B and c...will lose electron(s) in forming an Ion.
P is an Anion
b..Fe. and c...Pb form Cations (+) by losing electrons.
d. Se is an Anion.
The balanced combustion reaction of propane, C₃H₈, is
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Molar mass of propane: 44 g/mol
Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane
Molar mass of oxygen: 32 g/mol
Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen
Moles of oxygen needed to completely react propane:
0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen
Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.
Molar mass of water: 18 g/mol
Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)
Mass of water produced = 258.768 grams
N₂ + 3H₂ ⇒ 2NH₃
doesnt matterN₂ + 6.64H₂ ⇒ 2NH₃
(6.64H₂/3H₂) x (2NH₃) =4.4266667
rounded to sig figs= 4.43
One mole of a substance contains 6.02 × 10∧23 particles. Thus we first convert 89.2 g to moles. 1 mole of sodium contains 23 g
Hence 89.2 g = 89.2 / 23 g = 3.878 moles
Therefore, 3.878 × 6.02×10∧23 particles= 23.346 × 10∧23 particles
Hence 89.2 g of sodium contains 2.335 ×10∧24 particles
2KClO₃ → 2KCl + 3O₂
mole ratio of KClO₃ to O₂ is 2 : 3
∴ if moles of O₂ = 5 mol
then moles of KClO₃ =
= 3.33 mol
Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃
= 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol
= 407.93 g
