標題:

求一個未知化合物

發問:

一个化合物.K占有26.59%, Cr占35.36%,O占38.02%,求呢個化合物系咩也? ( 麻煩大家冩過程比我.THX)

最佳解答:

設有 100 g 的該化合物, 則: K 有 26.59 g, 即 26.59/39.1 = 0.680 摩爾 Cr 有 35.36 g, 即 35.36/52 = 0.680 摩爾 O 有 38.02 g, 即 38.02/16 = 2.376 摩爾 所以它們的摩爾比例為: K : Cr : O = 2 : 2 : 7 因此其方程式為 K2Cr2O7, 即重鉻酸鉀.

其他解答:

• 點樣可以BLOCK QQ
• 我想要英文詩歌歡欣(Give Thank)既笛譜@1@
• 全等與相似(作業)

• 點樣可以BLOCK QQ
• 我想要英文詩歌歡欣(Give Thank)既笛譜@1@
• 全等與相似(作業)

 

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mass in K: 26.59 mass in Cr: 35.36 mass in O: 38.02 no. of mole of K: 26.59 / 39.1 = 0.680 no. of mole of Cr: 35.36 / 52 = 0.68 no. of mole of O: 38.02 / 16 = 2.38 relative no. of moles of K: 0.680 / 0.680 = 1 relative no. of moles of Cr: 0.68 / 0.680 = 1 relative no. of moles of O: 2.38 / 0.680 = 3.5 therefore the ratio of K:Cr:O = 1:1:3.5 = 2:2:7 Therefore the empirical formula is K2Cr2O7 (correct to 3 significant figures) K2Cr2O7, Potassium dicromate, an orange compound which is toxic. For further information, please search wikipedia yourself. 2008-01-19 17:50:42 補充： correction:relative no. of moles of K: 0.680 / 0.68 = 1relative no. of moles of Cr: 0.68 / 0.68 = 1relative no. of moles of O: 2.38 / 0.68 = 3.50.680>0.68!!!|||||K2Cr2O7 Potassium Cromate