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user100 [1]
3 years ago
10

The average kinetic energy of water molecules is greatest in which of these samples? A) 100 g of water at45°C 8) 10 g of water a

t 55°C C) 10 g of water at 35°C D) l00 g of water at 25°C
Chemistry
1 answer:
Vadim26 [7]3 years ago
5 0

Answer:

10 g of water at 55^oC

Explanation:

According to the kinetic molecular theory, the kinetic energy of a particle depends only on temperature. This is described by the following equation: E_k = \frac{3}{2} kT. Notice that the kinetic energy term is directly proportional to the absolute temperature.

The average kinetic energy is independent of mass, meaning when we examine each of these samples, we are only looking for a sample that has the greatest temperature. Out of all the samples, the sample at 55^oC is at the highest temperature which implies it has the greatest average kinetic energy.

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na look it up please im not that smart

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Find ΔHrxn for the following reaction: <br><br> 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)
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Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

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5 0
3 years ago
What is the value of for this aqueous reaction at 298 K? <br><br>A+B↽⇀C+D ΔG°=12.86 kJ/mol<br><br>K=
agasfer [191]

Answer:

The equilbrium constant is 179.6

Explanation:

To solve this question we can use the equation:

ΔG = -RTlnK

<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>

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<em>T is absolute temperature = 298K</em>

<em>And K is equilibrium constant.</em>

Replacing:

12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK

5.19 = lnK

e^5.19 = K

179.6 = K

<h3>The equilbrium constant is 179.6</h3>

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