Suppose that in an ionic compound, "m" represents a metal that could form more than one type of ion. in the formula mf2 , the ch
1 answer:
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Explanation:
First, calculate the moles of using ideal gas equation as follows.
PV = nRT
or, n =
= (as 1 bar = 1 atm (approx))
= 0.183 mol
As, Density =
Hence, mass of water will be as follows.
Density =
0.998 g/ml =
mass = 3.25 g
Similarly, calculate the moles of water as follows.
No. of moles =
=
= 0.180 mol
Moles of hydrogen = = 0.36 mol
Now, mass of carbon will be as follows.
No. of moles =
0.183 mol =
= 2.19 g
Therefore, mass of oxygen will be as follows.
Mass of O = mass of sample - (mass of C + mass of H)
= 3.50 g - (2.19 g + 0.36 g)
= 0.95 g
Therefore, moles of oxygen will be as follows.
No. of moles =
=
= 0.059 mol
Now, diving number of moles of each element of the compound by smallest no. of moles as follows.
C H O
No. of moles: 0.183 0.36 0.059
On dividing: 3.1 6.1 1
Therefore, empirical formula of the given compound is .
Thus, we can conclude that empirical formula of the given compound is .
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
Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).