Answer:
Percent Composition
1. Find the molar mass of all the elements in the compound in grams per mole.
2. Find the molecular mass of the entire compound.
3. Divide the component's molar mass by the entire molecular mass.
4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.
I believe the answer is true. Hope this helps.
the calculated value is Ea is 18.2 KJ and A is 12.27.
According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.
At 500K, K=0.02s−1
At 700K, k=0.07s −1
The Arrhenius equation can be used to calculate Ea and A.
RT=k=Ae Ea
lnk=lnA+(RT−Ea)
At 500 K,
ln0.02=lnA+500R−Ea
500R Ea (1) At 700K lnA=ln (0.02) + 500R
lnA = ln (0.07) + 700REa (2)
Adding (1) to (2)
700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.
=ln [0.02/0 .07]
Ea= 2/35×100×8.314×1.2528
Ea =18227.6J
Ea =18.2KJ
Changing the value of E an in (1),
lnA=0.02) + 500×8.314/18227.6
= (−3.9120) +4.3848
lnA=0.4728
logA=1.0889
A=antilog (1.0889)
A=12.27
Consequently, Ea is 18.2 KJ and A is 12.27.
Learn more about Arrhenius equation here-
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Answer:
[Ag⁺] = 0.0666M
Explanation:
For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:
Ag⁺ + 2CN⁻ ⇄ Ag(CN)₂⁻
Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]
As initial concentrations of Ag⁺ and CN⁻ are:
[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M
[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M
The equilibrium concentrations of each compound are:
[CN⁻] = 9.7x10⁻⁴M - x
[Ag⁺] = 0.0676M - x
[Ag(CN)₂⁻] = x
<em>Where x is reaction coordinate</em>
Replacing in Kf formula:
1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]
1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0
Solving for x:
X = 9.614x10⁻⁴M
Thus, equilibrium concentration of Ag⁺ is:
[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>
The formula for Lithium Iodide is LIL