Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Explanation:
When magnesium metal burns is heated i the air it forms magnesium oxide.The balanced chemical reaction is given as:
2 moles of magnesium metal when reacts with 1 moles of oxygen it gives 2 moles of magnesium oxide which is white in color.
Some times along with formation of magnesium oxide small amount of magnesium nitride also produced due to which magnesium oxide appears grey in color .The balanced chemical reaction is given as:
3 moles of magnesium combines with 1 mol of nitrogen gas to to give 1 mol of magnesium nitride.
The average atomic weight is calculated by adding up the products of the percentage abundance and atomic weight. In this item, we have the equation,
A = (0.412)(21.016 amu) + (0.5012)(21.942 amu) + (0.0868)(23.974 amu)
Simplifying the operation will give us the answer of 21.74 amu.
<em>Answer: 21.74 amu</em>
Answer:
Fructose
Explanation:
A ketose is a monosaccharide containing one ketone group (RCOR) per molecule.
This means to identify a ketose, one has to look for the ketone group in the structural formuar of the compound.
Mannose - This is an aldose. It contains the aldehyde functional group.
Galactose - This is an aldose. It contains the aldehyde functional group.
Fructose - This is a keose. It contains the Ketone Functional group.
Glucose - This is an aldose. It contains the aldehyde functional group.
