一道化学问题
\u4e00\u9053\u5316\u5b66\u9898\uff0c\u6025\uff0c\u5728\u7ebf\u7b49\u8be5\u70c3\u7684\u6469\u5c14\u8d28\u91cf\u4e3a\uff1a3.75g/L*22.4L/mol=84g/mol
n(CO2)=5.8/44=0.13182mol
n(H2O)=2.16/18=0.12mol
n(C):C(H)=1:2 \uff08\u53d6\u6574\u7ea6\u7b49\u4e8e1\uff1a2\uff0c\u78b3\u6c22\u5747\u6765\u81ea\u4e8e\u70c3\uff09\u5206\u5b50\u5f0f\u4e3a\uff1a(CH2)n 14n=84 \u5f97n=6
\u6545\u5206\u5b50\u5f0f\u4e3aC6H12\uff0c\u53ef\u80fd\u7684\u7ed3\u6784\u7b80\u5f0f\u8f83\u591a\uff0c\u73af\u72b6\u4e3a\uff1a\u73af\u5df1\u70f7\uff1b\u76f4\u94fe\u578b\u4e3a\uff1a
CH2\uff1dCHCH2CH2CH2CH3\u3001CH3CH\uff1dCHCH2CH2CH3\u3001CH3CH2CH\uff1dCHCH2CH3\uff1b\u652f\u94fe\u578b\u8f83\u591a\uff0c\u4f60\u53ef\u4ee5\u81ea\u5df1\u6392\u4e00\u4e0b\u3002
\u6b64\u5916\uff0c\u6b64\u9898\u4e2d\u524d\u9762\u7b2c\u4e00\u6b65\u6c42\u51fa\u70c3\u7684\u6469\u5c14\u8d28\u91cf\u4e3a84\uff0c\u518d\u6839\u636e\u8d28\u91cf\u53ef\u4ee5\u6c42\u51fa\u70c3\u7684\u7269\u8d28\u7684\u91cf\u4e3a1.68/84=0.02mol.\u518d\u6839\u636e\u751f\u6210CO2\u7684\u7269\u8d28\u7684\u91cf0.132182mol\u548c\u751f\u6210H2O\u7269\u8d28\u7684\u91cf0.12mol.\u53ef\u6c42\u51faC\u548c\u6c22\u7684\u539f\u5b50\u6570\u5206\u522b\u4e3a6\u548c12\uff08\u53d6\u6574\u6570\uff09
1\u3001\u751f\u6210CO2\u4f7f\u6db2\u6001\u8d28\u91cf\u51cf\u5c11121-50-11-64.4\uff1d4.4g\uff0c\u78b3\u9178\u94a0\u7684\u8d28\u91cf\u4e3a\uff1a4.4*106/44\uff1d10.6g \u7eaf\u5ea6\u4e3a10.6*100%/11=96.36%
2.Na2SO4%=(4.4*142/44)*100%/(121-(11-10.6))=11.77%\uff1b NaCl%=\uff0811-10.6\uff09*100%/(121-14.2)=0.37%
旧键断裂需要吸收热量，新键的生成需要放出热量。吸收的多，反应就表现为吸热；放出的多，反应就表现为放热。
吸收能量：
断裂6个C-H，1个C-C，6*414.4kJ/mol+347.4kJ/mol =2833.8kJ/mol
放出能量：
生成4个C-H，1个C=C，1个H-H，4*414.4kJ/mol+615.3kJ/mol +435.3kJ/mol=2708.2kJ/mol
比较之后，吸收的多。因此，吸收热量=2833.8-2708.2=125.6kJ/mol
因为是吸热，所以△H=125.6kJ/mol
414.4kJ/mol*6+347.4kJ/mol *1-414.4kJ/mol *4-:615.3kJ/mol *1-435.3kJ/mol
一摩反应时 有一摩C-C 6摩C-H断裂 有一摩C=C 4摩C-H 和一摩H-H形成
125.2
断了两个C-H和一个C-C
释放能量
H=414.4+347.4=761.8KJ
生成了一个C=C和一个H-H
吸收能量
H=615.3+435.3=1050.6KJ
△H=-761.8+1050.6=-288.8KJ/mol
绛旓細(1) Si3N4 鍘熷瓙鏅朵綋 (2) 1s^2 2s^2 2p^6 3s^2 2p^2 C (3) 109掳28鈥 姝ｅ洓闈綋 sp^3 (4) Si3N4+4HF+9H2O锛3H2SiO3鈫+4NH4F (5) 8 鍒嗘瀽锛1锛夋瘡涓師瀛愰兘婊¤冻8鐢靛瓙绋冲畾缁撴瀯锛屾墍浠ュ叾鍖栧悎浠峰垎鍒槸锛4浠峰拰锛3浠凤紝鍥犳鍖栧寮忎负 Si3N4 銆傜敱浜庢櫠浣撲腑鐨勫井绮掓槸鍘熷瓙锛屾墍...
绛旓細鍥犵～閰告憾娑茬殑瀵嗗害澶т簬姘寸殑瀵嗗害锛岃屼箼閱囨按婧舵恫鐨勫瘑搴﹀皬浜庢按鐨勫瘑搴︼紝鏁呄1銆佅2蹇呭ぇ浜幭3鍜屜4锛闂鐨勫叧閿浆鍖栦负涓ょ纭吀婧舵恫鐨勫瘑搴ο1涓庛佅2鐨勬瘮杈冨拰涓ょ涔欓唶婧舵恫鐨勫瘑搴ο3涓幭4鐨勫ぇ灏忔瘮杈冦傚浜庡師鏉ョ殑纭吀婧舵恫锛岃川閲忓垎鏁拌秺澶у瘑搴﹁秺澶э紝璁惧師鏉ユ贩鍚堝墠涓ょ纭吀婧舵恫鐨勫瘑搴︿负蟻a鍜屜乥锛屼笖瀵嗗害蟻a锛炏乥...
绛旓細2HCl+CaCO3=CaCl2+CO2(姘斾綋锛+H2O 100 111 44 50g X Y 100/50g=111/X X=55.5g 100/50g=44/Y Y=22g 鎵浠ユ憾娑茬殑璐ㄩ噺=50g+9.5g-22g=37.5g 婧惰川鐨勮川閲=55.5g 婧惰川璐ㄩ噺澶т簬婧舵恫璐ㄩ噺锛岄鐩湁闂锛岀洂閰哥殑璐ㄩ噺搴旇鏄95g 婧舵恫鐨勮川閲=50g+95g-22g=123g 婧惰川鐨勮川閲=55.5g 婧...
绛旓細鑰冪偣锛氶噾灞炵殑鍖栧鎬ц川锛涙牴鎹寲瀛﹀弽搴旀柟绋嬪紡鐨勮绠楋紟涓撻锛氭帹鏂贩鍚堢墿鐨勫彲鑳界粍鎴愶紟鍒嗘瀽锛氬洜涓洪摑鐨勭浉瀵瑰師瀛愯川閲忔瘮閾併侀攲閮藉皬锛屾墍浠ョ浉鍚岃川閲忕殑閾濅笌鐩愰吀瀹屽叏鍙嶅簲鏃讹紝浜х敓鐨勬阿姘旇川閲忓锛庤В绛旓細瑙ｏ細鍋囪鍚堥噾涓搧鐨勮川閲忎负5.6 g锛屽垯鏍规嵁鍖栧鍙嶅簲鐨勬柟绋嬪紡锛欶e+2HCl=FeCl2+H2鈫戝彲璁＄畻鍑烘鏃朵骇鐢熸阿姘旂殑璐ㄩ噺涓0.2 g...
绛旓細0.2molHN3鍜0.2molNaOH鍙嶅簲锛岀敓鎴0.2molNaN3锛孨aN3娴撳害涓0.2mol梅0.5L=0.4mol/L銆侶N3涓嶤H3COOH鐩镐技锛屽垯NaN3鍦ㄦ按涓篃浼氭按瑙ｏ紝N3- + H2O = HN3 + OH-锛堝彲閫嗭級銆傚父娓╀笅锛屾憾娑瞤H = 伪 > 7銆侫銆佺敱浜嶩N3 = H+ + N3-锛堝彲閫嗭級锛孒2O = H+ + OH-锛堝彲閫嗭紝绋嬪害杈冨皬锛夛紝HN3婧舵恫涓細c(...
绛旓細鏍规嵁鏍规嵁璐ㄩ噺瀹堟亽鍙煡锛岀敓鎴怌 32g锛屽悓鏃朵篃鑳界‘瀹欰銆丅銆丆銆丏璐ㄩ噺姣旓紝璁続 鐨勭浉瀵瑰垎瀛愯川閲忎负16X锛屽垯B銆丆銆丏鐨勭浉瀵瑰垎瀛愯川閲忓垎鍒负20X銆32X 銆4X A + B = C + D 16X 20X 8g 10g銆15g 鐢变簬8gA涓15gB鍙嶅簲锛屽彲瑙丅鏈夊墿浣欙紝鍥犳鍙兘閫氳繃A鏉ヨ绠桟鐨勮川閲 A + B =...
绛旓細Al2O3鍜孎e2O3锛屾鏃舵墍寰楃殑婧舵恫涓篈l2(SO4)3銆傛矇娣涓淮鍔犳阿姘у寲閽狅紝鍥犱负姘у寲閾濅负涓ゆф哀鍖栫墿鍙互涓庢按鍙嶅簲鐢熸垚鍋忛摑閰搁挔鍜屾按銆傛鏃舵矇娣涓烘哀鍖栭搧銆傛憾娑蹭负鍋忛摑閰搁挔婧舵恫锛岄氬叆CO2浠ュ悗鍙戠敓锛2NaAlO2 + CO2 + 3H2O=2Al锛圤H)3娌夋穩 +Na2CO3,鑰屾阿姘у寲閾濆彈鐑垎瑙ｅ緱鍒版哀鍖栭摑鍜屾按锛屽浐浣揃涓篈l2O3銆傜涓姝ユ墍寰...
绛旓細棣栧厛鏄庣‘绛夋晥骞宠 鐨勪竴涓闂:瀵逛簬鍙嶅簲鍓嶅悗鎬昏鏁伴噺涓嶇浉绛夌殑鍙嗗弽搴旓紝濡傛灉鎸夊師姣斾緥鍚屽嶆暟鐨勫鍔犲拰闄嶄綆鍙嶅簲鐗╃殑娴撳害銆傜敱浜庡弽搴斾綋绯荤殑鍘嬪己鐩稿簲鐨勫澶ф垨鍑忓皬锛屼細閫犳垚骞宠 鐨勭Щ鍔ㄣ傝屽浜庡弽搴斿墠鍚庢昏鏁伴噺涓嶅彉鐨勫彲閫嗗弽搴旓紝鍘嬪己鐨勫彉鍖栦笉浼氶犳垚骞宠 鐨勭Щ鍔ㄣ傛墍浠ュ湪鍚屽嶆暟鐨勫鍔犳垨闄嶄綆鍙嶅簲鐗╃殑娴撳害鏄椂锛岃浆鍖栫巼涓嶅彉锛...
绛旓細璁撅細姘㈡哀鍖栭捑婧舵恫涓惈婧惰川X鍏嬶紝鐢熸垚姘寲閽綴鍏嬶紝鐢熸垚姘寸殑璐ㄩ噺涓篫鍏 KOH + HCl === KCl + H2O 56 36.5 74.5 18 X 10*36.5% Y Z X=5.6锛堝厠锛塝=7.46锛堝厠锛塟=1.8锛堝厠锛夋眰鍑烘阿姘у寲閽炬憾娑茬殑璐ㄩ噺锛堟憾璐ㄨ川閲/婧惰川璐ㄩ噺鍒嗘暟锛5.6/50%=11.2(鍏)姘㈡哀鍖栭捑婧舵恫涓憾鍓...
绛旓細2锛夌敤鏌愮搴熷純鐨勯噾灞炴槗鎷夌綈A銆丆銆丏缁勬垚鐨勫寲鍚堢墿婧舵恫鍙嶅簲锛屾墍浜х敓鐨勬皵浣撳彲濉厖姘旂悆锛屽啓鍑鸿鍙嶅簲鐨勭瀛愬弽搴旀柟绋嬪紡锛氾紙 2Al+2H2O+2OH- === 2AlO2- +3H2鈫 锛3锛堿CAE鍧囧彲褰㈡垚18鐢靛瓙鍒嗗瓙锛岃繖涓ょ鍒嗗瓙鍦ㄦ按婧舵恫涓弽搴旀湁娣￠粍鑹叉矇娣鐢熸垚锛屽啓鍑鸿鍙嶅簲鐨鍖栧鍙嶅簲鏂圭▼寮忥細锛 H2O2+H2S==S鈫+2H2O...
