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adelina 88 [10]
3 years ago
11

Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric

pressure is 101.3 kPa).
Chemistry
1 answer:
kaheart [24]3 years ago
7 0

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas  is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

pressure\propto \frac{1}{volume} (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure  is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

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<h2>Answer:</h2>

A). Increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object.

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When an excited electron in an atom moves to the ground state, the electron
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7 0
3 years ago
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Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?
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Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

  • For a weak acid like HF, the dissociation of HF will be:

<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>

[H₃O⁺] = [F⁻].

<em>∵ [H₃O⁺] = √Ka.C,</em>

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>

<em></em>

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>

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\text{Molarity}=\frac{\text{Moles of }KF}{\text{Volume of solution (in L)}}

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