Answer:
-10 m/s2
Explanation:
The acceleration is taken as −10 m/s2
Answer:
Ea = 177x10³ J/mol
ko = 
J/mol
Explanation:
The specific reaction rate can be calculated by Arrhenius equation:
Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.
k depends on the temperature, so, we can divide the k of two different temperatures:
Applying natural logathim in both sides of the equations:
ln(k1/k2) = Ea/RT2 - Ea/RT1
ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)
R = 8.314 J/mol.K
ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)
ln(0.052) = (Ea/8.314)x(-1.38x
-1.67x
xEa = -2.95
Ea = 177x10³ J/mol
To find ko, we just need to substitute Ea in one of the specific reaction rate equation:
ko = J/mol
Answer:
The volume of 8.625 g of sulphur dioxide at RTP is approximately 3,230.84 cm³
Explanation:
The question requires the determination of the volume occupied by the gas based on the molar volume of a gas at STP
The given parameters of the sulphur dioxide, SO₂, gas are;
The mass of the given SO₂ gas = 8.625 g
The molar mass of SO₂ = 64.07 g/mol
The number of moles, 'n', in the given sample of SO₂ gas = Mass of SO₂/(Molar Mass of SO₂)
∴ The number of moles of SO₂ in the gas sample = 8.625 g/(64.07 g/mol) ≈ 0.134618386 moles
The molar volume of a gas at RTP is approximately 24 dm³/mole
24 dm³ = 24,000 cm³
∴ The molar volume of a gas at RTP is approximately 24,000 cm³/mole
The volume occupied by a gas at RTP = (The number of moles of the gas) × (The Molar Volume of a gas at RTP)
∴ The volume occupied by the 8.625 g of SO₂ gas at RTP = 0.134618386 moles × 24,000 cm³/mole ≈ 3,230.84 cm³
The volume occupied by the 8.625 g of SO₂ gas at RTP ≈ 3,230.84 cm³.
Con: less energy
pro : helping the society
