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Swali 1 Ripoti
When a few drops of Millon reagents is added to egg-white solution in a test tube, the white precipitate changes to
Maelezo ya Majibu
When a few drops of Millon's reagent is added to an egg-white solution in a test tube and the solution is boiled, the white precipitate turns brick red. This indicates the presence of proteins.
Swali 2 Ripoti
For chemical reaction to be spontaneous, ∆G must be
Maelezo ya Majibu
In the context of chemical reactions, the spontaneity of a reaction is determined by the Gibbs Free Energy change, represented by the symbol ΔG. A chemical reaction is considered to be spontaneous if it proceeds on its own without needing continuous external input of energy.
For a reaction to be spontaneous, the value of ∆G must be negative. This is based on the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
Where:
A negative value for ΔG indicates that the process releases energy and will proceed spontaneously. This means the system is moving towards a lower energy and more stable state, naturally favoring the products over the reactants.
In contrast, a positive ΔG indicates that the reaction is non-spontaneous and requires energy input. If ΔG is zero, the system is at equilibrium, meaning there is no net change taking place, but this doesn't indicate spontaneity.
Therefore, in summary, for a reaction to be spontaneous, ∆G must be negative.
Swali 3 Ripoti
| COMPOUND | S | T | U | V | W |
| FORMULA | ROR' | RCOOH' | RCOR' | ROH' | RCOOR' |
From the table above, which of these two compounds can form functional group isomers?
Maelezo ya Majibu
ROH' and ROR' can form functional group isomers because they are the functional groups of alcohols and ethers, respectively.
Ethers have a pair of alkyl or aromatic groups attached to a linking oxygen atom. ROH is the functional group of alcohols, which are derivatives of water with one hydrogen atom replaced by an alkyl group.
Alcohols (ROH) and ethers (ROR') can form functional group isomers because they have the same chemical formula but different functional groups. E.g CH3 CH2 OH and CH3 OCH3
Swali 4 Ripoti
An example of a substance that does not change directly from solid to gas when heated is
Maelezo ya Majibu
When discussing the process of substances changing states, some substances can transition directly from a solid to a gas without passing through a liquid state. This process is called sublimation. However, not all substances exhibit this behavior. Let's examine the substances provided:
In conclusion, calcium carbonate (CaCO3) is the substance that does not change directly from a solid to a gas when heated, as it undergoes a decomposition process instead.
Swali 5 Ripoti
The volume occupied by 1 mole of an ideal gas at a temperature of 130 C and a pressure of 1.58 atm is
[ R = 0.082 atm dm3 K−1 mol−1 ]
Maelezo ya Majibu
According to the Ideal gas equation, PV = nRT
Given: P = 1.58 atm, V = ?, n = 1 mole, R = 0.082, T= 13 + 273K = 286K
Substituting all the given parameters,
V = nRTP
V = 1×0.082×2861.58
V = 14.84 dm3
Swali 6 Ripoti
Calculate the number of moles of Copper that will be deposited, if 2 Faraday of electricity is passed through the copper during the electrolysis of copper(II)tetraoxosulphate(VI)
[1F = 96500C ]
Maelezo ya Majibu
The electrolysis of copper(II) tetraoxosulphate(VI) involves the deposition of copper at the cathode. To understand how many moles of copper are deposited when 2 Faraday of electricity is passed through, we need to consider Faraday's first law of electrolysis. Faraday's first law states that the mass (or number of moles) of a substance deposited at an electrode is directly proportional to the quantity of electricity that is passed through the electrolyte.
A Faraday (or Faraday constant) is the charge of one mole of electrons, which is approximately **96500 coulombs** (C). During electrolysis, the chemical reaction occurring at the cathode for copper deposition can be represented by the following equation:
Cu2+ + 2e- → Cu
This equation shows that **2 moles of electrons** (represented by 2e-) are needed to deposit **1 mole of copper (Cu)**.
If we have **2 Faradays** of electricity, it means we have **2 x 96500 C = 193000 C**. Since **1 Faraday (96500 C)** is required to deposit **0.5 mole** of copper, **2 Faradays** will deposit twice that amount:
0.5 mole of copper deposited per Faraday x 2 Faradays = **1.0 mole** of copper
Thus, when **2 Faradays** of electricity are passed through copper(II) tetraoxosulphate(VI) solution, **1.0 mole** of copper will be deposited.
Swali 7 Ripoti
Kerosene is used as solvent for
Maelezo ya Majibu
Kerosene is commonly used as a solvent for paints. Let me explain why in a simple way:
Kerosene is a type of fuel that is composed of hydrocarbons, which are molecules made up of hydrogen and carbon atoms. These hydrocarbons give kerosene the ability to dissolve other similar substances.
Paints often contain oils and other hydrocarbon-based compounds. Since kerosene is also hydrocarbon-based, it can effectively dissolve and thin these compounds. This makes it suitable for use as a solvent in paints, allowing the paint to be thinned or cleaned up after use. This property makes kerosene a good choice for cleaning brushes and other painting tools or for dissolving dried paint.
On the other hand, sulphur, gums, and fats are typically not dissolved effectively by kerosene because of their different chemical properties. Therefore, kerosene as a solvent is primarily useful in the context of working with paints and similar hydrocarbon-based materials.
Swali 8 Ripoti
A type of isomerism that ClCH=CHCl can exhibit is
Maelezo ya Majibu
ClCH=CHCl can exhibit geometrical isomerism and positional isomerism. ClCH=CHCl can exhibit positional isomerism because the positions of the functional groups or substituent atoms are different. Positional isomerism occurs when compounds with the same molecular formula have different properties due to the difference in the position of a functional group, multiple bond, or branched chain.
Swali 9 Ripoti
The amount of water a substance chemically combined with is called water of
Maelezo ya Majibu
The amount of water that is chemically combined with a substance is referred to as water of crystallization. This is the water present in the crystalline form of a compound, necessary to maintain the structure of the crystals.
When certain substances crystallize from an aqueous solution, they incorporate a specific amount of water molecules into their crystal lattice structure. These water molecules are an integral part of the crystal and often affect its color, stability, and solubility. The water is combined in stoichiometric amounts, which means it is present in a fixed ratio relative to the rest of the molecule.
An example of this is copper(II) sulfate pentahydrate, which consists of copper(II) sulfate combined with five molecules of water per formula unit, represented as CuSO4·5H2O.
Swali 10 Ripoti
Which of the following represents an order of increasing reactivity?
Maelezo ya Majibu
To determine the order of increasing reactivity of the elements listed, it's important to understand the general trends in metal reactivity. Metals react by losing electrons, and their reactivity is often influenced by their ability to lose these electrons easily. In many cases, generally, alkali metals are the most reactive, and noble metals are the least reactive. Here's a basic description of the reactivity of the given metals:
With these considerations in mind, the order of increasing reactivity from the given options would be:
Gold (Au) < Copper (Cu) < Tin (Sn) < Iron (Fe) < Calcium (Ca)
This is the order where the least reactive element is first (gold), and the most reactive element is last (calcium). Hence, the correct option represents the order: Au < Cu < Sn < Fe < Ca.
Swali 11 Ripoti
Rust on the surface of a metal sheet contains
Maelezo ya Majibu
Rust on the surface of a metal, specifically on **iron**, is primarily composed of **hydrated iron(III) oxide**. The rusting process occurs when **iron** reacts with **oxygen** and **water** from the environment. This chemical reaction typically produces a compound called **iron(III) oxide**, which is then combined with water molecules, resulting in **hydrated iron(III) oxide**. This hydrated state gives rust its characteristic flaky and reddish-brown appearance.
Swali 12 Ripoti
An example of a compound that is acidic in solution is
Maelezo ya Majibu
Phosphoric acid is a weak acid that can donate three hydrogen ions in water. Phosphoric acid partially ionizes when dissolved in an aqueous solution.
Swali 13 Ripoti
The percentage of carbon(IV) oxide in air is
Maelezo ya Majibu
The air we breathe is made up of a mixture of gases. The most abundant gases in the atmosphere are nitrogen and oxygen, but there are other gases present in smaller amounts, one of which is carbon dioxide, chemically known as carbon(IV) oxide.
Carbon dioxide makes up approximately 0.03% of the Earth's atmosphere by volume. This value can also be expressed in different terms, such as 300 parts per million (ppm). Even though it is a small percentage, carbon dioxide plays a significant role in maintaining the Earth's temperature through the greenhouse effect.
In summary, the percentage of carbon(IV) oxide in air is 0.03%.
Swali 14 Ripoti
What would be the order of the electrolytic cell in an industry intending the production of silver plated spoons?
Maelezo ya Majibu
In the process of silver plating a spoon using an electrolytic cell, the correct configuration involves the following:
Cathode: The object to be plated, which in this case is the spoon. In an electrolytic cell, the cathode is where the reduction reaction occurs, and it is the surface on which the metal ions are deposited.
Anode: A rod made of silver. The anode is where oxidation occurs, meaning the silver rod will dissolve into the solution in the form of silver ions. These ions then move towards the cathode to be deposited as a thin layer on the spoon.
Electrolyte: A solution that contains a soluble silver salt (such as silver nitrate, AgNO3). The silver ions from this salt help in the process of transferring the silver from the anode to the cathode.
Thus, the proper order for silver plating a spoon in an electrolytic cell for industrial production is: "Cathode is the spoon; anode is a silver rod; electrolyte is a soluble silver salt."
Swali 15 Ripoti
The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is
Maelezo ya Majibu
The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is water.
Here's why:
Hence, the correct answer is water, as it is the substance that both reacts with sodium to form an alkali and changes the color of anhydrous copper(II) tetraoxosulphate (VI) to blue.
Swali 16 Ripoti
The IUPAC Nomenclature of CH3 CH2 C(CH3 )=C(CH3 )2 for the compound is
Maelezo ya Majibu
The compound in question is written as CH₃₃CH₂₂C(CH₃₃)=C(CH₃₃)₂₂, which seems to be intended as (CH3)3CH2CH=C(CH3)3. The IUPAC nomenclature of organic compounds follows specific rules to name the compound uniquely such that it is understood universally. Here is a comprehensive breakdown:
1. Select the longest carbon chain that includes the highest-order functional group, which, in this case, is the alkene group (double bond).
2. The longest chain consists of 5 carbons, which gives us the root name "pentene". We choose the carbon chain such that the double bond gets the lowest possible number, starting from the end of the chain closest to the double bond.
3. Number the carbon atoms in the chain from the end closest to the double bond. The numbering direction will determine the position of the double bond and substituents. The double bond starts on carbon 2.
4. Identify and name the substituents attached to the carbon chain. In this case, there are two methyl groups on carbon 3. This means it is dimethyl as there are two of them.
Thus, the complete name of the compound is 2,3-dimethylpent-2-ene. Here, "2,3-dimethyl" indicates the position and quantity of methyl groups, "pent" indicates the longest chain with 5 carbons, and "-2-ene" indicates a double bond starting at the second carbon.
Swali 17 Ripoti
An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5? [ C =12, H = 1, N = 14, O = 16].
Maelezo ya Majibu
To find the molecular formula of the compound, follow these steps:
1. Determine the Empirical Formula:
Start by assuming you have 100 grams of the compound. This means you have:
Now, convert these masses to moles using their atomic masses (C = 12, H = 1, N = 14, O = 16):
Next, divide each by the smallest number of moles to get the simplest ratio:
This gives us the empirical formula: C5H7NO2.
2. Determine the Molecular Formula:
The molecular formula is a multiple of the empirical formula. To determine this multiple, we need to find the empirical formula mass and compare it with the molar mass derived from the given vapor density.
Calculate the empirical formula mass:
The molar mass can be calculated from the vapor density:
Now, find the ratio of the molar mass to the empirical formula mass:
This ratio is approximately 1, indicating the molecular formula is the same as the empirical formula. Since empirical formulas typically should perfectly match the atomic proportions we derive from experiments, our calculations regarding the assumptions on the vapour and empirical formula mass remains our best match.
Therefore, the molecular formula is C5H7NO2.
Swali 18 Ripoti
The scientist that performed the experiment on discharged tubes that led to the discovery of the cathode rays as a sub-atomic particle is
Maelezo ya Majibu
The **scientist who performed the experiment on discharge tubes that led to the discovery of cathode rays as a sub-atomic particle** is J.J. Thomson.
In the late 19th century, J.J. Thomson conducted experiments using a cathode ray tube. This device involved an evacuated glass tube with electrodes at each end, through which an electric current was passed. **When a high voltage was applied, Thomson observed a stream of particles traveling from the negative electrode (cathode) to the positive electrode (anode).** These streams of particles were what he called "cathode rays."
Through his experiments, J.J. Thomson discovered that these cathode rays were composed of negatively charged particles. **He concluded that these particles were much smaller than atoms, and named them "electrons," which are now known to be sub-atomic particles.** His work was fundamental in advancing the atomic model and in understanding the structure of the atom.
Thomson's discovery was pivotal because it provided the first evidence that atoms are not indivisible, but rather consist of smaller subatomic particles. This **challenged the then-prevailing notion of atoms as indivisible units**, thus marking the birth of modern particle physics.
Swali 19 Ripoti
A gas when mixed with oxygen, it produces a very hot and early controllable flame. What is the name of the flame and where is it used?
Maelezo ya Majibu
The Oxy-ethylene flame is a type of flame produced when oxygen is mixed with a gas called ethylene. This mixture results in a flame that is extremely hot and can be easily controlled. Such a flame is often used in industrial applications related to cutting and welding metals. The heat generated by an oxy-ethylene flame is sufficient to melt metals, allowing them to be welded together or cut apart efficiently.
Swali 20 Ripoti
The highest isotope of hydrogen is
Maelezo ya Majibu
Hydrogen has three naturally occurring isotopes, and each of them contains the same number of protons but different numbers of neutrons. Let's briefly differentiate them:
The highest isotope of hydrogen is tritium because it has the most neutrons and, therefore, the greatest atomic mass compared to the other isotopes. It is also noteworthy that tritium is radioactive, while the other hydrogen isotopes are stable.
Swali 21 Ripoti
After breathing in a test tube that contains acidified K2 Cr2 O7 , a man noticed the change in the colour of K2 Cr2 O7 from orange to green. This suggests the presence of
Maelezo ya Majibu
When the acidified potassium dichromate (\(K_2Cr_2O_7\)) solution changes from orange to green, it indicates a chemical reaction is occurring where the chromium in the dichromate ion is being reduced. In this context, acidified \(K_2Cr_2O_7\) is commonly used as an oxidizing agent.
The change in color from orange (dichromate ion) to green (chromium ion) suggests that the dichromate ion is being reduced, and something in the person's breath is being oxidized.
The substances that can be oxidized in the breath are organic compounds, typically those containing functional groups with oxidizable hydrogen atoms or structures.
Therefore, when the color of acidified potassium dichromate changes from orange to green, it suggests the presence of an alkanol.
Swali 22 Ripoti
The composition of alloy permalloy is iron and
Maelezo ya Majibu
The alloy known as **permalloy** is composed primarily of **iron** and **nickel**. Permalloy is a well-known magnetic alloy that typically consists of about **80% nickel and 20% iron**. It is renowned for having high magnetic permeability, meaning it can become magnetized easily, which makes it extremely useful in a variety of electrical and magnetic applications, such as transformers, memory storage, and magnetic shielding. The nickel in permalloy enhances the magnetic properties of the iron, giving the alloy its unique characteristics.
Swali 23 Ripoti
Hydrochloric acid is regarded as a strong acid because it
Maelezo ya Majibu
Hydrochloric acid (HCl) is regarded as a strong acid because it ionizes completely in water. This means that when HCl is dissolved in water, it breaks down entirely into hydrogen ions (H+) and chloride ions (Cl-). In a solution, there are no molecules of HCl left; only its ions are present.
This complete ionization results in a high concentration of hydrogen ions, which is a key characteristic of strong acids. Because there are more hydrogen ions available, hydrochloric acid can readily participate in chemical reactions, particularly those involving proton transfers, like neutralization reactions with bases.
In summary, the reason HCl is considered strong is due to its ability to consistently and completely ionize in an aqueous solution, not because of its physical state, source, or reactive nature with bases. Therefore, the property that defines it as a strong acid is that it ionizes completely.
Swali 24 Ripoti
What method is suitable for the separation of gases present in air?
Maelezo ya Majibu
The suitable method for the separation of gases present in air is the fractional distillation of liquid air. This method is used due to the differing boiling points of the gases present in the air. Let me explain this in simple terms:
Air is a mixture of different gases, primarily nitrogen, oxygen, and argon, along with small amounts of other gases like carbon dioxide, neon, and krypton. Each of these gases turns into a liquid at different temperatures.
The process begins by cooling the air until it becomes a liquid. This is done at very low temperatures (around -200 degrees Celsius). Once the air is in liquid form, it is slowly warmed up in a distillation column. As it heats up, each gas boils off or evaporates at its respective boiling point and can be collected separately.
For example, nitrogen, which has a boiling point of about -196 degrees Celsius, will evaporate first and can be collected at the top of the distillation column. Following nitrogen, oxygen will evaporate at its boiling point of around -183 degrees Celsius. Finally, argon and other gases will do so at their respective temperatures.
In summary, fractional distillation of liquid air is effective because it takes advantage of the different boiling points to separate each gas from the air mixture.
Swali 25 Ripoti
Which of the following is an air pollutant?
Maelezo ya Majibu
An air pollutant is any substance in the air, introduced by natural or human activity, that causes harm or discomfort to living organisms, or damages the environment. Let's analyze the substances mentioned:
1. O2 (Oxygen)
Oxygen is the gas we need to breathe. It's not considered an air pollutant because it is essential for human and animal life, as well as many natural processes.
2. CO (Carbon Monoxide)
Carbon Monoxide is a colorless, odorless gas that is produced by burning fuel (like in cars and factories). This gas can be very dangerous if there is a lot of it, as it can prevent oxygen from entering the bloodstream. Because of its harmful effects, it is considered an air pollutant.
3. H2 (Hydrogen)
Hydrogen, while a flammable gas, is generally not harmful to the air or to organisms when it is released into the environment. Therefore, it is not considered an air pollutant.
4. O3 (Ozone)
Ozone is a bit tricky because it is both good and bad. Higher up in the atmosphere, it forms a layer that protects us from the sun’s UV radiation. However, at ground level, it is a harmful air pollutant. Ground-level ozone can cause health problems such as respiratory difficulties, so in this context, it is considered an air pollutant.
In conclusion, the substances that are considered air pollutants in this context are Carbon Monoxide (CO) and ground-level Ozone (O3).
Swali 26 Ripoti
Which of these is the most preferred separation technique for the isolation of solutes where the purity of the constituent is of utmost importance?
Maelezo ya Majibu
When the **purity of solutes** is of utmost importance, the most preferred separation technique is **recrystallization**. This method is widely used in chemistry for purifying solid compounds.
Here's a simple explanation of **recrystallization**:
1. **Dissolving the Impure Compound**: The impure solid is dissolved in a suitable hot solvent. The choice of solvent is crucial; it should dissolve the compound well at high temperatures but poorly at low temperatures.
2. **Cooling the Solution**: The solution is slowly cooled. As it cools, the solubility of the compound in the solvent decreases, causing the pure compound to form crystals and precipitate out of the solution.
3. **Collection and Drying of Crystals**: The pure crystals are collected through filtration and then allowed to dry, separating them from any remaining impurities that stay dissolved in the solvent.
The **advantage** of recrystallization is that it allows for the **removal of impurities** that are either more soluble than the desired compound at low temperatures or less soluble at high temperatures, resulting in a more purified product. Therefore, when achieving high purity is a priority, **recrystallization** is often the method of choice.
Swali 27 Ripoti
In the extraction of Aluminium, the silica impurity is removed by
Maelezo ya Majibu
Aluminum is extracted from bauxite by electrolysis. The extraction proceeds in two stages;
1. Purification of the Bauxite: The impure bauxite is heated with sodium hydroxide solution to form soluble sodium tetrahydroxy aluminate (iii). The impurities in the ore which are iron (iii) oxide and trioxosilicate (iv) compounds are not soluble in the alkali. They are therefore filtered off as a sludge.
Aluminum hydroxide crystals is then added to filtrate, NaAl(OH)4 solution to induce the precipitation of Aluminum hydroxide.
2. The electrolysis of the pure alumina
Swali 28 Ripoti
Hydrogen chloride gas and ammonia can be used to demonstrate the fountain experiment because they are
Maelezo ya Majibu
In the fountain experiment, hydrogen chloride gas (HCl) and ammonia (NH₃) are used to demonstrate the creation of a visible 'fountain' due to their high solubility in water. Here's a simple explanation:
When hydrogen chloride gas and ammonia gas come into contact with water, they dissolve very quickly and react vigorously. This is because both gases are very soluble in water. As they dissolve, a vacuum-like pressure is created inside the container where the gases are held, pulling water up into it, creating the 'fountain' effect.
Moreover, when HCl and NH₃ gases react with each other, they form a white, solid product known as ammonium chloride (NH₄Cl), which is a demonstration of how both gases can effectively dissolve and react with not just water, but also with each other.
Thus, the ability of these gases to create a fountain effect is primarily because they are very soluble in water, which allows them to dissolve rapidly and create the pressure differential necessary for the water to be pulled into the container dynamically.
Swali 29 Ripoti
The Van der waals forces of attraction operates between
Maelezo ya Majibu
The Van der Waals forces of attraction operate between molecules. These are weak forces of attraction that occur due to momentary changes in the electron distribution within molecules. Here's a simple explanation:
Therefore, the forces can affect the physical properties of molecular compounds, such as boiling and melting points, but do not generally involve charged particles like cations or anions.
Swali 30 Ripoti
The ions responsible for permanent hardness in water are sulphates of
Maelezo ya Majibu
Permanent hardness in water is mainly caused by the presence of certain metal ions, specifically the **sulfates (SO₄²⁻)** and **chlorides (Cl⁻)** of calcium (Ca) and magnesium (Mg). These compounds do not precipitate out when the water is boiled, which means they remain dissolved and continue to contribute to the hardness of the water.
Among the options you provided, the ions responsible for permanent hardness in water are the **sulfates of calcium (Ca²⁺)** and **magnesium (Mg²⁺)**. The presence of calcium sulfate (CaSO₄) and magnesium sulfate (MgSO₄) in water keeps it hard.
When compared to temporary hardness, which can be removed by boiling the water to precipitate bicarbonates, **permanent hardness cannot be removed by boiling**. Instead, methods such as ion exchange or the use of water softeners are required to remove these ions from the water.
In summary, the ions causing permanent hardness in water are the **sulfates of calcium (Ca²⁺)** and **magnesium (Mg²⁺)**. These ions remain dissolved and continue to make the water hard, despite boiling.
Swali 31 Ripoti
The electronic configuration of an atom of Nitrogen is 1s2 2s2 2p1x 2p1y 2p1z because the atom is
Maelezo ya Majibu
The electronic configuration of nitrogen is given as: 1s2 2s2 2px1 2py1 2pz1.
This configuration suggests that nitrogen has 7 electrons, as follows:
This is the **ground state** electron configuration of nitrogen, meaning that the atoms have electrons in the **lowest possible energy levels**. It demonstrates nitrogen's **stable configuration**, where it has half-filled p orbitals, each with a single electron. This configuration obeys Hund's Rule, which states that every orbital in a subshell gets one electron before any one orbital gets two (due to electron repulsion). It also obeys the Aufbau principle which suggests electrons fill orbitals starting from the lowest energy level.
Therefore, this configuration indicates that the atom is simply obeying rules governing electron configuration. The electrons are in their lowest energy orbitals, consistent with the principles that direct electron arrangement in an atom, ensuring stability without being excited or unstable. There are no **energy changes** being depicted nor is the atom in an **excited state**—it is showing the normal ground state.
Swali 32 Ripoti
Calculate the mass of Magnesium that will be liberated from its salt by the same quantity of electricity that liberated 16.0 g of Silver.
[Mg = 24.0, Ag = 108 ]
Maelezo ya Majibu
To solve this problem, we must consider the concept of electrochemistry and Faraday's laws of electrolysis. These laws are crucial for determining the mass of a substance liberated during electrolysis.
Faraday's first law states that the mass of a substance liberated is directly proportional to the quantity of electricity that passes through the electrolyte. The mass can be calculated using the formula:
m = (Q * M) / (n * F)
Where:
For silver (Ag), the chemical reaction at the cathode is:
Ag⁺ + e⁻ → Ag
This shows that **1 mole of electrons** is required to discharge **1 mole** of silver ions.
For magnesium (Mg), the chemical reaction at the cathode is:
Mg²⁺ + 2e⁻ → Mg
This means that **2 moles of electrons** are required to discharge **1 mole** of magnesium ions.
Given:
First, find the number of moles of Ag liberated:
Number of moles of Ag = 16 g / 108 g/mol = 0.1481 mol
The same quantity of electricity will be used to liberate an equivalent in moles of electrons for Mg.
0.1481 moles of Ag require 0.1481 moles of electrons, equivalent to:
0.1481 moles of electrons for Mg. Since Mg requires 2 moles of electrons for 1 mole of Mg:
Number of moles of Mg = 0.1481 / 2 = 0.07405 mol
Finally, calculate the mass of Mg liberated:
m = 0.07405 mol * 24 g/mol = 1.7772 g
Rounding this to the closest answer provided:
The mass of magnesium that will be liberated is approximately **1.78 g**.
Swali 33 Ripoti
H2 SO4
C2 H5 OH → C2 H4
1700 C
The reaction above illustrates
Maelezo ya Majibu
This reaction illustrates dehydration. In chemistry, dehydration refers to the process of removing water (H2O) from a compound. Let's break down the given reaction to understand this better.
The provided chemical equation is:
C2H5OH → C2H4 + H2O
This equation indicates that ethanol (C2H5OH) is being transformed into ethylene (C2H4) with the production of water (H2O).
The process involves the breaking of bonds in ethanol and the removal of a water molecule, as follows:
This reaction is typically carried out under certain conditions, in this case at a high temperature of 1700°C, to facilitate the dehydration process.
Therefore, this is indeed a dehydration reaction as it involves converting ethanol into ethylene by removing water.
Swali 34 Ripoti
CuOs + H2 (g ) ⇌ Cus + H2 O(g )
In the equation above, the effect of increased pressure on the equilibrium position is that
Maelezo ya Majibu
To understand the effect of increased pressure on the equilibrium position of the given reaction:
CuO(s) + H2(g) ⇌ Cu(s) + H2O(g)
We need to consider Le Chatelier's Principle. According to this principle, if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will adjust itself to counteract the effect of the change and re-establish equilibrium.
For the reaction in question, let's consider the number of gas molecules on each side of the equation:
Since both sides of the equation have the same number of gas molecules, an increase in pressure will not favor a shift to either the left or the right because the number of moles of gas on both sides of the equilibrium is the same.
Therefore, the effect of increased pressure on the equilibrium is that there is no effect. The position of the equilibrium remains unchanged, and pressure changes do not influence the production of more H2(g) or H2O(g) in this specific reaction.
Swali 35 Ripoti
25.0g of potassium chloride were dissolved in 80g of distilled water at 300 C. Calculate the solubility of the solute in mol dm3 . [K =39, Cl = 35.5]
Maelezo ya Majibu
To calculate the solubility of potassium chloride (KCl) in mol dm3, we need to follow these steps:
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Moles of KCl = Mass of KCl / Molar mass of KCl = 25.0 g / 74.5 g/mol = 0.3356 mol
Convert ml to liters: 80 ml = 0.080 L
Concentration = Moles of solute / Volume of solvent in liters = 0.3356 mol / 0.080 L = 4.195 mol/dm3
The solubility of potassium chloride at 30°C in mol/dm3 is therefore approximately 4.2 mol/dm3.
Swali 36 Ripoti
Na2 X ⇌ 2Na+ + X2−
The bond between Na and X is likely to be
Maelezo ya Majibu
The bond between Na and X is most likely to be ionic. Let's break this down simply:
In the equation provided:
Na2X ⇌ 2Na+ + X2−
The sodium (Na) atoms become positively charged ions (Na+), while X becomes a negatively charged ion (X2−). This change in charge occurs because sodium atoms donate electrons to the X atom. The donation of electrons by sodium to X indicates a transfer of electrons, which is a hallmark of an ionic bond.
In an ionic bond, electrons are transferred from one atom to another, resulting in a positively charged ion and a negatively charged ion. These oppositely charged ions attract each other, forming a strong ionic bond.
In summary, since sodium (Na) donates electrons to X forming ions, the bond between Na and X is most likely to be ionic.
Swali 37 Ripoti
An example of a physical change is
Maelezo ya Majibu
An example of a physical change is the boiling of water. Let me explain why this is considered a physical change:
A physical change is a change where the substances involved do not change their chemical composition, meaning they remain the same substance, just in a different form or appearance. In the case of boiling water, when water is heated to its boiling point, it changes from a liquid to a gas (steam), but it is still comprised of water molecules (H2O). The change is reversible, so the gas can condense back into liquid water without any new substance being formed.
On the other hand:
Thus, boiling water is an excellent example of a physical change as it involves only the change in the state of matter without altering the substance's identity.
Swali 38 Ripoti
The stability of atomic nucleus is determined by ratio of
Maelezo ya Majibu
The stability of an atomic nucleus is primarily determined by the neutron/proton ratio. This refers to the number of neutrons in relation to the number of protons within the nucleus. Let's break down why this ratio is crucial for nuclear stability:
The right balance between the number of neutrons and protons helps in achieving nuclear stability.
An imbalance in this ratio often results in an unstable nucleus, leading to radioactive decay as the nucleus attempts to reach a more stable form. This is why the neutron/proton ratio is a fundamental factor in the stability of the atomic nucleus.
Swali 39 Ripoti
The pH of a 0.001 mol dm−3 of H2 SO4 is
[Log10 2 = 0.3]
Maelezo ya Majibu
The question is asking about the pH of a 0.001 mol dm−3 solution of H2SO4 (sulfuric acid). To find the pH, we need to understand how sulfuric acid dissociates in water.
Step 1: Dissociation of H2SO4
Sulfuric acid, H2SO4, is a strong acid and dissociates completely in water in two steps:
1. The first dissociation: H2SO4 → H+ + HSO4-
2. The second dissociation: HSO4- → H+ + SO42-
For dilute solutions, particularly below 0.1 M, the first dissociation provides the major contribution to the H+ concentration. The second dissociation also contributes slightly to the acidity, but for simplicity and due to the dilute nature of this solution, the first step's contribution is primarily considered.
Step 2: Calculate the H+ Concentration
Since this is a strong acid and dissociates completely, for every 1 mole of H2SO4, we get 2 moles of H+. Therefore, for a 0.001 mol dm−3 solution of H2SO4, the concentration of H+ ions will be:
2 x 0.001 = 0.002 mol dm−3
Step 3: Calculate the pH
The pH is calculated using the formula: pH = -log[H+]
Substitute the H+ concentration:
pH = -log(0.002)
We know that log(10-2) = -2 and log(2) = 0.3 (as provided), so:
pH = -(log(2) + log(10-3))
pH = -(0.3 - 3)
pH = 3 - 0.3
pH = 2.7
Therefore, the pH of the 0.001 mol dm−3 H2SO4 solution is 2.7.
Swali 40 Ripoti
C2 H4(g) + 3O2(g) → 2CO2(g) + 2H2 O(g)
The above equation represents the combustion of ethene.If 10cm3 of ethene is burnt in 50cm3 of oxygen, what would be the volume of oxygen that would remain at the end of the reaction?
Maelezo ya Majibu
Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.
C2 H4(g) + 3O2(g) → 2CO2(g) + 2H2 O(g)
1 mole : 3 moles
Total volume required: 10 cm3 50 cm3
Reacted Volume: 10 cm3 30 cm3
Residual volume: 0 (50 - 30) = 20 cm3
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