Answer:
D. 18,800 J/mol
Explanation:
We need to use the Arrhenius equation to solve for this problem:
, where k is the rate constant, A is the frequency factor,
is the activation energy, R is the gas constant, and T is the temperature in Kelvins.
We want to find the value of
, so let's plug some of the information we have into the equation. The gas constant we can use here is 8.31 J/mol-K.
At 0°C, which is 0 + 273 = 273 Kelvins, the rate constant k is
. So:


At 20°C, which is 20 + 273 = 293 Kelvins, the rate constant k is
. So:


We now have two equations and two variables to solve for. We just want to find Ea, so let's write the first equation for A in terms of Ea:


Plug this in for A in the second equation:


After some troublesome manipulation, the answer should come down to be approximately:
Ea = 18,800 J/mol
The answer is thus D.
These are all the answers. Hope I helped
1. A
2. C
3. A
4. B
<span>Water soaks into it easily and during freeze-thaw cycles would be more likely to break apart because of the water freezing and thawing inside the rock. The water freezes and expands inside the rock, causing the surface to break off in flakes from the pressure of the expanding ice inside.</span>
Explanation:
A mixture in which there is uniform distribution of solute particles into the solvent is known as a homogeneous mixture.
For example, sugar dissolved in water is a homogeneous mixture.
On the other hand, a mixture in which there is uneven distribution of solute particles into the solvent is known as a heterogeneous mixture.
For example, sand present in water is a heterogeneous mixture.
Comment on given situations will be as follows.
(a) Air in a closed bottle - It is a homogeneous mixture because there will be even distribution of other gases that are present in air.
(b) Air over New York City - It is a heterogeneous mixture because there will be presence of some dust particles, fog or smoke into the air. Distribution of all these particles will be uneven. This will make air over New York City heterogeneous in nature.