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3 years ago

Converting the following mm Hg (torr) to standard

Chemistry

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

Explanation:

Given:

20.5 torr

78.6 torr

225 torr

Change into atm

Computation:

We know that;

1 torr = 0.00131579 atm

So,

1. 20.5 torr

= 20.5 x 0.00131579 atm

= 0.02697

= 0.027 atm (Approx)

2. 78.6 torr

= 78.6 x 0.00131579 atm

= 0.103090

= 0.103 atm (Approx)

3. 225 torr

= 225 x 0.00131579 atm

= 0.2960

= 0.296 atm (Approx)

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<h3>What is an experiment?</h3>

An experiment is a procedure that aims at probing or discovering something. For example, you can test if a plant grows faster/slower by using an experiment.

<h3>What causes different results in similar experiments?</h3>

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Based on this, if the metal content was different you should analyze if any of the factors changed in this experiment and find out how this change affected the general results.

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3 years ago

Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine

yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

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