False, because generally forensics collect the evidence to make sure the evidence doesn’t get damaged or tainted.... hope this helps
Answer:
Ethanol
Explanation:
Solvent extraction is the process in which a compound transfers from one solvent to another owing to the difference in solubility or distribution coefficient between these two solvents(Science Direct).
We have to remember that oxalic acid will be extracted better into a solvent in which it is more soluble. From the data given; oxalic acid is more soluble in ethanol than in ether.
This simply means that ethanol is a better solvent for extracting oxalic acid from water when compared to ether.
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.
Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol
Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.
PV = nRT PV
---- ------ ---> n = --------
RT RT RT
10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K
=106.29 mol Ar
So the moles of argon gas is 106.29 moles
Answer:
a) [H₃O⁺] = 1.8x10⁻⁵ M
b) pH = 4.75
c) % rxn = 3.5x10⁻³ %
Explanation:
a) The dissociation reaction of HCN is:
HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)
0.5 M - x x x
The dissociation constant from the above reactions is given by:
![Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH_%7B3%7DO%5E%7B%2B%7D%5D%5BCN%5E%7B-%7D%5D%7D%7B%5BHCN%5D%7D%20%3D%206.17%20%5Ccdot%2010%5E%7B-10%7D)
By solving the above quadratic equation we have:
x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]
Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.
b) The pH is equal to:
![pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%20-log%281.75%20%5Ccdot%2010%5E%7B-5%7D%20M%29%20%3D%204.75)
Then, the pH of the HCN solution is 4.75.
c) The % reaction is the % ionization:
![\% = \frac{x}{[HCN]} \times 100](https://tex.z-dn.net/?f=%20%5C%25%20%3D%20%5Cfrac%7Bx%7D%7B%5BHCN%5D%7D%20%5Ctimes%20100%20)
Therefore, the % reaction or % ionization is 3.5x10⁻³ %.
I hope it helps you!
