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3 years ago

What are some substances that from ion-ion forces?

Chemistry

1 answer:

Anika [276]3 years ago

6 0

Answer:

The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds.

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Fill in the blank questions usually require

TiliK225 [7]

The answer is a because a fill in the blank is always just one word

6 0

2 years ago

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Someone help please , I’ll give brain . Question 10 !

Rudik [331]

Answer:it’s c

Explanation:

I just know it is

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3 years ago

Phenols do not exhibit the same pka values as other alcohols; they are generally more acidic. Using the knowledge that hydrogen

nevsk [136]

Answer:

Phenols do not exhibit the same pka values as other alcohols;

They are generally more acidic.

Using the knowledge that hydrogen acidity is directly related to the stability of the anion formed, explain why phenol is more acidic than cyclohexane.

Explanation:

According to Bromsted=Lowry acid-base theory,

an acid is a substance that can release $H^{+}$ ions when dissolved in water.

So, acid is a proton donor.

If the conjugate base of an acid is more stable then, that acid is a strong acid.

In the case of phenol,

the phenoxide ion formed is stabilized by resonance.

$C_6H_5OH -> C_6H_5O^- +H^+$

The resonance in phenoxide ion is shown below:

Whereas in the case of cyclohexanol resonance is not possible.

So, cyclohexanol is a weak acid compared to phenol.

7 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

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3 years ago

Gas laws describe and predict the behavior of gases without attempting to explain why they happen

Nutka1998 [239]

The gas laws describe and predict the behavior of gases with an explanation and experimental data

So the given statement is False.

2) The volume of gas can be calculated based on Avagadro's law

It states that the volume of a gas is directly proportional or varies with the moles of the gas. Higher the moles more the volume, condition is the pressure and temperature are constants in the two conditions

Thus as here the pressure and temperature of nitrogen gas is kept constant

V α moles

$\frac{V1}{n1}=\frac{V2}{n2}$

Where

V1 = 6 l

n1 = 0.50 mol

V2 = ?

n2 = 0.75 mol

On putting values

V2 = 6 X 0.75 / 0.5 = 9 L

so resulting volume of the gas will be 9L

5 0

3 years ago

Read 2 more answers