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lisov135 [29]
3 years ago
15

Which of the following is not part of protozoa?

Chemistry
1 answer:
Serga [27]3 years ago
4 0

Answer: A. Cilla Is Correct.

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Which organism would have the most variation in the DNA of its offspring?
Arada [10]

The organism that would have the most variation in the DNA of its offspring is the cat (Option C). Meiosis is a type of cell division that generates more genetic variability than asexual types of reproduction.

Meiosis is a type of reductional cell division by which a parental cell produces 4 daughter cells (gametes), each containing half of the genetic material.

Animals (e.g., cats) generate gametes by meiosis which fuse during fertilization to produce new offspring.

Both amoeba and bacteria reproduce by a type of asexual reproduction called binary fission. Moreover, yeasts also reproduce asexually by a process called budding and fission.

Both asexual and sexual types of reproduction generate genetic variability by the emergence of new mutations in daughter cells.

Meiosis generates much more genetic variability than asexual types of reproduction due to two different processes:

  • Random assortment of chromosomes, which produces new allele combinations.

  • Recombination, i.e., by the exchange of genetic material (DNA) between non-sister chromatids during Prophase I.

Learn more in:

brainly.com/question/7002092

5 0
2 years ago
The thermochemical equation for the reaction of carbon dioxide with sulfur dioxide is shown below. How can the reaction be descr
prisoha [69]

Answer: d. has high activation energy

Activation energy is the energy that a system requires to start a certain process. Also, it <u>is the minimum energy necessary for a given chemical reaction to occur</u>. For a reaction to occur between two molecules, they must collide in the correct orientation and have a minimum amount of energy equal to the activation energy.

As the molecules approach, their electron clouds repel, so energy is required for the collision to occur and therefore the reaction. The activation energy comes from the heat of the system, that is, the translational, vibrational energy, etc. of each molecule. However, if this energy is not enough, the reaction will not be spontaneous.

<u>A reaction between two molecules can be favored by supplying energy to the system.</u> In the case raised in the question, <u>energy equal to 1104 kJ is provided to the system to favor the next reaction </u>

CO2 (g) + 2SO2 (g) → CS2 (g) + 3O2 (g)

<u>Since the energy equal to 1104 kJ is included in the reactants, it can be deduced that it is the energy that is provided to the system for the reaction to occur. </u>However, from the value of this energy it can not be said whether the system is endothermic or exothermic since it is a kinetic variable and the variables of this type do not allow predicting the thermodynamic behavior of a system.

Furthermore, it can be seen that the value of this energy is considerably high, therefore the reaction described has a high activation energy.

3 0
2 years ago
NEED HELP QUICKLY!!! How many moles are in each of the following?
oksano4ka [1.4K]

Answer: a. 0.26mol

b. 0.000479mol

c. 1.12mol

Explanation: Please see attachment for explanation

6 0
3 years ago
Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.
pogonyaev
I think it's balanced ?
There are two H's in the both side
And one P in the both side
7 0
1 year ago
g Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at
Alborosie

Answer :  The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

where,

\Delta G^o = standard Gibbs free energy  = -14.1 kJ/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

Q  = reaction quotient

[A]_{inside} = concentration inside the cell

[A]_{outside} = concentration outside the cell

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

-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

\frac{[A]_{inside}}{[A]_{outside}}=296.2

Thus, the ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

3 0
3 years ago
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