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The molarity of the potassium acetate solution given the data is 1.584 M
<h3>What is molarity? </h3>
This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:
Molarity = mole / Volume
<h3>How to determine the mole of CH₃COOK</h3>
- Mass of CH₃COOK = 19.4 g
- Molar mass of CH₃COOK = 98 g/mol
- Mole of CH₃COOK =?
Mole = mass / molar mass
Mole of CH₃COOK = 19.4 / 98
Mole of CH₃COOK = 0.198 mole
<h3>How to determine the molarity of CH₃COOK</h3>
- Mole of CH₃COOK = 0.198 mole
- Volume = 125 mL = 125 / 1000 = 0.125 L
- Molarity of CH₃COOK = ?
Molarity = mole / Volume
Molarity of CH₃COOK = 0.198 / 0.125
Molarity of CH₃COOK = 1.584 M
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Answer:
Kₐ = 6.7 x 10⁻⁴
Explanation:
First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:
HF + H₂O ⇄ H₃O⁺ + F⁻
Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]
Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.
pH = -log [ H₃O⁺ ]
1.68 = - log [ H₃O⁺ ]
Taking antilog to both sides of this equation:
10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]
[ F⁻ ] = 2.1 X 10⁻² M
Solving for Kₐ :
Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴
(Rounded to two significant figures, the powers of 10 have infinite precision )
Answer:
0.43
Explanation:
divide the given mass by molar mass from the periodic table
