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Ainat [17]
3 years ago
13

what are the relative strengths of intermolecular forces compared to the forces between ions in a salt or the forces between met

al atoms in a metallic bond?
Chemistry
1 answer:
Sophie [7]3 years ago
7 0
There are two kinds of forces, or attractions, that operate in a molecule—intramolecularand intermolecular. Let's try to understand this difference through the following example.



Figure of towels sewn and Velcroed representing bonds between hydrogen and chlorine atoms

We have six towels—three are purple in color, labeled hydrogen and three are pink in color, labeled chlorine. We are given a sewing needle and black thread to sew one hydrogen towel to one chlorine towel. After sewing, we now have three pairs of towels: hydrogen sewed to chlorine. The next step is to attach these three pairs of towels to each other. For this we use Velcro as shown above.

So, the result of this exercise is that we have six towels attached to each other through thread and Velcro. Now if I ask you to pull this assembly from both ends, what do you think will happen? The Velcro junctions will fall apart while the sewed junctions will stay as is. The attachment created by Velcro is much weaker than the attachment created by the thread that we used to sew the pairs of towels together. A slight force applied to either end of the towels can easily bring apart the Velcro junctions without tearing apart the sewed junctions.

Exactly the same situation exists in molecules. Just imagine the towels to be real atoms, such as hydrogen and chlorine. These two atoms are bound to each other through a polar covalent bond—analogous to the thread. Each hydrogen chloride molecule in turn is bonded to the neighboring hydrogen chloride molecule through a dipole-dipole attraction—analogous to Velcro. We’ll talk about dipole-dipole interactions in detail a bit later. The polar covalent bond is much stronger in strength than the dipole-dipole interaction. The former is termed an intramolecular attraction while the latter is termed an intermolecular attraction.

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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

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At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
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○ c. ○

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because a Research bias, also called experimenter bias, is a process where the scientists performing the research influence the results, in order to portray a certain outcome.

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