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Swali 1 Ripoti
Sulphur(IV)oxide can be used as a
Maelezo ya Majibu
Sulphur(IV) oxide has many uses including food preservation, refrigeration, laboratory reagent and solvent, sulphuric acid production, fumigant etc.Sulphur(IV) oxide is a good refrigerant because it has a high heat of evaporation and can be easily condensed.
Swali 2 Ripoti
Biodegradable pollutants are not safe in water systems because they can cause
Maelezo ya Majibu
Biodegradable pollutants are substances that can be broken down by natural processes and microorganisms. However, when they are present in water systems, they can lead to several environmental and health issues. One of the main concerns is their potential to cause ill health. Here's why:
When biodegradable pollutants such as organic waste are introduced into water bodies, they are decomposed by bacteria and other microorganisms. This process consumes dissolved oxygen in the water. As the oxygen levels decrease, aquatic life such as fish and plants may suffer or die due to a lack of oxygen, disrupting the entire aquatic ecosystem.
This situation is known as eutrophication, which can lead to the excessive growth of algae, commonly referred to as algal blooms. These blooms often produce toxins that are harmful to both aquatic life and humans. Furthermore, when this polluted water is used for drinking, agriculture, or recreational purposes, it poses serious health risks to humans. These risks may include gastrointestinal infections, neurological disorders, and skin problems.
In addition, as the pollutants decompose, foul smells may be released, which can affect air quality in the vicinity, although the primary concern with biodegradable pollutants in water is related to how they affect water quality and health.
Therefore, it is crucial to properly manage and treat biodegradable pollutants before they enter water systems to prevent these health hazards. Failure to do so can result in significant environmental and health issues.
Swali 3 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 4 Ripoti
The hybridization scheme in ethyne is
Maelezo ya Majibu
Ethyne, also known as acetylene, is a simple alkyne with the chemical formula C2H2. In ethyne, each carbon atom is bonded to two other atoms: one hydrogen atom and the other carbon atom. The molecular structure of ethyne is linear, with a triple bond between the two carbon atoms.
To determine the hybridization scheme in ethyne, we need to examine the arrangement of the electron pairs around each carbon atom. In ethyne, each carbon atom is forming two sigma (σ) bonds and two pi (π) bonds. Let's explain:
When we consider the hybridization of the carbon atoms, we focus on the formation of sigma bonds and lone pairs. In ethyne, each carbon atom utilizes two orbitals to form sigma bonds: one with the hydrogen atom and one with the other carbon atom. This implies that each carbon atom in ethyne must use two hybrid orbitals.
The two hybrid orbitals formed by each carbon atom in ethyne are a result of mixing one s orbital with one p orbital. This hybridization is referred to as sp hybridization, characterized by a linear electron geometry. The remaining two unhybridized p orbitals on each carbon atom are responsible for forming the two pi bonds in the triple bond.
In conclusion, the hybridization scheme in ethyne is sp.
Swali 5 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 6 Ripoti
A gas that turns lime water milky is likely to be from
Maelezo ya Majibu
The gas that turns lime water milky is **Carbon Dioxide**. This is because carbon dioxide reacts with calcium hydroxide, which is the main component of lime water, to form calcium carbonate. This chemical reaction can be represented by the equation:
Ca(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)
In this equation, calcium hydroxide ({Ca(OH)2}) in the lime water reacts with carbon dioxide ({CO2}) to produce calcium carbonate ({CaCO3}) and water ({H2O}).
The result is a milky or cloudy appearance due to the formation of insoluble calcium carbonate precipitate in the lime water. This reaction is a common test for the presence of carbon dioxide gas.
Among the options given, **Trioxocarbonate(IV)** is another name for the Carbonate group involving the gas carbon dioxide ({CO2}). Hence, the gas related to Trioxocarbonate(IV) is the one that turns lime water milky.
Swali 7 Ripoti
During the fractional distillation of crude oil, the fraction that distills at 200 - 2500 C is
Maelezo ya Majibu
The petroleum fractions that distill at 200–250°C are naphtha and kerosene,
Swali 8 Ripoti
What accounts for the low melting and boiling points of covalent molecules?
Maelezo ya Majibu
The low melting and boiling points of covalent molecules are primarily due to the presence of weak intermolecular forces between the molecules. While covalent molecules consist of atoms bonded together by strong covalent bonds, the forces between separate molecules, known as van der Waals forces or London dispersion forces, are much weaker. These weak forces require significantly less energy to overcome, which explains why covalent molecules tend to have lower melting and boiling points.
Although covalent molecules have definite shapes and possess shared electron pairs, these characteristics have little influence on the melting and boiling points. The focus is instead on how much energy is needed to separate the molecules from one another.
Covalent molecules are not typically three-dimensional structures like ionic compounds or metals which form intricate lattices and require more energy to disrupt. Thus, the primary reason for their lower melting and boiling points is the presence of weak intermolecular forces that can be more easily overcome with minimal energy input.
Swali 9 Ripoti
Solubility curve is a plot of solubility against
Maelezo ya Majibu
A solubility curve is a plot of solubility against temperature. Let me explain in a simple way:
Solubility refers to the amount of a substance (solute) that can dissolve in a given quantity of solvent to form a homogeneous solution at a specified condition. The most common factor that affects solubility is the temperature.
Here's why a solubility curve typically involves temperature:
Therefore, plotting solubility against temperature in a solubility curve allows us to visualize and understand how solubility changes with variations in temperature.
Swali 10 Ripoti
The chemical formula for potassiumhexacyanoferrate(II) is
Maelezo ya Majibu
The chemical formula for potassiumhexacyanoferrate(II) is K4Fe(CN)6.
Let's break down the name to understand why:
1. Potassium (K): The compound includes potassium ions. In this case, four potassium ions are present, indicated by the subscript 4 in K4.
2. Hexacyano: The prefix "hexa" means six, which signifies there are six cyanide ions (CN-) in the complex. This is represented as (CN)6.
3. Ferrate (II): The word "ferrate" suggests the presence of iron (Fe). The Roman numeral (II) indicates that the iron is in the +2 oxidation state.
Overall, the complex ion is [Fe(CN)6] with a charge of 4-, so to balance the charge, four potassium ions (each with a charge of +1) are needed, resulting in the formula K4Fe(CN)6.
Swali 11 Ripoti
Concentrated sodium chloride solution is electrolyzed using mercury cathode and graphite anode. The products at the anode and the cathode respectively are
Maelezo ya Majibu
When a concentrated sodium chloride solution is electrolyzed using a mercury cathode and graphite anode, the products are hydrogen gas at the cathode and chlorine gas at the anode
At the anode, 2Cl− → Cl2 + 2e−
At the cathode, 2H+ + 2e− → H2
During the electrolysis, hydrogen and chloride ions are removed from solution whereas sodium and hydroxide ions are left behind in solution. This means that sodium hydroxide is also formed during the electrolysis of sodium chloride solution.
Swali 12 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 13 Ripoti
Alkanoates are naturally found in
Maelezo ya Majibu
Alkanoates, also known as fatty acid esters, are primarily found in lipids. Lipids are a broad group of naturally occurring molecules that include fats, waxes, sterols, fat-soluble vitamins (such as vitamins A, D, E, and K), and others. One of the main components of lipids is fatty acids and their derivatives, such as alkanoates.
To be more specific, alkanoates can be found in the form of triglycerides, which are the main constituents of body fat in humans and animals, as well as vegetable fat. Triglycerides are composed of glycerol bound to three fatty acids, and these fatty acids are usually present in the form of alkanoates.
Unlike proteins and rubber, which are made up of amino acids and polymers of isoprene respectively, lipids are the primary class of biomolecules where these alkanoate compounds can be found in significant amounts.
Swali 14 Ripoti
The percentage of hydrogen in the sixth member of the class of the aliphatic alkanes is [H =1, C =12 ]
Maelezo ya Majibu
To determine the percentage of hydrogen in the sixth member of aliphatic alkanes, we first need to understand the general formula for alkanes. Aliphatic alkanes are a class of hydrocarbons with the general formula CnH2n+2, where 'n' is the number of carbon atoms.
The sixth member of this series will have n = 6. Therefore, the molecular formula for the sixth member is C6H14.
To find the percentage of hydrogen, we first calculate the molar mass of C6H14:
Total molar mass of C6H14 = 72 + 14 = 86
Next, we calculate the percentage of hydrogen:
Percentage of hydrogen = (Molar mass of hydrogen atoms / Total molar mass) × 100
Percentage of hydrogen = (14 / 86) × 100 = 16.28%
Therefore, the percentage of hydrogen in the sixth member of the aliphatic alkanes is 16.28%.
Swali 15 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 16 Ripoti
Water gas obtained from the gasification of coke is made up of
Maelezo ya Majibu
The gasification of coke to produce water gas involves reacting coke, which is primarily composed of carbon, with steam. The main chemical reaction that occurs is:
C (s) + H2O (g) → CO (g) + H2 (g)
From this reaction, the main constituents of water gas are hydrogen (H2) and carbon monoxide (CO), also known as carbon(II) oxide. Therefore, water gas obtained from the gasification of coke is made up of hydrogen and carbon(II) oxide.
Swali 17 Ripoti
The element which can combine with oxygen to form an acid anhydride of the form XO2 is
Maelezo ya Majibu
An Acid anhydride can be defined as a non-metal oxide which forms an acidic solution when reacted with water.
Sulphur is the element that can combine with oxygen to form an acid anhydride of the form XO2 .
An acid oxide is a compound that forms an acid when it reacts with water. Non-metals in groups 4–7 form acidic oxides.
Swali 18 Ripoti
H2 S(g) + Cl2 (g) → 2HCl(g) + S(s)
What is the change in oxidation state of sulphur from reactant to product?
Maelezo ya Majibu
To determine the change in oxidation state of sulfur, follow these steps:
In the given reaction:
H2S(g) + Cl2(g) → 2HCl(g) + S(s)
We observe:
Thus, the change in oxidation state of sulfur when moving from the reactants to the products is from **-2** to **0**. This indicates that sulfur is being oxidized.
The correct answer is that the oxidation state of sulfur changes from **-2 to 0**.
Swali 19 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 20 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 21 Ripoti
The reaction between alkanoic acids and alkanols in the presence of an acid catalyst is known as
Maelezo ya Majibu
The reaction between alkanoic acids and alkanols in the presence of an acid catalyst is known as esterification.
An alkanoic acid, also known as a carboxylic acid, is a type of organic acid that contains a carboxyl group (-COOH). An alkanol, commonly referred to as an alcohol, contains a hydroxyl group (-OH).
When an alkanoic acid reacts with an alkanol in the presence of an acid catalyst (commonly sulfuric acid), they combine to form an ester and water. This particular reaction is termed esterification. The acid catalyst speeds up the reaction by donating protons, which helps in breaking and forming new bonds.
Here's a simplified view of the reaction:
1. Alkanoic Acid (R-COOH) + Alkanol (R'-OH) -> Ester (R-COOR') + Water (H2O)
The key characteristics of esterification are:
Therefore, in summary, the process described is esterification.
Swali 22 Ripoti
In a chemical reaction, surface area of reactants can affect
Maelezo ya Majibu
The surface area of reactants affects the rate of a reaction between limestone and hydrochloric acid because it increases the number of collisions between the particles of the reactants. For example, if you have a large marble chip of calcium carbonate and hydrochloric acid, the acid can't reach all the calcium carbonate in the middle of the chip. If you break the marble chip into smaller pieces, you'll have a larger surface area for the acid to react with, and the reaction will happen faster.
Swali 23 Ripoti
The shape of the molecule of Carbon(IV) oxide is
Maelezo ya Majibu
The shape of the molecule of Carbon(IV) oxide, also known as carbon dioxide (CO2), is linear. This is because of the following reasons:
Due to this arrangement, carbon dioxide has a symmetric shape, making it non-polar despite having polar covalent bonds. The pulling forces of the two oxygen atoms on either side of the carbon atom cancel each other out, reinforcing its linear configuration.
Swali 24 Ripoti
When Calcium ethynide is decomposed by water, the gas produced is
Maelezo ya Majibu
When water reacts with calcium ethynide, the gas produced is ethyne (also known as acetylene), which is represented by the chemical formula C2H2.
The chemical reaction involved is as follows:
CaC2 + 2 H2O → C2H2 + Ca(OH)2
Let's break down this process to make it understandable:
The key point to remember here is that the gas produced is **ethyne (C2H2)**, which is useful in various industrial applications, such as welding and as a precursor for other chemicals.
Swali 25 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 26 Ripoti
Fats and oils are esters of fatty organic acids combined with a trihydric alkanol commonly referred to as
Maelezo ya Majibu
Fats and oils are types of lipids that belong to the category of esters of fatty acids. These are organic compounds formed when fatty acid molecules react with an alcohol. In the case of fats and oils, the alcohol involved is a trihydric alkanol, meaning it has three hydroxyl (-OH) groups.
The trihydric alkanol commonly found in fats and oils is glycerol. Glycerol, also known as glycerine, has the chemical formula C3H8O3 and has three carbon atoms, each of which is attached to a hydroxyl group, making it a perfect candidate to form esters with three fatty acid molecules.
When these fatty acids react with the hydroxyl groups of glycerol, they form compounds called triglycerides. These triglycerides are the primary constituents of both fats and oils. Therefore, the correct answer is that fats and oils are esters of fatty organic acids combined with glycerol as the trihydric alkanol.
Swali 27 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 28 Ripoti
Biuret test is a chemical test used for detecting the presence of
Maelezo ya Majibu
The Biuret test is a chemical test used for detecting the presence of proteins. When you perform a Biuret test, you are looking for peptide bonds, which are the connections between the amino acids in a protein. This is how it works:
The test is specifically tailored to proteins because carbohydrates, amines, and alkanoates do not exhibit the required peptide bonds necessary for this color change. Therefore, the Biuret test is not suitable for detecting these compounds.
Swali 29 Ripoti
Determine the half-life of a first order reaction with constant 4.5 x 10−3 sec−1 .
Maelezo ya Majibu
To determine the half-life of a first-order reaction, you can use the formula:
Half-life (\(t_{1/2}\)) = \(\frac{0.693}{k}\)
where \(k\) is the rate constant of the reaction. For the given problem, the rate constant (\(k\)) is 4.5 x 10-3 s-1.
Substituting the value of \(k\) into the formula, we have:
\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}}\)
Perform the division:
\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}} \approx 154\) s
Therefore, the half-life of the reaction is 154 seconds.
Swali 30 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
Swali 31 Ripoti
When n = 3, the quantum number of an element is
Maelezo ya Majibu
Quantum numbers are a set of numbers that describe the position and energy of an electron in an atom.
When the quantum number is equal to 3, the possible values for the azimuthal quantum number are 0, 1, and 2:
The three possible sub-shells when n=3 are 3s, 3p, and 3d.
Swali 32 Ripoti
Hydrochloric acid is not suitable in the preparation of ethanoic acid because it
Maelezo ya Majibu
Hydrochloric acid is not suitable for preparing ethanoic acid because it is too volatile.Being too volatile, means it has a low boiling point and is easily evaporated. Thus, HCl is not suitable because it cannot carry out the oxidation process required to convert alcohols into acids like ethanoic acid.
Ethanoic acid, also known as acetic acid, is a weak acid that doesn't fully dissociate in water, while hydrochloric acid is a strong acid that dissociates almost completely.
Swali 33 Ripoti
Alkylation of benzene is catalyzed by
Maelezo ya Majibu
Alkylation of benzene is a part of a reaction class called **Friedel-Crafts alkylation**. In this reaction, an alkyl group is transferred to the aromatic benzene ring, making it a more complex molecule. The catalyst used in this process is **aluminium chloride (AlCl3)**.
Here's how the reaction typically works:
In contrast, the other options wouldn't effectively catalyze alkylation of benzene for the following reasons:
Therefore, **aluminium chloride** is the catalyst used for the alkylation of benzene in Friedel-Crafts reactions.
Swali 34 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 35 Ripoti
Determine the empirical formula of an oxide of sulphur containing 60% of oxygen
[S = 32, O = 16 ]
Maelezo ya Majibu
To determine the empirical formula of an oxide of sulfur containing 60% of oxygen, we have to understand the concept of empirical formulas, which give the simplest whole-number ratio of atoms of each element in a compound.
Step 1: Assume 100g of the compound. In 100g of the compound:
Step 2: Convert masses to moles. Use the molar mass to find moles.
Step 3: Determine the simplest whole-number ratio.
To find the ratio, divide each mole value by the smallest number of moles calculated:
The simplest ratio of S:O is 1:3.
Thus, the empirical formula of the oxide is SO3.
Swali 36 Ripoti
An example of highly unsaturated hydrocarbon is
Maelezo ya Majibu
To determine a highly unsaturated hydrocarbon, we must first understand the concept of saturation in hydrocarbons. **Saturated hydrocarbons** are compounds that contain the maximum possible number of hydrogen atoms, single-bonded to carbon atoms, and they are alkanes. **Unsaturated hydrocarbons** have one or more double or triple bonds between carbon atoms, which reduces the number of hydrogen atoms that can be bonded.
Examining the given options:
Based on this analysis, **C2H2** (ethyne) is a highly unsaturated hydrocarbon due to the presence of a **triple bond**. The triple bond signifies a greater level of unsaturation compared to double bonds in hydrocarbons like ethene (C2H4).
Swali 37 Ripoti
The table above shows the formulae of some ions. In which of these compounds is the formula not correct?
Maelezo ya Majibu
To assess the correctness of the chemical formulae for the given compounds, let's break down each compound:
Aluminium Tetraoxosulphate(VI), Al2(SO4)3:
Aluminium ion is denoted as Al3+, and the sulphate ion is SO42-. To balance the charges between the positive and negative ions:
2 x (+3) from aluminium ions = +6
3 x (-2) from sulphate ions = -6
Thus, the charges balance out, making the formula correct.
Calcium Trioxonitrate(V), Ca(NO3)2:
Calcium ion is Ca2+, and the nitrate ion is NO3-. To balance the charges:
1 x (+2) from calcium ion = +2
2 x (-1) from nitrate ions = -2
The charges balance out, therefore, this formula is also correct.
Iron(III) Bromide, Fe3Br:
Iron(III) ion is Fe3+, and bromide ion is Br-. Each iron ion would pair with three bromide ions to balance the charges:
FeBr3, where:
1 x (+3) from iron = +3
3 x (-1) from bromide = -3
The charges balance out in the correct formula which should be FeBr3, making the given formula Fe3Br incorrect.
Potassium Sulphide, K2S:
Potassium ion is K+, and sulphide ion is S2-. To balance the charges:
2 x (+1) from potassium ions = +2
1 x (-2) from sulphide ion = -2
The charges balance out, making this formula correct.
Therefore, the compound with the incorrect formula is Iron(III) Bromide where the proper chemical formula should be FeBr3, not Fe3Br.
Swali 38 Ripoti
The term that is not associated with petroleum industry is ?
Maelezo ya Majibu
Cracking, saponification and polymerization are all terminologies associated with the petroleum industry but fermentation is associated with the brewery industry.
Cracking is a chemical process that breaks down heavy hydrocarbon molecules into lighter, more useful ones.
Saponification is a chemical reaction that converts fats and oils into soap. It's not directly involved in petroleum, but it can be used to analyze petroleum products.
Polymerization is a process in the petroleum industry that converts light olefin gases into higher molecular weight hydrocarbons.
Fermentation is the process in which a substance breaks down into a simpler substance. Microorganisms like yeast and bacteria usually play a role in the fermentation process, creating beer, wine, bread,yogurt and other foods.
Swali 39 Ripoti
The reaction of hydrogen and chlorine to produce hydrogen chloride gas is explosive in
Maelezo ya Majibu
The reaction between hydrogen and chlorine to produce hydrogen chloride gas is explosive in sunlight. This is because sunlight contains a broad range of electromagnetic radiation, including ultraviolet (UV) light, which is energetic enough to initiate the reaction.
Here is a simplified explanation:
In contrast, other forms of light like diffused light, infrared light, and Raman light do not provide enough energy to initiate this explosive reaction because they lack the necessary UV component found in sunlight.
Swali 40 Ripoti
If a salt weighs 2g and upon exposure to the atmosphere weighs 1.5g, this is as a result of
Maelezo ya Majibu
The observation that a salt initially weighs 2g, but reduces to 1.5g after exposure to the atmosphere is primarily due to the process called efflorescence.
Efflorescence occurs when a salt loses water molecules from its crystal structure when exposed to air, which is why the weight of the salt decreases over time. This loss of water is because some salts contain water of crystallization, and when such salts are exposed to the atmosphere, they can release this water, leading to a reduction in weight.
In this specific case, the salt has lost 0.5g of water, leading to the weight change from 2g to 1.5g. This process is different from hygroscopy, which involves absorbing moisture from the atmosphere, or deliquescence, where a substance absorbs moisture and eventually dissolves in it. It's also not related to effervescence, which is the escape of gas from an aqueous solution.
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