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Lorico [155]
3 years ago
5

16. DNA in a eukaryotic cell is contained in

Chemistry
2 answers:
Leto [7]3 years ago
6 0
A) a nucleus


The nucleus plays an important part within eukaryotic organelles due to it being the location of the cell’s DNA.


hope this helps,, have a great day :))
GrogVix [38]3 years ago
5 0

Answer:

A. Nucleus

Nucleus is the answe

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Which balanced equation represents an oxidation-reduction reaction?
Bingel [31]

Answer:

Oxidation states are used in chemistry solutions. It is a bond in which electron transfers easily from one nucleus to another nucleus.

Explanation:

  • Oxidation-reduction reactions have some rules.
  • The oxidation state is 0 at an uncombined bond.
  • The bond of oxidation reduction is +1 in alkeli metal.
  • The bond in two metal is +2
  • The oxidation reduction state at helogens is -1. It does not happened always.
  • The oxygen bond in oxidation and reduction is -2.
  • The sum of the oxidation state is equal to the compound charges.
  • In this process the changes occur for any elements. Redox could be occur. Its oxidized and reduction reaction can be seen in this process.
7 0
3 years ago
Arrange the colors of visible light, green, red, and blue, in order of increasing wavelength.
lukranit [14]

Answer: The increasing wavelength of colors:

Red > Green > Blue

Explanation:

Wavelength: This is the property of wave which includes the distance between two consecutive crests or trough. This is denoted by the Greek letter Lambda and it is found by dividing the velocity of the wave with its frequency.

Wavelength of colours are

Violet: 400 - 420 nm

Indigo: 420 - 440 nm

Blue: 440 - 490 nm

Green: 490 - 570 nm

Yellow: 570 - 585 nm

Orange: 585 - 620 nm

Red: 620 - 780 nm

8 0
3 years ago
Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reactio
vekshin1

Answer 1) In the given reaction of sulfuric acid


2NO_{(g)} + O_{2}_{(g)} ---->  2NO_{2}_{(g)}


2NO_{2}_{(g)} + 2SO_{2}_{(g)} ---->  2NO}_{(g)} + 2SO_{3}_{(g)}


On addition of nitrogen monoxide gas the reaction rate increases and more amount of product is formed.


So, it is clear that NO is the catalyst in this reaction.


Answer 2) This can be proven that NO is catalyst because it increases the rate of the reaction, but it is not consumed during the reaction, and it also gets regenerated at the end of reaction.


Hence, nitrogen mono oxide is considered as the catalyst in the given reaction.


Answer 3) It increases the rate of reaction by decreasing the activation energy of the reaction. Also it can be clearly seen in this reaction the NO is reacting with oxygen to lower the energy of activation. So, it is providing an alternative pathway for proceeding the reaction. This all confirms the assumptions of NO being the catalyst.


2NO_{(g)} + O_{2}_{(g)} ---->  2NO_{2}_{(g)}

4 0
3 years ago
Read 2 more answers
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0
3 years ago
Ok this is so weird but like how do you expose fake friends for talking about you? I know they did it but i dont know how to tal
murzikaleks [220]

Answer:

just do it

Explanation:

8 0
3 years ago
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