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Ajụjụ 1 Ripọtì
The combustion of candle under limited supply of air forms
Akọwa Nkọwa
When a candle burns under a limited supply of air, it doesn't get enough oxygen to completely burn the hydrocarbons in the wax. In complete combustion (with enough air), the candle would ideally produce water (H2O) and carbon dioxide (CO2). However, under limited air supply, the process is incomplete and results in the formation of soot and carbon monoxide (CO).
Here's why:
In summary, under limited air conditions, the combustion of a candle primarily forms soot and carbon monoxide (CO).
Ajụjụ 2 Ripọtì
An example of a physical change is
Akọwa Nkọwa
A physical change involves a change in the physical properties of a substance, without a change in its chemical composition. This means that the substance remains the same at the molecular level, despite how it might appear differently.
An example of a physical change from the given options is the liquefaction of liquids. In this process, a substance transitions from a solid or gas to a liquid state. This change is purely physical because the molecular structure of the substance does not change; only its state or form does. Importantly, such a change is usually reversible, meaning the substance can return to its original state. For instance, water can change into ice (frozen) or steam (vapor), and can still revert back to liquid water.
On the other hand, the other options involve chemical changes, where the original substances undergo chemical reactions to form new substances with different properties, thus altering the molecular structure depending on the option.
Ajụjụ 3 Ripọtì
H2 S(g) + Cl2 (g) → 2HCl(g) + S(s)
What is the change in oxidation state of sulphur from reactant to product?
Akọwa Nkọwa
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**.
Ajụjụ 4 Ripọtì
Scandium is not regarded as a transition metal because its ion has
Akọwa Nkọwa
Scandium is not regarded as a transition metal because its ion has no electron in the d-orbital.
To understand this, let's first define a transition metal. A transition metal is defined as an element that has an incomplete d-subshell in either its elemental form or in any of its common oxidation states.
When Scandium (Sc) loses electrons to form its most common ion (Sc3+), it loses three electrons. These electrons are removed from the 4s and 3d orbitals. The electron configuration for Scandium is [Ar] 3d1 4s2. Upon losing three electrons to form Sc3+, the resulting electron configuration is [Ar], which means there are:
As a result, there are no electrons in the d-orbital of the Scandium ion, which does not meet the criteria for a transition metal.
Ajụjụ 5 Ripọtì
Alkylation of benzene is catalyzed by
Akọwa Nkọwa
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.
Ajụjụ 6 Ripọtì
The product formed when ethyne is passed through a hot tube containing finely divided iron is
Akọwa Nkọwa
When **ethyne** (also known as acetylene) is passed through a hot tube containing finely divided iron, a process called decomposition occurs. The heat causes the ethyne molecules to break down, and under these conditions, they **re-combine** to form structures that result in more complex molecules.
The key transformation involves the conversion of these ethyne molecules into **aromatic compounds**. Aromatic compounds, such as **benzene**, have a distinct ring structure and are characterized by **stability** due to resonance (a phenomenon where electrons are delocalized over a certain structure, providing extra stability).
Thus, when ethyne is passed through a hot iron tube, it undergoes trimerization to form benzene, an **aromatic** compound. Therefore, the product formed is **aromatic**.
Ajụjụ 7 Ripọtì
The hybridization scheme in ethyne is
Akọwa Nkọwa
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.
Ajụjụ 8 Ripọtì
The compound of Copper used as a fungicide is
Akọwa Nkọwa
The compound of copper that is commonly used as a fungicide is **Copper(II) sulfate**, which is represented by the chemical formula **CuSO4**.
Let's break this down for better understanding:
The other compounds listed do not serve as common fungicides:
Therefore, the correct and widely used copper compound as a fungicide is Copper(II) sulfate (CuSO4).
Ajụjụ 9 Ripọtì
The percentage of hydrogen in the sixth member of the class of the aliphatic alkanes is [H =1, C =12 ]
Akọwa Nkọwa
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%.
Ajụjụ 10 Ripọtì
The reaction of hydrogen and chlorine to produce hydrogen chloride gas is explosive in
Akọwa Nkọwa
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.
Ajụjụ 11 Ripọtì
When n = 3, the quantum number of an element is
Akọwa Nkọwa
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.
Ajụjụ 12 Ripọtì
23892 U + 10 n → 23992 U
The process above produces
Akọwa Nkọwa
The process described appears to depict a nuclear reaction involving a nuclear transmutation. Let's break down the process:
1. The starting element is initially denoted as "23892", which represents Uranium-238. In nuclear notation, "23892" indicates an atomic mass number of 238 and an atomic number of 92.
2. The next step so happens with the element "238"; however, the numbers remain: "92" indicates that the atomic number is unchanged, suggesting no change in the element. This often means a step in between of hypothetical notation.
3. Then there's the occurrence of adding a "U + 10", which again leaves the original atomic number "92".
4. In subsequent steps, it seems that the number "n" transitions to become "23992". The mass number has increased by one unit, turning the initial isotope into "23992", which represents Uranium-239.
The key point here is the transition from Uranium-238 to Uranium-239, which typically happens through the process of a neutron absorption in which a neutron is added, resulting in a change of the mass number. Such a process often leads to the creation of a radioactive isotope.
Therefore, the process described is indicative of producing a radioactive isotope, specifically Uranium-239.
Ajụjụ 13 Ripọtì
Water gas obtained from the gasification of coke is made up of
Akọwa Nkọwa
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.
Ajụjụ 14 Ripọtì
When Calcium ethynide is decomposed by water, the gas produced is
Akọwa Nkọwa
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.
Ajụjụ 15 Ripọtì
The number of geometrical isomers of butene are
Akọwa Nkọwa
To understand the geometrical isomers of butene, we need to explore its structure. Butene has four carbon atoms, and there are various structural forms that butene can take. These structural forms include linear or branched chains, with a double bond present between carbon atoms.
Geometric isomerism is a type of stereoisomerism. It occurs due to restricted rotation around the double bond, leading to different spatial arrangements of groups attached to the carbons forming the double bond. The geometric isomerism primarily occurs in alkenes like butene where the positions of substituents can vary.
Let's consider the different types of butene, focusing on the possibility of geometrical isomerism:
In conclusion, for butene, only 2-butene has geometrical isomers (cis and trans). Therefore, the number of geometric isomers is 2.
Ajụjụ 16 Ripọtì
The empirical formula of an organic liquid hydrocarbon is XY. If the relative molar masses of X and Y are 72 and 6 respectively, it's vapour density is likely to be
Akọwa Nkọwa
To determine the vapor density of the organic liquid hydrocarbon with the empirical formula XY, we first need to determine the **molecular formula** of the compound, which represents the actual number of atoms of each element in a molecule.
The **relative molar masses** of X and Y are given as 72 and 6, respectively. To find the molar mass of XY, we can add these values together:
Molar mass of XY = Molar mass of X + Molar mass of Y = 72 + 6 = 78 g/mol
Vapor density is defined as half of the molar mass of the compound, since vapor density is often compared to hydrogen, where hydrogen is taken as the standard with a molar mass of 2 g/mol. Therefore, vapor density can be calculated using the formula:
Vapor Density = (Molar Mass of the Compound) / 2
Substituting the molar mass of XY:
Vapor Density of XY = 78 / 2 = 39
Therefore, the vapor density of the hydrocarbon with the empirical formula XY is **39**.
Ajụjụ 17 Ripọtì
An organic compound with general formula RCOR' is an
Akọwa Nkọwa
The general formula RCOR' represents a class of organic compounds known as ketones. In this formula, R and R' are alkyl groups, which are chains of carbon and hydrogen atoms. The CO in the middle is a carbonyl group, which consists of a carbon atom double-bonded to an oxygen atom. Therefore, with the presence of two alkyl groups on either side of the carbonyl group, the compound is categorized as a ketone, scientifically referred to as an alkanone.
Here is a simple breakdown of the terms:
Hence, by looking at the general formula RCOR', the organic compound in question is undoubtedly an alkanone.
Ajụjụ 18 Ripọtì
The reaction between alkanoic acids and alkanols in the presence of an acid catalyst is known as
Akọwa Nkọwa
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.
Ajụjụ 19 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 20 Ripọtì
A type of isomerism that ClCH=CHCl can exhibit is
Akọwa Nkọwa
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.
Ajụjụ 21 Ripọtì
How much of 5g of radioactive element whose half life is 50days remains after 200days?
Akọwa Nkọwa
To determine how much of a radioactive element remains after a certain period, we use the concept of half-life. The half-life of a substance is the time it takes for half of the initial amount of a radioactive element to decay. In this example, the half-life is given as 50 days.
We want to know how much of a 5g sample remains after 200 days. First, calculate how many half-lives occur in 200 days:
Number of half-lives = Total time elapsed / Half-life
= 200 days / 50 days
= 4 half-lives
Next, we calculate the remaining amount after each half-life period:
After 200 days, 0.31g of the radioactive element remains.
Ajụjụ 22 Ripọtì
When a few drops of Millon reagents is added to egg-white solution in a test tube, the white precipitate changes to
Akọwa Nkọwa
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.
Ajụjụ 23 Ripọtì
| 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?
Akọwa Nkọwa
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
Ajụjụ 24 Ripọtì
Fats and oils are esters of fatty organic acids combined with a trihydric alkanol commonly referred to as
Akọwa Nkọwa
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.
Ajụjụ 25 Ripọtì
When Sulphur(IV)oxide is passed into solution of acidified tetraoxomanganate(VII), the colour changes from
Akọwa Nkọwa
When Sulphur(IV) oxide (SO2) is passed into a solution of acidified tetraoxomanganate (VII) (KMnO4), it acts as a reducing agent. This reaction involves the reduction of potassium permanganate (KMnO4), which is characterized by a distinctive color change.
The tetraoxomanganate (VII) ion (MnO4-) is purple in color. During the reaction, SO2 gets oxidized while the MnO4- ion is reduced to Mn2+, which is almost colorless or pale pink, depending on the concentration.
Thus, the color of the solution changes from purple to almost colorless as the reaction progresses.
Ajụjụ 26 Ripọtì
CuOs + H2 (g ) ⇌ Cus + H2 O(g )
In the equation above, the effect of increased pressure on the equilibrium position is that
Akọwa Nkọwa
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.
Ajụjụ 27 Ripọtì
How many isomers has the organic compound represented by the formula C3 H8 O ?
Akọwa Nkọwa
The molecular formula C3H8O represents organic compounds that contain 3 carbon atoms, 8 hydrogen atoms, and 1 oxygen atom. Let's elucidate the possible isomers, which are molecules with the same molecular formula but different structural arrangements.
1. Alcohols: One class of compounds that can form isomers for this formula are alcohols, which include a functional group -OH.
a. Propan-1-ol: This is a straight-chain alcohol where the -OH group is on the first carbon. The structure is as follows:
CH3-CH2-CH2-OH
b. Propan-2-ol: This is another alcohol where the -OH group is on the second carbon, giving it a different structure and properties:
CH3-CH(OH)-CH3
2. Ethers: This is another class of possible isomers, where the oxygen atom is bonded to two alkyl groups.
c. Methoxyethane: Also known as ethyl methyl ether, it has a structure where the oxygen is in a bridge position between a methyl group and an ethyl group:
CH3-O-CH2-CH3
These are the possible structural isomers for this molecular formula. Therefore, the compound C3H8O has three isomers overall:
Thus, the answer is three distinct isomers.
Ajụjụ 28 Ripọtì
Hydrochloric acid is regarded as a strong acid because it
Akọwa Nkọwa
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.
Ajụjụ 29 Ripọtì
Akọwa Nkọwa
Silver and Gold are classified as noble metals. These metals are known for their resistance to corrosion and oxidation in moist air, unlike most other base metals. They can be found in the earth's crust as free, uncombined elements because they do not easily react with oxygen and other elements to form compounds. This property is what distinguishes noble metals from more reactive or corrosive ones. While the term "natural metals" seems applicable in that they occur naturally, the more precise and widely accepted term for metals like Silver and Gold is "noble metals".
Ajụjụ 30 Ripọtì
The main constituent of water-glass is
Akọwa Nkọwa
The main constituent of water-glass is sodium trioxosilicate(IV). Water-glass, also known as liquid glass, is common terminology for a mixture of sodium silicate and water. The primary chemical component in water-glass is sodium silicate, which includes sodium ions (Na+) bonded with silicate ions (SiO44-).
Essentially, when sodium silicate is dissolved in water, it results in a viscous liquid that can be utilized in various applications such as in cements, passive fire protection, textile and lumber processing, and as a sealant. Sodium trioxosilicate(IV) forms a significant part of this mixture as it reacts with other compounds to create a hardened, glass-like structure when it dries. Therefore, when water-glass is mentioned, it is mostly referring to solutions that have sodium trioxosilicate(IV) as their principal compound.
Ajụjụ 31 Ripọtì
The constituents of Alnico are Aluminium, Nickel and
Akọwa Nkọwa
Alnico is a type of alloy that is known for its strong magnetic properties. The name "Alnico" comes from the elements it is primarily composed of: Aluminum (Al), Nickel (Ni), and Cobalt (Co). These elements are combined to form an alloy that retains its magnetism well and can operate at high temperatures, making it ideal for applications like electric motors, sensors, and various electronic devices.
While there are different variations of Alnico, the presence of Cobalt (Co) is essential for enhancing the magnetic properties of the alloy. The other elements listed, such as Magnesium (Mg), Manganese (Mn), and Copper (Cu), are not typical core constituents of Alnico. Although trace amounts of other elements like copper may sometimes be included in specific formulations, the primary and most significant component responsible for Alnico's powerful magnetic characteristics is Cobalt (Co).
Ajụjụ 32 Ripọtì
The electronic configuration of an atom of Nitrogen is 1s2 2s2 2p1x 2p1y 2p1z because the atom is
Akọwa Nkọwa
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.
Ajụjụ 33 Ripọtì
Nitrogen obtained from air is not absolutely pure because it contains the following except
Akọwa Nkọwa
Nitrogen obtained from air is not absolutely pure because it contains other gases, including:
Ajụjụ 34 Ripọtì
What accounts for the low melting and boiling points of covalent molecules?
Akọwa Nkọwa
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.
Ajụjụ 35 Ripọtì
Akọwa Nkọwa
When a metal reacts with an acid, a chemical reaction takes place in which the metal displaces the hydrogen in the acid. This reaction produces a salt and hydrogen gas is liberated in the process.
Let's break it down further:
The general equation for the reaction is:
Metal + Acid → Salt + Hydrogen Gas
For example, when zinc (a metal) reacts with hydrochloric acid (an acid), the reaction is as follows:
Zn + 2HCl → ZnCl2 + H2
Here, zinc chloride (a salt) and hydrogen gas are produced. This illustrates that salt and hydrogen gas are formed when a metal reacts with an acid.
Ajụjụ 36 Ripọtì
When a specie undergoes oxidation, its
Akọwa Nkọwa
When a species undergoes oxidation, it experiences an increase in its oxidation number. Oxidation is a chemical process where a species loses electrons. In terms of oxidation number, electrons have a negative charge, so losing them results in an increase in charge. Thus, the oxidation number of the species becomes more positive or less negative.
To help understand, consider sodium (Na) reacting with chlorine (Cl2) to form sodium chloride (NaCl):
This change clearly shows that when sodium is oxidized, its oxidation number increases.
Therefore, the correct explanation is: a species undergoing oxidation will have its oxidation number increase.
Ajụjụ 37 Ripọtì
Na2 X ⇌ 2Na+ + X2−
The bond between Na and X is likely to be
Akọwa Nkọwa
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.
Ajụjụ 38 Ripọtì
In the graph above, y represents
Akọwa Nkọwa
To understand what y represents in the graph, we need to think about what graphs in chemistry, specifically regarding energy changes in reactions, generally show.
Chemical reaction energy diagrams often depict a reaction's energy change as a curve from the reactants to the products, showing different energy levels throughout the process. The energy required to start a reaction or to transform the reactants into an activated complex (also known as the transition state) is crucial.
The height of this energy barrier is called the activation energy. This is the minimum amount of energy required to start a chemical reaction. The activation energy is represented by the peak in the energy graph between the reactant energy level and the top of the curve.
Therefore, in this context, y represents the activation energy needed for the reaction to proceed. Understanding activation energy is vital as it determines how quickly a reaction will occur. Reactions with a high activation energy tend to happen more slowly because it is less probable that the necessary energy for the reaction to occur spontaneously will be present.
Ajụjụ 39 Ripọtì
In the treatment of water for municipal supply, chlorine is used to
Akọwa Nkọwa
In the treatment of water for municipal supply, chlorine is used to kill germs. This process is known as chlorination. Chlorine is a very effective disinfectant and is used to eliminate harmful microorganisms such as bacteria, viruses, and protozoans that may be present in the water. By doing so, chlorine helps to ensure that the water is safe for human consumption and protects public health by preventing waterborne diseases. It is important to note that **chlorine is not used to prevent tooth decay, prevent goitre, or to remove colour or odour** in water treatment for municipal supply.
Ajụjụ 40 Ripọtì
If 11.0g of a gas occupies 5.6 dm3 at s.t.p., calculate its vapour density (1 mole of a gas occupies 22.4 dm3 ).
Akọwa Nkọwa
The problem requires calculating the **vapor density** of the gas. Vapor density is defined as the mass of a certain volume of a gas compared to the mass of an equal volume of hydrogen, where the hydrogen standard is 2 g/mol (as the molecular weight of hydrogen gas, H₂, is 2).
Here's a step-by-step explanation:
The calculated vapor density of the gas is 22.
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