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3 years ago

Metal rings can be coated with a layer of copper using electricity.

Chemistry

1 answer:

Eduardwww [97]3 years ago

5 0

<u>First of all, what is electrolysis?</u>

Electrolysis is the process of breaking down ionic substances using direct current.

<u>Important points about electrolysis </u>

→ Ionic substances contain particles called ions.

→ Electricity is the flow of electrons or ions. For electrolysis to work, the compound must contain ions. The ions must be free to move for electrolysis to occur and it can happen by melting or dissolving an ionic substance in water.

→ Positively charged ions move to the negative electrode. They receive electrons and are <em>reduced</em>. The positive ions move towards the negative electrode because they want to cancel each other out.

→ Negatively charged ions move to the positive electrode. They lose electrons and are <em>oxidised</em>. The substance that is broken down is called the electrolyte <em>(an electrolyte is just a liquid or solution that can conduct electricity)</em> . The negative ions move towards the positive electrode because they want to cancel each other out.

<h3>Cathode = Negative electrode</h3><h3>Anode = Positive electrode</h3>

Metal ions form at the cathode and non-metal ions form at the anode

How I remember if an element is <em>oxidised</em> or <em>reduced</em> is by remembering OIL RIG

OIL = Oxidation is Loss (of electrons)

RIG = Reduction is Gain (of electrons)

<h2><em><u>The answer to your question</u></em></h2>

1) The first step would be to clean the metal ring and sand it down because when the metal atoms from the electrolyte are deposited onto the ring, they will form a weak bond and they may simply 'fall' off. Also this could affect conductivity and the whole experiment. The more things you do accurately now, the more accurate your result will be.

2) You want to put the solution you are given in to the tank your going to be using.

3) This is basically the main part, you want to set up the circuit, I have attached a diagram at the bottom to show you the circuit. The copper rod will be the anode and the metal ring will be a cathode (ignore the elements).

4) Now turn on the circuit and you will start to see the solution spilt with the the solution now being split some going to the anode and some going the cathode.

5) Then a thin layer should form on the electrode.

Hope this helps :)

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Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

A container with a volume 2. 0 L is filled with a gas at a pressure of 1. 5 atm. By decreasing the volume of the container to 1.

worty [1.4K]

The resulting pressure of the gas after decreasing the initial volume from 2 L to 1 L is 3 atm.

<h3>What is Boyle's Law?</h3>

According to the Boyle's Law at constant temperature, pressure of the gas is inversely proportional to the volume of that gas.

For the given question we use the below equation is:

P₁V₁ = P₂V₂, where

P₁ = initial pressure of gas = 1.5 atm

V₁ = initial volume of gas = 2 L

P₂ = final pressure of gas = ?

V₂ = final volume of gas = 1 L

On putting all these values on the above equation, we get

P₂ = (1.5atm)(2L) / (1L) = 3 atm

Hence required pressure of the gas is 3 atm.

To know more about Boyle's Law, visit the below link:
brainly.com/question/469270

5 0

2 years ago

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

$24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of 0.541284 moles of MgO:

$m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}$ .

7 0

3 years ago

use the internet to look up the sds for 2.0 m sodium hydroxide naoh to answer the following question a) list the potential acute

VashaNatasha [74]

A safety data sheet is informational data that states the properties of the various chemicals. According to the SDS of NaOH, it causes irritation, nausea, eye burn, burns in the digestive system, etc.

SDS is abbreviated for a safety data sheet which is a document that provides information on the chemicals including the properties, health hazards, measures, safety, prevention, etc.

According to SDS, NaOH results in various potential acute health effects including vomiting, diarrhea, nausea, irritation, etc. The chronic effect includes coma, burns in the eye, respiratory infection, digestive system, dermatitis, etc.

Therefore, sodium hydroxide has various acute and chronic health effects.

Learn more about SDS, here:

brainly.com/question/2471127

#SPJ4

Your question is incomplete, but most probably your full question was, Use the internet to look up the sds for 2.0 m sodium hydroxide, NaOH, to answer the following question: list the potential acute and chronic health effects.

6 0

1 year ago

HELP!!!!!!!!!!!!!!!!!!! 100 POINTSSSSSSSSSSSSSS

e-lub [12.9K]

Answer:

<em><u>The three-dimensional region of space that indicates where there is a high probability of finding an electron.</u></em>

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3 years ago

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