Answer:
Silver Nitrate is very soluble in water.
Answer:
Neutralization
Explanation:
When an acid react with base it form the salt and water. The reaction is also called neutralization reaction because both neutralize each other.
In neutralization reaction equal amount of acid and base react to neutralize each other and equal amount of water and salt are formed. When pH does not reach to 7 its means there is less amount of one of reactant which is not fully neutralize.
Neutralization reactions are also used as first aid. For example when someone is dealing with HCl for cleaning purpose of toilet and get touched. It is advised to neutralize it with soap, milk or egg white.
Example:
Hydrochloric acid when react with the sodium hydroxide, a salt sodium chloride and water are formed.
Chemical equation:
HCl + NaOH → NaCl + H₂O
Titration:
Neutralization reactions are also used to determine the concentration of solution. Titration is a quantitative technique in which acid or base is gradually added into the solution whose concentration is to be determine until the neutral point is reached.
Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ
A hydrate is a substance where in it contains water and other constituent elements. To know whether if that compound was a hydrate,you should record its mass, then put it in a test tube and heat it with a Bunsen burner. If the compound is a hydrate, the water in the compound will discharge in the form of water vapor. At the next 5-10 minutes, remove it in the test tube and weigh it up again. If the mass is now fewer, that means that there was water existing that has now evaporated, and the compound was a hydrate.