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3 years ago

7

A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.130 M AuNO 3 solution at 25 °C. The gold wire i

s connected to the positive end of a potentiometer, and the S.C.E. is connected to the negative end of the potentiometer. What is the half‑reaction that occurs at the Au electrode? Include physical states.

1 answer:

padilas [110]3 years ago

7 0

Answer:

The half reaction occurring at this electrode will be given as:

$Au^++e^-\rightarrow Au$

Explanation:

In electrochemical cell reduction occurs at anode.

But the gold wire is connected to the positive terminal of the battery which means that gold wire will act as a cathode nad reduction will take place

The half reaction occurring at this electrode will be given as:

$Au^++e^-\rightarrow Au$

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How many liters of o2 do you have if you have 5.8 moles of o2

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Answer:

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Explanation:

if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)

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Which form of energy is directly related to the measure of the average kinetic energy of the particles in a substance?

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

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Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

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$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

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2 years ago

Is O2 always a double bond?

Dima020 [189]

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H-O single bonds in $\text{H}_2\text{O}$ molecules.
O=O double bonds in $\text{O}_2$ molecules.

<h3>Explanation</h3>

How many valence electrons do atoms in each molecule need for them to be stable?

Each H atom needs two valence electrons to be stable.
Each atom of an element other than H needs eight valence electron to be stable.

There are two H atoms and two O atoms in an $\text{H}_2\text{O}_2$ molecule. Atoms in each $\text{H}_2\text{O}_2$ need $2 \times 1 + 2\times 2 = 6$ more electrons to be stable.
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How many chemical bonds in each molecule?

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