高一化学难题
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1\uff0e\u5df2\u77e5\u4e2d\u548c\u70ed\u7684\u6570\u503c\u662f57.3 kJ•mol\u20131\u3002\u4e0b\u5217\u7269\u8d28\u53cd\u5e94\u65f6\uff0c\u4ea7\u751f57.3 kJ\u70ed\u91cf\u7684\u662f\uff08 \uff09
A\uff0e\u7a00HCl\u548c\u7a00NaOH
B\uff0e1.0 mol•L-1 HCl\u548c1.0 mol•L \u20131 NaOH
C\uff0e500 mL 2.0 mol•L-1 HCl\u548c500 mL2.0 mol•L \u20131 NaOH
D\uff0e500 mL 2.0 mol•L-1 H2SO4\u548c500 mL 2.0 mol•L-1 Ba(OH)2
22.\u5b9e\u9a8c\u5ba4\u7528\u94c5\u84c4\u7535\u6c60\u4f5c\u7535\u6e90\u7535\u89e3\u9971\u548c\u98df\u76d0\u6c34\u5236\u53d6Cl2\uff0c\u5df2\u77e5\u94c5\u84c4\u7535\u6c60\u653e\u7535\u65f6\u53d1\u751f\u5982\u4e0b\u53cd\u5e94\uff1a
\u8d1f\u6781\uff1aPb+SO42\uff0d\uff0d2e\uff0d=PbSO4 \u6b63\u6781\uff1aPbO2+4H++SO42\uff0d+2e\uff0d= PbSO4+2H2O
\u4eca\u82e5\u5236\u5f970.050molCl2\uff0c\u8fd9\u65f6\u7535\u6c60\u5185\u6d88\u8017\u7684H2SO4\u7684\u7269\u8d28\u7684\u91cf\u81f3\u5c11\u662f\uff08 \uff09
A\uff0e0.025mol B\uff0e0.050mol C\uff0e0.10mol D\uff0e0.20mol
3\uff0e\u94c1\u68d2\u4e0e\u77f3\u58a8\u68d2\u7528\u5bfc\u7ebf\u8fde\u63a5\u540e\u6d78\u51650.01mol/L\u7684\u98df\u76d0\u6eb6\u6db2\u4e2d\uff0c\u53ef\u80fd\u51fa\u73b0\u7684\u73b0\u8c61\u662f\uff08 \uff09
A\uff0e\u94c1\u68d2\u9644\u8fd1\u4ea7\u751fOH\uff0d B\uff0e\u94c1\u68d2\u9010\u6e10\u88ab\u8150\u8680 C\uff0e\u77f3\u58a8\u68d2\u4e0a\u653e\u51faCl2 D\uff0e\u77f3\u58a8\u68d2\u4e0a\u653e\u51faO2
4\uff0e\u53cd\u5e943Fe(s) + 4H2O(g) = Fe3O4(s) + 4H2(g)\u5728\u4e00\u53ef\u53d8\u5bb9\u79ef\u7684\u5bc6\u95ed\u5bb9\u5668\u4e2d\u8fdb\u884c\uff0c\u4e0b\u5217\u6761\u4ef6\u7684\u6539\u53d8\u5bf9\u5176\u53cd\u5e94\u901f\u7387\u51e0\u4e4e\u65e0\u5f71\u54cd\u7684\u662f\uff08 \uff09
A\uff0e\u589e\u52a0Fe\u7684\u91cf B\uff0e\u5c06\u5bb9\u5668\u7684\u4f53\u79ef\u589e\u5927\u4e00\u500d
C\uff0e\u4fdd\u6301\u538b\u5f3a\u4e0d\u53d8\uff0c\u5145\u5165N2\u4f7f\u4f53\u7cfb\u538b\u5f3a\u589e\u5927 D\uff0e\u4f53\u79ef\u4e0d\u53d8\uff0c\u5145\u5165Ar\u4f7f\u5bb9\u5668\u4f53\u79ef\u589e\u5927
5\uff0e\u5728\u4e00\u5b9a\u6e29\u5ea6\u4e0b\u7684\u5bc6\u95ed\u5bb9\u5668\u4e2d\uff0c\u5f53\u4e0b\u5217\u7269\u7406\u91cf\u4e0d\u518d\u53d8\u5316\u65f6\uff0c\u8868\u660eA(s) + 2B(g) C(g) + D(g)\u5df2\u7ecf\u8fbe\u5230\u5e73\u8861\u72b6\u6001\u7684\u662f\uff08 \uff09
A\uff0e\u6df7\u5408\u6c14\u4f53\u7684\u538b\u5f3a B\uff0e\u6df7\u5408\u6c14\u4f53\u7684\u5bc6\u5ea6 C\uff0eB\u7684\u7269\u8d28\u7684\u91cf\u6d53\u5ea6 D\uff0e\u6c14\u4f53\u7684\u603b\u7269\u8d28\u7684\u91cf
6\uff0e\u5728\u5bc6\u95ed\u5bb9\u5668\u4e2d\u8fdb\u884c\u5982\u4e0b\u53cd\u5e94\uff1aX2( g) +3Y2( g) 2Z( g) \u5176\u4e2dX2\u3001Y2\u3001Z \u7684\u5e73\u8861\u6d53\u5ea6\u4f9d\u6b21\u662f0.1 mol•L\uff0d1\uff0c0.3 mol•L\uff0d1\uff0c0.2 mol•L\uff0d1 \u5219\u5404\u7269\u8d28\u7684\u8d77\u59cb\u7269\u8d28\u7684\u91cf\u6d53\u5ea6\u53ef\u80fd\u662f\uff08 \uff09
A\uff0ec(Z)=0.5 mol•L\uff0d1 B\uff0ec(Y2)=0.5 mol•L\uff0d1 C\uff0ec(X2)=0.5 mol•L\uff0d1 D\uff0ec(Z)=0.4 mol•L\uff0d1
21\uff0eA\uff5eH\u662f\u7531\u77ed\u5468\u671f\u5143\u7d20\u5f62\u6210\u7684\u5355\u8d28\u6216\u5316\u5408\u7269\uff0c\u5b83\u4eec\u4e4b\u95f4\u5b58\u5728\u5982\u4e0b\u8f6c\u5316\u5173\u7cfb\uff08\u90e8\u5206\u4ea7\u7269\u5df2\u7ecf\u7565\u53bb\uff09\uff1a
\u5df2\u77e5\uff1aA\u662f\u6de1\u9ec4\u8272\u56fa\u4f53\uff1bB\u5728\u901a\u5e38\u60c5\u51b5\u4e0b\u4e3a\u65e0\u8272\u6db2\u4f53\uff1bD\u662f\u6781\u6613\u6eb6\u4e8e\u6c34\u7684\u65e0\u8272\u6c14\u4f53\uff0c\u8be5\u6c14\u4f53\u901a\u5165AgNO3\u6eb6\u6db2\u4ea7\u751f\u96be\u6eb6\u4e8e\u6c34\u7684\u767d\u8272\u6c89\u6dc0\uff1bC\u3001F\u3001H\u3001G\u542b\u6709\u540c\u79cd\u5143\u7d20\u3002
\uff081\uff09\u82e5F\u4e3a\u65e0\u8272\u65e0\u5473\u6c14\u4f53
\u2460 \u5199\u51fa\u4e0b\u5217\u7269\u8d28\u7684\u5316\u5b66\u5f0f\uff1aG \u3001H \u3002
\u5199\u51fa\u4e0b\u5217\u7269\u8d28\u7684\u7535\u5b50\u5f0f\uff1aA \u3001F \u3001D \u3002
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\u2461 \u5199\u51fa\u4e0b\u5217\u53cd\u5e94\u7684\u79bb\u5b50\u65b9\u7a0b\u5f0f\uff1a
H\uff0bE\u6eb6\u6db2\u2192G\uff1a \u3000\u3000\u3000\u3000\u3000 \u3000\u3000 \uff1b
H\uff0bD\u6eb6\u6db2\u2192F\uff1a \u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000 \u3002
\uff082\uff09\u82e5F\u4e3a\u6613\u6eb6\u4e8e\u6c34\u7684\u56fa\u4f53
\u2460 F\u7684\u5316\u5b66\u5f0f \u3002
\u2461 \u5199\u51fa\u4e0b\u5217\u53cd\u5e94\u7684\u5316\u5b66\u65b9\u7a0b\u5f0f\uff1a
H\uff0bE\u6eb6\u6db2\u2192G\uff1a \u3000\u3000\u3000\u3000\u3000\u3000\u3000 \u3000\u3000\uff1b
H\uff0bD\u6eb6\u6db2\u2192F\uff1a \u3000\u3000\u3000\u3000\u3000\u3000\u3000 \u3000\u3002
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\uff08\u5047\u8bbe\u70e7\u74f6\u5185\u6eb6\u6db2\u6ca1\u6709\u6269\u6563\u5230\u6c34\u69fd\u4e2d\uff09\u3002\u5219\u6b64\u70e7\u74f6\u4e2d\u6240\u5f62\u6210\u6eb6\u6db2\u7684\u7269\u8d28\u7684\u91cf\u6d53\u5ea6\u4e3a \uff1b\u539f\u6df7\u5408\u6c14\u4f53\u7684\u5e73\u5747\u6469\u5c14\u8d28\u91cf\u4e3a \u3002
NaHSO3+HCL==NaCl+SO2(气体)+H2O
NaHCO3溶液
NaHCO3+SO2==NaHSO3+CO2(气体)+H2O
通CO2
Na2SiO3+CO2+H2O==Na2CO3+H2O+SiO2(沉淀)
之后在过滤
1.亚硫酸钠溶液 Na2SO3+2HCl=2NaCl+SO2↑+H2O(或生成NaHSO3而将HCL吸收)
2.碳酸氢钠溶液 NaHCO3+SO2=NaHSO3+CO2↑
3.二氧化碳 Na2SiO3+CO2+H2O=Na2CO3+H2SiO3↓(后过滤)
希望对您有帮助
1.亚硫酸钠溶液 Na2SO3+2HCl=2NaCl+SO2↑+H2O
2.楼上的,碳酸氢钠还要过滤.
不如直接用H2SO4吸收就好了,可以把SO2和吸收掉,而CO2直接出来了.
3.二氧化碳 Na2SiO3+CO2+H2O=Na2CO3+H2SiO3↓(硅酸不溶)
如果第一小题是浓盐酸应用饱和的NaCl溶液
第二小题可用酸性高锰酸钾除
绛旓細27.缁寸敓绱燙(鍙堝悕鎶楀潖琛閰,鍒嗗瓙寮忎负C6H8O6)鍏锋湁杈冨己鐨勮繕鍘熸,鏀剧疆鍦ㄧ┖姘斾腑鏄撹姘у寲,鍏跺惈閲忓彲閫氳繃鍦ㄥ急閰告ф憾娑蹭腑鐢ㄥ凡鐭ユ祿搴︾殑I2婧舵恫杩涜娴嬪畾銆傝鍙嶅簲鐨鍖栧鏂圭▼寮忓涓:C6H8O6+I2 C6H6O6+2HI,鐜版娴嬪畾鏌愭牱鍝佷腑缁寸敓绱燙鐨勫惈閲,鍏蜂綋鐨勬楠ゅ強娴嬪緱鐨勬暟鎹涓嬨傚彇10 mL 6 mol/L CH3COOH,鍔犲叆100 mL钂搁姘...
绛旓細2.璁鹃渶瑕乂 ml鐨勬祿鐩愰吀锛 V*1.19*0.37/36.5=HCl鐨勬懇灏旈噺=2L*0.2锛屾眰寰梀鍗冲彲銆
绛旓細1.鍥犱负NaHCO3鍐嶅姞鐑繃绋嬩腑NaHCO3涓殑C鏈変竴鍗婄敓鎴愪簡CO2锛屽洜涓篊O2鐨勭墿璐ㄧ殑閲忔槸100/100=1mol锛屾墍浠aHCO3鐨勭墿璐ㄧ殑閲忔槸2mol銆2.鏀堕泦鍒扮殑CO2鐗╄川鐨勯噺鏄88/44=2mol,鍏朵腑鏈夋潵鑷狽aHCO3涓殑1mol锛岄偅涔圢a2CO3涓殑C灏辨槸1mol浜嗭紝鏁匩a2CO3寰楃墿璐ㄧ殑閲忔槸1mol銆傛墍浠a2CO3璐ㄩ噺鏄106g锛孨aHCO3鐨勮川閲忔槸168g....
绛旓細O2鐨勮川閲=33.2g-28.4g =4.8g 璁惧弽搴旂敓鎴愮殑KCl璐ㄩ噺涓篨,鍒嗚В鐨凨ClO3璐ㄩ噺涓簓 2KClO3==2KCl+3O2 245^^^149^^^96 y^^^X^^^4.8 X=7.45 y=12.25 (1)KCl+AgNO3==AgCl+KNO3 ^^^74.5^^^143.5 ^^^7.45^^^z z=14.35g (2)MnO2+4HCL(娴)鈫扢nCl2+Cl2+2H2O ^^^87^^^1mo...
绛旓細c(HCl )=n(HCl)/V=0.00118a/(a*10-3)=1.18 mol/L 鎵浠 鈶犳槸1.18 mol/L 鈶℃湁C=n/V 鍙互绠楀嚭锛岄渶瑕佺殑娴撶洂閰革紱鐜荤拑妫掞紝瀹归噺鐡讹紝婊寸锛岀儳鏉 鈶 B (3)n(Na2CO3)=0.795/106=0.0075 mol Na2CO3+2HCl =2 NaCl +H2O +CO2鈫 1 2 0.0075mol x 鎵浠=0.0075mol ...
绛旓細4 8 0 X 3X 2X 4-X 8-3X 2X (2X)/(4-X+8-3X+2X)=0.2 瑙e緱X=1 鎵浠ュ墿浣欐阿姘斾负5mol銆傛荤墿璐ㄧ殑閲10mol.姘㈡皵鐧惧垎姣斾负50 锛2锛夊弽搴斿墠浣撶Н鐗╄川鐨勯噺12.鍙嶅簲鍚10.浣撶Н姣1姣1锛屽帇寮烘瘮5姣6 锛3锛塜/4=0.25锛堟爱姘旇浆鍖栫巼锛 3X/8=0.375锛堟阿姘旇浆鍖栫巼锛夛紙4锛...
绛旓細1.H O F Na Si 锛堝弽搴斿緢鍙伓 HF鍜孲iO2锛2.A 涓嶅 Be鍙嶄緥 B.I2鍙兘鍒癋e2+ C.杩欎釜瀵圭殑 D.绗笁鍛ㄦ湡鍙湁8涓厓绱 宸8 3. H C N O 鍙嶅簲寰堥珮绾 N2H4鍜孨2O4 鐢ㄤ綔鐏鐕冩枡
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绛旓細AI2<SO>3鐗╄川鐨勯噺鏄342 MgSO4 鐗╄川鐨勯噺鏄120 娣峰悎鐗╀负2mol 涓擜L2銆奡O4銆:MgSO4涓2锛3 鍒橝L2銆奡O4銆嬫湁0.8mol MgSO4鏈1.2mol 鎵浠 MgSO4 涓144鍏
绛旓細D锛500 mL 2.0 mol•L-1 H2SO4鍜500 mL 2.0 mol•L-1 Ba(OH)2 22.瀹為獙瀹ょ敤閾呰搫鐢垫睜浣滅數婧愮數瑙iケ鍜岄鐩愭按鍒跺彇Cl2锛屽凡鐭ラ搮钃勭數姹犳斁鐢垫椂鍙戠敓濡備笅鍙嶅簲锛氳礋鏋侊細Pb+SO42锛嶏紞2e锛=PbSO4 姝f瀬锛歅bO2+4H++SO42锛+2e锛= PbSO4+2H2O 浠婅嫢鍒跺緱0.050molCl2锛岃繖鏃剁數姹犲唴娑堣楃殑H...