Answer:
ΔU=-369.2 kJ/mol.
Explanation:
We start from the equation:
Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.
If Δ(PV) were calculated by ideal gas law,
PV=nRT
Δ(PV)=RTΔn.
Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.
So, for this case, ΔH=ΔU.
The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:
ΔU=-369.2 kJ/mol.
Answer:
Acetylene: -1,256 kJ/mol
Ethanol: -1,277 kJ/mol
The combustion of 0.25 mol of an unknown organic compound results in the release of 320 kJ of energy. Which of the compounds in the table could be the unknown compound?- Answer: Ethanol
Answer:
D
Explanation:
The fact is that the both elements belong to different groups in the periodic table. Mg is in group 12 while Al is in group 13. The outer most electron configuration for Mg2+=[Ne]3s0
Al+= [Ne]3s13p0
It is evident that in Al+, the second electron is to be removed from a filled 3s subshell which is energetically unfavourable. Unlike In Mg2+ where the two electrons are removed. There is always a tendency towards a quick loss of all the valence electrons in the valence shell. It will be more difficult to remove an electron from a filled shell.
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
3.25 kg in g = 3.25 * 1000 = 3250 g
Molar mass C₂H₆O₂ = 62.0 g/mol
Mass solvent = 7.75 kg
Number of moles:
n = mass solute / molar mass
n = 3250 / 62.0
n = 52.419 moles
Molality = moles of solute / kilograms of solvent
M = 52.419 / 7.75
M = 6.7637 mol/kg
hope this helps!
