Answer:
K = 2.7x10⁻⁵ at 25ºC
Explanation:
A way to write Arrhenius equation is:
ln K = - Ea/R × (1/T) + lnA
If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:
Y = -13815X +35.817
R² = 0.9927
(Taking the last k point as 0.0386) (ln 0.0386), <em>0.1386 has no sense</em>)
Your slope is -13815
-13815K = - Ea/R
-13815K×8.314J/molK = 114858J/mol = Ea
And your intercept =
lnA = 35.817
A = 3.59x10¹⁵
Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):
Y = -13815X +35.817
Y = -13815(1/298.15K) +35.817
Y = -10.5187
lnK = -10.5187
<h3>K = 2.7x10⁻⁵ at 25ºC</h3>
Answer:
1.5057×10^22 atom
Explanation:
As we
1 mole of argon = 40 g of argon
i.e 40 g of argon = 1 mole of argon
1 g of argon = 1/40 mole of argon
1 mole of argon = 6.023×10^23 atom of argon
1/40 mole if argon = 1/40 ×6.023×10^23
= 1.5057×10^22
NH3(g) will take the shape of and completely fill a closed 100.0 milliliter container.
Explanation:
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