A because the end result of this reaction is a radical created by the oxidation of an aromatic amine's or phenol's ring substituent. The hydroxyl group of a phenol acts as the ring substituent in this situation.
<h3>Which two enzyme types are required for the two-step process of converting cytosine to 5 hmC?</h3>
- The methyl group is transferred to cytosine in the first stage, and it is then hydroxylated in the second step.
- Therefore, a transferase and an oxidoreductase are the two groups of enzymes required.
<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 form an H-bond with any exposed 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 bind to a cation-exchange column, and this can only happen when the pH is lower than the pI.
- Proteins A and B would both be positively charged at pH 7.0.
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Answer : The volume given to the patient should be, 85.5 mL
Explanation :
As we are given that the suspension contains 10g/15mL.
Now we have to determine the volume should be given to the patient.
As, 10 grams of lactulose syrup needed 15 mL volume of solution
So, 57 grams of lactulose syrup needed 
volume of solution
Thus, the volume given to the patient should be, 85.5 mL
Answer:
-2.80 × 10³ kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.
Qcal + Qcomb = 0
Qcomb = - Qcal [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ
where,
C: heat capacity of the calorimeter
ΔT: change in the temperature
From [1],
Qcomb = - Qcal = -29.2 kJ
The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:
ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol
