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lawyer [7]
3 years ago
6

A chef finds a sealed container consisting of an ingredient that goes into his restaurant's secret sauce. The ingredient's molec

ules are moving in place. What will happen if the chef causes the ingredient to change phase by transferring energy into it? After the phase change, the ingredient's molecules will move . . .
Chemistry
2 answers:
LenaWriter [7]3 years ago
7 0

Answer:

faster, and the ingredient will be a liquid.

Explanation:

Anna11 [10]3 years ago
4 0

Answer:

... the ingredient's molecules would be more free and move faster, then the phase would change to liquid from solid.

Explanation:

This task seeks to examine the knowledge of the kinetic phase and characteristics of matters.

Here we see the chef described that the ingredient's molecules are moving in place – solid.

On the advent of causing the ingredient to change phase by transferring energy into it – the ingredient's molecules would be more free and move faster, then the phase would change to liquid from solid.

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3 years ago
Value of Δ H ∘ rxn for the equation NH 3 ( g ) + 2 O 2 ( g ) ⟶ HNO3 ( g ) + H 2 O ( g )
koban [17]

Answer: - 894.6 kJ/mol.

Explanation:

Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.

∆H is the change in the sum of the internal energy of a system.

We are to find the Value of ΔH°(rxn) for the equation:

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).

From the series of equations given;

==> 4NH3 (g) + 5O2 (g) -------->4 NO(g) + 6H2O (l). ∆H = -1166.0 kJ/mol.--------------------------------------(1).

===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).

===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).

The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.

6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).

6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).

Next, add equations (4) and (5) to give;

4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).

Add this equation to the equation (1) from above, we have;

4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).

Then, divide the equation (7) above by 2 to give us back the equation (**).

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.

Δ H^∘ (rxn)= - 894.6 kJ/mol.

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3 years ago
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KonstantinChe [14]

Answer:

Gastropoda.

Explanation:

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