Probably very negative or sightless negative
Answer:
The answer is: +7
Explanation:
Oxidation state or oxidation number of an element is the hypothetical charge on an element that forms completely ionic bonds. The oxidation number represents the number of electrons lost or gained by that element.
Perchlorate ion is a molecule with a chemical formula: ClO₄⁻
The oxidation state of oxygen in ClO₄⁻ = -2,
the total charge on the ClO₄⁻ molecule = -1,
let the oxidation state of chlorine be x
<u><em>As the sum of oxidation states of all elements in a molecule is equal to the total charge on the molecule.</em></u>
⇒ oxidation state of chlorine + oxidation state of oxygen × 4 = total charge on the molecule
⇒ x + (-2) × 4 = -1
⇒ x + (-8) = -1
⇒ x = -1 + 8 = +7
⇒<u> </u><u>x = +7</u>
<u>Therefore, the oxidation state of chlorine in the perchlorate ion (ClO₄⁻): x = +7</u>
Answer:
Ka = 1.39x10⁻⁶
Explanation:
A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
<em>Where Ka is:</em>
Ka = [H⁺] [X⁻] / [HX]
<em>Where [] is the molar concentration in equilibrium of each specie.
</em>
The equilibrium is reached when some HX reacts producing H+ and X-, that is:
[HX] = 1.64M - X
[H⁺] = X
[X⁻] = X
As pH is 2.82 = -log [H⁺]:
[H⁺] = 1.51x10⁻³M:
[HX] = 1.64M - 1.51x10⁻³M = 1.638M
[H⁺] = 1.51x10⁻³M
[X⁻] = 1.51x10⁻³M
And Ka is:
Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]
<h3>Ka = 1.39x10⁻⁶</h3>
Answer:
Average density for method A = 2.4 g/cm³
Average density for method B = 2.605 g/cm³
Explanation:
In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):
Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6
Average = 9.6/4 = 2.4 g/cm³
Σ = 2.603 + 2.601 + 2.605 + 2.611 = 10.420
Average = 10.420/4 = 2.605 g/cm³
