Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
<span>The minimum amount of energy needed to get a chemical reaction started is called the activation energy.
</span>
<span>According to the question-
1 mol C3H8O = 60.096 g C3H8O
2 mol C3H8O = 9 mol O2
1 mol O2 = 31.998 g O2
[(3.00 g C3H8O)/1][(1 mol C3H8O)/(60.096)][(9 mol O2)/(2 mol C3H8O)][(32.998 g O2)/(1 mol O2)] = 7.1880435 g O2
Since 7.1880435 g of O2 is needed, and 7.38 g of O2 is available, 0.199565 g of O2 will be left over and oxygen is present in excess.
Next, we need to convert 0.199565 g of O2 into moles of O2:
[(0.199565 g O2)/1][(1 mol O2)/(31.998 g O2)] = 0.005999 mol O2, or 0.006 mol O2</span>
Your answers would 35/3, 11 2/3, and 11.6 !! :D
Answer:
PCl₅ = 0.03 X 208 = 6.24g
PCl₃ = 0.05 X 137 =6.85 g
Cl₂ = 0.03X71 = 2.13 g
Explanation:
The equilibrium constant will remain the same irrespective of the amount of reactant taken.
Let us calculate the equilibrium constant of the reaction.
Kc=![\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BPCl_%7B3%7D%5D%5BCl_%7B2%7D%5D%7D%7B%5BPCl_%7B5%7D%5D%7D)
Let us calculate the moles of each present at equilibrium
molar mass of PCl₅=208
molar mass of PCl₃=137
molar mass of Cl₂=71
moles of PCl₅ = 
moles of PCl₃= 
moles of Cl₂ = 
the volume is 5 L
So concentration will be moles per unit volume
Putting values
Kc = 
Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow
Initial 0.02 0.06 0.04
Change -x +x +x
Equilibrium 0.02-x 0.06+x 0.04+x
Conc. (0.02-x)/2 (0.06+x)/2 (0.04+x)/2
Putting values
0.024 = 
Solving
x = -0.01
so the new moles of
PCl₅ = 0.02 + 0.01 =0.03
PCl₃ = 0.06-0.01 = 0.05
Cl₂ = 0.04-0.01 = 0.03
mass of each will be:
mass= moles X molar mass
PCl₅ = 0.03 X 208 = 6.24g
PCl₃ = 0.05 X 137 =6.85 g
Cl₂ = 0.03X71 = 2.13 g
