The answer for the following problem is mentioned below.
- <u><em>Therefore the final temperature of the gas is 740 K</em></u>
Explanation:
Given:
Initial pressure of the gas (
) = 1.8 atm
Final pressure of the gas (
) = 4 atm
Initial temperature of the gas (
) = 60°C = 60 + 273 = 333 K
To solve:
Final temperature of the gas (
)
We know;
From the ideal gas equation;
we know;
P × V = n × R × T
So;
we can tell from the above equation;
<u> P ∝ T</u>
(i.e.)
<em> </em>
<em> = constant</em>
= 
Where;
= initial pressure of a gas
= final pressure of a gas
= initial temperature of a gas
= final temperature of a gas
= 
=
= 740 K
<u><em>Therefore the final temperature of the gas is 740 K</em></u>
Answer:
Explanation:
Mole = no. Molecules/6.02×10^23
Mole = (1.40×10^24)/(6.02×10^23)
Mole = 2.33mole
Explanation:
Reaction:
Cu + 2AgC₂H₃O₂ → Cu(C₂H₃O₂)₂ + 2Ag
The problem is to split the reaction into oxidation and reduction halves:
The oxidation half is the sub-reaction that undergoes oxidation
The reduction half is the one that undergoes reduction:
The ionic equation:
Cu + 2Ag⁺ + 2C₂H₃O₂⁻ → Cu²⁺ + 2C₂H₃O₂⁻ + 2Ag
Oxidation half:
Cu → Cu²⁺ + 2e⁻
Reduction half:
2Ag⁺ + 2e⁻ → 2Ag
C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.
learn more:
Oxidation state brainly.com/question/10017129
#learnwithBrainly
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 