Step 1:
Divide mass of each element with its M.mass in order to find out moles.
C = 63.2 g / 12 g/mol = Moles = 5.26 moles
H = 5.26 g / 1.008 g/mol = Moles = 5.21 moles
C = 41.6 g / 16 g/mol = Moles = 2.6 moles
Step 2:
Select moles of the element with least value and divide all moles of element by it,
C H O
5.26/2.6 : 5.21/2.6 : 2.6/2.6
2.02 : 2.00 : 1
Result:
Empirical Formula = C₂H₂O
Answer:
mass (g) needed = 710.2 grams Na₂SO₄(s)
Explanation:
Needed is 2.5 Liters of 2.0M Na₂SO₄; formula wt Na₂SO₄ = 142.04g/mol.
mass (grams) of Na₂SO₄(s) = Molarity needed x Volume needed in Liters x Formula Wt of solute
mass (grams) of Na₂SO₄(s) = (2.5L)(2.0M)(142.04g/mol) = 710.2 grams Na₂SO₄(s)
Mixing: Transfer 710.4 grams Na₂SO₄ into mixing vessel and add water-solvent up to but not to exceed 2.5 Liters total volume. Mix until dissolved.
Gives 2.5 Liters of 2.0M Na₂SO₄(aq) solution.
Answer:
6.25%
Explanation:
Given data:
Half life of lutetium-117 = 6.75 days
Percentage remaining after 27 days = ?
Solution;
Number of half lives = Time elapsed / half life
Number of half lives = 27 days / 6.75 days
Number of half lives = 4
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
At 3rd half life = 25%/2 = 12.5%
At 4th half life = 12.5%/2 = 6.25%
Answer:
Explanation:
1. Miles travelled in an average month
2. Using a gasoline powered vehicle
(a) Moles of heptane used
(b) Equation for combustion
C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O
(c) Moles of CO₂ formed
(d) Volume of CO₂ formed
At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.
3. Using an electric vehicle
(a) Theoretical energy used
(b) Actual energy used
The power station is only 85 % efficient.
(c) Combustion of CH₄
CH₄ + 2O₂ ⟶ CO₂ +2 H₂O
(d) Equivalent volume of CO₂
The heat of combustion of methane is -802.3 kJ·mol⁻¹
4. Comparison
