Hg + O2 --> HgO + O
Note that Mercury has a variable valency.
Answer:
Explanation:
Pipet is used to dispense a very small amount of liquid.
Test tube rack is used to hold multiple test tubes at the same time.
Test Table is used to view chemical reactions or hold or heat small amounts of substance.
Scoopula is used to dispense chemicals from a larger container.
Graduated cylinder is used to measure volume very precisely.
Bunsen burner is used to heat objects.
Beaker is used to transport heat or store substance.
Spot plate is used to observe the color changes of small quantities of a reacting mixture.
Goggles are used to protect the eyes from flying objects or chemical splashes.
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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
Answer:
More Energy
Explanation:
Energy is required to break bonds
In Lewis dot structures, you draw the atom in the center and then surround the atom with its valence electrons. The Lewis structure for O is as shown in the attached diagram.
<h3>What is the Lewis structure of O ?</h3>
Lewis Structure of an atom of oxygen contains 6 electrons in the valence shell. Four valence electrons exist in lone pairs. It implies that oxygen atom must participate in two single bonds or one double bond in order to have an octet configuration.
A simplified representation of the valence shell electrons in a molecule is called Lewis Structure. It shows how electrons are arranged around individual atoms in the molecule.
To know more about Lewis structure, refer
brainly.com/question/1525249
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