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Zepler [3.9K]
3 years ago
6

How do models help scientists predict the polarity of molecules?

Chemistry
2 answers:
yuradex [85]3 years ago
8 0

Answer:There are three main properties of chemical bonds that must be considered—namely, their strength, length, and polarity. The polarity of a bond is the distribution of electrical charge over the atoms joined by the bond. Specifically, it is found that, while bonds between identical atoms (as in H2) are electrically uniform in the sense that both hydrogen atoms are electrically neutral, bonds between atoms of different elements are electrically inequivalent. In hydrogen chloride, for example, the hydrogen atom is slightly positively charged whereas the chlorine atom is slightly negatively charged. The slight electrical charges on dissimilar atoms are called partial charges, and the presence of partial charges signifies the occurrence of a polar bond.

Explanation:

Svetach [21]3 years ago
7 0

Answer:

The first activity was developed in collaboration with our student teacher, Dr. Chris Sosa. In this activity, the students model the pull of electrons in a bond between two elements, demonstrating covalent bonding. In particular, they practice differentiating between polar and nonpolar bonds. Most compounds, however, have polar covalent bonds, which means that electrons are shared unequally between the bonded atoms. Figure 9.3.1

9.3.

1

compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall that a lowercase Greek delta (δ

δ

) is used to indicate that a bonded atom possesses a partial positive charge, indicated by δ+

δ

+

, or a partial negative charge, indicated by δ−

δ

−

, and a bond between two atoms that possess partial charges is a polar bond.

Read more on Brainly.com - brainly.com/question/13921838#readmore

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Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

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And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

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Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

Learn more:

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