C. air, fire, water, earth. 4 elements
Alfheim
In Norse mythology it means home of Elves and it is known as home to the light elves
Answer:
Explanation:
Mole = no. Molecules/6.02×10^23
Mole = (1.40×10^24)/(6.02×10^23)
Mole = 2.33mole
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g
