Answer:
2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)
Explanation:
Chlorine is a diatomic halogen gas known for its greenish-yellow colour. It has a pungent smell and is only moderately soluble in water.
It is a very reactive gas and is never found in free state in nature.
Chlorine can be prepared in the laboratory by oxidation of hydrochloric acid using KMnO4 as follows;
2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)
The set up does not need to be heated.
Answer: 44
Explanation: Because its easy
Answer:
CH3CH2CH2Cl
CH3CH2CH2CH2CH2SH
Br2
Explanation:
Dispersion forces increases with increase in relative molecular mass. The specie having the greater relative molecular mass definitely has greater dispersion forces. A rough estimation of the relative molecular masses of the species stated in the answer will reveal this fact.
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