The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
<h3>Position of 18o-labeled methanol in the products</h3>
The 18O label will appear at position b in the product as indicated in the image.
This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).
While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.
Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
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Answer:
162.7miles/hr
Explanation:
Given parameters:
Distance covered by the train = 813.5miles
Time taken = 5hours
Unknown:
Speed of the train = ?
Solution:
Speed is a physical quantity.
It is mathematically expressed as;
Speed =
So, input parameters and solve;
Speed = = 162.7miles/hr
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.
