Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
Answer:
No
Explanation:
as it would stay sticked on them forerver amd a time of 2 months it will kill them
We are given with
Cobalt phosphate - CoPO4
We are asked for the net ionic equation for the phosphate dissolving in H3O+
The net ionic equation is
CoPO4 (s) + H3O+ (aq) -----> HPO42- (aq) + Co3+ (aq) + H2O *(l)
At constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.
<h3>What is Boyle's law?</h3>
Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.
Boyle's law is expressed as;
P₁V₁ = P₂V₂
Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.
Given that;
- Initial volume of the gas V₁ = 22.5L
- Initial pressure of the gas P₁ = 0.98atm
- Final pressure of the gas P₂ = 0.95atm
- Final volume of the gas V₂ = ?
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = (0.98atm × 22.5L) / 0.95atm
V₂ = 22.05Latm / 0.95atm
V₂ = 23.2L
Therefore, at constant temperature, if the pressure is compressed to the given value, the volume of the nitrogen gas increases to 23.2L.
Learn more about Boyle's law here: brainly.com/question/1437490
#SPJ1
