Answer :
121.5 <span>
μCi
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
</span>
3.8 /
=
((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5 </span>μci
The answer is zero!
let me know if you need help with anything else! :)
A because the outcome of this reaction exists a radical formed by the oxidation of an aromatic amine's or phenol's ring substituent. The hydroxyl group of a phenol serves as the ring substituent in this condition.
<h3>Which two enzyme types are required for the two-step process of converting cytosine to 5 hmC?</h3>
- The methyl group exists moved to cytosine in the first step, and it exists then hydroxylated in the second stage.
- Thus, a transferase and an oxidoreductase exist as the two groups of enzymes needed.
<h3>Which kind of interaction between proteins and the dextran column material is most likely to take place?</h3>
- Hydrogen bonding because the glucose's OH would create an H-bond with any disclosed polar side chains on a protein surface.
<h3>Two out of the four proteins would adhere to a cation-exchange column at what buffer pH?</h3>
- Only positively charged proteins can attach to a cation-exchange column, and this can only occur when the pH exists lower than the pI.
- Proteins A and B would both be positively charged at pH 7.0.
To learn more about hydroxyquinoline refer to:
brainly.com/question/26102339
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Which question the top part, middle or the 7 covalents?
