<span>26.833 liters
Aluminum oxide has a formula of Al</span>₂O₃,<span> which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).
Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.
Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.
Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815
Finally, you need to calculate the volume of </span>1.1815 <span>moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,
1.1815 * 22.7 = </span>26.8 liters <span>of oxygen gas.
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To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy.
The amount of energy required by an isolated, gaseous molecule in the electronic state of the ground to absorb in order to discharge an electron and produce a cation has been known as the ionization energy. The amount of energy required for every atom in a mole to drop one electron is most often given as kJ/mol.
Anything that causes electrically neutral atoms and molecules to gain or lose electrons in order to become electrically charged atoms as well as molecules .
Therefore, the "To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy."
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Answer:
<h3>The answer is 235.29 mL</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question.
density = 0.017 g/mL
mass = 4 g
We have
We have the final answer as
<h3>235.29 mL</h3>
Hope this helps you
