Using this formula, to find K, what does E0 represent?

2 answers:

6 0

The correct answer would be A. The symbol Eo would represent the cell potential of an electrolytic cell. This potential is being created by two metals that possess different properties. The energy per charge that is available from the reaction of the metals is the measure of this potential and is related to the equilibrium constant, K.

natita [175]3 years ago

6 0

The correct answer is a

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Ethanol (C2H5OH) and water (H2O) are both liquids at or near room temperature. When a sample of ethanol is poured into a beaker

BaLLatris [955]

Answer:

a. Are miscible because each can hydrogen bond with the other.

Explanation:

Both ethanol and water are miscible. The reason why they can both mix freely is due to the hydrogen bonds that will form between their molecular structure.

Hydrogen bonds are special dipole-dipole attraction between polar molecules in which hydrogen atoms are directly joined to an electronegative atom.

Ethanol has an hydroxyl group which will bond to form an intermolecular bond with the oxygen and hydrogen on the water molecule. This attraction makes them miscible.

5 0

3 years ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$