I believe it means the imitations on what you did what were you not able to do to the object. I'm not too sure but I believe that's what it means. I hope I was able to help
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
Start with the process of elimination. Obviously, color is not a specific component to a wave, so we can cancel that out. While there are frequencies and measurable speeds of waves, those are not considered parts of the wave; nor is the wavelength or the base considered to be a part of the wave. The crest is the highest point of the wave, and is considered a part of the wave, as well as the trough, which is the the lowest point on the wave.
<span>People with protanopia are unable to sense any ‘red’ light, people with deuteranopia do not sense ‘green’ light and people with tritanopia cannot sense ‘blue’ light. If a person perceives the color green, then the yellow sensitive nerves must work somewhat effectively since green is a combination of yellow and blue. Red-sensitive nerves are most likely not responding properly for this person. The answer is C.</span>
Number 2 is the correct answer
