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solong [7]
2 years ago
7

If ∆G◦ = 27.1 kJ at 25◦C for the reaction CH3COOH (aq) + H2O (l) → CH3COO− (aq) + H3O+ (aq), calculate Ka for this reaction at 2

98 K.1. 1.012. 5.63 × 1043. 1.15 × 10−114. 9.89 × 10−15. 1.78 × 10−5
Chemistry
1 answer:
frosja888 [35]2 years ago
5 0

Answer:

Ka = 1.78 × 10⁻⁵

Explanation:

Let's consider the following thermochemical equation.

CH₃COOH(aq) + H₂O(l) → CH₃COO⁻(aq) + H₃O⁺(aq)   ∆G° = 27.1 kJ/mol

At 25°C (298 K), we can find the equilibrium constant (Ka) using the following expression.

∆G° = - R × T × lnKa

where,

R: ideal gas constant

T: absolute temperature

27.1 × 10³ J/mol = - (8.314 J/K.mol) × 298 K × lnKa

Ka = 1.78 × 10⁻⁵

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Given:
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Refer to the periodic table tool and write the electron configurations of the following elements in both long and short terms
ahrayia [7]

Answer:-

Carbon

[He] 2s2 2p2

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Explanation:-

For writing the short form of the electronic configuration we look for the nearest noble gas with atomic number less than the element in question. We subtract the atomic number of that noble gas from the atomic number of the element in question.

The extra electrons we then assign normally starting with using the row after the noble gas ends. We write the name of that noble gas in [brackets] and then write the electronic configuration.

For carbon with Z = 6 the nearest noble gas is Helium. It has the atomic number 2. Subtracting 6 – 2 we get 4 electrons. Helium lies in 1st row. Starting with 2, we get 2s2 2p2.

So the short term electronic configuration is [He] 2s2 2p2

Similarly, for potassium with Z = 19 the nearest noble gas is Argon. It has the atomic number 18. Subtracting 19-18 we get 1 electron. Argon lies in 3rd row. Starting with 4, we get 4s1.

So the short electronic configuration is [Ar] 4s1.

For long term electronic configuration we must write the electronic configuration of the noble gas as well.

So for Carbon it is 1s2 2s2 2p2.

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5 0
2 years ago
How many moles of compound are there in the following?
slavikrds [6]

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8 0
3 years ago
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sergeinik [125]

Answer:

The empirical formula is = C_4H_8O

The formula of Valproic acid = C_8H_{16}O_2

Explanation:

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2 moles of hydrogen atoms are present in 1 mole of water. So,

<u>Moles of H = 2 x 0.00922 = 0.01844 moles </u>

Molar mass of H atom = 1.008 g/mol

<u>Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g </u>

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of CO_2 = 0.403 g  /44.01 g/mol = 0.009157 moles

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<u>Moles of C = 0.009157 moles </u>

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<u>Given that the Valproic acid only contains hydrogen, oxygen and carbon. </u>So,

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<u>Moles of O  = 0.036412  / 15.999  = 0.002276 moles</u>

<u></u>

<u>Taking the simplest ratio for H, O and C as: </u>

<u>0.01844 : 0.002276 : 0.009157</u>

<u> = 8 : 1 : 4</u>

The empirical formula is = C_4H_8O

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,  

Molecular mass = n × Empirical mass

144 = n × 72

<u>⇒ n = 2</u>

The formula of Valproic acid = C_8H_{16}O_2

7 0
3 years ago
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