1.31 × 10⁴ grams.
<h3>Explanation</h3>
Assume that oxygen acts like an ideal gas. In other words, assume that the oxygen here satisfies the ideal gas law:
,
where
the pressure on the gas,
;
the volume of the gas,
;
the number of moles of the gas, which needs to be found;
the absolute temperature of the gas,
.
the ideal gas constant,
if P, V, and T are in their corresponding SI units: Pa, m³, and K.
Apply the ideal gas law to find
:
.
In other words, there are 410.3 moles of O₂ molecules in that container.
There are two oxygen atoms in each O₂ molecules. The mass of mole of O₂ molecules will be
. The mass of 410.3 moles of O₂ will be:
.
What would be the mass of oxygen in the container if the pressure is approximately the same as STP at
or
instead?
Answer:
0.465
Explanation:
To find the volume of a substance, divide the mass by the density.
M/D = V
10.0 / 21.5 = 0.4651163
Then round to 3 significant figures: and the density is 0.465
Since HClO4 is a strong acid it will completely dissociate in solution.
HClO4 + H2O --> H3O+ + ClO4-
Therefore, your concentration of HClO4 will be equal to your amount of H3O+, the conjugate acid in solution.
The Ka expression is as follows:
Ka= [H3O+][ClO4-] / [HClO4]
So you know that -log [H3O+] gets you pH, so just plug in [2.2] as your concentration of H3O+ and your answer should be -.34 pH
The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g
<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
- Volume = 250 mL = 250 / 1000 = 0.25 L
- Molarity = 0.3 M
Mole = Molarity x Volume
Mole of Na₂SO₄ = 0.3 × 0.25
Mole of Na₂SO₄ = 0.075 mole
<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
- Molar mass of Na₂SO₄ = 142.05 g/mol
- Mole of Na₂SO₄ = 0.075 mole
Mass = mole × molar mass
Mass of Na₂SO₄ = 0.075 × 142.05
Mass of Na₂SO₄ = 10.65 g
Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.
Learn more about molarity:
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Answer:
(O = 16) , 2H2O, and (C = 12)
Explanation: