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Alisiya [41]
2 years ago
14

Aryan want to deposit silver on an iron spoon using silver nitrate as electrolyte. Which terminal of the battery he should conne

ct the spoon? What material should the other electrode be made of?
Chemistry
1 answer:
34kurt2 years ago
8 0

Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.

<h3>What is electroplating?</h3>

Electroplating is a way of electrolysis, where a thin layer of metal is used to plate a determined object. It is a kinda process to purify a material you want.

The anode contains the metal you want to plate with, in this case, the Ag.

The anode contains the half-reaction of oxidation so:

Ag(s) → Ag ⁺ (aq)  +  e⁻

In the cathode, you have the spoon, which it takes place the half-reaction of reduction:

Ag ⁺ (aq)  +  e⁻  →  Ag(s)

The electrolytic cell, where the redox reaction takes place, must be filled with a AgNO₃ solution.

Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.

Learn more about the electroplating here:

brainly.com/question/20112817

#SPJ1

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One method used commercially to peel potatoes is to soak them in a solution of NaOH for a short time and then remove them from t
sergij07 [2.7K]

Answer:

b. 2.28 M

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The reaction of neutralization of NaOH with H2SO4 is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

<em>Where 2 moles of NaOH react per mole of H2SO4</em>

<em />

To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:

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45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4

<em>Moles NaOH:</em>

0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH

<em>Molarity NaOH:</em>

0.0457moles NaOH / 0.020L =

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Right option:

<h3>b. 2.28 M</h3>
7 0
3 years ago
Fe2O3(s) + 3CO(g) ---&gt; 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast
baherus [9]

Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles

\text{Moles of} CO=\frac{260}{28}=9.3moles

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)

According to stoichiometry :

1 mole of Fe_2O_3 require 3 moles of CO

Thus 2.8 moles of Fe_2O_3 will require=\frac{3}{1}\times 2.8=8.4moles  of CO

Thus Fe_2O_3 is the limiting reagent as it limits the formation of product and CO is the excess reagent.

As 1 mole of Fe_2O_3 give = 2 moles of Fe

Thus 2.8 moles of Fe_2O_3 give =\frac{2}{1}\times 2.8=5.6moles of Fe

Mass of Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%

Therefore, the percent yield is, 91.8%

6 0
3 years ago
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