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4 years ago

Identify the anode and cathode when plating an iron nail with zinc.

Chemistry

2 answers:

Lunna [17]4 years ago

5 0

The anode and cathode when plating an iron nail with zinc :

anode : zinc

cathode : iron nail

<h3>Further explanation</h3>

Electrolysis utilizes electrical energy to carry out non-spontaneous redox reactions. The ions in the solution flowed by an electric current

The electrolyzed substance is an electrolyte which can be in the form of a melt or a solution.

In the electrolytic cell the positive pole - the anode is the site of oxidation, whereas the negative pole - the cathode is the site of the reduction

Electrons (electricity) enter the electrolysis cell through the negative pole (cathode)

Electroplating is one of the applications of electrolysis. In the electroplating process, the metal to be plated as a cathode and the plating metal as an anode. Both electrodes are dipped in an electrolyte saline solution from the positive metal plating ion. In this electroplating process, the metal at the anode will oxidize and dissolve into metal ions. Then this ion will experience a reduction reaction and is deposited on the surface of the cathode.

The reaction that occurs in plating iron nail with zinc.

Cathode (Fe): Zn²⁺ (aq) + 2e⁻ ---------> Zn (s)

Anode (Zn): Zn (s) ----------> Zn²⁺ (aq) + 2e⁻

-------------------------------------------------- -----------

Zn (s) Anode -----> Zn (s) Cathode

The electrolyte used can be ZnCl₂

<h3>Learn more </h3>

an electrolytic cell

brainly.com/question/1319774

During electrolysis

brainly.com/question/11868966

the substance is produced at the cathode during the electrolysis

brainly.com/question/6854891

gtnhenbr [62]4 years ago

3 0

The anode is zinc, and the cathode is the nail

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A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given

m_a_m_a [10]

Answer:

$A=69.8\frac{bar}{m^6}\\\\ B=0.00302bar*m^6$

2. $W=-8.2kPa$

Explanation:

Hello,

1. In this case, for the given p-V equation, one could use the two states to form a 2x2 linear system of equations in terms of A and B:

$\left \{ {{0.1^2A+0.1^{-2}B=1} \atop {0.04^2A+0.04^{-2}B=2}} \right.$

$\left \{ {{0.01A+100B=1} \atop {0.0016A+625B=2}} \right$

Whose solution by any method for solving 2x2 linear system of equations (elimination, reduction or substitution) is:

$A=69.8\frac{bar}{m^6}\\\\ B=0.00302bar*m^6$

2. Now, for us to compute the work, we must first compute n, as the power relating the pressure and volume for this process:

$P_1V_1^n=P_2V_2^n\\\\\frac{P_1}{P_2}=(\frac{V_2}{V_1} )^n\\\\\frac{1bar}{2bar}= (\frac{0.04m^3}{0.1m^3} )^n\\\\0.5=0.4^n\\\\n=\frac{ln(0.5)}{ln(0.4)} =0.7565$

Now, we compute the work:

$W=\frac{P_2V_2-P_1V_1}{1-n} =\frac{2bar*0.04m^3-1bar*0.1m^3}{1-0.7565} \\\\W=-0.082bar*m^3*\frac{1x10^2kPa}{1bar}\\ \\W=-8.2kPa$

Regards.

8 0

4 years ago

1. A student needs to make 80mL of a 4.5 x 10^-3 M solution of H_3PO_4 (mm = 98) from solid. Describe how the student would do t

slamgirl [31]

Answer:

1. 35 mg of H₃PO₄

2. 27 mol AlF₃; 82 mol F⁻

3. 300 mL of stock solution.

Explanation:

1. Preparing a solution of known molar concentration

Data:

V = 80 mL

c = 4.5 × 10⁻³ mol·L⁻¹

Calculations:

(a) Moles of H₃PO₄

Molar concentration = moles of solute/litres of solution

c = n/V

n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol

(b) Mass of H₃PO₄

moles = mass/molar mass

n = m/MM

m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg

(c) Procedure

Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,

2. Moles of solute.

Data:

V = 4900 mL

c = 5.6 mol·L⁻¹

Calculations:

Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃

Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.

3. Dilution calculation

Data:

V₁= 750 mL; c₁ = 0.80 mol·L⁻¹

V₂ = ? ; c₂ = 2.0 mol·L⁻¹

Calculation:

V₁c₁ = V₂c₂

V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL

Procedure:

Measure out 300 mL of stock solution. Then add 500 mL of water.

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4 years ago

What is the mass of 8.0 x 10^26 UF6 molecules?

gulaghasi [49]

First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23

Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles

Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol

Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams

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3 years ago

Which of the following best describes the valence electrons in a polar

Ghella [55]

Answer:

C. The electrons are closer to the atom with the higher

electronegativity.

Explanation:

In a covalent bond, the electrons are shared equally when the atoms involved have the same electronegativity. However, when there are differences of electronegativity between those atoms, the electrons preferred are closer to the atom with the higher electronegativity.

5 0

4 years ago

Which of the following molecules or ions has a bent shape?

Dovator [93]

It’s c I believe but I don’t know

4 0

3 years ago

Read 2 more answers