Answer:
1.37 × 10²³ Atoms of Mercury
Solution:
Step 1: Calculate Mass of Mercury using following formula,
Density = Mass ÷ Volume
Solving for Mass,
Mass = Density × Volume
Putting values,
Mass = 13.55 g.cm⁻³ × 3.4 cm³ ∴ 1 cm³ = 1 cc
Mass = 46.07 g
Step 2: Calculating number of Moles using following formula;
Moles = Mass ÷ M.mass
Putting values,
Moles = 46.07 g ÷ 200.59 g.mol⁻¹
Moles = 0.229 mol
Step 3: Calculating Number of Atoms using following formula;
Number of atoms = Moles × 6.022 ×10²³
Putting value of moles,
Number of Atoms = 0.229 mol × 6.022 × 10²³
Number of Atoms = 1.37 × 10²³ Atoms of Hg
The free electrons available and the electrons participating in making the bond
The percent yield for this reaction is 91%.
<u>Explanation</u>:
Ba(NO3)2 + Na2SO4 — > BaSO4 + 2NaNO3
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Assume that 0.45 mol of Ba(NO3)2 reacts with excess Na2SO4 to create 0.41 mol of BaSO4.
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Convert to grams of BaSO4 utilizing its molar mass (233.38 g/mol), which will be giving the actual yield.
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Actual Yield = 0.41 mol BaSO4 x (233.38 g BaSO4/1 mol BaSO4)
= 96 g BaSO4
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Beginning with 0.45 mol of Ba(NO3)2, compute the theoretical yield of BaSO4. Mole proportion of Ba(NO3)2 to BaSO4 is 1:1.
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0.45 mol Ba(NO3)2 x (1 mol BaSO4/1 mol Ba(NO3)2) x (233.38 g BaSO4/1 mol BaSO4) = 105 g BaSO4
% Yield = (Actual Yield/Theoretical Yield) x 100%
= (96/105) x 100%
= 91%.
