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ollegr [7]
2 years ago
9

What is pressure in relation to a gas?

Chemistry
1 answer:
labwork [276]2 years ago
7 0

The pressure of a gas is the force that a gas exerts per unit area of the container.

Pressure is defined as force per unit area. Gas molecules are constantly colliding against the walls of the container. The pressure of the gas is the force the gas is exerting on its container.

Since temperature is defined as the average kinetic energy of the molecules of a gas then the higher the temperature, the faster the particles move.

The volume of a container refers the size if the container.

The pressure of a gas is inversely proportional to its volume according to Boyle's law. Thus implies that if the pressure of the gas goes up, the volume has to go down.

For a compound to be called an acid, it must contain H+ and H3O+ when dissolved in water.

For a compound to be called a base, the compound must dissolve in water to yield hydroxide ions.

Learn more: brainly.com/question/11543614

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

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

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