標題:
chem~~~
發問:
6.85 g of a compund Z consisting of only nitrogen, hydrogen and chlorine gives 2240 cm^3 of nitrogen gas a s.t.p. with suitable treatment. Addition of a slight excess of acidified silver nitrate solution to an aqueous solution containing 1.37 g of Z precipotates 2.87g of silver chloride. Calculate the empirical... 顯示更多 6.85 g of a compund Z consisting of only nitrogen, hydrogen and chlorine gives 2240 cm^3 of nitrogen gas a s.t.p. with suitable treatment. Addition of a slight excess of acidified silver nitrate solution to an aqueous solution containing 1.37 g of Z precipotates 2.87g of silver chloride. Calculate the empirical formula of compund Z.
最佳解答:
For normal gas, 1 mole of gas will occupy a volume of 22,400 cm3 at stp. If 6.85g of compound Z has a volume of 2,240cm3, it means that there is 0.1 mole of gas. Therefore, the molecular weight of compound Z will be: 6.85/0.1 = 68.5g/mol Besides, by addition of excess AgCl, all the chloride from compound will precipitate out teh siverto form silver chloride. If the mass of silver chloride (MW = 143.5) is 2.87g, the mole of chloride will be: 2.87g/143.5 = 0.02mol. Since there is only 1.37g of compound Z, the MW of Z will be: 1.37/0.02 = 68.5g/mol. The AW of Cl = 35.5, N = 14 and H= 1. The precipitation reaction shows that the compound Z contains only 1 Cl atom in the molecule. So, 33 of the MW will attributted to N and H. There are two possbilities: N2H5 and NH19. Since it is unlikely to have such a small molecule with 19 H atoms. So, N2H5 will be the case. Therefore, the empirical formula of the compound Z will be: N2H5Cl I hope the information can help.
其他解答:
chem~~~
發問:
6.85 g of a compund Z consisting of only nitrogen, hydrogen and chlorine gives 2240 cm^3 of nitrogen gas a s.t.p. with suitable treatment. Addition of a slight excess of acidified silver nitrate solution to an aqueous solution containing 1.37 g of Z precipotates 2.87g of silver chloride. Calculate the empirical... 顯示更多 6.85 g of a compund Z consisting of only nitrogen, hydrogen and chlorine gives 2240 cm^3 of nitrogen gas a s.t.p. with suitable treatment. Addition of a slight excess of acidified silver nitrate solution to an aqueous solution containing 1.37 g of Z precipotates 2.87g of silver chloride. Calculate the empirical formula of compund Z.
最佳解答:
For normal gas, 1 mole of gas will occupy a volume of 22,400 cm3 at stp. If 6.85g of compound Z has a volume of 2,240cm3, it means that there is 0.1 mole of gas. Therefore, the molecular weight of compound Z will be: 6.85/0.1 = 68.5g/mol Besides, by addition of excess AgCl, all the chloride from compound will precipitate out teh siverto form silver chloride. If the mass of silver chloride (MW = 143.5) is 2.87g, the mole of chloride will be: 2.87g/143.5 = 0.02mol. Since there is only 1.37g of compound Z, the MW of Z will be: 1.37/0.02 = 68.5g/mol. The AW of Cl = 35.5, N = 14 and H= 1. The precipitation reaction shows that the compound Z contains only 1 Cl atom in the molecule. So, 33 of the MW will attributted to N and H. There are two possbilities: N2H5 and NH19. Since it is unlikely to have such a small molecule with 19 H atoms. So, N2H5 will be the case. Therefore, the empirical formula of the compound Z will be: N2H5Cl I hope the information can help.
其他解答:
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