It isn't a materialistic object.
Answer:
Moon rocks contain few volatile substances (e.g. water), which implies extra baking of the lunar surface relative to that of Earth. The relative abundance of oxygen isotopes on Earth and on the Moon are identical, which suggests that the Earth and Moon formed at the same distance from the Sun.
Explanation:
<h2>
Hello!</h2>
The answer is:
The new volume will be 1 L.

<h2>
Why?</h2>
To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.
To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.
Boyle's Law equation is:

So, we are given the information:

Then, isolating the new volume and substituting into the equation, we have:



Hence, the new volume will be 1 L.

Have a nice day!
Answer:
2

Explanation:
Half-life


Concentration
![{[A]_0}_A=1.2\ \text{M}](https://tex.z-dn.net/?f=%7B%5BA%5D_0%7D_A%3D1.2%5C%20%5Ctext%7BM%7D)
![{[A]_0}_B=0.6\ \text{M}](https://tex.z-dn.net/?f=%7B%5BA%5D_0%7D_B%3D0.6%5C%20%5Ctext%7BM%7D)
We have the relation
![t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%5Cpropto%20%5Cdfrac%7B1%7D%7B%5BA%5D_0%5E%7Bn-1%7D%7D)
So
![\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%7Bt_%7B1%2F2%7D%7D_A%7D%7B%7Bt_%7B1%2F2%7D%7D_B%7D%3D%5Cleft%28%5Cdfrac%7B%7B%5BA%5D_0%7D_B%7D%7B%7B%5BA%5D_0%7D_A%7D%5Cright%29%5E%7Bn-1%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B2%7D%7B4%7D%3D%5Cleft%28%5Cdfrac%7B0.6%7D%7B1.2%7D%5Cright%29%5E%7Bn-1%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B2%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7Bn-1%7D)
Comparing the exponents we get

The order of the reaction is 2.
![t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%5E%7Bn-1%7D%7D%5C%5C%5CRightarrow%20k%3D%5Cdfrac%7B1%7D%7Bt_%7B1%2F2%7D%5BA%5D_0%5E%7Bn-1%7D%7D%5C%5C%5CRightarrow%20k%3D%5Cdfrac%7B1%7D%7B2%5Ctimes%201.2%5E%7B2-1%7D%7D%5C%5C%5CRightarrow%20k%3D0.4167%5C%20%5Ctext%7BM%7D%5E%7B-1%7D%5Ctext%7Bmin%7D%5E%7B-1%7D)
The rate constant is 