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3 years ago

14

After performing a western blot analysis, you obtain a blot showing a single band. Plotting electrophoretic mobility of the mark

er proteins and the protein of interest, you determine that the molecular weight of the protein is 213 kD . Estimate the number of amino acids in the protein.

1 answer:

saveliy_v [14]3 years ago

3 0

Answer:

The estimate is $N = 1936$

Explanation:

From the question we are told that

The molecular weight of the protein is $MW_p = 213 kD = 213 *10^{3} \ D$

Generally the average molecular weight of amino acid is $MW_a = 110 \ D$

Generally the number of amino acids in the protein is mathematically represented as

$N = \frac{MW_p}{MW_a}$

$N = \frac}{213 *10^{3}{110}$

=> $N = 1936$

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A 50.0 mL sample of 12.0 M HCl is diluted to 200 mL. What is true about the diluted solution?

kobusy [5.1K]

The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

V1 = 50mL
C1 = 12.0M
V2 = 200mL
C2 = ?

<h3>Facts about the diluted solution</h3>

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

$c_1 v_1 = c_2 v_2\\12 * 50 = c_2 * 200\\c_2 = \frac{600}{200} \\c_2 = 3M$

The concentration of the solution reduces.

2. The number of moles remains the same.

When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.

Learn more on concentration of a solution here;

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3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

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

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

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4 years ago

After reading the list of physical properties above, you realize that they are describing a substance that is a

Alecsey [184]

The answer is B

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4. An 82 kg hiker climbs Mt. Humphrey near Flagstaff. During a two hour period, the hiker's

bazaltina [42]

Answer:

$\Delta E = 434253.96\,J$ , $\Delta P = 60.313\,W$

Explanation:

The change in energy is given by the change in gravitational potential energy:

$\Delta E = m\cdot g \cdot \Delta h$

$\Delta E = (82\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (540\,m)$

$\Delta E = 434253.96\,J$

The average rate of change in terms of time is approximately this:

$\Delta P = \frac{\Delta E}{\Delta t}$

$\Delta P = \frac{434253.96\,J}{(2\,h)\cdot (\frac{3600\,s}{1\,h} )}$

$\Delta P = 60.313\,W$

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what mass of carbon dioxide will be produced when 12.9 g of butane reacts with an excess of oxygen in the following reaction?

Nadya [2.5K]

39.1 gCO2
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3 years ago