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
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\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
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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
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\u2460 F\u7684\u5316\u5b66\u5f0f \u3002
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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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质量为pv=1.18a 克，那么m(HCl )=1.18a*36.5%=0.04307a 克
n(HCl)=m(HCl )/M=0.04307a/36.5=0.00118a (mol)
c(HCl )=n(HCl)/V=0.00118a/(a*10-3)=1.18 mol/L 所以 ①是1.18 mol/L
②有C=n/V 可以算出，需要的浓盐酸；玻璃棒，容量瓶，滴管，烧杯
③ B
(3)n(Na2CO3)=0.795/106=0.0075 mol
Na2CO3+2HCl =2 NaCl +H2O +CO2↑
1 2
0.0075mol x
所以x=0.0075mol *2=0.015mol 所以250ml HCl 溶液中有
n(HCl)=(250/30.60)*0.015=0.1225mol
C=n/V=0.1225/（250*10-3）=0.490 mol/L
（4）Na2CO3+2HCl =2 NaCl +H2O +CO2↑
1 2
y 0.490*0.036mol
所以y =0.00088mol
m(Na2CO3)=0.00088*106=0.0934 g
Na2CO3%=(0.0934/0.825)*100%=11.33%
21、 CO2会先与Ca(OH)2，碱性 Ca(OH)2>NaOH
（1）第一阶段：Ca(OH)2+CO2=CaO3↓+H2O ；2NaOH + CO2 =Na2CO3+H2O
第一阶段反应后，还有余下的CO2为0.7-（0.2+0.1）=0.4 mol
第二阶段 Na2CO3+H2O+CO2=2NaHCO3; CaO3+H2O +CO2=Ca(HCO3)2
（2）看反应式，知道，第一阶段离子是减少了，减少的是Ca2+,（0.2mol）CO32-也是0.2mol，OH-也0.2mol。当CO2为0.4mol后，又继续增多，一直会回到反应前的起点：0.2*（2OH- + Ca2+)+ 0.2*（Na+OH-)=1 mol
CO2为0.2mol，0.3mol，均填空0.4mol；CO2为0.5mol ，填空0.5mol， CO2为0.6mol ,填1.0
看不清
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绛旓細姘у寲閽犲拰姘达細Na2O + H2O = 2NaOH 杩囨哀鍖栭挔鍜屾按锛2Na2O2 + 2H2O = 4NaOH + O2 鎴戜滑鍋囪棰樼洰涓繃姘у寲閽犲惈鏈 Xg锛岄偅涔堟哀鍖栭挔鍚湁锛70-X锛塯.鎴戜滑鍙戠幇杩欎袱涓弽搴斾腑锛屼娇寰楁憾娑茶川閲忔崯澶辩殑鍙嶅簲鍙湁涓涓紝灏辨槸杩囨哀鍖栭挔鍜屾按銆傜敱鏂圭▼鐭ワ紝x g杩囨哀鍖栭挔鍜屾按鍙嶅簲锛岀敓鎴愭阿姘у寲閽狅紙40/39锛墄 g锛屾崯澶辫川閲...
绛旓細棣栧厛閰嶅钩鍖栧寮忥細 xCuCO3*yCu(OH)2+2(x+y)HCl=(X+Y)CuCl2+(2y+x)H2O+xCO2(姘斾綋涓婂崌绗﹀彿锛 鎽╁皵鏁帮細 0.100mol 0.560/22.4L/mol 鐩稿鍒嗗瓙璐ㄩ噺锛 2(x+y)*36.5 x*44 鍙互寰楀嚭姣斾緥 x/y=1/1 閭ｄ釜鐭抽潚鐨勪竴鏍 绗簩闂 閭ｄ釜鍔犵洂閰哥殑杩樻槸鍙互鐢ㄧ涓姝ョ殑鏂圭▼寮 浠涔堥兘涓嶇敤鏀 涓婇潰...
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...
绛旓細涓銆佷笉鍙兘 浜屻佽浜屾哀鍖栫⒊ x mol 姘 y mol 鍒欐湁 44x+18y=8 28x+2y=3.2锛堝樊閲忔硶锛夊洜姝や簩姘у寲纰宠川閲忎负 44x g ,姘磋川閲忎负 18y g 銆備笁銆佽姹傛按鐨勭墿璐ㄧ殑閲忚寖鍥达紝鍥犱负褰撴按瓒呰繃涓婇檺鏃讹紝鍥轰綋澧為噸浼氬皬浜3.2g,鑰屽皬浜庝笅闄愬垯姹傚嚭鐨勬皵浣撴婚噸浼氬浜8g.鍥犳鏈夋瀬闄愭儏鍐碉細璁 浜屾哀鍖栫⒊ x mol 姘 y ...
