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Dafna1 [17]
3 years ago
7

Calculate the energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 the centers of which ar

e separated by a distance of 7.5 nm
Chemistry
1 answer:
Marrrta [24]3 years ago
8 0

Answer:

The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J

Explanation:

Let the charge on the cation be q₁

Also let the charge on the anion be q₂

A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C

An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C

The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m

Energy of attraction = \frac{Kq_1q_2}{r^2}

Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂

Energy of attraction = \frac{8.99 X 10^9 (1.602X10^{-19})(4.806 X10^{-19})}{(7.5X10^{-9})^2}

Energy of attraction = 1.231 X 10⁻¹¹ J

Therefore, the energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J

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Answer:

1. The gas law used: Dalton's law of partial pressure.

2. Pressure of nitrogen = 331 mmHg

Explanation:

From the question given above, the following data were obtained:

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Answer: The correct answer is -297 kJ.

Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

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Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

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Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

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Hope this helps!

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