How to convert 0.000984 into scientific notation?
1 answer:
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Answer:
Mass of rock = 4533.4 kg
Explanation:
Percent composition is percentage by the mass of element present in the compound.
Given that the is 82 % by mass.
It means that 82 kg of is present in 100 kg of rock
To find the mass of rock which contains 2600 kg of iron
2600 kg = 2600*1000 g
2 moles of iron are present in 1 mole of
Molar mass of iron = 55.845 g/mol
Mass of 2 moles = 55.845 * 2 = 111.69 g/mol
Molar mass of = 159.69 g/mol
It means,
111.69 g of iron are present in 159.69 g of
1 g of iron is present in 159.69/111.69 g of
2600*1000 g is present in (159.69/111.69)*2600*1000 g of
Mass of = 3717378.47 g =3717.38 kg
Thus,
82 kg of is present in 100 kg of rock
1 kg of is present in 100/82 kg of rock
3717.38 kg of is present in (100/82)*3717.38 kg of rock
Mass of rock = 4533.4 kg
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