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1 year ago

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

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1 answer:

dusya [7]1 year 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:

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Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T

shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

C = $\frac{P}{K_{h}}$

where, pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at $25^{o}C$ are as follows.

$p(O_{2})$ = 0.21 atm = 0.21 bar

$p(N_{2})$ = 0.78 atm = 0.78 bar

$K_{h}$ for $O_{2}$ = $7.9 \times 10^{2} bar/mol$

$K_{h}$ for $N_{2}$ = $1.6 \times 10^{3} bar /mol$

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

$C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L$

$C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L$

Now, we will calculate the number of moles as follows.

$n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3$ mol

$n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3$ mol

As the molar mass of $O_2$ = 32 g/mol

Hence, mass of oxygen will be as follows.

Mass of $O_2 = 32 \times 1.995 \times 10^-3 \times 1000$

= 63.84 mg

As the molar mass of $N_{2}$ = 28

Mass of $N_{2} = 28 \times 3.66 \times 10^-3$ = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg and nitrogen is 102.5 mg.

5 0

3 years ago

Read 2 more answers

What is a hypothesis?

gladu [14]

Answer: a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation.

Explanation: definition

3 0

2 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2Sr(s)+O_2(g)\rightarrow 2SrO(s)$ $\Delta H_1=-1184kJ$

(2) $SrO(s)+CO_2(g)\rightarrow SrCO_3(s)$ $\Delta H_2=-234kJ$ ( × 2)

(3) $CO_2(g)\rightarrow C(s)+O_2(g)$ $\Delta H_3=394kJ$ ( × 2)

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

4 0

2 years ago

What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°C to 70.0°C when 2,520 J of hea

Ksju [112]

Answer:

A. 0.00420 J/(gi°C)

Explanation:

i got it rigt

8 0

2 years ago

Concentrated nitric acid is 15.6M and has a density of 1.41g/mL. What is the weight percent of concentrated HNO3?

tankabanditka [31]

Molarity= mol/ liters

since the molarity is given, we can assume that we have 1.0 Liters of solution

15.6 M= mol/ 1 liters---> this means that we have 15.6 moles of HNO3

we need to convert these moles to grams using the molar mass of HNO3

molar mass HNO3= 1.01 + 14.0 + (3 X 16.0)= 63.01 g/mol

15.6 mol HNO3 (63.01 g/ mol)= 983 grams HNO3

now we have to determine the grams of solution using the assumption of 1 liters of solution and the density

1 liters= 1000 mL

1000 mL (1.41 g/ ml)= 1410 grams solution

mass percent= mass of solute/ mass of solution x 100

mass percent= 63.01/ 1410 x 100= 4.47 %

6 0

3 years ago