涓€銆侀€夋嫨棰橈紙鏈ぇ棰樺寘鎷?6灏忛锛屾瘡棰樺彧鏈変竴涓€夐」绗﹀悎棰樻剰銆傛瘡灏忛3鍒嗭紝鍏?8鍒嗭級銆侟br>1銆佷笅鍒楀寲鍚堢墿鐨勬牳纾佸叡鎸阿璋变腑鍑虹幇涓夌粍宄扮殑鏄紙 锛堻br>A锛庣敳鑻 B锛 1,3-涓佷簩鐑 C 2-鐢插熀涓欑兎 D锛 2,4-浜岀敳鍩烘垔鐑饵br>2銆佸皢鈶燞+銆佲憽Cl锕ｃ€佲憿Al3+銆佲懀K+銆佲懁S2锕ｃ€佲懃OH锕ｃ€佲懄NO3锕ｃ€佲懅NH4+ 鍒嗗埆鍔犲叆H2O涓紝鍩烘湰涓婁笉褰卞搷姘寸殑鐢电骞宠 鐨勬槸
A锛庘憼鈶モ懄 B锛庘憼鈶⑩懁鈶︹懅 C锛 鈶♀懀鈶 D锛庘憽鈶ｂ懃鈶?br>3銆佷笅鍒楃儹鍖栧鏂圭▼寮忔垨鍙欒堪姝ｇ‘鐨勬槸( )
A. 1 mol娑叉€佽偧鍦ㄨ冻閲忔哀姘斾腑瀹屽叏鐕冪儳鐢熸垚姘磋捀姘旓紝鏀惧嚭642 kJ鐨勭儹閲忥細
N2H4(l)+O2(g)= N2(g)+2H2O(g) 螖H锛?642 kJ?mol-1
B. C2H6鐨勭噧鐑х儹涓篞 kJ?mol-1锛屽垯琛ㄧずC2H6鐕冪儳鐑殑鐑寲瀛︽柟绋嬪紡涓裹br>2C2H6(g)+7O2(g)= 4CO2(g)+6H2O(l) 螖H锛?2Q kJ?mol-1
C.宸茬煡锛欻2(g)+ O2(g)= H2O(l) 螖H锛?286 kJ?mol-1锛屽垯锛欬br>2H2O(l)= 2H2(g)+O2(g)鐨勎擧锛?572 kJ?mol-1
D.宸茬煡N2(g)+3H2(g) 2NH3(g) 螖H锛?92.4 kJ?mol-1锛屽垯鍦ㄤ竴瀹氭潯浠朵笅鍚戝瘑闂鍣ㄤ腑鍏呭叆0.5 mol N2(g)鍜?.5 mol H2(g)鍏呭垎鍙嶅簲鏀惧嚭46.2 kJ鐨勭儹閲廃br>4銆佸凡鐭sp(AgCl)锛?.8脳10锛?0锛孠sp(AgI)锛?.0脳10锛?6銆備笅鍒楄娉曢敊璇殑鏄?br>A锛嶢gCl涓嶆憾浜庢按锛屼笉鑳借浆鍖栦负AgI
B锛庡湪鍚湁娴撳害鍧囦负0.001 mol?L锛?鐨凜l锛嶃€両锛嶇殑婧舵恫涓紦鎱㈠姞鍏gNO3绋€婧舵恫锛岄鍏堟瀽鍑篈gI娌夋穩
C锛嶢gI姣擜gCl鏇撮毦婧朵簬姘达紝鎵€浠ワ紝AgCl鍙互杞寲涓篈gI
浼氬憳鐧诲綍
鍏嶈垂娉ㄥ唽
璁块棶PC缃戠珯