Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol
moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2
Formula of substance:
C₂H₄
Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
<span>KE = 1/2mv^2
KE = 1/2(8)4 m/s^2
KE = 4*4
KE = 16 Joules
Kinetic energy would equal 16 J </span>
Answer:
Explanation:
Hello there!
In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:
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Answer:
An object becomes electrically charged when <em>electrons</em><em> </em><em>are</em><em> </em><em>transferred</em><em> </em><em>to</em><em> </em><em>it</em>
Explanation:
<em>PLEASE</em><em> </em><em>DO MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>
Answer:
4.43 g of Oxygen
Explanation:
As shown in Chemical Formula, one mole of Aluminium Sulfate [Al₂(SO₄)₃] contains;
2 Moles of Aluminium
3 Moles of Sulfur
12 Moles of Oxygen
Also, the Molar Mass of Aluminium Sulfate is 342.15 g/mol. It means,
342.15 g ( 1 mole) of Al₂(SO₄)₃ contains = 192 g (12 mole) of O
So,
7.9 g of Al₂(SO₄)₃ will contain = X g of O
Solving for X,
X = (7.9 g × 192 g) ÷ 342.15 g
X = 4.43 g of Oxygen
