69.9%
Explanation:
To find the mass percentage of iron in the compound in Fe₂O₃, we would go ahead to express the given molar mass of the iron to that of the compound.
Mass percentage = x 100
Molar mass of Fe = 55.85g/mol
Molar mass of O = 16g/mol
Molar mass of Fe₂O₃ = 2(55.85) + 3(16) = 159.7g/mol
Mass percentage = = 69.94% = 69.9%
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The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
mass of CO₂ formed is 591.8 g
The image represents A COMPOUND because the molecules are BONDED CHEMICALLY.
A compound is a substance formed when two or more elements combine together chemically. In the process of chemical combination, the chemical bonds that were present in the participating elements will be broken and new chemical bonds will be formed in the product.
Answer:
V = 11.21 L
Explanation:
Given data:
Volume of helium = ?
Number of moles = 0.500 mol
Temperature = 273.15 K
Pressure of gas = 1 atm
Solution:
Formula:
PV = nRT
R = general gas constant = 0.0821 atm.L/ mol.K
V = nRT/P
V = 0.500 mol × 0.0821 atm.L/ mol.K × 273.15 K / 1 atm
V = 11.21 L / 1
V = 11.21 L