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Ilia_Sergeevich [38]
3 years ago
14

Chymotrypsin is a serine protease enzyme that we talked about in class. In order for chymotrypsin to function, His-57 in the act

ive site must be deprotonated. At pH = 7, approximately what percentage of chymotrypsin proteins are active? Use the Henderson-Hasselbalch equation provided and assume the pKa of His is 6.
Chemistry
1 answer:
Marizza181 [45]3 years ago
3 0

Answer:

Percentage of chymotrypsin proteins that are active is 90,91%

Explanation:

The active site, His-57, is in equilibrium, thus:

His-57H⁺ ⇄ His-57 + H⁺

Where His-57H⁺ is the active site in its protonated form.

For this equilibrium, Henderson-Hasselbalch formula is:

pH = pka + log₁₀ (His-57/His-57H⁺)

If pH is 7 and pka is 6:

7 = 6 + log₁₀ (His-57/His-57H⁺)

1 = log₁₀ (His-57/His-57H⁺)

10 = His-57/His-57H⁺

10 (His-57H⁺) = His-57 <em>(1)</em>

If total active sites are 100%

100 = His-57H⁺ + His-57 <em>(2)</em>

Replacing (2) in (1)

His-57H⁺ = 9,09%

<em>His-57 = 90,91%</em>

Thus, percentage of chymotrypsin proteins that are active is 90,91%

I hope it helps!

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Answer: The number of neutrons will increase as we move from left to right in a periodic table.

Explanation:

Atomic number is equal to the number of protons.

Mass number is the sum of number of neutrons and number of protons.

As we move from left to right, both the atomic number and mass number increases.

For example: As we move from Lithium to berrylium to boron to carbon to nitrogen to oxygen to fluorine to neon , the number of neutrons increase from 4 to 5 to 6 to 6 to 7 to 8 to 10 to 10.

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2 years ago
mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
anygoal [31]

<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

  • <u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

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A fuel tank holds 22.3 gallons of gasoline. If the density is 0.8206 g/mL, what is the mass in kilograms of gasoline in a full t
7nadin3 [17]

Answer:

m=69.3kg

Explanation:

Hello!

In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:

V=22.3gal*\frac{3.78541L}{1gal}=84.415L

Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:

m=0.8206g/L*84.415L\\\\m=69.3kg

Best regards!

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