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3 years ago

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On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of ac

etone (56.2 °c) and 1-propanol (97.4 °c), which have similar molar masses.

1 answer:

Inga [223]3 years ago

7 0

Go on google bro you can get more help there and hope you get what your looking for and good luck

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The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion fo

Anna007 [38]

Answer:

NaCl: ionic, HF: hydrogen bond, HCl: dipole dipole , F2: dispersion force

Explanation:

complete question is:

The four major attractive forces between particles are ionic bonds, dipole-dipole attractions, hydrogen bonds, and dispersion forces. Consider the compounds below, and classify each by its predominant attractive or intermolecular force among atoms or molecules of the same type.Identify each of the following ( NaCl, HF, HCl, F2) as Ionic, H Bonding, Dipole or Dispersion.

3 0

3 years ago

Read 2 more answers

Calculate the concentration of all species in a 0.160 m solution of h2co3.

enot [183]

H₂CO₃ ⇔ HCO₃⁻ + H⁺

I 0.160 0 0

C -x +x +x

E 0.160-x +x +x

Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]

4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)

x² = 6.88 x 10⁻⁸

x = 2.62 x 10⁻⁴

HCO₃⁻ ⇔ CO₃⁻² + H⁺

I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴

C -x +x +x

E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x

Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]

5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)

x = 5.6 x 10⁻¹¹

Thus,

[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M

[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M

[CO₃⁻²] = 5.6 x 10⁻¹¹ M

[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M

[OH⁻] = 3.8 x 10⁻¹¹

8 0

3 years ago

Picture shown!<br><br><br> What is the melting point of this substance at 1 ATM of pressure?

Oksana_A [137]

Answer:

The melting point of this substance at 1 ATM of pressure is 110°.

5 0

3 years ago

[03.04]When one atom of potassium (K) combines with one atom of fluorine (F), an ionic bond forms, resulting in potassium fluori

dybincka [34]

Heyo, I believe that the answer would be D. The valence electrons are given up by the K and gained by the F.

3 0

3 years ago

If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass pe

NNADVOKAT [17]

First we need to calculate the number of moles of NaOH titrated.

molar concentration = number of moles / solution volume (liter)

number of moles = molar concentration × solution volume

number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles

Then we look at the chemical reaction:

CH $_{3}$ -COOH + NaOH = CH $_{3}$ COONa + H $_{2}$ O

We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.

Now we can get the mass of acetic acid:

number of moles = mass (grams) / molecular mass (g/mol)

mass = number of moles × molecular mass

mass of acetic acid = 0,01125 × 60 = 0.675 g

We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:

concentration of acetic acid = (mass of acetic acid / mass of vinegar) × 100

concentration of acetic acid = (0.674 / 7) × 100 = 9.6 %

3 0

3 years ago