When drawing a covalent lewis dot diagram its important to ensure that:

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

saveliy_v [14]

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

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

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

=> $N = 1936$

3 0

4 years ago

A 27.3 g marble sliding to the right at 21.0 cm/s overtakes and collides with a 11.7 g marble moving in the same direction at 12

NISA [10]

Answer:

The answer to this is

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

Explanation:

To solve the question, let us list out the given variables and their values

Mass of first marble m1 = 27.3g

Velocity of the first marble v1 = 21.0 cm/s

Mass of second marble m2 = 11.7g

Velocity of the second marble v2 = 12.6 cm/s

After collision va1 = unknown and va2 = 23.7 cm/s

From Newton's second law of motion, force = rate of change of momentum produced

Hence m1v1 + m2v2 = m1va1 + m2va2 or

va1 = (m1v1 + m2v2 - m2va2)÷m2 or (720. 72-277.29)÷m1 → va1 = 16.24 cm/s

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

4 0

3 years ago

Find the potential difference required to accelerate protons from rest to 10% of the speed of light. (at this point, relativisti

elixir [45]

I don't know much about the potential difference but I certainly can calculate the energy of the proton <span>from rest to 10% of the speed of light.

Special relativity tells us that "</span><span>the energy needed to accelerate a particle (with mass) grow super-quadratically when the speed is close to c, and is ∞ when it is c."

Expressed the theory in equation is as follows:
</span>
ENERGY = rest mass × speed of light squared / (1−(“percent of speed of light”)squared ) <span> = mc2/(1-("percent of speed of light")2)
</span>
Just plug-in the values and you can now have the answer what you are looking for!

5 0

4 years ago

A solution containing 20.0 g of sodium sulfite reacts with 7.0 ml of phosphoric acid. The concentration of the acid solution is

stiv31 [10]

a) The mass of the excess reactant remaining at completion is 7.36 grams.

b) mass of water produced is 3.51 grams.

c) moles of sodium phosphate produced is 0.0186.

d) grams of sulphur dioxide produced is 12.49 grams.

Explanation:

Data given:

mass of sodium sulphite = 20 grams

volume of phosphoric acid = 7 ml

concentration of the phosphoric acid is = 1.83 grams/ml

mass of excess reactant =?

grams of water produced =?

moles of sodium phosphate =?

grams of sulphur dioxide =?

balance equation for the reaction:

2 $H_{3}$ P $O^{4}$ + 3 $Na_{2}$ $SO_{3}$ ⇒ 2 $Na_{3} PO_{4}$ +3 $SO_{2}$ +3 $H_{2}$ O

Now the number of moles will be calculated by using the formula:

molarity = $\frac{number of moles}{volume of solution}$

For phosphoric acid number of moles (atomic mass of H3PO4 = 97.99 grams/mole), mass is 1 ml has 1.83 grams so 7 ml wil have 12.81 grams

number of moles = $\frac{mass}{atomic mass of 1 mole}$

= $\frac{12.81}{97.99}$

= 0.13 moles of H3PO4 in 7 ml

number of moles of sodium sulphite (atomic mass = 126.04 grams/mole)

= $\frac{20}{126.04}$

= 0.15 moles

from the stoichiometry:

limiting reagent is the one which yield low amount of product.

3 moles of phosphoric acid gave 2 moles of Na3PO4

0.13 moles give x moles

$\frac{2}{3}$ = $\frac{x}{0.13}$

3x = 0.26

x = 0.086 moles of Na3PO4

2 moles of phosphoric acid gave 3 moles of sulphur dioxide and 3 moles of water

0.13 moles of phosphoric acid will give x moles of water and sulphur dioxide.

$\frac{3}{2}$ = $\frac{x}{0.13}$ (SO2) and $\frac{3}{2}$ = $\frac{x}{0.13}$

so 0.195 moles of water and 0.195 moles of SO2 is formed.

Now the second reactant:

3 moles of 3 $Na_{2}$ $SO_{3}$ gave 2 moles of Na3PO4

So, 0.15 moles will give

$\frac{2}{3}$ = $\frac{x}{0.15}$

x = 0.1 moles of Na3P04

3 moles $Na_{2}$ $SO_{3}$ gives 3 mole of water and SO2

so 0.15 mole of water and SO2 formed

so the limiting reagent is H3P04

mass of the products:

mass = number of moles x atomic mass

b) mass of water = 0.195 x 18

= 3.51 grams

c) mass of SO2 = 0.195 x 64.06

= 12.49 grams

d ) mass of sodium phosphate = 0.086 x 163.94

= 14.09 grams

a) mass of excess reactant

Sodium sulphite is excess reactant,

so we had started with 0.15 sodium sulphite

ratio of excess and limiting reagent as moles of sodium phosphate from sodium sulphite is 0.1 and from H3P04 is 0.086 moles so excess is

0.1 x 126.04

= 12.64 grams

we started with 20 grams and used 12.64 grams

excess reagent = 20 - 12.64

= 7.36 grams is the excess reagent.

4 0

3 years ago

What coefficients are needed to balance the equation for the complete combustion of methane? enter the coefficients in the order

Art [367]

The coefficients in the order CH₄, O₂, CO₂, and H₂O, are 1,2,1,2

The equation becomes:

CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)

<h3><em>Further explanation</em></h3>

Complete combustion of hydrocarbon with oxygen will be obtained by CO₂ and H₂O compounds.

If O is insufficient there will be incomplete combustion produced by CO and H and O

Hydrocarbon combustion reactions (specifically alkanes)

$\large {\boxed {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}}$

Equalization of chemical reaction equations can be done using variables.

Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C atoms, H and last atoms O atoms

Methana combustion reaction

CH₄ (g) + O₂ (g) -> CO₂ (g) + H₂0 (g)

We give the most complex compounds, namely Methane, we give the number 1

So the reaction becomes

CH₄ (g) + aO₂ (g) -> bCO₂ (g) + cH₂0 (g)

C atom on the left 1, right b, so b = 1

H atom on the left 4, right 2c, so 2c = 4 ---> c = 2

Atom O on the left 2a, right 2b + c, so 2a = 2b + c

2a = 2.1 + 2

2a = 4

a = 2

The equation becomes:

CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)

<h3><em>Learn more</em></h3>

the combustion of octane in gasoline

brainly.com/question/8175791

brainly.com/question/897044

hydrogen and excess oxygen

brainly.com/question/1405182

Keywords: methane, combustion, hydrocarbons, equalization of reaction equations

3 0

4 years ago

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