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3 years ago

Identify the arrows that show removal of thermal energy when matter changes state.

Chemistry

1 answer:

BARSIC [14]3 years ago

6 0

When matter loses heat, one of the following happens:
Liquid changes to solid (freezing), or
Gas changes to liquid (condensation), or
Gas changes to solid (deposition)

1- Freezing: when the temperature of a liquid matter drops below its freezing point, the liquid usually solidifies forming crystals. Example of this is when water changes to ice
2- Condensation: when the temperature of gas is reduced, the molecules slow down and come together changing the matter into liquid. Example of this is when water vapor condenses into water droplets.
3- Deposition: Sometimes when the temperature of a gas drops, it solidifies directly without passing through the liquid phase. An example of this is when water vapor changes directly to ice in sub-freezing areas without passing through the liquid phase. This is the process through which clouds are formed.
The diagram below shows these three changes in matter.

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3 years ago

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lina2011 [118]

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3 years ago

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

3 years ago

Explain in your own words the interplay of temperature (heat) and intermolecular forces and the effect of these two variables on

DochEvi [55]

The forces between particles are called intermolecular forces. A strong intermolecular force means that the particles are tightly paced and is associated with the solid phase. Moderate intermolecular force is associated with the liquid state and little to no intermolecular force is associated with the gaseous state. Temperature has a direct effect on the state of matter in which the substance exists has. Generally speaking, a rise in tempreature changes a substance from the solid to liquid phase and from liquid to gaseus phase. The reverse is true, if the temperature lowers then the substance will go from gas to liquid and liquid to solid. It is important to not that temperature affects intermolecular forces. As the temperature increases then the individual particles become excited and gain enough energy to over the intermolecular forces and so the particles seperate from each other.

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1 year ago