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crimeas [40]
3 years ago
7

What is the oxidation state of each element in febr2?

Chemistry
2 answers:
Sauron [17]3 years ago
5 0
<span>To determine the oxidation state of a metal, like Fe (Iron) or anything else really, you generally won't know. But look at a periodic table and you'll see that most metals fall into the sort of "valley" section, and this part of the periodic table is called the "d-block" and it has 10 columns. These guys in the d-block of the periodic table have a special way of filling their orbitals, but everything on the left and right are a little simpler to understand. So we see that Br (Bromine) is one of the easier ones to find the oxidation state of, since it's on the right part of the periodic table (where there are 6 columns) and we know that they follow the octet rule, and since bromine is a halogen (That just means it's in the second to last column, which is next to the noble gasses) and things at the top right of the periodic table are highly electronegative, so they'll tend to gain another electron to have a complete octet to fill its orbitals and have a negative charge. Since it only had to gain one to have a complete octet, it's charge is now -1. We can now look at FeBr2 and see that overall the charge is 0. But it has two bromines which each have a charge of -1, so we see that we can set up a little algebra equation: (Charge of Iron) + (Charge of Bromines)=(Total Charge) So, we have: (Charge of Iron) + 2 (-1)=0 So then we can see that Iron is a +2 oxidation state, Iron (II) Bromide.     
</span>
Hope that helps 
Happy Studying!

Oksi-84 [34.3K]3 years ago
4 0
FeBr2 
1*z + 2*(-1) = 0 
z = 2 is the o<span>xidation state of each element in FeBr2</span>
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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

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Na+ = 0.01M

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pOH = 2

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