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marta [7]
2 years ago
5

Based on the passage, what is the structure of the product of the reaction between 8-hydroxyquinoline-5 sulfonate and HRP mcat r

eddit
Chemistry
1 answer:
dusya [7]2 years ago
4 0

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

#SPJ4

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What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

  • i: van 't Hoff factor (1 for non-electrolytes)
  • Kf: cryoscopic constant
  • m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

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