高中化学必修一有关物质的量的计算 求：高中化学必修一物质的量你一节的所有计算公式与相关例题。
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\u4e2d\u542b\u6709NA\u4e2aH2O\u5206\u5b50\uff0c\u7ea6\u4e3a6.02\u00d71023 \u6c34\u5206\u5b50\u30021mol H2SO4 \u4e2d\u542b\u67092molH ,4molO ,1.204\u00d71024\u4e2aCO2\u5206\u5b50\u7684\u7269\u8d28\u7684\u91cf\u662f 2mol1mol\u4efb\u4f55\u7c92\u5b50\u6240\u542b\u7684\u7c92\u5b50\u6570\u76f8\u7b49\uff0c\u90fd\u662fNA\u4e2a\uff0c3\u516c\u5f0f\u7269\u8d28\u7684\u91cf\u3001\u963f\u4f0f\u52a0\u5fb7\u7f57\u5e38\u6570 \u4e0e\u7c92\u5b50\u6570\uff08\u7b26\u53f7\u4e3aN\uff09\u4e4b\u95f4\u7684\u5173\u7cfb n=N/NA 9.03*10^22\u4e2aH+\u7684\u7269\u8d28\u7684\u91cf\u662f \uff1f \u516c\u5f0f\u53d8\u5f62 N=n*NA \u4e8c\u3001\u6469\u5c14\u8d28\u91cf1\u6469\u5c14\u8d28\u91cf\u5355\u4f4d\u7269\u8d28\u7684\u91cf\u7684\u7269\u8d28\u6240\u5177\u6709\u7684\u8d28\u91cf\u53eb \uff0c\u7b26\u53f7M\uff0c\u5355\u4f4d g/mol 1mol12C \u7684\u8d28\u91cf\u662f 12g , \u90a3\u4e481molH\u7684\u8d28\u91cf\u662f\uff1f\u7535\u5b50\u7684\u8d28\u91cf\u5ffd\u7565\u4e0d\u8ba1\uff0c\u90a3\u4e48 1molH+\u7684\u8d28\u91cf\u662f\uff1f 1molNa+\u7684\u8d28\u91cf\u662f\uff1f1molH2O\u7684\u8d28\u91cf? 1molH2SO4 \u7684\u8d28\u91cf\u901a\u8fc7\u4ee5\u4e0a\u8ba1\u7b97\uff0c\u6211\u4eec\u53ef\u4ee5\u5f97\u51fa\u7ed3\u8bba 1mol\u4efb\u4f55\u7269\u8d28\u7684\u8d28\u91cf\u4ee5\u514b\u4e3a\u5355\u4f4d\uff0c\u6570\u503c\u4e0a\u7b49\u4e8e\u5176\u76f8\u5bf9\u539f\u5b50\u8d28\u91cf(Ar)\u6216\u8005\u76f8\u5bf9\u5206\u5b50\u8d28\u91cf(Mr)\u3002\u4f46\u662f\u6469\u5c14\u8d28\u91cf\u7684\u8868\u8fbe\u5f0f\u662f\u7279\u5b9a\u7684\uff0cH\u7684\u6469\u5c14\u8d28\u91cf 1g/mol , NaCl\u7684\u6469\u5c14\u8d28\u91cf\u662f 58.5g/molSO42-\u7684\u6469\u5c14\u8d28\u91cf\u662f 96g/mol\u601d\u8003\uff1a 1molH2O \u7684\u8d28\u91cf\u662f\u591a\u5c11\uff0c \u6c34\u7684\u6469\u5c14\u8d28\u91cf\u662f\u591a\u5c11\uff0c 36g\u6c34\u7684\u7269\u8d28\u7684\u91cf\u662f\u591a\u5c11 2\u516c\u5f0f\u901a\u8fc7\u4e0a\u9762\u7684\u6f14\u7ec3\uff0c\u53ef\u4ee5\u5f97\u51fa\u7269\u8d28\u7684\u8d28\u91cf\uff0c\u7269\u8d28\u7684\u91cf\u4ee5\u53ca\u6469\u5c14\u8d28\u91cf\u4e4b\u95f4\u7684\u516c\u5f0fM=m/n \u516c\u5f0f\u53d8\u5f62 m=n*M n=m/M \u4f53\u4f1a\u7269\u8d28\u7684\u91cf\u7684\u6865\u6881\u4f5c\u7528\uff1a \u628a\u5fae\u89c2\u4e0e\u5b8f\u89c2\u8054\u7cfb\u8d77\u6765\uff1a\u7c92\u5b50\u6570 N ========NA======== \u7269\u8d28\u7684\u91cfn =========M======= \u8d28\u91cfm\u5fae\u89c2 ====================== \u6865\u6881 ================== \u5b8f\u89c2 \u8fd9\u6837\u5c31\u53ef\u4ee5\u628a\u7269\u8d28\u7684\u91cf\u8fd0\u7528\u5230\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e2d\uff0c2H2+O2=\u70b9\u71c3==2H2O\u5316\u5b66\u8ba1\u91cf\u6570 2 1 2\u5206\u5b50\u6570\u76ee 2 1 2*NA 2NA NA 2NA\u7269\u8d28\u7684\u91cf\u5173\u7cfb 2mol 1mol 2mol\u8d28\u91cf\u5173\u7cfb 4g 32g 36g
2012-03-23 20:581.溶质的物质的量=溶质的物质的量浓度x溶液的体积c=n / v2.物质的量=微粒数/阿伏伽德罗常数(n=N/Na)
3.物质的量=物质的质量/物质的摩尔质量（n=m/M)
4.物质的量=气体的体积/气体的摩尔体积（n=V/Vm)
5.c=1000ρ(密度) w% / M
注：n（mol）： 物质的量 ；V(L)：物质的体积 ；M(mol/L）：摩尔质量；w%：溶液中溶质的质量分数 密度单位：g/cm^3
n=m/M(M是摩尔质量）
n=c*v（C是物质的量浓度）
n=V/22.4(V是STP下的体积）
我的百度空间有复习资料
从单位的统一性理解
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绛旓細婧惰川鐨鐗╄川鐨勯噺锛坢ol锛夛紳鐗╄川鐨勯噺娴撳害锛坢ol/L锛壝楁憾娑蹭綋绉紙L锛夈傛湁鍏虫憾娑茬殑璁＄畻鍏紡:锛1锛夋祿婧舵恫鐨勮川閲徝楁祿婧舵恫婧惰川鐨勮川閲忓垎鏁帮紳绋婧舵恫鐨勮川閲徝楃█婧舵恫婧惰川鐨勮川閲忓垎鏁帮紙鍗虫憾璐ㄧ殑璐ㄩ噺涓嶅彉锛夈傦紙2锛夋祿婧舵恫鐨勮川閲徝楁祿婧舵恫婧惰川鐨勮川閲忓垎鏁帮紳绋婧舵恫鐨勮川閲徝楃█婧舵恫婧惰川鐨勮川閲忓垎鏁帮紙鍗虫憾璐ㄧ殑璐ㄩ噺涓嶅彉锛夈鍖...
绛旓細楂樹腑鍖栧鍚堥泦鐧惧害缃戠洏涓嬭浇 閾炬帴锛歨ttps://pan.baidu.com/s/1znmI8mJTas01m1m03zCRfQ ?pwd=1234 鎻愬彇鐮侊細1234 绠浠嬶細楂樹腑鍖栧浼樿川璧勬枡涓嬭浇锛屽寘鎷細璇曢璇曞嵎銆佽浠躲佹暀鏉愩佽棰戙佸悇澶у悕甯堢綉鏍″悎闆嗐
绛旓細鍦ㄩ珮涓寲瀛﹀涔犱腑锛屾懇灏旇繖涓寮犵殑鐭ヨ瘑鐐瑰彲浠ョ畻浣滄牳蹇冨熀纭鐭ヨ瘑鐐逛簡銆傛兂瑕佸涔犲ソ鍖栧锛屽氨棣栧厛闇瑕佹墦濂藉熀纭銆備笅闈㈡垜涓哄ぇ瀹跺甫鏉ラ珮涓寲瀛︽懇灏旂煡璇嗙偣锛屽笇鏈涘彲浠ュ澶у鐨勫寲瀛︽垚缁╂湁鎵甯姪銆楂樹腑鍖栧蹇呬慨涓鎽╁皵锛氫竴銆 1.鐗╄川鐨勯噺(n)鏄〃绀哄惈鏈変竴瀹氭暟鐩矑瀛愮殑闆嗕綋鐨勭墿鐞嗛噺.2.鎽╁皵(mol): 鎶婂惈鏈6.02 脳1023涓...
绛旓細2mol 瑙ｉ鎬濊矾SiO2 +CaCO3 =CaSiO3 +CO2鈫 CaCO3 =CaO +CO2鈫戙愬弽搴旀潯浠跺潎涓哄姞鐑瓟妗 V(CO2)=4.48L n(鍥 )=0銆併佹垜鎵撲笉鍑哄姞鐑鍙.36L 5g 0.05mol 1.12L .浣5g V(CO2)=4銆傛湜瑙佽皡銆60g 100g 1mol 22.4L 100g 1mol 22.4L 9g 20g(杩囬噺)0.15mol 3 ...
绛旓細楂樹腑鍖栧甯哥敤璁＄畻鍏紡 1.鏈夊叧鐗╄川鐨勯噺锛坢ol锛夌殑璁＄畻鍏紡 锛1锛夌墿璐ㄧ殑閲忥紙mol锛= (g)(g/mol)鐗╄川鐨勮川閲忕墿璐ㄧ殑鎽╁皵璐ㄩ噺 锛2锛夌墿璐ㄧ殑閲忥紙mol锛= ()(/mol)23 寰矑鏁颁釜6.0210 涓 锛3锛夋皵浣撶墿璐ㄧ殑閲忥紙mol锛= (L)22.4(L/mol)
绛旓細鐗╄川鐨勯噺鏄〃绀虹墿璐ㄦ墍鍚井绮掓暟(N)(濡:鍒嗗瓙,鍘熷瓙绛)涓庨樋浼忓姞寰风綏甯告暟(NA)涔嬫瘮,鍗硁=N/NA銆傞樋浼忎冀寰风綏甯告暟鐨勬暟鍊间负0.012kg C鎵鍚⒊鍘熷瓙鐨勪釜鏁,绾︿负6.02脳10銆傚畠鏄妸涓瀹氭暟鐩殑寰绮掑瓙涓庡彲绉伴噺鐨勫畯瑙傜墿璐ㄨ仈绯昏捣鏉ョ殑涓绉嶇墿鐞嗛噺銆
绛旓細鍏堢敱Cu39.50% N8.64 寰梟(Cu):n(N)=(N鎬*39.50%/64)锛(N鎬*8.64%/14)=1 鍗矯u锛歂=1:1 鎺掗櫎BD 鎺ョ潃鍒╃敤Cu鎴朜鐨勪竴涓垎瀛愰噺姣斾緥鍗冲彲 A涓64+14+4+32+16*3=N鎬=162 64/162=39.5 鑰孊涓64*2+14*2+4*2+32*3+16*12=N鎬=452 128/452=28.32%涓嶆纭 鏁呴堿 ...
绛旓細3 7.8/78=0.1mol 鎵浠ョ～鏈0.1mol锛岄偅涔堢～閰搁摑灏辨湁1/3mol锛屾墍浠ュ氨鏈夛紙1/3锛*342=114g 4 136*0.86=116.96g 116.96/58.5=2mol 136-116.96=19.4 19.4/95=0.2mol 鎵浠ユ隘绂诲瓙鐨勬鐗╄川鐨勯噺涓2+0.2*2=2.4mol 鍗崇閰搁摱2.4mol锛屾墍浠ュ氨鏈2.4*170=408g 5...
绛旓細鎴戣寰楂樹竴鍖栧姘斾綋绫荤殑鍏紡绮惧崕灏卞湪鐞嗘兂姘旀佹柟绋嬶紙鍏嬫媺浼緳鏂圭▼锛塒V=NRT P鏄帇寮猴紝V鏄綋绉紝N鏄鐗╄川鐨勯噺锛孯鏄竴涓父鏁帮紝鍙互鐩存帴璁版垚8.314锛孴鏄俯搴 鐢ㄨ繖涓柟绋嬪彲浠ヨВ鍐宠澶氶棶棰橈紝濡傦紙鍦ㄧ浉鍚屾俯搴︿笅锛岀瓑浣撶Н鐨勪竴瀹氳川閲忕殑姘斾綋锛屽帇寮烘槸涓嶆槸涓鏍凤級绛 杩樻湁灏辨槸鍩虹鏂圭▼浜嗭紝閫傜敤鎵鏈夌殑鍏堟潵 N=m/M ...
