Isotopes of any given factor all incorporate the equal variety of protons, so they have the identical atomic wide variety (for example, the atomic wide variety of helium is usually 2). Isotopes of a given factor include exceptional numbers of neutrons, therefore, special isotopes have special mass numbers.
Answer: Solubility
Explanation:
Solubility is the ability or property of a solute which can be a solid, liquid, or gaseous chemical substance to dissolve in a solvent (solid, liquid or gas) and form a solution. The factors that affect how soluble a substance is depends on the solvent ,temperature and pressure.
Here, Mario trying to dissolve certain substances (solutes) in water (the solvent shows he is trying to find out the solubility of the substances which can be insoluble, sparingly soluble or very soluble.
<u>Answer:</u> The correct option is Option D.
<u>Explanation:</u>
Significant figures are defined as the figures that represent the digits of a number which carry an important contribution to the numerical value, starting with the first non-zero digit. For Example: 105.268 has 6 significant figures because every digit carry its own contribution towards the numerical value.
Whenever there is multiplication, the answer will contain the same number of significant figures as there are in the least precise numerical value.
Significant figures in 8.01 are 3 and that in 4.1 are 2.
So, the answer of these two numerical value will contain 2 significant figures.
Hence, the correct option is Option D.
Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³
