Answer:
2 NO (g) → N2 (g) + O2 (g)
2 NOCl (g) → 2 NO (g) + Cl2 (g)
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2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)
ΔH = [90.3 kJ x 2 x -1] + [-38.6 kJ x -1 x 2] = -103.4 kJ
The ΔH for the reaction is -103.4 kJ
Answer:
D. an orbital notation of the atom
Explanation:
Orbital notiation uses lines and arrows to show shells, subshells, and orbitals for electrons in an atom. Since it shows arrows being paired up in this diagram it would be the best model for Chuck to use.
To know the answer, you either know what is really the
nature and chemistry of a sugar solution. You can also know the answer by
knowing the meaning of entropy. Entropy is often interpreted as the degree of
disorder or randomness in the system. So the correct statement is that the
system becomes more disordered and has an increase in entropy.
Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

Regards!
Although you have not provided the circled electron, I can help you with a wide explanation.
1) Atomic number of manganese is 25. That means that it has 25 protons and 25 electrons.
2) Those 25 electrons are distributed (electron configuration) as per the quantum rules:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
3) The most reasonable is that you have been asked to give the possible quantum numbers for an electron in the 4s or 3d.
4) Those are 7 electrons and these are their possible sets of quantum numbers:
i) For the two electrons in 4s:
n is the main energy level so n = 4
l tells the kind of orbital, which is s, so l = 0
ml is also 0 (it can be from -l to + l, so given that l i s0, ml is 0)
ms: one is +/12 and the other is -1/2 (this is the spin number).
ii) For the 5 electrons in 3d
n = 3
l can be 0, 1, or 2
if l = 0, then ml = 0
if l = 1, then ml can be -1, 0 , or 1 (from - l to + l)
ms can be either +1/2 or - 1/2 (spin)