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3 years ago

Use mathematics to prove that the 5.50 grams of salicylic acid (C7H6O3) used in this experiment is equal to 0.04 mol.

Chemistry

1 answer:

g100num [7]3 years ago

7 0

Answer:

Please find the answer to the question below

Explanation:

In chemistry, the following mathematical formula is used to calculate the number of moles contained by a substance:

mole = mass of substance (g)/molar mass of substance (g/mol)

Molar mass of salicylic acid (C7H6O3) = 12(7) + 1(6) + 16(3)

= 84 + 6 + 48

= 138g/mol

Mass = 5.50grams

mole = 5.5/138

mole = 0.039

Approximately, the number of moles of 5.5grams of salicylic acid is 0.04moles. This is in accordance with the mole value (0.04) given in this question.

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Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

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Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$