Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
Answer:
A
Explanation:
A formal charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity(Wikipedia).
The formal charge on an atom in a molecule reflects the electron count associated with the atom compared to the isolated neutral atom(University of Calgary).
Looking at all the structures listed A-E for SeO2F2, only structure A minimizes the formal charges for each atom in SeO2F2.
240 g NaOH
<em>Step 1</em>. Calculate the moles of NaOH
Moles of NaOH = 0.750 L solution × (8 mol NaOH/1 L solution) = 6 mol NaOH
Step 2. Calculate the mass of NaOH
Mass of NaOH = 6 mol NaOH × (40 g NaOH/1 mol NaOH) = 240 g NaOH
B) calciummmmmmmmmmmmmmmm
Answer:
See explanation
Explanation:
The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;
Bismuth;
[Xe]4f14 5d10 6s2 6p3
Chromium;
[Ar]4s1 3d5
Strontium;
[Kr]5s2
Phosphorus;
[Ne]3s2 3p3
2.
Bi
6p- n=6, l= 1, ml= 1, ms= 1/2
Cr
3d- n=3, l=2, ml=2,ms=1/2
Sr
5s- n=5, l=0, ml=0, ms=1/2
P
3p- n=3, l= 1, ml= 1, ms=1/2
3.
a) Tin (Sn) - [Kr] 5s2 4d10 5p2
b) Caesium (Cs)- [Xe] 6s1
c) Copper (Cu)- [Ar] 4s1 3d10
