What is the isoelectric point of proteins?

2 answers:

7 0

The isoelectric point<span> (</span>pI, pH(I),IEP<span>), is the pH at which a particular molecule carries no net electrical charge in the statistical mean.</span>

lakkis [162]3 years ago

6 0

Isoelectric point. The isoelectric point (pI, pH(I), IEP), is the pH at which a particular molecule carries no net electrical charge in the statistical mean. The standard nomenclature to represent the isoelectric point is pH(I), although pI is also commonly seen, and is used in this article for brevity.

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If body fats has a density of 0.94 g/mL and 3.0 liters of fat are removed, how many pounds of fat were removed from the patient?

-Dominant- [34]

Answer:

6.217 pounds

Explanation:

We are given;

Density of body fats 0.94 g/mL
Volume of fats removed = 3.0 L

We are required to determine the mass of fats removed in pounds.

We need to know that;

Density = Mass ÷ volume

1 L = 1000 mL, thus, volume is 3000 mL

Rearranging the formula;

Mass = Density × Volume

= 0.94 g/mL × 3000 mL

= 2,820 g

but, 1 pound = 453.592 g

Therefore;

Mass = 2,820 g ÷ 453.592 g per pound

= 6.217 pounds

Thus, the amount of fats removed is 6.217 pounds

4 0

3 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.