Calculate the change in energy of an atom that emits a photon of wavelength 2.21 meters. (Planck’s constant is 6.626 x 10-34 jou
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The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.
<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction
<u>Explanation:</u>
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.
For the given chemical reaction:
<u>On the reactant side:</u>
Oxidation number of H = +1
Oxidation number of N = +5
Oxidation number of O = -2
Oxidation number of S = 0
<u>On the product side:</u>
Oxidation number of H = +1
Oxidation number of N = +2
Oxidation number of O = -2
Oxidation number of S = +6
As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.
The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.
Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction