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geniusboy [140]
2 years ago
8

What kind(s) of intermolecular forces exist in the compounds given below? (a) CH3CH2CH2CH2CH3(l) dispersion forces dipole-dipole

forces hydrogen bonding (c) H2CO(l) dispersion forces dipole-dipole forces hydrogen bonding (b) CH3CH2OH(l) dispersion forces dipole-dipole forces hydrogen bonding (d) O2(l) dispersion forces dipole-dipole forces hydrogen bonding
Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
7 0

Answer:

(a) CH3CH2CH2CH2CH3(l)

  • dispersion

(b) CH3CH2OH(l)

  • Dipole-dipole interaction
  • Dispersion forces
  • Hydrogen bonding

(c) H2CO(l)

  • Dipole-dipole
  • Dispersion

(d) O2(l)

  • Dispersion

Explanation:

Dispersion forces are those forces that occur between two non polar molecules.They form the weakest bonds.Here electrons of one molecule is attracted to the nucleus of the other molecule. Example are;

  • Interaction of two methyl (-CH₃) group
  • Interaction between nitrogen gas , N₂ molecules
  • Interaction between oxygen gas ,O₂ molecules

Dipole-Dipole interaction happens when two polar molecules interact.Positive charges in one molecule is attracted to negative charge of another molecule.Examples

  • Chloroform (CHCl₃)
  • Ammonia (NH₃)

Hydrogen bonding are created when an hydrogen atom bonded to an electronegative atom is attracted to a another electronegative atom.Example is the hydrogen bond between oxygen atom and hydrogen is water (H₂O).

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5 0
2 years ago
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
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24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

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3 years ago
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Answer:

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Answer:

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Explanation:

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