explanation:we will suppose the fuel has the formula nxoy.the molar mass of the fuel = (x . atomic mass of n) + (y . atomic mass of o) = 92.0 g/mole.
the % composition of n = 30.43 %the % composition of n = (x.atomic mass of n)/(molar mass of fuel) x 100.then x = (% composition of n)(molar mass of fuel) / 100(atomic mass of n) = (30.43 %)(92.0 g/mole) / 100(14.00 g/mole) = 1.9996 ≅ 2.0.
by the same way;
the % composition of o = 69.57 %the % composition of o = (y.atomic mass of o)/(molar mass of fuel) x 100then y = (% composition of o)(molar mass of fuel) / 100(atomic mass of o) = (69.57 %)(92.0 g/mole) / 100(16.00 g/mole) = 4.0.
so, the formula of the compound is n₂o₄.
inital velocity (u) should be less
the formula of the original halide is srcl₂.
explanation:the balanced equation of this reaction is:
srx₂ + h₂so₄ → srso₄ + 2 hx, where x is the halide.from the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of srso₄.the number of moles of srso₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.the number of moles of srx are 4.11 x 10⁻³ moles from the stichiometry of the balanced equation.n = mass / molar mass, n = 4.11 x 10⁻³ moles and mass = 0.652 g.the molar mass of srx₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole. the molar mass of srx₂ (158.62 g/mole) = atomic mass of sr (87.62 g/mole) + (2 x atomic mass of halide x).the atomic mass of halide x = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.this is the atomic mass of cl.so, the formula of the original halide is srcl₂.
the answer is e.sodium