Answer:
[O₂(g)] = 0.0037M
Explanation:
2SO₂(g) + O₂(g) => 2SO₃(g)
Conc: [SO₂(g)] [O₂(g)] [SO₃(g)] and [SO₂(g)] = [SO₃(g)]
Kc = [SO₃(g)]²/[O₂(g)][SO₂(g)]² => Kc = 1/[O₂(g)] = 270 if [SO₂(g)] = [SO₃(g)]
∴ [O₂(g)] = (1/270)M = 0.0037M
Answer:
HNO₃
Explanation:
Data given
Nitrogen = 9.8 g
Hydrogen = 0.70 g
Oxygen = 33.6 g
Empirical formula = ?
Solution:
Convert the masses to moles
For Nitrogen
Molar mass of N = 14 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 9.8 g/ 14 g/mol
no. of mole = 0.7
mole of N = 0.7 mol
For Hydrogen
Molar mass of H = 1 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 0.70 g/ 1 g/mol
no. of mole = 0.7
mole of H = 0.7 mol
For Oxygen
Molar mass of O = 16 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 33.6 g / 16 g/mol
no. of mole = 2.1
mole of O = 2.1 mol
Now we have values in moles as below
N = 0.7
H = 0.7
O = 2.1
Divide the all values on the smallest values to get whole number ratio
N = 0.7 / 0.7 = 1
H = 0.7 / 0.7 = 1
O = 2.1 / 0.7 = 3
So all have following values
N = 1
H = 1
O = 3
So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.
Group 2 because the outer electrons ( also known as valence electrons) are 2 and that’s how we know which group it is in
A. the number dots shown on the lewis dot diagrams for the element in the period
I believe what you are trying to ask here is the balanced chemical reaction. We first identify the chemical formula of the substances.
Lead(II) sulfide = PbS
Oxygen = O2
Lead(II) oxide = PbO
Sulfur dioxide = SO2
The balanced chemical reaction is written as:
PbS + 3/2O2 = PbO + SO2
