Answer:
Explanation:
H₂SO₄ is a strong acid, which means that most of it ionizes in aqueous solution.
Since it is a diprotic acid (two hydrogen ions) its ionization occurs in two steps:
- H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
- HSO₄⁻ (aq) → H⁺(aq) + SO₄²⁻(aq)
Thus, almost all H₂SO₄ has ionized and its final concentration is almost nothing.
After the first ionization, the conentrations of H⁺(aq) and HSO₄⁻ are equal but by the second ionization more H⁺ ions are produced along with SO₄⁻.
You can show it as one step dissociation, assuming 100% dissociation (given this is a strong acid):
By the stequiometry you can build this table:
H₂SO₄ (aq) → 2H⁺(aq) + SO₄²⁻(aq)
Initial A 0 0
Change - x +2x +x
Equilibrium A - x 2x x
As explained, A - x is very low, and 2x is twice x. Thus,
The rank of the concentrations from highest to lowest is:
States that the properties of elements are periodic or recurring and are correlated to their atomic number.
After one half-life, one half is remaining. After two half-lives one fourth is remaining. So that would be 25%
The answer is, 699g
The idea here is that you need to use the mole ratio<span> tha exists between </span>ferric oxide<span>, </span><span><span>Fe2</span><span>O3</span></span><span>, and iron metal, </span>Fe<span>, to determine how many moles of the latter will be produced when </span>all the given mass<span> of the ferric oxide reacts...
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