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Makovka662 [10]
3 years ago
13

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

a 8 A current? Recall the density of gold is 19.3 g/cm^3.
Chemistry
1 answer:
konstantin123 [22]3 years ago
7 0

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = a=31 cm^2

Thickness of the gold plating  = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3

Density of the gold = d=19.3 g/cm^3

m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g

Moles of gold = \frac{29.915 g}{197 g/mol}=0.152 mol

Au^{3+}+3e^-\rightarrow Au

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

\frac{3}{1}\times 0.152 mol=0.456 mol of electrons

Number of electrons = N =0.456\times \times 6.022\times 10^{23}

Charge on single electron = q=1.6\times 10^{-19} C

Total charge required = Q

Q=N\times q

Amount of current passes = I = 8 Ampere

Duration of time  = T

I=\frac{Q}{T}

T=\frac{N\times q}{I}

=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

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The temperature at which vapor pressure of a substance becomes equal to the atmospheric pressure is called boiling point.

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In CH_{3}OH (methanol), there is hydrogen bonding present which is a stronger force. So, it will have highest boiling point as compared to CH_{3}Cl and CH_{4}.

In CH_{3}Cl (chloroform), there is more electronegative atom attached (Cl) is attached to less electronegative atom (C and H). So, electrons are more pulled towards the chlorine atom. So, boiling point of  CH_{3}Cl is more than methane (CH_{4}).

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To measure the amount of calcium carbonate in a seashell, an analytical chemist crushes a sample of the shell to a fine powder a
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The given question is complete, the complete question is:

To measure the amount of calcium carbonate (CaCO) in a seashell, an analytical chemist crushes a 4.80 g sample of the shell to a fine powder and titrates it to the endpoint with 515. mL of 0.140 M hydrogen chloride (HCl) solution. The balanced chemical equation for the reaction is: 2HCI(a)Co (a) H2Co,(aq) + 2Cl (aq)

What kind of reaction is this?

If you said this was a precipitation reaction, enter the chemical formula of the precipitate.

If you said this was an acid-base reaction, enter the chemical formula of the reactant that is acting as the base.

If you said this was a redox reaction, enter the chemical symbol of the element that is oxidized

Calculate the mass percent of CaCO in the sample. Be sure your answer has the correct number of significant digits.

Answer:

It is an acid-base reaction and the mass percent of CaCo3 is 75.2%.

Explanation:

A chemical reaction in which an insoluble salt produces from two soluble salts is termed as precipitation reaction.

A reaction in which atleast exchange of one proton takes place between the two species is termed as an acid-base reaction.

A reaction in which any of the element has a change in oxidation state is termed as redox reaction.

In the mentioned reaction, there is a transfer of H⁺, no precipitation is forming, and no change in oxidation state taking place, thus, it is an acid-base reaction.

In the acid-base reaction, the base refers to the species that accepts hydrogen ion or proton. In the given case, CO₃²⁻ is accepting H+ ion to become H₂CO₃. Hence, CO₃²⁻ is the base.

In order to calculate mass percent of CaCO₃, first there is a need to find the moles of HCl reacted for a solution,

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Now from the balanced equation, one mole of CaCO₃ needs two moles of HCl.

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The mass of calcium carbonate taking part in reaction will be,

= Moles × Molar mass

= 0.03605 × 100 gram/mole

= 3.6086 gram

Mass% of CaCO₃ = Mass of CaCO₃/Mass of sample × 100

= 3.6086 grams/4.80 grams × 100

Mass % = 75.2%

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