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3 years ago

Heating curve shows temperature verses energy gain. Which parts of the curve represent a gain in potential energy?

Chemistry

1 answer:

Brilliant_brown [7]3 years ago

6 0

Answer:

Those two horizontal lines.

Explanation:

Hello there!

In this case, when focusing on these heating curves, it is important to say they tend to have two constant-temperature sections and three variable-temperature sections. Thus, from lower to higher temperature, the first constant-temperature section corresponds to melting and the second one vaporization, whereas the three variable-temperature sections correspond to the heating of the solid until melting, the liquid until vaporization and the gas until the critical point.

In such a way, we infer that the boxes referred to constant temperature are referred to a gain in potential energy, that is, the two horizontal lines.

Regards!

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In a solution of pure water, the dissociation of water can be expressed by the following: H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) T

kakasveta [241]

Answer:

[H₃O⁺] = 2.90 × 10⁻¹⁰ M

Explanation:

1) Ionization equilibrium equation: given

H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

2) Ionization equilibrium constant, at 25°C, Kw: given

Kw = 1.0 × 10⁻¹⁴

3) Stoichiometric mole ratio:

As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:

[H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M

⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M

4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C and you need to calculate what the [H₃O⁺(aq)] is.

Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:

Kw = 1.0 × 10⁻¹⁴ M², and

Kw = [H3O⁺] [OH⁻]

Then you can substitute the known values and solve for the unknown:

1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M

⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M

As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.

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