An abyssal plain......................................
Divide by the molar mass of HCl which is 36.5g/mol grams cancels out and you are left with 1.99 mol.
55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
Answer:
year 1 is 5.5%
year 2 is 7.5%
year 3 is 10.2%
Explanation:
since,
length of the transect covered in seaweed / total lenth of transect x 100
then,
0.55 / 10.0 x 100 = 5.5
and
0.75 / 10.0 x 100 = 7.5
and
1.02 / 10.0 x 100 = 10.2
you could also just move the decimal to the right once
:)
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
