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2 years ago

How does energy flow through the food chain?

Chemistry

2 answers:

Delicious77 [7]2 years ago

6 0

Answer:

So, energy is really all one big circle. Think of it like this:

When an organism dies, it's body is always inevitably taken back by the earth. For the purpose of the example, we'll use one specific scavenger as the main contributor here. When the body soaks back into the soil, the nutrients act as a fertilizer and aid in the growth of a patch of grass, that grass is eaten by a grasshopper, then, a frog comes along and is eaten by a snake, and eventually, that snake is eaten by an eagle. Let's say that eagle dies of natural causes, and once it passes away, the earth reclaims it's body. Now, we have a whole new fresh patch of grass and the cycle continues on, and on, and on. In this way, all energy is made eternal through the cycle of life and the food chain.

yuradex [85]2 years ago

3 0

Answer:

2nd order.

Explanation:

gras, grasshopper, frog, snake, eagle

Pleas mark branliest if you are satisfied with the answer. Thanking you in anticipation.

You might be interested in

How many particles are in 47.7 g of Magnesium? (Round the average

Lynna [10]

Answer:

1.18 × 10²⁴ particles Mg

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

47.7 g Mg

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Mg - 24.31 g/mol

Step 3: Convert

$47.7 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg} )(\frac{6.022 \cdot 10^{23} \ particles \ Mg}{1 \ mol \ Mg} )$ = 1.18161 × 10²⁴ particles Mg

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

1.18161 × 10²⁴ particles Mg ≈ 1.18 × 10²⁴ particles Mg

3 0

3 years ago

. Why a chlorine atom with an atomic number of 17 will be negatively charged when its valance shell gains an electron.

sesenic [268]

Explanation:

When chlorine bonds and becomes stable, (had a full outer electron shell), because we know that chlorine has 7 electrons in it’s outer shell, it requires 1 to gain noble gas structure. So, when if bonds with an element and then becomes stable, the gaining of an electron means it becomes a negatively charged ion. (an ion being a charged particle.)

3 0

3 years ago

The contents of two beakers are described in the table.

velikii [3]

The answer is D. The final substance in Beaker X is a mixture and in Beaker Y is a pure substance.

The question tells us that the sugar is not visible. This is because it is soluble. The sugar has dissolved into the water, making the two a mixture. The sand, on the other hand, is not dissolved at all. The water and sand are two separate pure substances.

3 0

3 years ago

Write a brief explanation about why a plant extract is a mixture

Musya8 [376]

Extraction is the crucial first step in the analysis of medicinal plants, because it is necessary to extract the desired chemical components from the plant materials for further separation and characterization

3 0

4 years ago

Considering the limiting reactant, what is the mass of manganese produced from 25.0 g of manganese(iv) oxide (86.94 g/mol) and 2

Evgen [1.6K]

Answer: 15.749g of manganese will be produced from the given reaction.

Explanation: We are given a chemical reaction:

$3MnO_2(l)+4Al(l)\rightarrow 3Mn(l)+2Al_2O_3(s)$

To know which of the reactant is the limiting reagent, we will have to first find out the moles of each reactant. Formula used for calculating the moles is:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Moles of Manganese(IV) oxide:

Given mass = 25 g

Molar mass = 86.94 g/mol

$Moles=\frac{25g}{86.94g/mol}=0.2875moles$

Moles of aluminium:

Given mass = 25 g

Molar mass = 26.982 g/mol

$Moles=\frac{25g}{26.982g/mol}=0.9265moles$

By stoichiometry of the reaction:

3 moles of manganese (IV) oxide reacts with 4 moles of aluminium.

So, 0.2875 moles of manganese (IV) oxide will react with = $\frac{4}{3}\times 0.2875$ = 0.3833 moles of aluminium.

As, the moles of aluminium is more than the required. Hence, it is the excess reagent. Therefore, manganese (IV) oxide is the limiting reagent because it limits the formation of the product.

By Stoichiometry of the reaction:

3 moles of manganese (IV) oxide produces 3 moles of manganese.

So, 0.2875 moles of manganese (IV) oxide will produce = $\frac{3}{3}\times 0.2875$ = 0.2875 moles of manganese.

To calculate the mass of manganese, we use equation 1:

Molar mass of manganese = 54.938 g/mol

Putting values in equation 1, we get:

$0.2875mol=\frac{\text{Mass of manganese}}{54.938g/mol}$

Mass of manganese = 15.794g

7 0

4 years ago