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jeka57 [31]
3 years ago
8

What is the ground-state electron configuration of a neutral atom of cobalt?

Chemistry
1 answer:
Andre45 [30]3 years ago
8 0
4d7 , 3s2 will be the valence shell electronic configuration of a neutral atom of cobalt!
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A plant living in the environment shown in the picture would have to be adapted not to loose to much water through its___
Ray Of Light [21]

Answer:

Many plants have thorns on their stems or leaves. What is the MOST likely explanation for the evolution of thorns?

A) Thorns help plants produce more food from photosynthesis.

B) Thorns are an example of a mutation that arises in the genetic code of plants.

C) Thorns help plants to conserve resources like water and soil nutrients that may be used by other organisms.

D) Thorns are an adaptation that some plants have evolved in order to discourage herbivores from eating the plant.

2)

Explanation:

5 0
2 years ago
Read 2 more answers
What is the difference between the number of electrons in an atom of selenium?
azamat

Explanation:

Supposing we have to find the difference between the number of electrons in Selenium and Aluminium.

Well, electrons can be transformed to an atomic number, so if SE is 34, that means it has 34 electrons. AI has an atomic number of 13, which means it has 13 electrons. So the difference is that more electrons than AI are available from SE.

3 0
2 years ago
Which material has carbon atoms interspersed with iron atoms in its molecular structure?
gogolik [260]

Answer:

steel should be correct

4 0
2 years ago
The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0
3 years ago
Identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent for the following equation
bazaltina [42]

The element that gains electrons, becomes reduced.

While the one which loses electrons, becomes oxidized.

In this equation,

CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.

By balancing the equation, we will get:

3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O

Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.

Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.

So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .

4 0
3 years ago
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