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3 years ago

Why Is It Important To Classify The Millions Of Species On Earth

2 answers:

5 0

Because without classification there wouldn't be a way to decide what is what and which is which. Classification allows the justification of someone or something to have it's own. For ex : This monkey is red, and this monkey is blue. They have something different between them, they aren't just a "monkey" they are specified under what you can specify about them or what others specify about them. That is a Red Monkey, that is a Blue Monkey. And so forth.

salantis [7]3 years ago

5 0

Answer:

Why Is It Important To Classify The Millions Of Species On Earth?

The importance mainly is the Taxonomy

Explanation:

The importance mainly is the Taxonomy, which gives an order to living beings, the taxonomy facilitates the study and compression of the species on Earth. Also taxonomy gives the scientific names and classify by their kingdom, family and species, in addition to including their characteristics

One of the important teachings that the different ways of classifying give us have been the great diversity of living beings that inhabit the Earth.

Even the most recent classifications are provisional, because new species are discovered every year. The main value of all classifications is that they have allowed us to know the different organisms that exist much better.

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What mass, in grams, of CO2 and H20 N is formed from 2.55 mol of propane?

Black_prince [1.1K]

Answer: The mass of $CO_2$ and $H_2O$ produced are 336.6 g and 183.6 g respectively.

Explanation:

The combustion reaction between propane and oxygen leads to formation of carbon dioxide and water.

Law of Conservation of mass states that the mass will remain constant for a balanced equation. This is carried out when the total number of atoms on reactant side is same as the total number of atoms on the product side. Thus the equation must be balanced.

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

a) 1 mol of propane produces = 3 moles of $CO_2$

Thus 2.55 mol of propane produces = $\frac{3}{1}\times 2.55=7.65 moles of [tex]CO_2$

mass of $CO_2=moles\times {\text {molar mass}}=7.65mol\times 44g/mol=336.6g$

b) 1 mol of propane produces = 4 moles of $H_2O$

Thus 2.55 mol of propane produces = $\frac{4}{1}\times 2.55=10.2 moles of [tex]H_2O$

mass of $H_2O=moles\times {\text {molar mass}}=10.2mol\times 18g/mol=183.6g$

The mass of $CO_2$ and $H_2O$ produced are 336.6 g and 183.6 g respectively.

8 0

3 years ago

What is the solution to the problem rounded to the correct number of significant figures? 8.01 × 4.1 = ? A. 32.8 B. 32.84 C. 32.

kicyunya [14]

Answer: The correct option is Option D.

Explanation:

Significant figures are defined as the figures that represent the digits of a number which carry an important contribution to the numerical value, starting with the first non-zero digit. For Example: 105.268 has 6 significant figures because every digit carry its own contribution towards the numerical value.

Whenever there is multiplication, the answer will contain the same number of significant figures as there are in the least precise numerical value.

Significant figures in 8.01 are 3 and that in 4.1 are 2.

So, the answer of these two numerical value will contain 2 significant figures.

$8.01\times 4.1=32.841\approx 33$

Hence, the correct option is Option D.

7 0

3 years ago

Read 2 more answers

What does your heart do for your body?

e-lub [12.9K]

Answer:

The heart sends blood around your body. The blood provides your body with the oxygen and nutrients it needs. It also carries away waste. Your heart is sort of like a pump, or two pumps in one

3 0

3 years ago

Read 2 more answers

I need help with number 8 and fast it's due tomorrow. :)

EastWind [94]

Nuetrons+protons. and number 7 is electrons and protons

3 0

3 years ago

Using the following thermochemical data: 2Y(s) + 6HF(g) → 2YF3(s) + 3H2(g) ΔH° = –1811.0 kJ/mol 2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2

Luba_88 [7]

Answer:

ΔH° = 182.4 kJ/mol

Explanation:

The ΔH wanted is for the reaction :

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g)

This is a Hess Law problem where e will have to algebraically manipulate the first and second equations , add them together, and arrive at the desired equation above.

Notice if we reverse the first equation and divide it by 2 and add to the the second only divided by two, we will arrive to the desired equation:

2YF3(s) + 3H2(g) → 2Y(s) + 6HF(g) ΔH° = 1811.0 kJ/mol (change sign)

dividing by two :

YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1

2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2(g) ΔH° = –1446.2 kJ/mol

dividing this one by two,

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

Now adding 1 and 2

YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

________________________________________________________

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g). ΔH° = 905.5 + (-723.1) kJ/mol

ΔH° = 182.4 kJ/mol

Notice how the Y(s) and H2 cancel nicely and the coefficients are the right ones.

8 0

4 years ago