<span>Fry's Reagent a chemical used on metals such as firearms, it corrodes them and allows you to see the etching that could have been scratched off by the criminal, such as the serial number of the gun. It is </span><span>a chemical that can restore a serial number that has been removed from a gun by dissolving the area that has been flattened when the number was stamped on.</span>
For this reaction to proceed, the following bond breaking should occur:
*one C-O bond* one H-Cl bond
After, the following bond formations should occur:*one C-Cl bond*one O-H bond
Now, add the bond energies for the respective bond energies which can be found in the attached picture. For bond formations, energy is negative. For bond breaking, energy is positive.
ΔHrxn = (1)(358) + (1)(431) + 1(-328) + 1(-463) =
<em> -2 kJ</em>
The answer should be D. Light
Answer:
The correct answer is 0.92 g
Explanation:
The density is defined as the mass per unit of volume:
Density= mass/volume
From the data provided:
volume= 5.4 L
density= 0.17 g/L
Thus, to calculate the mass of helium:
mass= density x volume = 0.17 g/L x 5.4 L= 0.918 g ≅ 0.92 g
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
