Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
1) V(CH₄) = 0,376 L.T(CH₄) = 304 K.p(CH₄) = 1,5 atm 101325 Pa/atm = 151987,5 Pa = 151,9875 kPa.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(CH₄) = p · V ÷ R · T.n(CH₄) = 151,9875 kPa · 0,376 L ÷ 8,314 J/K· mol · 304 K.n(CH₄) = 0,0226 mol.V(CH₄) = n(CH₄) · Vm.V(CH₄) = 0,0226 mol · 22,4 dm³/mol.V(CH₄) = 0,506 dm³ = 0,506 L.
2) V(SO₂) = 5,2 L.p(SO₂) = 45,2 atm = 45,2 atm · 101,325 kPa/atm = 4579,89 kPa.T(SO₂) = 293 K.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(SO₂) = p · V ÷ R · T.n(SO₂) = 4579,89 kPa · 5,2 L ÷ 8,314 J/K· mol · 293 K.n(CH₄) = 9,77 mol.There is not enogh SO₂, 225 mol - 9,77 mol = 215,23 mol is needed.
3) p(He) = 3,50 atm · 101,325 kPa/atm = 354,63 kPa.V(He) = 4,00 L.n(He) = 0,410 mol.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.T = p · V ÷ R · n.T(He) = 354,63 kPa · 4,00 L ÷ 8,314 J/K· mol · 0,410 mol.T(He) = 416,14 K.n - amount of substance.
4) p(Ar) = 1,00 atm · 101,325 kPa/atm = 101,325 kPa.V(Ar) = 3,4 L.T(Ar) = 263 K.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(Ar) = p · V ÷ R · T.n(Ar) = 101,325 kPa · 3,4 L ÷ 8,314 J/K· mol · 263 K.n(Ar) = 0,157 mol.n(Ar) = 0,157 mol + 2,5 mol = 2,657 mol.p(Ar) = 2,657 mol · 8,314 J/K· mol · 263 K ÷ 3,4 L.p(Ar) = 1708,74 kPa.
This interaction must be one of the intermolecular forces, particularly, Van der Waals forces. From the description given, this force is called <em>induced dipole-induced dipole forces</em>. Dipole is defined as the separation of positive and negative charges. This type of intermolecular force is very weak compared to hydrogen bonding.
Answer:
✓no change in the composition of the platinum could be detected
