a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp AgBr = s²
c. 5 x 10⁻¹³ mol/L
<h3>Further explanation</h3>
Given
solubility AgBr = 7.07 x 10⁻⁷ mol/L
Required
The dissolution reaction
Ksp
The solubility product constant
Solution
a. dissolution reaction of AgBr
AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp
Ksp AgBr = [Ag⁺] [Br⁻]
Ksp AgBr = (s) (s)
Ksp AgBr = s²
c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L
The flame test is a qualitative test used in chemistry to help determine the identity or possible identity of a metal or metalloid ion found in an ionic compound. If the compound is placed in the flame of a gas burner, there may be a characteristic color given off that is visible to the naked eye. And for the proof. The flame test provided evidence that specific atoms are present in compounds by the color of the flame. The metal atoms are what is responsible for the colors during the flame test. The color of the flame will be yellow-orange because Sodium (Na) is present in all the compounds that have a yellow-orange flame. Hope this helps! Mark brainly please!
You would need exactly 50 molecules of glucose.
Answer:
The number of neutrons is entirely dependent on the Mass number of the particular atom. The standard mass for potassium is 39.
Potassium is element number 19, so it has 19 protons and 19 electrons in the neutral atom. It has therefore 39-19 = 20 Neutrons.
Explanation:
