All categories

2 years ago

Can y’all please help me answer?

Chemistry

1 answer:

valentina_108 [34]2 years ago

5 0

Answer:

It really depends on the situation. If you have 1000 elk and 2 wolves, they really would play a better role in killing off elk so that the elk population doesn't get over populated and diseased, causing a major population collapse. But if there were only 10 elk and 50 wolves, the wolves would play a very negative role on the population.

Hope this helps!

Explanation:

You might be interested in

CO(g) + 12 O2(g) → CO2(g)The combustion of carbon monoxide is represented by the equation above.(a) Determine the value of the s

devlian [24]

Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The combustion of $CO$ will be,

$CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)$ $\Delta H_1=-110.5kJ/mol$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ/mol$

Now we are reversing reaction 1 and then adding both the equations, we get :

(1) $CO(g)\rightarrow C(s)+\frac{1}{2}O_2(g)$ $\Delta H_1=110.5kJ/mol$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ/mol$

The expression for enthalpy change for the reaction will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2$

$\Delta H_{rxn}=(110.5)+(-393.5)$

$\Delta H_{rxn}=-283kJ/mol$

Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

6 0

4 years ago

Location E is located at the Tropic of Cancer, while location F is at the equator. Which location is likely to be warmer at the

Stells [14]

The answer is "Location E, because Earth's axis is tilted towards the sun" :D

6 0

2 years ago

Which scientist developed the 1st model of the atom that showed the structure of the inside of an atom?

frez [133]

C. Rutherford (1911)

7 0

3 years ago

Which does not affect chemical equilibrium

Amanda [17]

An inert gas will not react with either the reactants or the products, so it will have no effect on the product/reactant ratio, and therefore, it will have no effect on equilibrium.

5 0

3 years ago

Read 2 more answers

A sample of 23.2 g of nitrogen gas is reacted with

slavikrds [6]

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

5 0

3 years ago