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3 years ago

How does evolution relate to tectonic plate movement?

Chemistry

1 answer:

Reil [10]3 years ago

7 0

Answer:

Below!

Explanation:

A planet with oceans, continents, and plate tectonics maximizes opportunities for speciation and natural selection, whereas a similar planet without plate tectonics provides fewer such opportunities. Plate tectonics exerts environmental pressures that drive evolution without being capable of extinguishing all life.

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It is primarily used as energy storage found in the cell membrane

Gnesinka [82]

Answer:

ATP—Adenosine triphosphate, a nucleotide which is the most important short-term energy storage compound in cells. It is the “energy currency” of the cell, necessary for practically all metabolic activities. Carbohydrate—A type of organic molecule made of carbon, hydrogen, and oxygen.

4 0

2 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Nimfa-mama [501]

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

$a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8$ cm

for a BCC structure, the atomic radius is equal to

$r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3} }{4}=2.3103x10^{-8}cm$

7 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

5 0

3 years ago

Which is one advantage of using a reflecting telescope instead of a refracting telescope?

prohojiy [21]

Answer: C

Explanation:

4 0

3 years ago

Read 2 more answers

The pH at the midpoint in the titration of an acid with a base is A) equal to the pK of the corresponding base. B) equal to the

iVinArrow [24]

Answer:

The pH at the midpoint in the titration of an acid with a base is

A) equal to the pK of the corresponding base.

B) equal to the pK of the corresponding acid.

C) equal to 14 minus the pK of the corresponding acid.

D) equal to 14 plus the pK of the corresponding base.

E) none of the above

Explanation:

When a weak acid is titrated with a strong base, then a buffer solution is formed.

pH of a buffer solution can be calculated by using the formula:

$pH=pKa+log\frac{[salt]}{[acid]}$

Exactly at the mid point,

[conjugate base of the salt]=[acid]

So, log [salt]/[acid] =0

Hence, pH of the solution will be equal to pKa of the weak acid.

Answer is option B.

6 0

3 years ago