An electron in the 3s orbital. The order of electron orbital energy levels starting from lowest to highest is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
Answer:
3,964 years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of the element is 5,730 years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 100%).
[A] is the remaining concentration of the sample ([A] = 61.9%).
∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.
To determine the mass of oxygen per gram of sulfur for sulfur dioxide, we simply obtain the ratio of the mass of oxygen and the mass of sulfur produced from the decomposition of sulfur dioxide. All other values given in the problem statement above are just to confuse us that the question is a difficult one. We do as follows:
mass of oxygen per gram sulfur = 3.45 g / 3.46 g
mass of oxygen per gram sulfur = 0.9971 g O2 / g S
Sulfur smells like rotten egg and is yellow...
The equation that relates both energy and wavelength is:
where e is the energy and lambda is the wavelength.
Therefore, as we can see from this equation, the energy of an electromagnetic wave is inversely related to the wavelength of the electromagnetic wave.
