What does the subscript 2 indicate in the formula o2?

1 answer:

6 0

The subscript 2 in the formula o2 indicates that there are 2 atoms of the element oxygen, represented by the symbol "O." The subscript to the right of the symbol tells how many atoms are present of that element.

You might be interested in

The uses of sodium phosphate include clarifying crude sugar, manufacturing paper, removing boiler scale, and washing concrete. W

garri49 [273]

2.4 × 10⁻² is the pH of a solution

What is the short definition of pH?

The pH scale determines how acidic or basic water is. The range is 0 to 14, with 7 representing neutrality. Acidity is indicated by pH values below 7, whereas baseness is shown by pH values above 7.
In reality, pH is a measurement of the proportion of free hydrogen and hydroxyl ions in water.

Equation representing the i onization of Na₃PO₄ is as fallow

Na₃PO₄ → PO₄³⁻ + 3Na⁺

Concentration of Na₃PO₄ = $\frac{26.9 * \frac{1 mol}{164 g} }{1 L} = 0.164 M$

Since 1 mol of Na₃PO₄ forms 1 mol of PO₄³⁻ [ PO₄³⁻ ] = [ Na₃PO₄ ] = 0.164 M

Equation representing the i onization of HPO²⁻₄ is as fallow

HPO²⁻₄ → PO₄³⁻ + H⁺

Acidionization constant Ka₃ of HPO²⁻₄ = 4. 17 × 10⁻¹³

Baseionization constant Kb₃ of PO₄³⁻ = Ka₃/ Kb₃

= 1 .0 × 10⁻¹⁴/ 4. 17 × 10⁻¹³ = 2.4 × 10⁻²

Learn more about pH

brainly.com/question/15289741

#SPJ4

4 0

1 year ago

If 17. 6 g of hcl are used to produce a 14. 5 l solution, what is the ph of the solution?.

kati45 [8]

This problem is providing us with the mass of hydrochloric acid and the volume of solution and asks for the pH of the resulting solution, which turns out to be 1.477.

<h3>pH calculations</h3>

In chemistry, one can calculate the pH of a solution by firstly obtaining its molarity as the division of the moles of solute by the liters of solution, so in this case for HCl we have:

$M=\frac{17.6g*\frac{1mol}{36.46g} }{14.5L} \\\\M=0.0333 M$

Next, due to the fact that hydrochloric acid is a strong acid, we realize its concentration is nearly the same to the released hydrogen ions to the solution upon ionization. Thereby, the resulting pH is:

$pH=-log(0.0333)\\\\pH=1.477$