1 mole --------------- 6.02 x 10²³ atoms
( moles iron) -------- 5.0 x 10²⁵ atoms
( moles iron ) = 5.0 x 10²⁵ x 1 / 6.02 x 10²³
moles iron = 5.0 x 10²⁵ / 6.02 x 10²³
= 83.05 moles of iron
hope this helps!
Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
B is the answer I think better sure
Answer: 10.9 mol.
Explanation:
- To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.
Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
- Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.
Using cross multiplication:
1.0 mole of PbO₂ → 2.0 moles of H₂O
5.43 moles of PbO₂ → ??? moles of water
The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.
