Answer:
- 10.555 kJ/mol.
Explanation:
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).
∆H°rxn is the standard enthalpy change of the reaction (J/mol).
T is the temperature of the reaction (K).
∆S°rxn is the standard entorpy change of the reaction (J/mol.K).
∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants
<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.
∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants
<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>
<em></em>
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>
This may seem confusing because they give you two masses, but all you have to do is pick one to do the calculations. Personally, I would pick O2, since the molar mass is easier to calculate. The answer would be 3.3 g (rounded for sig figs). To get this, first take the 5.9 grams of O2 and convert it to moles by dividing by the molar mass of oxygen gas, which is 32. Then, multiply both by the mole-mole ratio, which is 2:2, or simply 1:1. After that, multiply that by 18g, which is the molar mass of water to get grams of water.
REMEMBER, you have to write and balance the chemical equation before you can do any of that work.
That happens to be CH4 + 2O2 => CO2 + 2H2O
Here are some disadvantages, is that nitrogen dioxide is a toxic gas and it can still be harmful when ingested by human, also critics of hydrogen fuel cells argue that although these cells do not emit carbon after burning, they give out nitrogen dioxide and other emissions.
Hope this helps
Answer:
53 meters = 53000 millimeters
Explanation:
In this question we have to convert meters into millimeters .
By metric conversion,
Since, 1 meter = 1000 mm
Therefore, 53 meters = 53 × 1000
= 53000 millimeters
53 meters = 53000 millimeters is the answer.
and 
.
Assuming complete decomposition of both samples,
First compound:
; 
of the first compound would contain
Oxygen and mercury atoms seemingly exist in the first compound at a 
ratio; thus the empirical formula for this compound would be where the subscript "1" is omitted.
Similarly, for the second compound
; 
of the first compound would contain
and therefore the empirical formula
.
