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gogolik [260]
3 years ago
6

What is Darwin's name for species that do not appear to have changed for millions of years?

Chemistry
1 answer:
ioda3 years ago
4 0

Since, the options have not been given the question is incomplete.

What is Darwin's name for species that do not appear to have changed for millions of years?

a.

Dinosaurs

b.

Living fossils

c.

Old souls

d.

Ancient moderns

Answer: b. Living fossil

Explanation:

In 1859 Charles Darwin proposed the term living fossil, that means a species or group of species that had not changed in terms of evolutionary context thus can be useful in tracing the extinct or previously existing forms of life. The examples of the living fossils are horseshoe crabs, ginkgo (Conifers) and tuatara. These group of animals were existed unchanged in the Ordovician, Permian, and Triassic periods respectively with few surviving species.

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The following reaction is an example of what type of equilibrium reaction?
Mrac [35]

Hello, Brainly User!

A heterogeneous equilibrium reaction is when there are different phases in the reaction. An equilibrium reaction is a reaction that can go in both the forward and reverse directions, and the concentrations of the reactants and products remain the same.

So therefore, your answer is A. Heterogeneous!

Hope i helped and i was honored to answer your question!

Have a great day..

-Philaeagles14

6 0
3 years ago
What happens to the electron configuration when you move down a family?
stiv31 [10]

Answer:

The number of energy levels increases as you move down a group as the number of electrons increases.

Explanation:

Each subsequent energy level is further from the nucleus than the last. Therefore, the atomic radius increases as the group and energy levels increase. 2) As you move across a period, atomic radius decreases.

4 0
3 years ago
The volume of a fixed amount of gas is double and the absolute temperature os doubled according to the ideas gas law how has the
olya-2409 [2.1K]
Given that <span>P=<span><span>nRT/</span>V</span></span><span>, for an ideal gas, the pressure would be unchanged.</span>
5 0
3 years ago
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

brainly.com/question/5976752

brainly.com/question/13707449

brainly.com/question/13707449

brainly.com/question/6263007

brainly.com/question/14641878

brainly.com/question/2912965

#SPJ1

7 0
2 years ago
QUESTION 1
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B. Electricity
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