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Molodets [167]
3 years ago
14

Decide if the following statements regarding Intermolecular forces are True or False. H2S will have a higher boiling point than

F2. HF will have a lower vapor pressure at -50oC than HBr. CH3CH2OH will have a higher melting point than CH3OCH3. HF will have a higher vapor pressure at 25oC than KCl. As the polarity of a covalent compound increases, the solubility of the compound in CCl4 decreases.
Chemistry
1 answer:
igomit [66]3 years ago
4 0

Answer:

1. True

2. False

3. True

4. False

5. True

Explanation:

The intermolecular forces are the forces that put together the molecules in a substance. When the molecule is polar, it will have partial charges in the atoms, so the molecules will be attracted by electrostatic force, such as an ionic compound (but less effective). The force is called dipole-dipole.

When the molecule is nonpolar, to be together, a partial charged is induced, and then they will be attracted. Because it is induced, the force is weaker then dipole-dipole, and it's called dipole-induced-dipole-induced.

When a polar molecule is formed between hydrogen and an extremely electronegativity element(N, O, and F), the dipole-dipole bond will be stronger, so it has a special name: hydrogen bond.

As stronger is the force, as difficult is to break it, so more difficult will be to boil it. Then:

1. H₂S is polar, so it has dipole-dipole forces, and F₂ is nonpolar, so it has dipole-induced-dipole-induced forces, then H₂S has a higher boiling point.

2. HF has a hydrogen bond, and HBr a dipole-dipole force, so HF has a higher boiling point, because of that, at the same temperature, HF will need a higher pressure to boil, so the vapor pressure (the pressure necessary to boil it) of it will be higher than at HBr.

3. CH₃CH₂OH has a hydrogen bond because there's a bond between hydrogen and oxygen, so it will be stronger than the forces in CH₃OCH₃. Then CH₃CH₂OH has a higher melting point.

4. KCl is an ionic compound, which has stronger forces than the covalent compounds, so KCl will have a higher pressure vapor.

5. In solubilization, "like dissolves like", it means that polar solvents dissolve polar solutes, and nonpolar solvents dissolve the nonpolar solute. CCl₄ is a nonpolar solvent (the tetrahedral geometry makes the dipole forces cancel), so it dissolves nonpolar solutes.

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4 0
3 years ago
Read 2 more answers
In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,
Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -              -

At t =equilibrium      (x-s)                s           2s          

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

K_{sp}=s\times {2s}^2

K_{sp}=4s^3

Given  s = 7.4×10⁻³ M

So, Ksp is:

K_{sp}=4\times (7.4\times 10^{-3})^3

K_{sp}=4\times (7.4\times 10^{-3})^3

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -            0.20

At t =equilibrium      (x-s')             s'         0.20+2s'         

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.</u>

3 0
4 years ago
A book with a weight of 12 N rests on its back cover. If the back cover measures 0.21 m by 0.28 m, how much pressure does the bo
erik [133]

Answer:

option D = 204 Pa

Explanation:

Given data:

Weight of book = 12 N

Area of book =  0.21 m × 0.28 m = 0.0058 m²

Pressure of book = ?

Solution:

formula

pressure = force / area

P = f/ A

now we will put the values in formula,

P = 12 N/ 0.0058  m²

P= 204 Nm⁻²      ( Nm⁻²= Pa)

P = 204 Pa

4 0
3 years ago
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