Answer:
Explanation:
As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)
ΔG = ΔG° + RTInQ
Q = 1
ΔG = ΔG°
ΔG = =nFE°
n=no of electrons transfered.
E° = 1.1v
ΔG° = -2 * 96500 * 1.10
= -212300J
ΔG° =-212.3kJ/mol
<h3>Therefore, the ΔG° = -212.3kJ/mol</h3>
Answer : The pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
If it is a big open wound you apply pressure
<span>1. Which variable is the independent variable and which is the dependent variable? Density vs. ethylene glycol
The independent variable would be ethylene glycol and dependent variable would be density.
A. A 25-mL volumetric flask with its stopper has a mass of 32.6341 g. The same flask filled to the line with ethylene glycol (C2H6O2, automotive antifreeze) solution has a mass of 58.0091 g. What is the density of the ethylene glycol solution?
Density = 58.0091 - 32.6341 / .025 = 1015 g/L
B. What is the molarity of the ethylene glycol solution, if the mass of ethylene glycol in the solution is 12.0439 g?
Molarity = 12.0439 ( 1 mol / 62.07 g) / 0.025 = 7.8 M</span>
Answer:
A
Explanation: A wolf is a predator and a hen is prey.
