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gavmur [86]
3 years ago
10

What kind of bond does these have?

Chemistry
1 answer:
RUDIKE [14]3 years ago
7 0
<span>1. Tap water has a small concentration of H+ & OH- ions as well as water molecules, hence there would be permanent dipole-permanent dipole (p.d.-p.d.) forces of attraction between the water molecules (aka H-bonds) as well as ionic bonds between the H+ & OH- ions.

2. Distilled water does not have H+ & OH- ions, hence only H-bonds exist between the water molecules.

3. There are covalent bonds between the individual sugar molecules.
 
4. There are ionic bonds between the Na+ & Cl- ions in NaCl.
 
5. There are p.d.-p.d. forces of attraction between the Na+ ions and the O2- partial ions of the water molecules as well as between the Cl- ions and the H+ partial ions of the water molecules. There are also H-bonds between the individual water molecules and ionic bonds between the Na+ & Cl- ions (although these are in much lower abundance than in unsolvated solid NaCl).

6. There are i.d.-i.d. as well as p.d.-p.d. forces of attraction between the sugar molecules and the water molecules. There are also H-bonds between the individual water molecules and covalent bonds within the sugar molecules.</span>
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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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