1道高一化学难题,希望各位帮帮忙啊!!!!!!
1\u9053\u9ad8\u4e00\u5316\u5b66\u96be\u9898,\u5e0c\u671b\u5404\u4f4d\u5e2e\u5e2e\u5fd9\u554a!!!!!!-\u8fd9\u662f\u9ad8\u4e00\u7684\u9898\u76ee\u561b \u6211\u600e\u4e48\u89c9\u5f97\u662f\u9ad82\u7684
\u9996\u5148C2H4\uff08\u4e59\u70ef\uff09\u548c\u6eb4\u6c34\u662f\u53cd\u5e94\u7684 \u4e5f\u5c31\u662f\u6eb4\u6c34\u5438\u6536\u4e86C2H4 \u90a37g\u5c31\u662fC2H4\u7684\u8d28\u91cf
\u90a3CH4\uff08\u7532\u70f7\uff09\u5c31\u662f15-7=8g
\u6545\uff1avCH4\uff1avC2H4=8/16\uff1a7/28=2\uff1a1
\u53c2\u52a0\u53cd\u5e94\u7684AgNO3\u7684\u7269\u8d28\u7684\u91cf\u662f
(15.46-4.66)/108 = 0.1(mol)
BrClx \u2014\u2014\u2014 (x+1)AgNO3
80+35.5x x+1
4.66g 0.1 mol
\u5217\u5f0f: (80+35.5x):4.66 = (x+1):0.1
\u89e3\u5f97 x = 3
\u8fd9\u6837\u53ef\u4ee5\u4e86\u5417 \u6211\u53ef\u662f\u7b2c\u4e00\u4e2a\u554a~ ..\u7ed9\u6211\u5206\u5427....
溴水可以和CH4发生取代反应,和C2H4发生加成反应.取代反应时可生成气体,而加成反应时溴气全被吸收.所以Br2+CH4=HBr+CH3Br...1分有4分HBr生成消耗4分溴气.1分C2H4有4分HBr生成消耗5分溴气4*81X+4*81Y-4*160X+5*160Y+332X+504Y-16X+30Y=7
X+Y=15所以X/Y就是体积CH4/C2H4=1/2
2:1
1:2
绛旓細閫堿B 鍥犱负鑷哀鐨勭浉瀵瑰垎瀛愯川閲忎负48锛岃岀墿璐ㄧ殑閲=鐩稿鍒嗗瓙璐ㄩ噺锛48g鑷哀=3mol锛屾墍浠ヤ负3闄樋浼忎冀寰风綏甯搞傚洜姝瀵广傚悓鏍種a鐨勭浉瀵瑰垎瀛愯川閲忎负23锛屽洜璐ㄩ噺涓2.3g锛屾墍浠ュ惈鐢靛瓙鏁颁负0.1闄樋浼忎冀寰风綏甯告暟銆傚洜姝瀵癸紒甯告俯甯稿帇涓锛1mol鐨勬哀姘=22.4L锛屽洜11.2=0.5mol锛屾墍浠閿欍侱绛旀閿欙紝绯诲洜涓1mol浠讳綍绮掑瓙...
绛旓細锛圓锛夋阿姘斾笌姘ф皵鐨勫弽搴旀柟绋嬪紡涓猴細2H2+O2=2H2O.渚夸簬鐞嗚В鐨勬柟娉曪細浠ゆ哀姘旀阿姘斿潎涓2mol.鍒欐哀姘旇繃閲,鏈鍚庡墿浣欐皵浣撲负1mol姘ф皵.涓哄墠姘斾綋鐨1/4.
绛旓細鐢变簬姘斾綋浣撶Н姣旂瓑浜庣墿璐ㄧ殑閲忕殑姣旓紝涓嶅Θ璁剧┖姘旀湁1 mol锛屽垯鍙充晶姘斾綋搴旇鏈3 mol锛岀偣鐕冨悗鍋滅暀鍦ㄤ腑澶鏄庣噧鐑у悗鍓╀綑鐨勬皵浣撲篃鏄1 mol锛岄偅涔堝弽搴斾腑娑堣楃殑H2銆丱2鎬婚噺灏辨槸2 mol锛屾牴鎹柟绋嬪紡2H2 + O2 = 2H2O鍙煡锛屾秷鑰楃殑H2 = 4/3 锛汷2 = 2/3 鍓╀綑鐨1 mol姘斾綋鑻ヤ负H2锛屽垯H2銆丱2鐨勪綋绉瘮涓(1+4/3)...
绛旓細1.D 鍥犱负纰抽吀姘㈡牴绂诲瓙浼氫笌姘㈡哀鏍圭瀛愬弽搴 2.B 鑻ユ湁缃崲鍙嶅簲锛屽垯浼氭湁鍖栧悎浠风殑鍗囬檷锛堟哀鍖栬繕鍘熷弽搴旓級锛岃岄涓笉瀛樺湪鍖栧悎浠峰崌闄嶏紝鎵浠ヤ笉浼氬寘鍚疆鎹㈠弽搴 3.B 鐩愬彲鐪嬫垚鏄吀纰变腑鍜屽悗鐨勪骇鐗 锛屾棦鏈夐槼绂诲瓙涔熸湁閰告牴绂诲瓙 銆傛哀鍖栫墿婧朵簬姘村悗鍙兘褰㈡垚鍏朵腑鐨勪竴绉嶇瀛愭垨绂诲瓙鍥紝鍥犳涓嶆瀯鎴愮洂銆
绛旓細1銆2H2SO4+C=CO2+2SO2+2H2O 2銆佸惛鏀舵按锛岀櫧鑹插彉涓鸿摑鑹茬~閰搁摐 3銆佹楠屼簩姘у寲纭紝鍝佺孩瑜壊 4銆佸惛鏀朵簩姘у寲纭玈O2+Br2+2H2O=H2SO4+2HBr 5銆佺‘瀹氫簩姘у寲纭畬鍏ㄦ哀鍖 6銆佹楠屼簩姘у寲纰筹紝鐭崇伆姘村彉娴戞祳銆
绛旓細鏍峰搧涓惈鏈夌⒊閰告阿閽 纰抽吀閽 鍜屽皯閲忕殑姘 鏁呬笉鑳界敤宸噺娉 21.2g鍥轰綋閮芥槸纰抽吀閽 0.2mol 鏀堕泦鍒版皵浣撴槸浜屾哀鍖栫⒊ 0.325mol 璁剧⒊閰告阿閽 鍜岀⒊閰搁挔鐨勭墿璐ㄧ殑閲忓垎鍒槸xmol ymol 鏍规嵁閽犲師瀛愬畧鎭掑緱 x + 2y = 0.2 * 2 鏍规嵁纰冲師瀛愬畧鎭掑緱 x + y = 0.325 寰楀嚭 x = 0.25mol y = 0.075 mol...
绛旓細绛夋俯绛変綋绉殑鏃跺欏姞鍏ユ儼鎬ф皵浣擄紝鍙嶅簲閫熺巼涓嶅彉锛屽洜涓哄師姘斾綋鐗╄川鐨勯噺涓嶅彉锛屼綋绉笉鍙橈紝鍔犲叆鎯版ф皵浣撳悗鍘熸潵姘斾綋娴撳害娌℃湁鍙橈紝鎵浠ラ熷害灏辨病鏈夊彉銆傝岀瓑娓╃瓑鍘嬫椂锛屽姞鍏ユ儼鎬ф皵浣撳悗锛屼负淇濊瘉鍘嬪姏涓嶅彉锛屽氨蹇呴』澧炲ぇ浣撶Н锛岃繖鏄師姘斾綋鐗╄川鐨勯噺涓嶅彉锛屼綋绉彉澶э紝娴撳害鍙樺皬锛屽弽搴旈熺巼涓嬮檷 ...
绛旓細KMgCl3•6H2O 瑙f瀽锛5.55g鍔犵儹澶辨按鍚庝负3.39g锛屾湁缁撴櫠姘2.16g锛屽悎0.12mol 灏3.39g鍓╀綑鍥轰綋鍒嗕负涓や唤锛屾瘡鍒嗕负1.695g 绗竴涓1.695g锛氬姞鍏gNO3浜х敓涓嶆憾浜庣閰哥殑娌夋穩鏄剧劧涓篈gCl锛屽彲鐭ワ細涓0.03mol 鍗冲惈Cl鍏冪礌0.03mol锛屽悎1.065g 绗簩涓1.695g锛氭皑姘翠负纰辨х墿璐紝鏁呯敓鎴愮殑娌夋穩蹇呬负姘...
绛旓細璁剧洂閰稿拰姘㈡哀鍖栭挔婧舵恫鐨勭殑鐗╄川鐨勯噺娴撳害鍒嗗埆涓簒鍜寉,鍐嶇敱CaCO3-2HCl 鏈: 0.025x = 0.020y 0.025x- (0.1/100)*2 = 0.004y 瑙e緱锛歺锛0.1mol/L y= 0.125mol/L
绛旓細绗竴棰橈紝浣犳妸B閫夐」鍐欓敊浜嗭紝AD娌℃湁閰嶅钩锛孋涓⒊閰搁挋鏄浐浣擄紝涓嶆憾浜庢按 绗簩棰橈紝閫塁锛屽洜涓鸿鍙嶅簲涓哄彲閫嗗弽搴旓紝涓嶈兘瀹屽叏杩涜锛屼笖娌℃湁璇存槑鏄湪浠涔堟儏鍐典笅 绗笁棰 3Cu+8H(+)+2NO3(-)+=3Cu(2+)+4H2O+2NO,鐢变簬閾滅殑鐗╄川鐨勯噺涓0.03鎽╁皵锛屾阿绂诲瓙鐨勭墿璐ㄧ殑閲忎负0.1鎽╁皵锛岀敱鍙嶅簲绯绘暟鍙煡姘㈢瀛愯繃閲...