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nalin [4]
3 years ago
11

Based on these ideas, what are the pro's and con's for both Sexual Reproduction and Asexual reproduction?

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
6 0
Asexual
Pro:
1. inexpensive to make offspring (usually make a lot at a time and not invest a lot of time in raising them).
2. Do not need a mate to reproduce.
3. Can rapidly expand a population
Con:
1. genetically identical- prone to extinction because once a parasite has evolved to attack a specific genotype, it can kill them all.
2. Lineages usually don't last longer than a couple thousand years

Sexual:
Pros:
1. Genetically unique- so more likely to create a "successful" offspring
2. Lineages more likely to last hundreds of thousands of years

Cons:
1. More effort into creating offspring- require more parental effort
2. STD's- easily to pass
3. need to find a mate or else won't be successful as an organism.

Hope this helps you.
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un estudiante introduce un corcho cuya densidad es 0.95 g/cm en cada uno de los liquidos que figura representa mejor la situacio
bagirrra123 [75]

Answer:

La respuesta correcta es la opción A.

Explicación:

La densidad del corcho es 0.95g / cm3 por lo que se hundirá en la solución de acetona porque el corcho es más denso que la solución de acetona mientras que por otro lado, el corcho flotará en otros dos líquidos porque la densidad del corcho es menor que en otros dos líquidos o, en otras palabras, los dos líquidos son más densos que el corcho, por eso el corcho flotará en estos dos líquidos.

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3 years ago
Why can iron filings be used to visualize a magnetic field?
MrMuchimi

Answer:

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3 years ago
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

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

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
Read 2 more answers
How many molecules of nitrogen gas are in 22.4 L?
Serhud [2]
The answer is letter (B).
3 0
3 years ago
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Calculate how much heat is absorbed by a sample that weighs 12 kilograms, has a specific heat of 0.231 kg/CJ, and is heated from
Andrei [34K]

Answer:

97 J

Explanation:

Step 1: Given data

  • Mass of the sample (m): 12 kg
  • Specific heat capacity (c): 0.231 J/kg.°C (this can also be expressed as 0.231 J/kg.K)
  • Initial temperature: 45 K
  • Final temperature: 80 K

Step 2: Calculate the temperature change

ΔT = 80 K - 45 K = 35 K

Step 3: Calculate the heat required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.231 J/kg.K × 12 kg × 35 K = 97 J

4 0
3 years ago
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