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2 years ago

Based on its type of chemical bond, which substance has the highest boiling point?

Chemistry

2 answers:

Aleonysh [2.5K]2 years ago

8 0

Answer: Option (b) is the correct answer.

Explanation:

When a compound contains a strong bond between its combining atoms or molecules then the compound will have a high boiling point.

An ionic bond is a strong bond because it contains strong force of attraction between two oppositely charged atoms.

For example, LiF is an ionic compound in which lithium is $Li^{+}$ ion and fluorine is $F^{-}$ ion.

Therefore, LiF compound will have strong force of attraction.

On the other hand, covalent compounds have weak bond due to sharing of electrons. Hence, they have low boiling point.

For example, $CO_{2}$ , $CCl_{4}$ , and $H_{2}$ etc are covalent compounds.

Thus, we can conclude that LiF substance has the highest boiling point.

Ber [7]2 years ago

7 0

Answer: Lithium Fluoride

Explanation: Let’s see,,,
1. Obviously not hydrogen or carbon dioxide because they’re gas at room temp.
2. Lithium Fluoride is an ionic compound which means they have higher intermolecular forces and require more energy to break them, so it probably has a higher boiling point.

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What is the mass of 0.572 moles of Al?

kozerog [31]

Answer: 15.433704

Use the periodic table to check the atomic mass, this is the number of grams per mole → 1 mole of Aluminum is 26.982 g

5 0

3 years ago

What would the molecule CH₄ be classified as?

larisa [96]

Answer:

Alkane

Explanation:

Definition of Alkane "any of the series of saturated hydrocarbons including methane, ethane, propane, and higher members. (google dictionary)"

CH4 is methane.

6 0

3 years ago

Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole

kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where CH₃COOH is the acid and CH₃COO⁻ is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ $\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$

a) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[2 mol]}$

pH = 4,75

b) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[1mol]}$

pH = 5,05

I hope it helps!

7 0

3 years ago

The latent heat of melting is 80 calories for 1 gram of ice. if 1 gram of ice absorbs 60 calories what do i get?

Eva8 [605]

Latent heat of melting is the energy that a solid absorbs to change its phase as its liquid. During this process, since all energy is used to change the phase, the temperature is constant.

Here the latent energy of melting for 1 g of ice is 80 calories and that 1 g of ice only absorbed 60 calories. hence the phase is not changed because it requires more 20 calories to melt.

Hence 1 g of ice remains as its solid phase (ice).

3 0

3 years ago

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 = [A⁻]/[HA] = 2.69

[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0

3 years ago