Questions 1. If a student used a sample of metal that was wet, a significant per

Question

Questions 1. If a student used a sample of metal that was wet, a significant percentage of the apparent mass of the metal would be water. Would the resulting measured value of the heat capacity be too high or too low? Explain 2. A particular metallic element, M, has a heat capacity of 0.36 J.g.t forms an oxide that contains 2.90 g of M per gram of oxygen. (a) Using the law of Dulong and Petit (Cp x M 25 J mol, estimate the molar mass of the metal M. g/mol (b) From the composition of the oxide (2.90 g M/g O) and the molar mass of oxygen (16.0 g/mol), determine the mass of M that combines with each mole of oxygen. g M/mol O (c) Use your estimate of the atomic mass from (a) and the information from (b) to determine the empirical formula of the oxide (the moles of M that combine with each mole of O) (d) Now that you know the composition of the oxide and its formula, what is the accurate value of the atomic mass of the metal M? What is the identity of M? g/mol

Explanation / Answer

Q1.The smple of metal contains water as impurity and water has higher heat capacity. Therefpre more heat will be absorbed. to rise the temperature of the sample. Consequently the measured heat capacity of the sample will be higher thn the actual value .

Q2.

(a) Cp X M = 25J/K mol

M = 25 J/ K mol / 0.36g/K g = 69.44 g/mol

(b) Let there is 1g of Oxygen in the compound

2.9 gram of M react with 1g of oxygen,

Therefore mass of M that will react with one mole (16g ) of oxygen = 2.9 gX 16 = 46.4g

(c) Let there be 16g of Oxygen present in the oxide

Moles of oxygen = mass of oxygen / atomic mass of oxygen = 1 6g / 16g /mol = 1 mole

For 16g gram of oxygen there are 46.4g of M

Moles of M = 46.4g / 69.44 g/mol = 0.67 moles

Devide the moles of M and oxygen by 0.67

Hence empirical formulla = M0.67O = M2O3

(d) The element is Gallium and accurate atomic mass is 69.7g/mol

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