<span>The elements that form the compound don't compare because it doesn't matter about the elements when they are combined. The elements that formed it together is completely different than the outcome compound. </span>
Answer:
option d
Explanation:
Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.
Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.
Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
Answer:
16.5 days
Explanation:
Given that:
Half life = 26.5 days
Where, k is rate constant
So,
The rate constant, k = 0.02616 days⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given:
35.0 % is decomposed which means that 0.35 of is decomposed. So,
![\frac {[A_t]}{[A_0]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D)
= 1 - 0.35 = 0.65
t = 7.8 min
![\frac {[A_t]}{[A_0]}=e^{-k\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-k%5Ctimes%20t%7D)
<u>t = 16.5 days.</u>
Answer:
I think it's C
Explanation:
i think its c b3cause when you guess c always feels right
