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2 years ago

Why is it important to turn off the power supply when the loading buffer gets toward the bottom of the gel?.

1 answer:

4 0

At this condition, it is necessary to to stop the current as the molecules may pass through the gel if the current is allowed to continue even after the loading buffer gets towards the bottom of the gel.

A narrow dye front is seen when the gel percentage is low, whereas a diffuse dye front is seen when the acrylamide percentage is large in the gel. To prevent a circuit, the power should be turned off before removing the gels and then the cords. When an electric current is sent through a gel, charged molecules move throughout the substance. One end of the gel receives a positive charge, and the other receives a negative charge due to the application of an electric current across the gel. Migration describes how charged molecules migrate. To the opposing charge, molecules move. In the sample wells at the top of the gel, protein samples are introduced.

Learn more about molecules here:

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If you were about 10 kilometers above earth, what would the pressure approximately be?

Sliva [168]

Answer:

About 25 kPa

Explanation:

Pressure decreases with height above sea level.

The calculation is rather complicated, so I will refer to the figure below.

It shows that the pressure at 10 km is about 25 kPa.

7 0

4 years ago

A hamburger containing 335.4 kcal of energy was combusted in a bomb calorimeter with an unknown heat capacity. The temperature o

Zielflug [23.3K]

Answer:

$Cv_{calorimeter}=18.4kcal/K$

Explanation:

Hello!

In this case, since the combustion of the hamburger released 335.4 kcal of energy and that energy is received by the calorimeter, we can write:

$Q_{hamburguer}=-Q_{calorimeter}$

And the heat of the calorimeter is written in terms of the temperature change and the calorimeter constant:

$Q_{hamburguer}=-Cv_{calorimeter}\Delta T$

Thus, given the released heat by the hamburger due to its combustion and the temperature change, Cv for the calorimeter turns out:

$Cv_{calorimeter}=\frac{-Q_{hamburguer}}{\Delta T} =\frac{-(-335.4kcal)}{18.2K}\\\\Cv_{calorimeter}=18.4kcal/K$

Best regards!

4 0

3 years ago

In 1939, at the age of 20, a man stood 270 cm tall (over 8 1/2 feet) and wore size 37 shoes! The hormonal disorder from which he

rodikova [14]

Answer:

Somatotropin(Growth Hormone)

Explanation:

-Uncontrolled growth in a person is usually caused by the excessive secretion of growth hormone.

-This hormone is also known as Somatotropin.

-This hormone is produced in the pituitary gland.

3 0

4 years ago

In the writing of ionic chemical formulas, what factor is "crossed over" in the crossover rule?

Levart [38]

Ionic compounds are formed by the complete transfer of electrons between the atoms. The atom which gains electron(s) forms anion whereas loss of electron(s) results in the formation of cation. They are bonded to each other by electrostatic force of attraction between the negatively and positively charged atoms.

While writing the ionic chemical formula for binary ionic compound the rules are:

Writing the chemical symbol of the metals and non-metals involved in the compound formation.
The charge i.e. the absolute value of oxidation number of each metal is written on respective atoms.
The charge i.e. the absolute value of oxidation number are cross-multiplied that is the charge of first ion becomes the subscript of second ion and vice versa.

Hence, in the writing of ionic chemical formulas, the value of each ion's charge is "crossed over" in the crossover rule.

5 0

3 years ago

Three Stoichiometry Questions

andrezito [222]

Answer:

Explanation:

Given data:

Mass of aluminium = 2.5 g

Mass of oxygen = 2.5 g

Mass of aluminium oxide = 3.5 g

Percent yield = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 27 g/mol

Number of moles = 0.09 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 32 g/mol

Number of moles = 0.08 mol

Now we will compare the moles of aluminium oxide with aluminium and oxygen.

Al ; Al₂O₃

4 : 2

0.09 : 2/4×0.09 = 0.045

O₂ : Al₂O₃

3 : 2

0.08 : 2/3 ×0.08 = 0.053

The number of moles of aluminium oxide produced by Al are less so it will limiting reactant.

Mass of aluminium oxide:

Mass = number of moles × molar mass

Mass = 0.045 × 101.96 g/mol

Mass = 4.6 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 3.5 g / 4.6 ×100

Percent yield = 76.1%

Given data:

Mass of copper produced = 3.47 g

Mass of aluminium = 1.87 g

Percent yield = ?

Solution:

Chemical equation:

2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 1.87 g/ 27 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of copper with aluminium.

Al ; Cu

2 : 3

0.07 : 3/2×0.09 = 0.105

Mass of copper:

Mass = number of moles × molar mass

Mass = 0.105 × 63.55 g/mol

Mass = 6.67 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 3.47 g / 6.67 × 100

Percent yield = 52%

4 0

3 years ago