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marta [7]
1 year 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]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:

brainly.com/question/26102339

#SPJ4

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16

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4 nitrogen (N) atoms and 12 (3×4) atoms in Hydrogen (H)

Total = 4 + 12 = 16 molecules

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Calculate the percentage mass of hydrogen in hydrochloric acid, HCI​
PIT_PIT [208]

Answer:

To find the mass percent of hydrogen in hydrogen chloride, we must divide the weight of the hydrogen atom alone by the weight of the entire molecule. Then we multiply by 100% to find the percentage. Thus, 2.77% of the mass of hydrogen chloride is hydrogen.

Explanation:

i hope you understand better

4 0
2 years ago
What is the density of air at 75 F and latm pressure? Express in lbm/ft3 and kg/m3
ser-zykov [4K]

Answer : The density of air in lbm/ft^3 and kg/m^3 is, 0.0743lbm/ft^3 and 1.19kg/m^3 respectively.

Explanation :

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

where,

P = pressure of air = 1 atm

V = volume of air

T = temperature of air = 297 K

The conversion used for the temperature from Fahrenheit to degree Celsius is:

^oC=(^oF-32)\times \frac{5}{9}

^oC=(75-32)\times \frac{5}{9}=24^oC

The conversion used for the temperature from degree Celsius to Kelvin is:

K=273+^oC

K=273+24=297K

n = number of moles

m = mass of air

M = average molar mass of air = 28.97 g/mole

\rho = density of air = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the above formula, we get:

\rho=\frac{PM}{RT}

\rho=\frac{(1atm)\times (28.97g/mole)}{(0.0821L.atm/mol.K)\times (297K)}

\rho=1.19g/L

Now we have to calculate density in lbm/ft^3.

Conversion used :

1g/L=0.0624lbm/ft^3

So,

1.19g/L=\frac{1.19g/L}{1g/L}\times 0.0624lbm/ft^3=0.0743lbm/ft^3

The density of air in lbm/ft^3 is, 0.0743lbm/ft^3

Now we have to calculate density in kg/m^3.

Conversion used :

1g/L=1kg/m^3

So,

1.19g/L=1.19kg/m^3

The density of air in kg/m^3 is, 1.19kg/m^3

8 0
3 years ago
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