Answer: 405.3 minutes
Explanation: In order to explain this problem we have to use the following:
Fisrtly we calculate the volume of the wire, this is given by:
Vwire=π*r^2*L where r and L are the radius and L the length of teh wire, respectively.
Vwire=π*1.25*10^-3*0.26=1.27*10^-6 m^3
then the number of the total electrons in tthe wire volume is given by;
n° electrons in the wire=ρ*Vwire=8.4*10^28*1.27*10^-6 m^3=1.07 *10^23
Finally, considering the current in the wire equal to 4.4*10^18 electrons/s
the time consuming to extract all the electrons from the wire is given by:
t= total electrons in the wire/ current=1.067*10^23/4.4*10^18=24,318 s
equivalent to 405.3 minutes
Answer: Frequency factor A = 8 × 10⁹
activation energy Ea = 15.5 KJ/Mol
Explanation: to begin, let us first define the parameters given;
K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹
K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹
K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹
also T₁ = 300.3 K
T₂ = 341.2 K
T₃ = 392.2 K
we know that;
㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]
where R is given as 8.314 J/mol-k
Ea = activation energy
K₁, K₂ = rate constant
T₁, T₂ = Temperature
therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]
this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol
∴ Ea = 15.5 KJ/ Mol
also given that K = A e⁻∧Ea/RT
here A = frequency factor
∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)
A = 7.99 × 10⁹ = 8 × 10⁹
Where Gravity rely's on only mass and distance and nothing else, so the weight on the planets will vary like you have stated. However Mars is smaller than Mercury, so the weight on Mars will be less, and the weight on Mercury will be more. Think this way.
More Mass = More Gravity = More Weight
Less Mass = Less Gravity = Less Weight
It depends on how long the ball stays in the air
1 s=20m
10 s=200m
