Answer:
I think its iron(II) oxide
Explanation:
The substance is black but when it's mixed with water it turns green solution so when mixed with sodium Hydroxide it turns green
Answer:
I don't know if you can directly prove it with evidence if you haven't observed it but you can maybe take an educated guess by the aftermath of it?
For example, you see a burnt log. At this time, people don't know what fire is. After we study the log, we could see that it takes extreme temperature in order to burn the log and that would help people see that there is a force like fire that can cause this. In a way, finding out that extreme temperatures burns stuff is another step closer to the discovery and proof of fire
I hope that makes sense
Answer:
1.66 × 10⁻¹⁸ Moles
Explanation:
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Atoms and Avogadro's Number is given as,
Number of Moles = Number of Atoms ÷ 6.022 × 10²³ Atoms/mol
Putting values,
Number of Moles = 1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol
Number of Moles = 1.66 × 10⁻¹⁸ Moles
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
