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.
Answer:
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
Explanation:
Step 1: Data given
Mass of Al2O3 = 3.50 grams
Molar mass of Al2O3 = 101.96 g/mol
Number of Avogadro = 6.022 * 10^23 /mol
Step 2: Calculate moles Al2O3
Moles Al2O3 = mass Al2O3 / molar mass Al2O3
Moles Al2O3 = 3.50 grams / 101.96 g/mol
Moles Al2O3 = 0.0343 moles
Step 3: Calculate moles Aluminium
In 1 mol Al2O3 we have 2 moles Al
in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al
Step 4: Calculate aluminium atoms
Aluminium atoms = moles aluminium * Number of Avogadro
Aluminium atoms = 0.0686 * 6.022 * 10^23
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
Answer:
C. The species became more resistant to the insecticide.
Explanation:
The more of something you receive, the more immune you are to it. For example, every year, many people get a flu shot. Flu shots contain small amounts of the flu, and it is inserted into your blood stream so that you may become more immune to the disease when it comes around. The same happens for many living things, including insects.
Cu = 63.546
N= 14.001 g/mol
O= 15.999 g/mol * 3 = 47.997
Copper (II) Nitrate has a MW of 125.544 g/mol
6.25 x 125.544
= 784.65 <--- is your answer, if there were was a multiple choice or not :)
Answer:
C. effusion because there is a movement of a gas through a small opening into a larger volume
Explanation:
Effusion makes fluid/gas molecules move to the container with less pressure or larger volume. In diffusion, the movement should work two ways even though one side might receive more. But in effusion, the movement is rather one way.
This case shows how effusion work because its not the concentration that makes the balls moving to the bottom part of the container. No ball moving from bottom container to top either.
