Solution:
The molecular formula is PbSO₄, indicating lead sulfate
Option c.
Explanation:
The percentage makeup shows that in 100 g of this compound, there are:
68.3 g of Pb, 10.6 g of S, and (100 - 68.3 - 10.6) = 21.1 g of O
To find the moles of each element, we divide by their molar masses:
68.3 g Pb / 207.2 g/mol = 0.329 moles Pb
10.6 g S / 32.06 g/mol = 0.331 moles S
21.1 g O / 16 g/mol = 1.32 moles O
Next, we find the mole ratio by dividing each by the smallest number of moles:
0.329 / 0.329 = 1 Pb
0.331 / 0.329 = 1 S
1.32 / 0.329 = 4 O
Thus, the molecular formula is PbSO₄, representing lead sulfate.
Cu(NO3)2 --> MM187.5558
NiNO3 *COEF2* --> 120.6983
(a)
Write the balanced half-reactions for the overall process:
Oxidation: Se^2- (aq) → Se
(s) + 2e-
Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- →
S2O3^2- + 6OH- (aq)
(b)
Assuming E sulfite is 0.57 V, compute E selenium:
E anode = E cathode – E cell
= -0.57 – 0.35
=
-.092
Answer: The number of sulfur dioxide molecules present is 1.27·10²³.
Calculating: m(SO₂) equals 13.5 g.
Using the formula n(SO₂) = m(SO₂) ÷ M(SO₂).
This gives n(SO₂) = 13.5 g ÷ 64 g/mol.
Resulting in n(SO₂) = 0.21 mol.
Subsequently, N(SO₂) = n(SO₂) ·Na.
Therefore, N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
Ultimately, N(SO₂) equals 1.27·10²³.
Where n represents amount of substance.
M refers to molar mass.
Na is Avogadro's number.