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1 year ago

Based on the passage, what is the structure of the product of the reaction between 8-hydroxyquinoline-5 sulfonate and hrp?.

1 answer:

4 0

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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Water is the solvent

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2 years ago

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Answer:

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

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$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
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$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
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$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
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Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
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$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
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$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

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2 years ago

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