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masya89 [10]
2 years ago
11

1. In what ways does an organism’s mode of reproduction impact the diversity of its offspring? (Asexual vs. Sexual)

Chemistry
2 answers:
NikAS [45]2 years ago
8 0

Answer:

Sexual reproduction uses two organisms in order to create another organism (reproduce). The organisms' offspring will have different traits than it's parent organisms. For example, if the offspring's female parent organism has green eyes and it's male parent organism has brown eyes, the offspring will have either green or brown eyes. If it has green eyes, the trait is known as the dominate trait while the brown eye trait is known as the recessive trait (or vise versa).

During asexual reproduction an organism, quite literally, splits itself in two. Thus creating a literal clone of itself. There is no difference between the original organism and it's cloned offspring.

Explanation:

Hope this helps. Have a nice day.

Katen [24]2 years ago
7 0

Answer:

<u>Asexual reproduction</u> = An exact copy of the reproducer

<u>Sexual reproduction</u> = Genetic variation into the organism if a random mutation in the organism's DNA is transmitted to offspring.

Explanation:

Example of Asexual Reproduction diversity - A red apple has apple seeds. You plant the red apple seeds, and it grows up to be no different than the apple before it.

Sexual reproduction Example of diversity -Two parents are of different ethnicities. The female gets pregnant, but because of the different genetics from both of the parents, the child will inherit their genetics + some random mutation in the DNA.  

This impacts the diversity of the offspring because it can be an exact copy of the reproducer (Asexual), or significantly different with some similarities if the mode of reproduction was sexual.

I really hope this helps you! Please tell me if it did or not. Good luck with your assignment/exam/quiz!

(OMG TYSM FOR BRAINLIEST!! :D)

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Answer:

(a) 10.62

(b) 2.82

(c) 1.95

Explanation:

The neutralization reaction in this question is

(CH3CH2)3N + HCl   ⇒ (CH3CH2)3NH⁺ + Cl⁻

The problem  can be  solved by calculating the number of moles of triethylamine after  addition of the portions of HCl. Since it is a weak base if it is not consumed completely, that is in excess we will have a buffer of a waek base. If its consumed completely the pH will be determined by the strong acid HCl.

The pOH for a buffer of a weak base is gven by

pOH = pKb + log [(CH3CH2)3NH⁺] / [(CH3CH2)3N]

(a) 11 mL of 0.100 M HCl

mol HCl = 0.011 L x 0.100 mol/L = 0.0011 mol HCl

mol  (CH3CH2)3N reacted = 0.0011 mol

mol (CH3CH2)3NH⁺ produced = 0.0011 mol

mol (CH3CH2)3N  initially = 0.020 L x 0.1000 mol/L 0.0020 mol

mol (CH3CH2)3N left = 0.0020 mol - 0.0011 = 0.0009 mol

pKb = - log Kb = - log (5.2 x 10⁻⁴) = 3.284

Now we can compute pOH,

pOH = 3.284 + log ( 0.0011 / 0.0009 ) = 3.37

pH = 14 - pOH = 14 - 3.37 = 10.62

(b) 20.60 mL HCl

mol HCl = 0.0206 L x 0.100 mol/L = 0.00206

mol  (CH3CH2)3N consumed = 0.0020 mol

This is so  because the acid will consume completely the 0.0020 mol of the weak base  we had originally present.

Now the problem circumscribes to that of calculating the pH of the unreacted HCl

Total Vol = 0.0206 L + 0.02 L = 0.0406 L

mol HCl = 0.0206 L x .100 = 0.00206 mol

mol HCl left = 0.00206 mol - 0.0020 mol = 0.00006 mol

[HCl] = 0.00006 mol / 0.0406 L = 0.0015 M

Since HCl is a strong acid ( 100 % ionization) :

pH = - log [H⁺] = - log ( 0.0015 ) = 2.82

(c) We will compute the pH in  the same way we did for part (b)

mol HCl = 0.025 L x 0.100 mol/L = 0.0025 mol

mol HCl left = 0.0025 mol  - 0.0020 mol = 0.0005

Total Volume = 0.020 L + 0.025 L = 0.045 L

[HCl] = 0.0005 mol / 0.045 L = 0.111

pH = - log ( 0.111) = 1.95                                            

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