If you are given the
standard potential for the reduction of X^2+ is +0.51 V, and the standard
potential for the reduction of A^2+ is -0.33, just add the two. The standard
potential for an electrochemical cell with the cell is 0.18V
342.15 g/mol is the molar mass of Al2(SO4)3 Aluminium sulfate, This is what I found I hope this is right. Hope this helps;)
We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
1 mm : 10⁻³ m
We require a conversion from m³ to mm³, so we must take the cube of the ratio we have made:
1 mm³ = (10⁻³)³ m³
Therefore, the conversion used will be:
(1 mm / 10⁻³ m)³
When we multiply by this conversion, we will get:
32 m³ = 32 x 10⁹ mm³
Answer:
Soap decreases surface tension by changing the way water behaves on the surface. Hard and soft water react differently when soap is added to them.
Explanation: i think it is this
The mass of ammonium chloride that must be added is : ( A ) 4.7 g
<u>Given data :</u>
Volume of water ( V ) = 250 mL = 0.25 L
pH of solution = 4.85
Kb = 1.8 * 10⁻⁵
Kw = 10⁻¹⁴
Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :
NH₄CI + H₂O ⇄ NH₃ + H₃O⁺
where conc of H₃O⁺
[ H₃O⁺ ] =
and Ka = Kw / Kb
∴ Ka = 5.56 * 10⁻¹⁰
Next step : Determine the concentration of H₃O⁺ in the solution
pH = - log [ H₃O⁺ ] = 4.85
∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵
Next step : Determine the concentration of NH₄CI in the solution
C = [ H₃O⁺ ]² / Ka
= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰
= 0.359 mol / L
Determine the number of moles of NH₄CI in the solution
n = C . V
= 0.359 mol / L * 0.25 L = 0.08979 mole
Final step : determine the mass of ammonium chloride that must be added to 250 mL
mass = n * molar mass
= 0.08979 * 53.5 g/mol
= 4.80 g ≈ 4.7 grams
Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g
Learn more about ammonium chloride : brainly.com/question/13050932