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Gelneren [198K]
3 years ago
12

3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 1

9.2°C. Calculate ∆H for the dissolution of the unknown solid. (The specific heat of the solution is 4.18 J/g・°C and the density of the solution is 1.20 g/mL).
Chemistry
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

Enthalpy change for the dissolution of the unknown solid = 4.6 Kj/mole

Explanation:

Using Q = m x Cs x ΔT ................................(1)

where Q = Amount of heat absorbed

            m = Mass of solution

           Cs = Specific heat capacity of solution

          ΔT = Change in temperature

Given Density of solution = 1.20 g/ml

And volume of solution = 150 ml

Mass of solution = density x volume

                           = 1.2 x 150

                           = 180 g

From equation (1)

              Q = 180 x 4.18 x 19.2 = 14446.08 = 14.4 Kj

So, ΔH  of the dissolution of the unknown solid = \frac{Q}{n} = \frac{14.4}{3.15} = 4.6 kj/mole

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Answer:

The molecular formule for this unknow molecule is C2H4O2

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The empirical formula is CH2O  ( or better said CnH2nOn)

This means there are 3 elements in the formula of this molecule

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⇒ Hydrogen (H) with a Molar mass of 1 g/mole

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We can also notice that the amount of hydrogen should 2x the amount of carbon ( also 2x the amount of oxygen).

The mass of the empirical formule = 12g/ mole + 2* 1 g/mole + 16 g/mole = 30 g/mole

To know what number is n in CnH2nOn we should divide the molecular mass by the empirical mass:

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this means n = 2

and this will give a molecular formule of C2H4O2  

We can control this to calculate the molecular mass:

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Explanation:

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hola, esta pregunta es bastante difícil pero está bien, no lo sé, lo siento :) :)

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