Answer:
8.45 moles are produced
Explanation:
CaCl₂ + Na₂CO₃ → CaCO₃ + 2 NaCl
From the equation, we can see that for every 1 mole of CaCl₂ and 1 mole Na₂CO₃ will give 1 mole of CaCO₃ and 2 moles of NaCl
to calculate how many moles of CaCO₃ ,we simply multiply multiply each by the 8.45 moles of CaCl₂ which will reacts
these is because for every 1 mole of CaCl₂ and 1 mole Na₂CO₃ will give 1 mole of CaCO₃ and 2 moles of NaCl
therefore we have every 1x8.45(8.45) mole of CaCl₂ and 1x8.45(8.45) mole Na₂CO₃ will give 1x8.45(8.45) mole of CaCO₃ and 2x8.45(16.9) moles of NaCl
8.45 moles are produced in the reaction
The atomic mass of the element would simply be equal to
the sum of the weighted average of each isotope, that is:
atomic mass = 59.015 amu * 0.717 + 62.011 amu * (1 – 0.717)
<span>atomic mass = 59.863 amu</span>
Answer:
the answer is distillation
The answer to your question is,
An Arrhenius acid is a type of substance that separates in water to form Hydrogen ions.
-Mabel <3
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