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Dominik [7]
3 years ago
5

In part 1 of lab 2 you will make and standardize a solution of naoh(aq). suppose in the lab you measure the solid naoh and disso

lve it into 100.0 ml of water. you then measure 0.2005 g of khp (kc8h5o4, 204.22 g/mol) and place it in a clean, dry 100-ml beaker, and then dissolve the khp in about 25 ml of water and add a couple of drops of phenolphthalein indicator. you titrate this with your naoh(aq) solution and find that the titration requires 9.82 ml of naoh(aq).
Chemistry
2 answers:
Nostrana [21]3 years ago
3 0
Shusbbcjdkbdjekenbfhfkdndhd
Agata [3.3K]3 years ago
3 0

Answer:

0.100 M

Explanation:

KHP is used to standardize the NaOH solution, according to the following neutralization reaction.

KC₈H₅O₄ + NaOH → NaKC₈H₄O₄ + H₂O

We can establish the following relations.

  • The molar mass of KHP is 204.22 g/mol.
  • The molar ratio of KHP to NaOH is 1:1.

The moles of NaOH that reacted with 0.2005 g of KHP are:

0.2005gKHP.\frac{1molKHP}{204.22gKHP} .\frac{1molNaOH}{1molKHP} =9.818 \times 10^{-4} molNaOH

The molarity of NaOH is:

M=\frac{9.818 \times 10^{-4} mol}{9.82 \times 10^{-3}L} =0.100M

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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)

7 0
3 years ago
A spectator ion is (Select all that apply.) a. an ionic component of a reactant that is unchanged by the reaction your eye, b. c
DENIUS [597]

Correct Question:

A spectator ion is (Select all that apply.)

- a piece of french fry contaminating the reaction mixture

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Answer:

- an ionic component of a reactant that is unchanged by the reaction

Explanation:

A spectator ion is an ion that exists as a reactant and a product in a chemical equation. A spectator ion is one that exists in the same form on both the reactant and product sides of a chemical reaction.

Spectator ions are ions that are present in a solution but don't take part in the reaction. When reactants dissociate into ions, some of the ions may combine to form a new compound. The other ions don't take part in this chemical reaction and are therefore called spectator ions.

The correct option is therefore the option;

- an ionic component of a reactant that is unchanged by the reaction

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3 years ago
Why don't we use meters or miles when discussion distances between stars or galaxies?
taurus [48]
If we use mile or something else it will be hard to measure in sky .
Because the mile or meters too short for the space,
Galaxy and stars are too long far away from us if we use mile or meter,
It till be take like miliion year to measure.


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Answer:

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Answer:

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3 years ago
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