H20. 2 of hydrogen and oxygen
Sodium Hydroxide (NaOH) is also known as lye which is a base (very high ph; Alkaline)
Now, in chemistry, equilibrium is what affects the reaction rate of a reaction. If they are in equilibrium, the concentrations of them will not change (both reactants and products).
Now, lets say that to synthesize a certain chemical, we need it to be in an acidic environment with HCL or some other acid as the catalyst for the reaction.
Well, if we were to add Sodium Hydroxide to this which is very alkaline, the ph would change greatly which affects the reaction rate. If we do not have enough energy to overcome the activation barrier, the reaction will not occur (atleast for a very long time).
However, a common mistake is thinking that a catalyst will affect the equilibrium. This is not true. The reaction will still take place but it will have a very slow reaction rate.
TLDR; Adding a catalyst (like NaOH or Sodium Hydroxide) will not change the equilibrium but instead change the reaction rate. The reaction can still occur, although it can take a very, very long time (like diamonds turning into graphite)
410g Ag
2.3*10^24 atoms
1 molcule Ag- 6.02g*10^3
Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
