We are given that the specific heat of water is 4.18 J / g
°C. We know that the molar mass of water is 18.02 g/mol, therefore the molar
heat capacity is:
molar heat capacity = (4.18 J / g °C) * 18.02 g / mol
<span>molar heat capacity = 75.32 J / mol °C</span>
Answer:
- Add AgNO₃ solution to both unlabeled flasks: based on solubility rules, you can predict that when you add AgNO₃ to the NaCl solution, you will obtain AgCl precipitate, while no precipitate will be formed from the NaClO₃ solution.
Explanation:
<u>1. Adding AgNO₃ to NaCl solution:</u>
- AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)
<u>2. Adding AgNO₃ to NaClO₃ solution</u>
- AgNO₃ (aq) + NaClO₃ (aq) → AgClO₃ (aq) + NaNO₃ (aq)
<u />
<u>3. Relevant solubility rules for the problem.</u>
- Although most salts containing Cl⁻ are soluble, AgCl is a remarkable exception and is insoluble.
- All chlorates are soluble, so AgClO₃ is soluble.
- Salts containing nitrate ion (NO₃⁻) are generally soluble and NaNO₃ is not an exception to this rule. In fact, NaNO₃ is very well known to be soluble.
Hence, when you add AgNO₃ to the NaCl solution the AgCl formed will precipitate, and when you add the same salt (AgNO₃) to the AgClO₃ solution both formed salts AgClO₃ and NaNO₃ are soluble.
Then, the precipiate will permit to conclude which flask contains AgCl.
The molar heat of vaporization of ammonia is 23.3 kJ/mol. The molar heat of condensation of ammonia is - 23.3 kJ/mol.
The molar heat of condensation is the opposite of the molar heat of vaporization. The molar heat of vaporization of ammonia is given :
ΔH evaporation = - ΔH condensation
Therefore the molar heat of condensation of ammonia is given by:
ΔH condensation = - 23.3 kJ / mol
That's right. The molar heat of vaporization of ammonia is 23.3 kJ/mol. The molar heat of condesation of ammonia is - 23.3 kJ/mol.
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Steam, being in gas form, has the highest energy as particles have more kinetic energy and they move more vigorously than in ice or water that are in solid and liquid form respectively.
