Cryo-EM is used to preserve and characterize cycled positive electrodes. Under regular cycling conditions, there isn't an intimate coating layer like CEI.A small electrical short can cause a stable conformal CEI to form in place. The conformal CEI's chemistry is revealed by EELS and cryo-(S)TEM.
It has been assumed that the intimate coating layer generated on the positive electrode, known as cathode electrolyte interphase (CEI), is crucial. However, there are still numerous questions about CEI. This results from the absence of useful instruments to evaluate the chemical and structural characteristics of these delicate interphases at the nanoscale. Here, using cryogenic electron microscopy, we establish a methodology to maintain the natural condition and directly see the interface on the positive electrode.
Learn more about Cathode electrolyte interphase here:
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This combination in non polar.
Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 - 
Solving
-
=75.5 kJ/mol - 2*90.25 kJ/mol
-
=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
Answer:
XY₂Z₄
2.35 mol Z
Explanation:
A sample of the compound contains 0.221 mol X, 0.442 mol Y, and 0.884 mol Z. We can find the simplest formula (empirical formula) by <em>dividing all the numbers of moles by the smallest one</em>.
X: 0.221/0.221 = 1
Y: 0.442/0.221 = 2
Z: 0.884/0.221 = 4
The simplest formula is XY₂Z₄.
The molar ratio of X to Z is 1:4. The moles of Z in a sample that contained 0.588 moles of X is:
0.588 mol X × (4 mol Z/1 mol X) = 2.35 mol Z