl-£©¡£½«ËùµÃ»ìºÏÒºÅä³É250mL´ý²âÈÜÒº£¬È¡25.00 mL´ý²âÒº£¬ÓÃ2.000 mol£®L-lNa2S203±ê×¼ÒºµÎ¶¨£¨I2+2S2O32-= 2I-+S4O62-£©£¬²âµÃÏûºÄNa2SO3ÈÜҺƽ¾ùֵΪ16.40mL¡£¸ÃÑùÆ·ÖÐNaClO2µÄÖÊÁ¿·ÖÊýΪ ¡£

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

¾«Ó¢¼Ò½ÌÍø