Answer:
There are many equations for oxidation and reduction, but the most relevant ones are the equations for the half-reactions. In general, the equation for oxidation is:
Oxidation: Reducing Agent + O2 → Oxidizing Agent
And the equation for reduction is:
Reduction: Oxidizing Agent + e- → Reducing Agent
These equations show the transfer of electrons that occurs during oxidation and reduction.
Explanation:
Cesium.
Groups are the vertical columns that run up and down while periods are the horizontal rows. So to find the answer to this, go to the first column (Group 1) and find the sixth period (row 6) which will land you on Cesium, element 55.
Hope this helps!
The question is incomplete , complete question is:
Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:
Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.
Answer:
The ΔH of the reaction is -626 kJ/mol.
Explanation:
We are given with:
ΔH = (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)
The ΔH of the reaction is -626 kJ/mol.
Answer:
Positive; negative
Explanation:
When electrons are lost, a positive ion is formed. A positive ion is called a cation.
When electrons are gained, a negative ion is formed. A negative ion is called an anion.
Answer:
5 × 10^-4 L
Explanation:
The equation of the reaction is;
2KClO3 = 2KCl + 3O2
Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles
From the stoichiometry of the reaction;
2 moles of KClO3 yields 3 moles of O2
0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas
From the ideal gas equation;
PV= nRT
P= 85.4 × 10^4 KPa
V=?
n= 0.165
R= 8.314 J K-1 mol-1
T= 40+273 = 313K
V= 0.165 ×8.134 × 313/85.4 × 10^4
V=429.4/85.4 × 10^4
V= 5 × 10^-4 L
