It isn't a materialistic object.
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,
CO + 2 H₂ → CH₃OH
Calculating Moles of CO:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO
Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g
X = 11.25 Moles of CO
Calculating Moles of H₂:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂
Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g
X = 22.5 Moles of H₂
Result:
3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
Answer:
Moles of H₂S needed = 6.2 mol
Moles of SO₂ produced = 6.2 mol
Explanation:
Given data:
Number of moles of O₂ = 9.3 mol
Moles of H₂S needed = ?
Moles of SO₂ produced = ?
Solution:
Chemical equation:
2H₂S + 3O₂ → 2SO₂ + 2H₂O
Now we will compare the moles of oxygen with H₂S.
O₂ : H₂S
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
Now we will compare the moles of SO₂ with both reactant.
O₂ : SO₂
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
H₂S : SO₂
2 : 2
6.2 : 6.2 mol
So 6.2 moles of SO₂ are produced.
The color emitted be larger atoms is lower in energy then the light emitted by smaller atoms
B because inner core has metal and it’s solid
