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2 years ago

Prepare a list of objects around you that are electroplated

Chemistry

1 answer:

DedPeter [7]2 years ago

8 0

Answer:

Hello! Some household items that are electroplated include kitchenware, such as metal pots and pans, door handles, mobile phones, and coins. An electroplated object is coated by electrolytic deposition with chromium, silver, or another metal.

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A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode c

Andre45 [30]

Answer: The standard potential of the cell is 0.77 V

Explanation:

We know that:

$E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V$

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq)+2e^-$

Reduction half reaction: $Cu^{+}(aq)+e^-\rightarrow Cu(s)$ ( × 2)

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation follows:

$E^o_{cell}=0.52-(-0.25)=0.77V$

Hence, the standard potential of the cell is 0.77 V

5 0

3 years ago

How can you tell that a precipitate has formed?

ladessa [460]

Answer:

if a solid appears when mixing two liquids

Explanation:

i remember this question :D

7 0

3 years ago

Read 2 more answers

Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 10.0 kJ or greater.

nikdorinn [45]

Answer:

The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ

Explanation:

From Arrhenius equation showing the temperature dependence of reaction rates.

$K = Ae^{\frac{Ea}{RT} }$ where

k = rate constant

A = Frequency or pre-exponential factor

Ea = energy of activation

R = The universal gas constant

T = Kelvin absolute temperature

we have

$f = e^{\frac{Ea}{RT} }$

Where

f = fraction of collision with energy higher than the activation energy

Ea = activation energy = 10.0kJ = 10000J

R = universal gas constant = 8.31 J/mol.K

T = Absolute temperature in Kelvin = 400K

In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature

Plugging in the values into the equation relating f to activation energy we get

$f = e^{\frac{10000J}{(8.31J/((mol)(K)))(400K)} }$ or f = $e^{3.01}$ = 20.22 moles of argon have an energy of 10.0 kJ or greater