Which statement describes the general trends in electronegativity and first ionization energy as the elements in Period 3 are co
2 answers:
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Answer:
0.185moles of Al₂O₃
Explanation:
Mass of Al = 10g
Mass of O₂ = 19g
Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃
This is the balanced reaction equation.
Solution
From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.
Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.
Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:
Number of moles of Al =
=
= 0.37mol
From the equation:
4 moles of Al produced 2 moles of Al₂O₃
0.37 mole will yield: = 0.185moles of Al₂O₃
Explanation:
The given data is as follows.
T = = (120 + 273.15)K = 393.15 K,
As it is given that it is an equimolar mixture of n-pentane and isopentane.
So, = 0.5 and
= 0.5
According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.
(393.15 K) = 9.2 bar
(393.15 K) = 10.5 bar
Hence, we will calculate the partial pressure of each component as follows.
=
= 4.6 bar
and,
=
= 5.25 bar
Therefore, the bubble pressure will be as follows.
P =
= 4.6 bar + 5.25 bar
= 9.85 bar
Now, we will calculate the vapor composition as follows.
=
= 0.467
and,
=
= 0.527
Calculate the dew point as follows.
= 0.5,
= 0.5
= 0.101966
P = 9.807
Composition of the liquid phase is and its formula is as follows.
=
= 0.5329
=
= 0.467
Answer:
50 mg
Explanation:
First, we have to calculate the partial pressure of O₂ (pO₂) using the following expression.
pO₂ = P × X(O₂) = 1.13 atm × 0.21 = 0.24 atm
where,
P: total pressure
X(O₂): mole fraction of oxygen
Then, we can calculate the concentration of O₂ in water (C) using Henry's law.
C = k × pO₂ = (1.3 × 10⁻³ M/atm) × 0.24 atm = 3.1 × 10⁻⁴ M
where,
k: Henry's constant for O₂
The mass of oxygen in a 5.00 L bucket with a concentration of 3.1 × 10⁻⁴ M is: (MM 32.0 g/mol)