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Ibeere 1 Ìròyìn
A liquid hydrocarbon obtained from fractional distillation of coal tar that is used in the pharmaceutical industry is
Awọn alaye Idahun
Benzene is a liquid hydrocarbon that is obtained from the fractional distillation of coal tar, and it is extensively used in the pharmaceutical industry. Let me break this down for you:
That's why benzene plays an important role in the pharmaceutical industry, making it a highly valued product obtained through the distillation of coal tar.
Ibeere 2 Ìròyìn
The amount of water a substance chemically combined with is called water of
Awọn alaye Idahun
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.
Ibeere 3 Ìròyìn
What method is suitable for the separation of gases present in air?
Awọn alaye Idahun
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.
Ibeere 4 Ìròyìn
When Calcium ethynide is decomposed by water, the gas produced is
Awọn alaye Idahun
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.
Ibeere 5 Ìròyìn
An organic compound with general formula RCOR' is an
Awọn alaye Idahun
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.
Ibeere 6 Ìròyìn
The combustion of candle under limited supply of air forms
Awọn alaye Idahun
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).
Ibeere 7 Ìròyìn
The reaction of hydrogen and chlorine to produce hydrogen chloride gas is explosive in
Awọn alaye Idahun
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.
Ibeere 8 Ìròyìn
A type of isomerism that ClCH=CHCl can exhibit is
Awọn alaye Idahun
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.
Ibeere 9 Ìròyìn
Heat of solution involves two steps that is accompanied by heat change. The energies involved in this steps are
Awọn alaye Idahun
The heat of solution refers to the overall energy change that occurs when a solute dissolves in a solvent. This process involves breaking and making of intermolecular forces, and it can be broken down into two main steps that are each accompanied by heat change. The energies involved in these steps are:
Lattice energy: This is the energy required to break the bonds between the ions in the solid crystal lattice of the solute. Breaking these bonds requires energy, and this step is usually endothermic, meaning it absorbs heat from the surroundings. The more energy needed to break the lattice, the higher the lattice energy.
Hydration energy: Once the lattice is broken, the ions are surrounded by solvent molecules, typically water, in a process known as hydration. The energy released when the solvent molecules interact with and stabilize the ions is called the hydration energy. This step is usually exothermic, meaning it releases heat into the surroundings.
In conclusion, the two energies involved in the heat of solution are lattice energy and hydration energy. The balance between these two energies determines whether the overall process of dissolving a solute in a solvent is endothermic or exothermic.
Ibeere 10 Ìròyìn
Cx Hy O + 5O2 → 4CO2 + 4H2 O
Cx Hy O in the equation is
Awọn alaye Idahun
Cx Hy O + 5O2 → 4CO2 + 4H2 O
On balancing the equation, we should have
X = 4 , y = 8 and O = 2 ⇒ C4 H8 O2
Since 2 is a common factor to the three atoms, we can divide through by 2, considering the fact that that formula is not in the option.
We finally have C2 H4 O
Ibeere 11 Ìròyìn
Which of the following is an air pollutant?
Awọn alaye Idahun
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).
Ibeere 12 Ìròyìn
The law which states that a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is
Awọn alaye Idahun
The law that states a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is the law of definite proportion.
To explain this simply, let's consider water as an example. Water is made up of hydrogen and oxygen. According to the law of definite proportion, a sample of pure water taken from anywhere in the world will always contain the same ratio of hydrogen to oxygen by mass. Specifically, water will always have approximately 88.8% oxygen and 11.2% hydrogen by mass.
This is because a chemical compound has a fixed composition, regardless of the process used to create it or the source from which it is derived. The law of definite proportion, also known as the law of constant composition, is fundamental in chemistry because it supports the idea that chemical compounds are composed of elements in specific and fixed ratios. This does not change regardless of how the compound is prepared or where it is found.
Ibeere 13 Ìròyìn
The compound of Copper used as a fungicide is
Awọn alaye Idahun
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).
Ibeere 14 Ìròyìn
CH3 -CH2 -OH and CH3 -O-CH3
The relationship between the two compounds above, is that they are
Awọn alaye Idahun
The relationship between the two compounds is that they are isomers.
To understand why these compounds are isomers, let's break down their structures and definitions:
1. Structures of the Compounds:
2. Definitions:
Both compounds have the same molecular formula: C2H6O. However, they have different arrangements of their atoms. Ethanol has a hydroxyl group (-OH) attached to an ethyl group (CH3-CH2-), while dimethyl ether involves two methyl groups (CH3-) bonded to an oxygen atom (O). This difference in structure leads to different chemical and physical properties, despite having the same molecular formula. Hence, these two compounds are classified as isomers.
Ibeere 15 Ìròyìn
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 ]
Awọn alaye Idahun
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.
Ibeere 16 Ìròyìn
The element which can combine with oxygen to form an acid anhydride of the form XO2 is
Awọn alaye Idahun
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.
Ibeere 17 Ìròyìn
H2 SO4
C2 H5 OH → C2 H4
1700 C
The reaction above illustrates
Awọn alaye Idahun
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.
Ibeere 18 Ìròyìn
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]
Awọn alaye Idahun
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.
Ibeere 19 Ìròyìn
Boyle's law can be expressed mathematically as
Awọn alaye Idahun
Boyle's Law describes the relationship between the volume and pressure of a given amount of gas held at a constant temperature. It states that the pressure of a gas is inversely proportional to its volume. In simpler terms, if you decrease the volume of a gas, its pressure increases, provided the temperature remains constant, and vice versa.
The mathematical expression of Boyle's Law is PV = K, where:
This relationship implies that if you multiply the pressure by the volume, the result will always be the same constant as long as no other variables are changed. This is the classic formulation of Boyle's Law, illustrating the inverse relationship between pressure and volume for a gas at constant temperature.
Ibeere 20 Ìròyìn
Solubility curve is a plot of solubility against
Awọn alaye Idahun
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.
Ibeere 21 Ìròyìn
The IUPAC name of the compound above is
Awọn alaye Idahun
To determine the IUPAC name of a compound, follow these steps:
Hence, by following these steps, if the bromo and methyl groups are both attached to the second carbon (lowest numbering possible), the IUPAC name of the compound is "2-bromo, 2-methyl butane."
Ibeere 22 Ìròyìn
The shape of the molecule of Carbon(IV) oxide is
Awọn alaye Idahun
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.
Ibeere 23 Ìròyìn
If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are
Awọn alaye Idahun
To answer this question, let's break it down step by step:
Mass Number: The mass number is the total number of protons and neutrons in an atom's nucleus. In this case, the mass number is given as 31.
Stable Neutral Atom: A stable neutral atom has no overall electrical charge, meaning the number of protons (positively charged) must equal the number of electrons (negatively charged).
If we symbolize the number of protons by the atomic number (Z), we can say:
1. **Protons = Electrons** in a neutral atom.
2. **Mass Number (A) = Protons + Neutrons**.
Given that the mass number is 31, we have the equation:
A = Protons + Neutrons = 31.
Assuming a commonly known stable element like Phosphorus, which has an atomic number (Z) of 15, it means:
1. **Protons = 15**.
2. **Electrons = 15** (because it's a neutral atom).
3. To find Neutrons: Neutrons = Mass Number - Protons = 31 - 15 = 16.
So, in this scenario, the number of electrons is 15 and the number of neutrons is 16. This combination is found in the first option given.
Ibeere 24 Ìròyìn
In the conductance of aqueous CuSO4 solution, the current carriers are the
Awọn alaye Idahun
In the conductance of aqueous CuSO4 solution, the current carriers are the hydrated ions.
Here's why:
The other options can be understood as follows:
The correct answer is therefore hydrated ions because they enable the conduction of electricity through the aqueous solution.
Ibeere 25 Ìròyìn
Benzene formed nitrobenzene at temperature of 600 C when it reacts with mixture of concentrated trioxonitrate(V) acid and concentrated
Awọn alaye Idahun
The reaction described is the nitration of benzene to form nitrobenzene. This is an example of an electrophilic aromatic substitution reaction. **Nitration** involves replacing a hydrogen atom on a benzene ring with a nitro group (NO2). This reaction requires a nitrating mixture composed of concentrated nitric acid (trioxonitrate(V) acid) and concentrated sulfuric acid (tetraoxosulphate(VI) acid). Let me explain why:
Nitration is typically carried out using a mixture of **concentrated nitric acid and concentrated sulfuric acid** at a temperature of around **60°C**. The role of sulfuric acid in this mixture is to act as a catalyst and a dehydrating agent. It helps generate the nitronium ion (NO2+), which is the active electrophile that attacks the benzene ring.
Here's a simplified mechanism for this reaction:
None of the other options listed (hydrochloric acid, phosphoric acid, and hydrogen iodide) contain the necessary combination of properties to generate the nitronium ion and facilitate the nitration of benzene.
Therefore, the correct mixture to carry out the nitration of benzene, forming nitrobenzene at a temperature of 60°C, is a combination of **concentrated nitric acid and concentrated sulfuric acid (tetraoxosulphate(VI) acid)**.
Ibeere 26 Ìròyìn
Esterification reaction is analogous to
Awọn alaye Idahun
The **esterification reaction** is analogous to a **condensation reaction**. In chemistry, a **condensation reaction** is a type of chemical reaction where two molecules or functional groups combine to form a larger molecule, with the simultaneous loss of a small molecule, usually water. **Esterification** specifically involves the reaction between an acid (often a carboxylic acid) and an alcohol, resulting in the formation of an **ester** and the release of a molecule of water.
To explain this further, in an esterification reaction:
Conversely, the other types of reactions you've mentioned have different mechanisms:
Therefore, given the nature of how molecules join and release water, it's clear that the **esterification reaction** is analogous to a **condensation reaction**.
Ibeere 27 Ìròyìn
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 ).
Awọn alaye Idahun
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.
Ibeere 28 Ìròyìn
The stability of atomic nucleus is determined by ratio of
Awọn alaye Idahun
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.
Ibeere 29 Ìròyìn
Determine the empirical formula of an oxide of sulphur containing 60% of oxygen
[S = 32, O = 16 ]
Awọn alaye Idahun
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.
Ibeere 30 Ìròyìn
The basicity of tetraoxophosphate(V) acid is
Awọn alaye Idahun
The term basicity of an acid refers to the number of hydrogen ions (H⁺) that an acid can donate when it dissociates in water. In simpler terms, it's the number of replaceable hydrogen ions in one molecule of the acid.
Tetraoxophosphate(V) acid is another name for phosphoric acid, which has the chemical formula H₃PO₄. In this molecule, there are three hydrogen (H) atoms bonded to the phosphate group (PO₄).
When H₃PO₄ dissolves in water, it donates hydrogen ions in three steps:
Therefore, phosphoric acid, or tetraoxophosphate(V) acid, can donate a total of three hydrogen ions. Hence, the basicity of tetraoxophosphate(V) acid is 3.
Ibeere 31 Ìròyìn
The number of molecules of helium gas contained in 11.5g of the gas is
Awọn alaye Idahun
To find the number of molecules of helium gas in a given mass, we can use Avogadro's number and the molar mass of helium.
Step 1: Determine the molar mass of helium.
Helium is a noble gas with an atomic mass of approximately 4 grams per mole (g/mol).
Step 2: Calculate the number of moles in 11.5 grams of helium.
The formula to find the number of moles is:
Number of moles = Mass (g) / Molar Mass (g/mol)
So for helium:
Number of moles = 11.5 g / 4 g/mol = 2.875 moles
Step 3: Use Avogadro's number to find the number of molecules.
Avogadro's number is 6.022 x 1023 molecules per mole.
The formula to find the number of molecules is:
Number of molecules = Number of moles x Avogadro's Number
Number of molecules = 2.875 moles x 6.022 x 1023 molecules/mole
Number of molecules ≈ 1.73 x 1024 molecules
Therefore, the number of molecules of helium gas in 11.5g of helium is approximately 1.73 x 1024.
Ibeere 32 Ìròyìn
The percentage of carbon(IV) oxide in air is
Awọn alaye Idahun
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%.
Ibeere 33 Ìròyìn
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?
Awọn alaye Idahun
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.
Ibeere 34 Ìròyìn
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?
Awọn alaye Idahun
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
Ibeere 35 Ìròyìn
Strong acids can be distinguished from weak acids by any of the following methods, EXCEPT
Awọn alaye Idahun
To distinguish between strong acids and weak acids, we can employ several methods based on their chemical properties:
Conductivity Measurement: Strong acids dissociate completely in water, releasing more ions. Because ion concentration is directly related to electrical conductivity, strong acids exhibit higher conductivity than weak acids, which only partially dissociate.
Litmus Paper: This method helps determine if a solution is acidic or basic but does not provide detailed information about the strength (strong or weak) of an acid. Both strong and weak acids turn blue litmus red. Therefore, **litmus paper cannot effectively distinguish between a strong and a weak acid.**
Measurement of pH: Strong acids have a lower pH because they fully dissociate to release more hydrogen ions (H+), whereas weak acids have a relatively higher pH as they do not dissociate completely. Thus, pH measurement can distinguish the extent of acidity.
Measurement of Heat of Reaction: The heat of reaction can give insights into the strength of an acid because it involves the degree of ionization and the energetics associated with it. A strong acid will exhibit a different calorimetric response compared to a weak acid.
In summary, **litmus paper is not suitable for distinguishing between a strong and a weak acid**, as it only indicates acidity but does not reveal the strength of the acid.
Ibeere 36 Ìròyìn
At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that
Awọn alaye Idahun
To solve the problem, we can use **Graham's law of effusion**. This law states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is represented as:
Rate of diffusion of Gas X / Rate of diffusion of Gas Y = sqrt(Molar mass of Gas Y / Molar mass of Gas X)
According to the given information, gas X diffuses **twice as fast** as gas Y. This implies:
2 = sqrt(Molar mass of Gas Y / Molar mass of Gas X)
To eliminate the square root, square both sides of the equation:
(2)^2 = Molar mass of Gas Y / Molar mass of Gas X
This simplifies to:
4 = Molar mass of Gas Y / Molar mass of Gas X
Rearranging the equation, we find:
Molar mass of Gas Y = 4 * Molar mass of Gas X
This means that **Gas Y is four times as heavy as Gas X**. Therefore, the correct statement is:
Ibeere 37 Ìròyìn
The IUPAC Nomenclature of CH3 CH2 C(CH3 )=C(CH3 )2 for the compound is
Awọn alaye Idahun
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.
Ibeere 38 Ìròyìn
Aqueous solution of sodium hydroxide can be used to test for the presence of : I. Ca2+ , II. Zn2+ , III. Cu2+
Awọn alaye Idahun
Aqueous solution of sodium hydroxide (NaOH) is a versatile reagent in chemistry, often used to test for the presence of metal ions. When sodium hydroxide is added to solutions containing certain metal ions, it forms precipitates that are characteristic of those ions. Here's how it interacts with each of the mentioned ions:
Calcium ions (Ca2+): When NaOH is added to a solution containing calcium ions, a white precipitate of calcium hydroxide (Ca(OH)2) can form. However, the precipitate is only slightly soluble in water, and this reaction is not the most definitive test for calcium ions.
Zinc ions (Zn2+): When sodium hydroxide is added to a solution containing zinc ions, a white gelatinous precipitate of zinc hydroxide (Zn(OH)2) forms. This precipitate is soluble in excess NaOH, leading to a clear, colorless solution. This reaction is used to test for zinc ions.
Copper ions (Cu2+): When NaOH is added to a solution containing copper ions, a pale blue precipitate of copper(II) hydroxide (Cu(OH)2) forms. This precipitate is insoluble even in excess NaOH, and the formation of this blue precipitate is a common test for copper ions.
Therefore, an aqueous solution of sodium hydroxide can be used to test for the presence of all three ions: calcium (Ca2+), zinc (Zn2+), and copper (Cu2+). The reaction and precipitate formation with each ion serve as indicators of their presence. Thus, the correct answer is:
I, II and III.
Ibeere 39 Ìròyìn
The pH of a 0.001 mol dm−3 of H2 SO4 is
[Log10 2 = 0.3]
Awọn alaye Idahun
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.
Ibeere 40 Ìròyìn
When a few drops of Millon reagents is added to egg-white solution in a test tube, the white precipitate changes to
Awọn alaye Idahun
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.
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