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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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Papessa [141]
I know CL is chlorine and NA is sodium.
8 0
3 years ago
Write the name and molecular formula of an organic compounds having its name suffixed with -ol and having 2 carbon atoms in the
777dan777 [17]

Ethanol C₂H₆O

Explanation:

When ethanol (CH₃-CH₂-OH) is heated in the presence of the sulphuric acid (H₂SO₄) it will produce ethylene (CH₂=CH₂ ) and water (H₂O).

CH₃-CH₂-OH → CH₂=CH₂ + H₂O

Learn more about:

sulphuric acid

brainly.com/question/867125

#learnwithBrainly

4 0
3 years ago
Ammonia and oxygen react to produce nitric oxide and water. Which of the following chemical equations describes this reaction an
aksik [14]

Answer:

Option B. 4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Explanation:

The reaction between ammonia, NH3 and oxygen, O2, will produce nitric oxide and water as shown below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

Now, let us balance the equation.

This is illustrated below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

There are 3 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of NH3 and 6 in front of H2O as shown below:

4NH3(g) + O2(g) —> NO(g) + 6H2O(g)

There are 4 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 4 in front of NO as shown below:

4NH3(g) + O2(g) —> 4NO(g) + 6H2O(g)

Now, the are 2 atoms of O on the left side and a total of 10 atoms on the right side. It can be balance by putting 5 in front of O2 as shown below:

4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Now, we can see that the equation is balanced.

Therefore, option B gives the balanced chemical equation for the reaction.

5 0
3 years ago
How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC?
vodomira [7]

Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J

Explanation:

1) Data:

Water ⇒ C = 1 cal/g°C

m = 65.8 g

Ti = 31.5°C

Tf = 36.9°C

Heat, Q = ?

2) Formula:

Q = mCΔT

3) Calculations:

Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal

4) You can convert from calories to Joules using the conversion factor:

1 cal = 4.18 J

⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J

3 0
3 years ago
A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

                                           = 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

             x = 0.0097 mol

              y = (0.0294 - 0.0097)

                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

                               = \frac{0.0197}{0.0294}

                              = 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0
3 years ago
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