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Vanyuwa [196]
4 years ago
9

Basics of solution formation learning goal: to learn how the solvent water forms hydrogen bonds, solvates polar molecules, and w

eakens ionic bonds. water is one of the most essential substances on earth. one of the most important properties of water is its ability to dissolve many different substances. the bent nature of the molecule and the unequal charge distribution play pivotal roles in the solution process. polarity: polarity is directly proportional to the dipole moment of a chemical bond. ionic nature: the ionic nature of a compound depends on the difference in electronegativity between atoms. hydrogen bonding: hydrogen bonding is a weak electrostatic attractive force developed between partly polar hydrogen and a highly electronegative atom, usually oxygen, nitrogen, or fluorine. part a check all that apply. water is known as a polar molecule because
Chemistry
1 answer:
exis [7]4 years ago
6 0
Water is deemed as a polar molecule because of the electrostatic attraction occurring between molecules. One major key characteristic of knowing polar molecules is knowing their atomic geometry. Now, we know that water has two hydrogens and one oxygen atom covalently bonded to each other. This means that they share electrons. However, this sharing is unequal because oxygen is more electronegative compared to hydrogen. So, the electrons tend to be nearer to the oxygen atom making it partially negative. On the other hand, the hydrogen would be partially positive. Polarity arises when there is an imbalance in the pull of electrostatic forces between individual atoms. If you look at the molecular geometry of water, its shape is bent. As a result, it creates a dipole moment which attracts solute to bond with water. In chemistry, the primary rule of solubility is that 'like dissolves like'. This means that polar solutes will dissolve in polar solvents. Similarly, nonpolar solutes will dissolve in nonpolar solvents.
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Answer:

\large \boxed{\text{-2043.96 kJ/mol}}

Explanation:

Assume the reaction is the combustion of propane.

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\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}

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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)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\=  \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}

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