Effect of Concentration on the rate of reaction of magnesium ribbon in hydrochloric acid

Effect of concentration on rate of reaction of magnesium ribbon in hydrochloric acid

Aim / Objective: 

To investigate the effect of concentration on rate of reaction of magnesium ribbon in hydrochloric acid.

Abstract:  

The rate of reaction was determined by measuring the time required for a given amount of magnesium metal to be consumed by Hydrochloric acid (HCl) solution of varying concentrations.

Introduction: 

The rate of a chemical reaction is the time required for a given quantity of reactant(s) to be changed to product(s). The unit of time may be seconds, minutes, hours, days or years.

The rate is affected by several factors, some of which are listed as follows:

(1) Nature of the reactants, i.e., one metal may react vigorously with acid while another does not react.

(2) The particle size of the reactants, i.e., a lump of coal burns slowly but powdered coal may explode.

( 3 ) Temperature increases in general increase the rate of reaction, i.e., a 2O°C rise in temperature doubles the reaction rate.

( 4 ) Catalysts affect the rate by using or allowing a different pathway for the reaction to follow.

(5) Concentration affects the rate of reaction, i.e., if the concentration of one of the reactants is doubled and is an integral part of the reaction then rate increases appropriately.

Some reactions are fast and other reactions are slow. The rate of a specific reaction can be found only by experiment.

Apparatus/Materials:

Magnesium ribbon , ruler, scissors, analytical balance, sandpaper, hydrochloric acid, measuring cylinder, graduated cylinder, distilled water, glass stirring rod.

Method / Procedure:

  1. Clean a 25cm length magnesium ribbon lightly by using sandpaper to remove the surface oxide layer. Cut the clean ribbon into five equal pieces of 6cm using a ruler.
  2. Weigh the five pieces together and determine the mass of one piece assuming all six are the same.
  3. Make up 30mL of each of the following five solutions of hydrochloric acid (HCl) with water: 2.0M, 1.5M, 1.0M, 0.5M and 0.25M. To do this, calculate the volumes of 3.0 mol L-1 stock hydrochloric acid solution and water that must be mixed using the dilution equation: (number mL stock HCL solution) x [HCL]= (3.0mL diluted solution) x [HCL] diluted
  4. Measure the calculated volume of stock hydrochloric acid solution using 10mL graduated cylinder. Pour this acid into a 50-mL graduated cylinder and then dilute to the 30-mL mark by carefully adding water from a bottle.
  5. Make up the other for solutions in a similar way. Pour each diluted solution into a 50-mL beaker.
  6. Drop a piece of magnesium into the 2.0 mol L-1 acid solution and start timing the reaction. Stir gently at first using the glass stirring rod to make sure the metal does not stick to the sides of the beaker.
  7.  Measure the time elapsed when the reaction stops and record the time.
  8. Repeat the procedure with the other four acid solutions.
  9. **Your instructor may require you to plot a graph

Suggested Results:

Table showing the effect of varying concentration of HCl on rate of reaction of Magnesium ribbon

Table showing the effect of varying concentration of HCl on rate of reaction of Magnesium ribbon

Discussion:

Write a balanced equation for the reaction

>>> Mg(s) + 2HCl(aq)      =         MgCl2(aq)   + H2(g)

 

Write the ionic equation for the reaction

>>>Mg(s) + 2H+(aq)           =        Mg2+(aq)  + H2(g)

 

Identify the variables in the experiment

  • The manipulated variables were the HCl and the water
  • The responding variable was time
  • The controlled variable was the magnesium ribbon

Based on your experimental data, make a general statement about the effect of concentration of reactants on time and reaction rate

>>>Concentration affects the rate of reaction. Therefore over time as the concentration of HCl increased then the rate of the reaction also increased.

Source of Error/ Limitations/ Assumptions:

  •  Inaccurate timing. Time may be lost during the experiment between the times taken to notice the cross has disappeared to the actual stopping of the watch.
  • Inaccurate measurement of reactants will affect the overall rate of reaction.

Reactivity series of metals|Reaction with hydrochloric acid and sulphuric acid (Mg, Zn. Al, Fe, Pb, Cu)

Reactivity series of metals|Reaction with hydrochloric acid and sulphuric acid (Mg, Zn. Al, Fe, Pb, Cu)

Aim / Objective: 

To investigate the reaction of some metals: Mg, Zn. Al, Fe, Pb, Cu, with hydrochloric acid and sulphuric acid.

Apparatus/ Materials:

metals of magnesium (Mg) turnings, zinc (Zn) granules, aluminum (Al) turnings, iron (Fe) filings, lead (Pb) foil, copper (Cu), 6 test-tubes, test-tube rack, dilute hydrochloric acid (HCl), dilute sulphuric acid (H2SO4), splints, Bunsen burner and spatula.

Method / Procedure:

  1. Half fill a test- tube with dilute HCl.
  2. Add a spatula full of magnesium turnings to the acid, place cork stopper in the mouth of test-tube and observe for effervescence of gas.
  3. Record your observations. Heat gently under the Bunsen burner if no reaction is taking place or if it is too slow.
  4. Remove cork-stopper and place a lighted splint at the mouth of test-tube. Note the reaction.
  5. Repeat steps 1-4 with the remaining metals and dilute HCl. Then repeat the same procedure with dilute sulphric acid.
  6. Tabulate your results.

Suggested Results: 

reactivity table 1 reactivity table 2

 

Discussion:

reactivity series equations

reactivity series results

 

 

 

 

Source of Error/ Limitations/ Assumptions:  

  • Overheating of test-tube.
  • Allowing some gas to escape when HCl and H2SO4 is added

Oxidizing and Reducing Agents

 Identification of oxidizing and reducing agents using solutions of acidified potassium dichromate (VI), hydrogen peroxide, acidified potassium iodide, acidified potassium manganate (VII), bleach

Aim / Objective: 

To identify oxidizing and reducing agents using solutions of acidified potassium dichromate (VI), hydrogen peroxide, acidified potassium iodide, acidified potassium manganate (VII), bleach.

Introduction: 

An oxidizing agent, or oxidant, is one that gains electrons and is reduced in a chemical reaction. They are also known as electron acceptors, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced. Some examples of oxidizing agents include potassium nitrate, halogens and nitric acid.

A reducing agent, or reductant, loses electrons and is oxidized in a reaction.  A reducing agent is usually in one of its lower possible oxidation states and is the electron donor. A reducing agent is oxidized because it loses electrons in the redox reaction. Examples of reducing agents include formic acid, sulfite compounds and earth metals.

Materials/ Apparatus:

Hydrogen peroxide (H2O2), acidified iron (II) Sulphate (FeSO4), acidified potassium iodide (KI), acidified potassium dichromate(VI) (K2Cr2O7) acidified potassium manganate (VII) (KMnO4), sulphuric acid (H2SO4), bleach (NaClO), 7 test tubes, test tube rack, glass stirring rod, metal boiling tube holder, droppers, sodium hydroxide (NaOH).

Method / Procedure:

  1. Label test tubes 1-7 and place in test tube rack
  2. Half fill test tube 1 with acidified KI and use a dropper to add acidified potassium dichromate (VI) (K2Cr2O7) to the solution. Note color change.
  3. Half fill test tube 2 with acidified KI and use a dropper to add acidified potassium manganate (VII) (KMnO4) to the solution. Note color change.
  4. Half fill test tube 3 with acidified KI and use a dropper to add NaClO to the solution. Note color change.
  5. To test tube 4 half fill with hydrogen peroxide (H2O2) and use a dropper to add acidified KMnO4 to the solution. Note color change.
  6. Half fill test tube 5 with KI and use a dropper to add H2O2 to the solution. Note color change.
  7. Half fill test tube 6 with H2O2 and use a dropper to add acidified K2Cr2O7 to the solution. Note color change.
  8. To the final test tube half fill with FeSO4 and use a dropper to add acidified K2Cr2O7 to the test tube. Add sodium hydroxide (NaOH) to the test tube and note color change.
  9. Tabulate results

Suggested Results:

 

oxidizing and reducing agents 1 oxidizing and reducing agents 2 oxidizing and reducing agents 3 oxidizing and reducing agents 4

 

Hope this one helps, remember to keep reading and studying as the weeks go by and your exams will be great.

Qualitative Analysis on Compound X in order to identify cations and anions present

Qualitative Analysis on Compound X in order to identify cations and anions present

Aim / Objective: 

To perform qualitative analysis tests on compound X in order to identify cations and anions present.

Introduction: 

Qualitative analysis is a technique that is used to separate and detect cations and anions in a sample. Anions are atoms or groups of atoms that have gained an electron or electrons. The atoms that form ions most easily are the Group 17 or VII atoms, also called halides: Fluorine (F), Chlorine (Cl), Bromine (Br) and Iodine (I). These form anions with a -1 charge. Oxygen (O), Sulphur (S), Nitrogen (N) and Phosphorus (P) also form anions. Most anions are composed from multiple atoms, and are called polyatomic ions.

Cations are atoms that have lost an electron to become positively charged. For example: Sodium (Na)  has one valence electron, one electron in its outer energy level, so it tends to lose one electron, and to become an ion with a +1 charge

Materials/ Apparatus:

Bunsen burner, red litmus paper, blue litmus paper, distilled water, test tube rack, test tubes, glass stirring rod, metal test tube holder, dilute nitric acid (HNO3), sodium hydroxide (NaOH), potassium iodide (KI), barium chloride (BaCl2), ammonia solution, lead nitrate (Pb(NO3)2), silver nitrate (AgNO3), dilute hydrochloric acid (HCl), splint, dropper, compound X.

Method / Procedure:

  1. Heat solid compound X in a dry test tube.
  2. Test the gas evolved with red and blue litmus paper and a glowing splint.
  3. Record the observations.
  4. To a solution of compound X in a test tube add distilled water and dilute HNO3, then add AgNO3 and note observations.
  5. In a test tube mix solution X with distilled water, then add dilute HNO3 and BaCl2 to the mixture. Record the observations.
  6. In a test tube mix solution X with distilled water and heat the test tube moderately. Add dilute HNO3 and Pb(NO3) to the heated mixture. Record the observations.
  7. Mix a small portion of solution X with distilled water in a test tube. Use a dropper to add NaOH in excess to the mixture and heat test tube moderately. Record your observations.
  8. Mix a small portion of solution X with distilled water in a test tube. Add excess ammonia to the mixture and note the results.
  9.  Half fill a test tube with solution X and add KI(aq). Record the observations.
  10. Half fill a test tube with solution X, add NaOH(aq) and heat the test tube moderately. Dip the glass stirring rod with the mixture in a beaker of HCl. Test the gas evolved with litmus paper. Record the observations.
  11. Tabulate results

Suggested Results: 

qualitative analysis 1 qualitative analysis 2

 

Conclusion:

  1. The cations present In compound X were Al3+ and NH4+
  2. The anions present were SO42- and Cl

 

Properties & Characteristics of Matter

Properties & Characteristics of Matter

Any characteristics that can be used to describe or identify matter are called a property. Some well-known properties are mass color melting point, boiling point, temperature etc.

Properties are often classified as either physical or chemical:

Physical properties- identify the substance without causing a change in its chemical composition. Alternatively it’s a property that can be measured without change of identity or composition. Examples are color melting point, mass density, electrical conductivity atomic or ionic size evaporation, condensation etc.

Chemical properties – one that can be observed only when the substance changes its chemical composition. In other words how a substance reacts with one another e.g. combustion, explosion, rusting.

Intensive / Extensive Properties of Matter 

Properties of matter can also be classified as either extensive or intensive.

Intensive- has values that do not depend on sample size. This property helps to identify substances e.g. color b.p m.p taste odor, temperature, density.

Extensive- have values that depend on the amount of sample present e.g. mass, volume length etc.

Physical and Chemical Changes

Physical change- a change in the form or state of matter but in not in its chemical make-up.

Examples are:

  • Dissolving salt in water. The water and the salt in the solution retain their chemical identities and can be separated.
  • Melting ice to liquid water.

Chemical change – a chemical reaction, a change in the composition of the substances that takes place and a new and different substance is formed.

Examples are:

  • Hydrogen and oxygen gas explode with a big bang when ignited producing water
  • Enzymatic digestion of food by different enzymes

Chemical Composition of Matter 

A sample of matter can also be classified based on its chemical composition.

Any sample of matter can be classified as either a pure substance or a mixture. A pure substance can either be an element or a compound.

properties of matter

 

Pure substance- has a fixed composition. It cannot be separated or purified into other kinds of substances by physical means. E.g. NaCl, H2O

Element- a substance that is comprised of only one type of atom. It cannot be chemically broken down into different simpler substances e.g. all the elements in the periodic table

Molecules – a substance formed when two or more atoms join together chemically e.g. iodine I2 or Water H2O

Compound – a substance composed of two or more different elements chemically combined or any pure substance that can be broken down by chemical means into two or more different, simpler substances e.g. NH3, H2O, Glucose.

A compound is a molecule that contains at least two different elements all compounds are molecules but not all molecules are compounds. Molecular hydrogen and molecular oxygen are not compounds because each is composed of a single element and cannot be broken down chemically into anything simpler. Water H2O, carbon dioxide CO2 and methane CH4 are compounds because each is made from more than one element. Water can be decomposed through electrolysis into hydrogen and oxygen gas.

Mixture – formed when two or more substances are mixed together in some random proportion without chemically changing the individual substances in the process, e.g. sugar in water, coke, fruit juices etc.

Law of Conservation of Mass – mass can neither be created nor destroyed; the total mass of the substances involved in a physical or chemical reaction remains constant.

e.g. During the rusting of a nail, the initial mass of the iron plus the mass of the oxygen combines to produce a mass of iron oxide (rust) equal to the sum of masses of iron and oxygen consumed.

Law of conservation of energy – energy changes can neither be created nor destroyed in a chemical process or physical change. It can only be converted from one form to another

Energy is the capacity or ability to do work or to transfer heat. E.g. light energy, electrical energy, and heat energy, mechanical and chemical energy.

Anytime a chemical reaction takes place, there is also a change in energy. Either energy is released by the reaction -exothermic reaction (e.g. metabolism) or energy is required to keep the reaction going – endothermic reaction (photosynthesis)

 

Separating a mixture of Oil and Water

Separating a mixture of Oil and Water using a Separating Funnel Aim / Objective: To separate a mixture of cooking oil and water. Abstract: In this experiment, an immiscible mixture of oil and water was separated using a separating funnel. Materials/ Apparatus: measuring cylinder, retort stand, beaker, clamp, separating funnel, conical flask, cooking oil and …

Continue reading ‘Separating a mixture of Oil and Water’ »

30 FACTS YOU MUST KNOW TO PASS CHEMISTRY (PART 1)

 30 FACTS YOU MUST KNOW TO PASS CHEMISTRY (PART 1)

  1. An atom is defined as the basic unit of matter which consist of a nucleus and a cloud of negatively charged electrons.
  2. A proton is a positively charged particle and is found in the nucleus of the atom
  3. An electron had a negative charge and surrounds the nucleus
  4. A neutron has no electric charge.
  5. Atomic number is the same as the number of protons in the nucleus
  6. Mass number is equal to the number of protons plus the number of neutrons
  7. Number of neutrons is found by subtracting the atomic number from the mass number
  8. Number of protons is equal to number of electrons in a neutral atom.
  9. The number of particles in one mole of a substance is called Avogadro’s number and is calculated to be 6.02 x 1023
  10. When electrons become excited they move from one energy level to a higher energy level. As the fall back to ground state they emit energy as light and a bright line spectra is produced.
  11. The distribution of an electron in an atom is referred to as the electronic configuration.
  12. Anions are negative ions and is formed when atoms gain electrons
  13. Cations are positive ions and are formed when atoms lose electrons
  14. Elements are pure substances consisting of one or more classes of substances that cannot be broken down into simpler substances by chemical processes.
  15. Isotopes are any two or more forms of an element with same number of protons but different number of neutrons
  16. A solution is a homogenous (same) mixture of two or more substances
  17. Heterogeneous mixtures consist of different, discernable substances and are not uniformed throughout.
  18. A solute is a substance that is being dissolved e.g. salt, while a solvent is the substance (usually liquid) that dissolves the solute e.g. water.
  19. Chemical formulas are written so that the charges on the cation cancel one another.
  20. Coefficients refer to those numbers written in front of products and reactants in order to balance a chemical equation.
  21. Reactants are found on the left side of the reaction arrow and the products are found on the right.
  22. Endothermic reactions absorb the energy from the surrounding while exothermic reactions release energy.
  23. Physical changes refer to the appearance of the material and do no form new substances
  24. Chemical changes result in the formation of new substances
  25. The “Law of Conservation” of energy states that the masses of the reactants in a chemical equation will always be equal to the masses of the products.
  26. There are three states of matter: solid, liquids and gases.
  27. Solids have a definite shape and volume.
  28. The particles in liquids are closely spaced and can easily move alongside each other. They have no definite shape however they have a definite volume.
  29. Gases have widely spaced particles that are in constant random motion and will collide with the walls of the container to create pressure.
  30.  “STP” stands for Standard, Temperature and Pressure.

Now that you are familiar with or have reminded yourself of some of these interesting “MUST KNOWS” in chemistry, wait for Part 2. So while you wait, don’t forget to read a chemistry book, read a chapter, a half of a chapter or even two pages. Just keep the brain fresh with chemistry and your exams will be a breeze!

Let us know how the reading went, join the conversation on the Facebook page and Twitter.

 

Acids, Bases and Indicators

Acids, Bases and Indicators

An acid was defined n the early times as a substance that tastes sour. Vinegar, which contains acetic acid, is sour, and so is lemon juice, which contains citric acid. Acids were also known to change the colors of organic dyes. The dye present in red cabbage, for example, turns from reddish purple to brighter red color when the cabbage is marinated with vinegar. A common organic acid dye used to identify acids is litmus, which is derived from various lichens. Litmus turns from blue to red when mixed with an acidic solution.

Bases, or alkalis, also known from ancient times, have a bitter taste and slippery feel. Bases turn red litmus back to blue, and turn the otherwise colorless organic dye phenolphthalein a bright pink.

The Arrhenius theory of acids and bases defines an acid as a substance which, when placed in water, generates protons (H+ ions), and base as a substance which, when placed in water, generates hydroxide ions (OH-).

Many compounds may be reorganized from their formulas. In the case of acids, the available hydrogen atoms which are responsible for the acidity are written first, followed by the symbols of their other elements in the formula. Similarly, some bases (but not all) may be recognized by their formulas as an hydroxide ion combined with a positive metal ion.

Examples:                                                          ACIDS

Hydrochloric acid              HCL

Nitric acid                            HNO3

Sulphuric acid                    H2SO4

Acetic acid                           HC2H3O2

Oxalic acid                           H2C2O4

Carbonic acid                     H2CO3

                        BASES

Sodium hydroxide           NaOH

Potassium hydroxide     KOH

Calcium hydroxide           Ca(OH)2

Ammonia                            NH3

 

According to Arrhenius theory, the solutions of hydrochloric acid and sodium hydroxide may be represented by their respective equations:

 

HCl(aq)     =         H+(aq) + Cl-(aq)

NaOH(aq)         =           Na+(aq) + OH-(aq)

 

The hydrogen ion, present in all aqueous acid solutions is species responsible for all the acidic properties from the sour taste to the color of litmus. Likewise, the hydroxide ion is responsible for all the basic properties of a solution.

Water behaves as both acid and as a base, as illustrated by the equation:

H2O +H2O                =           H3O+     +    OH-

 One water molecule has donated a proton (acted as an acid) and another water molecule has accepted a proton (acted as a base). Since hydronium ions and hydroxide ions are produced in equal concentrations, water is neutral. Aqueous solutions contain both hydronium and hydroxide ions. In acidic solutions, the concentration of hydronium ions is greater than that of hydroxide ions. In basic solutions, the concentration of hydroxide ions is greater than that of hydronium ions.

Indicators

Substances used to determine whether a solution is acidic or basic are known as indicators. Indicators are usually organic compounds whose color depends on the concentration of H+ ions. The table below lists several common indicators, their color in acid and in base solutions, and the approximate pH (given as a range) at which the color changes.

Since:   pH = -log (H+ concentration

An indicator can also be used to determine the approximate pH of a given solution.

INDICATOR Color IN ACID SOLUTION CHANGE RANGE pH Color IN BASIC SOLUTION
Methyl orangeMethyl redBromothymol bluePhenolphthalein

Litmus paper

OrangeRedYellowColorless

Red

2-43-66-88-10

5-8

Yellow

Yellow

Blue

Pink

Blue

 

There are several characteristic reactions of acids and bases:

Neutralization is a reaction of an acid and a base to form a salt and water:

 

e.g:                                                                        HCl(aq) + NaOH(aq)     =          H2(g) + NaCl(aq)

In neutralization reaction, the hydrogen ion of the acid solution unites with the hydroxide from the basic solution to form water molecule. The net ionic equation is:

H+(aq) + OH(aq)      =         H2O(l)

For the above reaction, when NaOH solution is added to HCL solution containing a few drops of phenolphthalein, the solution turned from colorless to pink when the acid is neutralized and one more drop of the base is abed.

The reaction of an active metal with acids forms a salt and hydrogen gas (a SINGLE REPLACEMENT REACTION):

Mg(s) + 2HCl(aq)        =            MgCl2(aq) + H2(g)

The reaction of a carbonate with an acid forms a salt, carbon dioxide and water:

CaCO3(s) + 2HCL(aq)         =          CO2(g) + H2O(l) + CaCl2(aq)

The reaction of a strong base with ammonium salts forms ammonia gas, water and salt:

2KOH(aq) + (NH4)2 SO4 (aq)       =         2NH3(g)  + 2H2(l) + K2SO4(aq)

The reaction of a transition metal salt with a base results in precipitation of the transition metal hydroxide and another salt:

3 Sr(OH)2(aq_ + 2FeCl3(aq)        =        2Fe(OH)3(s) +3SrCl2(aq)

Be sure to see our lab on indicators in the Labs and Experiments page.

Solubility and Electrical Conductivity of molecular and ionic solids | Sugar| Sodium Chloride| Iodine|

Solubility and Electrical Conductivity of molecular and ionic solids | Sugar| Sodium Chloride| Iodine|

Aim / Objective:

To investigate the solubility and electrical conductivity of molecular and ionic solids.

Introduction: 

Ionic solids are formed by ionic bonding where the particles in the solids are ions. These solids are crystals because of their structures. The regular structure of a crystal tells you that the ions are arranged in an orderly manner. This orderly arrangement is a crystal lattice.

Most substances that contain covalent bonds are also described as simple molecular substances, because they consist of separate molecules. Sugar is a molecular crystal which has a low melting and boiling point.

Sodium Chloride is an ionic compound which has a high melting and boiling point due to strong forces which exist between the ions.

Iodine, like sugar is described to be a molecular crystal for it consists of separate molecules.

Materials/ Apparatus:

Iodine and Sugar solid (molecular solids), sodium chloride (ionic solid), distilled water (polar solvent), tetrachloromethane (non-polar solvent), beaker, electrolytic cell, test tubes, measuring cylinder.

Method / Procedure:

  1. Label three test-tubes A, B and C. In test tube A put one crystal of iodine solid.
  2. In test-tube B, put a spatula full of sugar and finally in test-tube C put a spatula full of sodium chloride.
  3. Add 5cm3 of distilled water to the three test tubes. Record your observations.
  4. Repeat steps 1 and 2, then add 5cm3 of tetrachloromethane and record your observations.
  5. Tabulate results.
  6. Label three (3) beakers A, B and C. In beaker A, mix one crystal of iodine and 25cm3 of tetrachloromethane.
  7. In beaker B, mix three spatulas full of sugar and 25cm3 of water and in beaker C, mix three spatulas full of sodium chloride and 25cm3 of water.
  8. Place electrodes into the solutions and close the switch. Watch the ammeter for readings and record the registered reading to determine if the solution conducted an electric current.
  9. Tabulate results.

Suggested Results: 

Table showing the electrical conductivity of iodine, sugar and sodium chloride

Table showing the electrical conductivity of iodine, sugar and sodium chloride

Table showing the solubility of sugar, iodine and sodium chloride in water and solvent

Table showing the solubility of sugar, iodine and sodium chloride in water and solvent

Discussion/Answers:

1. State the reason(s) why sugar did not conduct electricity when dissolved in water

>>>This is due to the absence of ions, for when ions are set free they move to an oppositely charge electrode when a voltage is applied.

2. State the reason for the differences in solubility of sugar in water and tetrachloromethane

>>>The reason why sugar dissolves in water and not in tetrachloromethane is because sugar has polar hydroxyl groups and polar substances dissolve in polar solvent (water), whereas they do not dissolve in nonpolar organic solvent (tetrachloromethane).

3. Explain why sodium chloride was the only substance that conducts electricity

>>>The sodium chloride solutions conducts electricity because the ions are set free upon melting and these ions will furthermore move to an oppositely charge electrode when a voltage is applied.

4. State the reason for the differences in solubility of NaCl in water and tetrachloromethane

>>>The reason why sodium chloride dissolves in water and not in tetrachloromethane is because NaCl has polar hydroxyl groups and polar substances dissolve in polar solvent (water), whereas they do not dissolve in nonpolar organic solvent (tetrachloromethane).

5. Why did iodine not conduct electricity?

>>> Iodine did not conduct electricity when molten or dissolved due to the absence of ions from its structure.

6.  State the reason for the differences in solubility of iodine in water and tetrachloromethane

>>>Iodine consists of molecules and dissolves in tetrachloromethane and is insoluble in water. This is because tetrachloromethane is a non-polar organic solvent and water is a polar solvent, and with iodine being a non-polar molecular substance it will readily dissolve in solvents that are non-polar. This means essentially that the atoms of the element are identical and has the same attraction to the tetrachloromethane.

Source of Error/ Limitations/ Assumptions: 

–          Incorrect reading from the ammeter

–          Contamination of solids resulting in incorrect voltage readings

–          Parallax error

–          Incorrect measurement of solutions

–          Not dissolving solids completely

Remember to always consult your textbook for any other useful information that would really impress your teacher.

Rate of Diffusion of two gases – Ammonia (NH3) and Hydrochloric Acid (HCl)

Rate of Diffusion of two gases – Ammonia (NH3) and Hydrochloric Acid (HCl)

Aim / Objective:

To compare the rates of diffusion of two gases- Ammonia (NH3) and Hydrochloric Acid (HCl)

Introduction: 

The particles in a fluid are in continuous random motion. In light of this, gases and gaseous mixtures spread out to occupy any space available to them, eventually acquiring a uniformed composition. The process by which fluid mixtures kept a constant temperature becomes uniformed is known as diffusion.

In this experiment, a comparison on the rate of diffusion of ammonia and hydrogen chloride will be done.

Materials/ Apparatus:

Glass tubing, rubber bung, concentrated ammonia solution, concentrated hydrochloric acid, cotton and tape, stop-watch

Method / Procedure:

  1. Set up apparatus as shown below.
  2. Soak one cotton wool in the concentrated ammonia solution and a next piece in the concentrated hydrochloric acid.
  3. Insert both cotton wools simultaneously at points A (one end of the tube) and B (the other end of the tube) respectively and then quickly insert rubber bungs at both ends of the as shown in the diagram.
  4. Begin timing and record the time taken for the white ring to for in the glass tube.
  5. Record the following distances:
    •  From End A to the center of the white ring and label this distance as X
    •  From End B to the center of the white ring, and label this distance as X2
Diagram of apparatus showing the diffusion of ammonia and hydrochloric acid

Diagram of apparatus showing the diffusion of ammonia and hydrochloric acid

 

Suggested Results:

Diagram showing the reaction of  the diffusion of ammonia and hydrochloric acid

Diagram showing the reaction of the diffusion of ammonia and hydrochloric acid

  1. After the cotton wools were placed in the tube, it took approximately 13mins and 22 seconds for a reaction to start taking place.
  2. When the two gases reacted there was a white ring formed 2.5cm from the cotton wool with the HCl and 29.5cm from the cotton wool with the NH3
  3. The length of the tube was estimated to be 32cm and from the experiment, one can deduce that HCl is more dense than NH3 because the NH3 traveled a farther distance than the HCl in the same period of time , hence the white ring formed closer to the cotton with the HCl (Hydrogen Chloride)

Discussion/ Answers:

1. Define diffusion.

>>>> Diffusion is the process by which molecules or ions move from an area of high concentration to a region of low concentration along a concentration gradient.

2. Use the values of X1 and X2 and time (t1) to calculate the relative rates of movements of ammonia gas and hydrogen chloride under similar conditions.

>>> Rate = distance / time

Time = 13m 22s gives 802 seconds

(a) Distance travelled by HCl = 0.25m

   Rate of HCl = Distance/ Time = 0.25m/ 802 seconds

                                                         = 3.12 x10-4 ms-1

(b) Distance travelled by NH3 = 0.295m

Rate of NH3 = Distance/ Time = 0.295m/ 802 seconds

                                                         = 3.67 x10-4 ms-1

3. Calculate the relative molecular masses of ammonia Mr(NH3) and Mr(HCl)

>>>  Relative molecular mass of HCl =

H                                   Cl                      = HCl

1                                   35.5                   = 36.5

>>>Relative molecular mass of NH3 =

N                                   H3                      = NH3

14                                 (3×1)                   = 17

Explain why the rates of diffusion for the two gases are different.

>>> The rate of diffusion for the gases were different even though they were under the same conditions was a result of ammonia being less dense and moving at a further distance than the hydrogen chloride. The formula given for rate is distance/ time thus ammonia’s rate of diffusion higher than that of HCl as it moved a further distance.

In one short sentence explain why concentration affects rate of diffusion.

>>> The steeper the concentration gradient is the faster the rate of diffusion.

Write a balanced equation for the formation of the white ring.

>>> NH3(g)       +                HCl(g)                      =         NH4Cl

Ammonia    +      hydrogen chloride       =      ammonium chloride.

 

Sources of Error/ Limitations/ Assumptions:

  1.  Inaccuracy in timing due to strips not changing entirely and the time is recorded too soon.
  2. Inconsistency with volume of ammonia used.

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