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Tambaya 1 Rahoto
The indicator used in a titration between strong acid and weak base is
Bayanin Amsa
A titration is a process used to determine the concentration of an unknown solution by adding a solution of known concentration. The indicator used in a titration is a substance that changes color at the specific pH level of the solution, which usually happens at the equivalence point.
For a titration between a strong acid and a weak base, the solution at the equivalence point is slightly acidic. This is because the salt formed as a result of the neutralization reaction can undergo hydrolysis, producing an excess of hydronium ions (H₃O⁺) which makes the solution acidic.
Among the provided indicators, methyl orange is the most suitable for indicating this type of reaction because it changes color in an acidic pH range of about 3.1 to 4.4. It shifts from red at a pH below 3.1 to yellow at a pH above 4.4.
Therefore, for a titration involving a strong acid and a weak base, methyl orange is the appropriate indicator as it can show the end point effectively when the solution is slightly acidic. The pH at the equivalence point falls within the color change range of methyl orange.
Tambaya 2 Rahoto
The volume occupied by 1 mole of an ideal gas at a temperature of 130 C and a pressure of 1.58 atm is
[ R = 0.082 atm dm3 K−1 mol−1 ]
Bayanin Amsa
According to the Ideal gas equation, PV = nRT
Given: P = 1.58 atm, V = ?, n = 1 mole, R = 0.082, T= 13 + 273K = 286K
Substituting all the given parameters,
V = nRTP
V = 1×0.082×2861.58
V = 14.84 dm3
Tambaya 3 Rahoto
One of the following is not a water pollutant?
Bayanin Amsa
Water pollutants are substances that, when introduced into the water, cause harm to ecosystems, human health, and the overall quality of the water. Each of the options provided has the potential to be considered a water pollutant, except for one. Let's explain them:
1. Inorganic fertilizers: These are substances mainly composed of synthetic chemicals, including nitrates and phosphates. When these fertilizers enter water bodies, they can lead to nutrient pollution, which causes excessive growth of algae (eutrophication), leading to a decrease in oxygen levels in the water, harming aquatic life.
2. Warm water affluent: This refers to the discharge of heated water into natural water bodies. This heat contamination can change the temperature of the water, affecting the metabolism of aquatic life and leading to thermal pollution.
3. Oxygen gas: Oxygen gas is a fundamental component of the Earth's atmosphere and is not considered a water pollutant. In fact, dissolved oxygen is crucial for the survival of aquatic organisms. Rather than causing any harm, adequate levels of dissolved oxygen in water bodies are essential for maintaining healthy aquatic ecosystems.
4. Biodegradable waste: These are organic materials that decompose in the environment. When introduced in large quantities into water bodies, they can consume a significant amount of dissolved oxygen as they decompose, which can lead to depletion of oxygen levels and cause harm to aquatic life, making them pollutants in aquatic ecosystems.
Given the explanations above, oxygen gas is the option that is not a water pollutant. It is vital for the health of aquatic ecosystems, unlike the other options, which can all lead to some form of pollution in water bodies.
Tambaya 4 Rahoto
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
Bayanin Amsa
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.
Tambaya 5 Rahoto
During the fractional distillation of crude oil, the fraction that distills at 200 - 2500 C is
Bayanin Amsa
The petroleum fractions that distill at 200–250°C are naphtha and kerosene,
Tambaya 6 Rahoto
The highest isotope of hydrogen is
Bayanin Amsa
Hydrogen has three naturally occurring isotopes, and each of them contains the same number of protons but different numbers of neutrons. Let's briefly differentiate them:
The highest isotope of hydrogen is tritium because it has the most neutrons and, therefore, the greatest atomic mass compared to the other isotopes. It is also noteworthy that tritium is radioactive, while the other hydrogen isotopes are stable.
Tambaya 7 Rahoto
The pH of a 0.001 mol dm−3 of H2 SO4 is
[Log10 2 = 0.3]
Bayanin Amsa
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.
Tambaya 8 Rahoto
Which of the following is an air pollutant?
Bayanin Amsa
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).
Tambaya 9 Rahoto
The reaction between alkanoic acids and alkanols in the presence of an acid catalyst is known as
Bayanin Amsa
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.
Tambaya 10 Rahoto
The scientist that performed the experiment on discharged tubes that led to the discovery of the cathode rays as a sub-atomic particle is
Bayanin Amsa
The **scientist who performed the experiment on discharge tubes that led to the discovery of cathode rays as a sub-atomic particle** is J.J. Thomson.
In the late 19th century, J.J. Thomson conducted experiments using a cathode ray tube. This device involved an evacuated glass tube with electrodes at each end, through which an electric current was passed. **When a high voltage was applied, Thomson observed a stream of particles traveling from the negative electrode (cathode) to the positive electrode (anode).** These streams of particles were what he called "cathode rays."
Through his experiments, J.J. Thomson discovered that these cathode rays were composed of negatively charged particles. **He concluded that these particles were much smaller than atoms, and named them "electrons," which are now known to be sub-atomic particles.** His work was fundamental in advancing the atomic model and in understanding the structure of the atom.
Thomson's discovery was pivotal because it provided the first evidence that atoms are not indivisible, but rather consist of smaller subatomic particles. This **challenged the then-prevailing notion of atoms as indivisible units**, thus marking the birth of modern particle physics.
Tambaya 11 Rahoto
In the graph above, y represents
Bayanin Amsa
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.
Tambaya 12 Rahoto
Which of the following is used in forming slag in the blast furnace for the extraction of iron?
Bayanin Amsa
In the process of extracting iron in a blast furnace, CaCO3, or calcium carbonate, plays a crucial role in forming slag. Here is a simple and comprehensive explanation of how it works:
1. Role of Calcium Carbonate (CaCO3):
Calcium carbonate is commonly used as a flux in the blast furnace. When it is introduced into the furnace, it undergoes a decomposition reaction due to the high temperatures, breaking down into calcium oxide (CaO) and carbon dioxide (CO2).
2. Formation of Slag:
The calcium oxide (CaO) produced then reacts with silicon dioxide (SiO2) present in the iron ore. This reaction forms a liquid slag of calcium silicate. The slag serves two main functions:
Thus, calcium carbonate (CaCO3) is crucial for forming slag by providing the necessary calcium oxide (CaO) that reacts with impurities to form slag during the extraction of iron in a blast furnace.
Tambaya 13 Rahoto
The IUPAC name of the compound above is
Bayanin Amsa
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."
Tambaya 14 Rahoto
An example of an amphoteric oxide is
Bayanin Amsa
An example of an amphoteric oxide is Al2O3 (aluminum oxide).
Amphoteric oxides are special because they can act as both acidic and basic oxides. This means they can react with both acids and bases to form salts and water, showcasing their dual behavior.
Here is how it works:
In contrast, oxides like CuO (copper(II) oxide) are basic oxides, and K2O (potassium oxide) is a basic oxide as well. They don't exhibit both acidic and basic properties.
Therefore, the amphoteric nature of Al2O3 is what distinguishes it from common oxides that are strictly acidic or basic. This property is crucial in various chemical processes and applications.
Tambaya 15 Rahoto
How much of 5g of radioactive element whose half life is 50days remains after 200days?
Bayanin Amsa
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.
Tambaya 16 Rahoto
For chemical reaction to be spontaneous, ∆G must be
Bayanin Amsa
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.
Tambaya 17 Rahoto
The main constituent of water-glass is
Bayanin Amsa
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.
Tambaya 18 Rahoto
Water gas obtained from the gasification of coke is made up of
Bayanin Amsa
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.
Tambaya 19 Rahoto
A liquid hydrocarbon obtained from fractional distillation of coal tar that is used in the pharmaceutical industry is
Bayanin Amsa
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.
Tambaya 20 Rahoto
A major effect of oil pollution in coastal water is
Bayanin Amsa
One of the major effects of oil pollution in coastal water is the destruction of aquatic life.
When oil spills into a water body, it forms a thin layer called a sheen on the surface of the water. This oil layer blocks sunlight from reaching aquatic plants and phytoplankton, inhibiting their ability to perform photosynthesis. As a result, these plants and microorganisms suffer, impacting the entire food chain.
Moreover, oil can coat the feathers of birds and the fur of marine mammals, which affects their insulation and buoyancy, leading to hypothermia, drowning, or inability to fly. Additionally, the toxic components in oil are harmful if ingested, causing internal damage to fish and other marine organisms. These combined effects can lead to significant mortality in aquatic ecosystems, threatening biodiversity and the natural balance of coastal waters.
Therefore, oil pollution can severely affect the health and survival of aquatic life, creating disruptions that can persist for many years.
Tambaya 21 Rahoto
Esterification reaction is analogous to
Bayanin Amsa
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**.
Tambaya 22 Rahoto
Fats and oils are esters of fatty organic acids combined with a trihydric alkanol commonly referred to as
Bayanin Amsa
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.
Tambaya 23 Rahoto
Determine the half-life of a first order reaction with constant 4.5 x 10−3 sec−1 .
Bayanin Amsa
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.
Tambaya 24 Rahoto
When Sulphur(IV)oxide is passed into solution of acidified tetraoxomanganate(VII), the colour changes from
Bayanin Amsa
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.
Tambaya 25 Rahoto
The product formed when ethyne is passed through a hot tube containing finely divided iron is
Bayanin Amsa
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**.
Tambaya 26 Rahoto
127g of sodium chloride was dissolved in 1.0dm3 of distilled water at 250 C . Determine the solubility in moldm−3 of sodium chloride at that temperature. [Na = 23, Cl = 35.5]
Bayanin Amsa
To determine the solubility of sodium chloride (NaCl) in mol/dm3 at the given temperature, you need to first calculate the number of moles of NaCl dissolved.
Step 1: Calculate the molar mass of NaCl.
The molar mass of a compound is found by adding the atomic masses of its constituent elements:
- Sodium (Na) has an atomic mass of 23.
- Chlorine (Cl) has an atomic mass of 35.5.
Thus, the molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.
Step 2: Calculate the number of moles of NaCl.
The formula to calculate moles is:
Number of moles = Mass (g) / Molar mass (g/mol)
Given mass of NaCl = 127 g,
Number of moles = 127 g / 58.5 g/mol ≈ 2.17 mol
Step 3: Calculate the solubility in mol/dm3.
Since the sodium chloride is dissolved in 1.0 dm3 of water, the solubility is the same as the number of moles, since the volume is already 1.0 dm3.
Therefore, the solubility of sodium chloride at that temperature is 2.17 mol/dm3.
Rounded to the options given, 2.17 mol/dm3 is approximately equal to 2.2 mol/dm3.
Tambaya 27 Rahoto
In the treatment of water for municipal supply, chlorine is used to
Bayanin Amsa
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.
Tambaya 28 Rahoto
The hybridization scheme in ethyne is
Bayanin Amsa
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.
Tambaya 29 Rahoto
An example of a substance that does not change directly from solid to gas when heated is
Bayanin Amsa
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.
Tambaya 30 Rahoto
How many moles of CO2 are produced when ethanol is burnt with 6g of oxygen
Bayanin Amsa
To determine how many moles of carbon dioxide (CO2) are produced when ethanol is burnt with 6g of oxygen, we need to understand the balanced chemical equation for the combustion of ethanol. The reaction is as follows:
C2H5OH + 3O2 → 2CO2 + 3H2O
This equation tells us that 1 mole of ethanol (C2H5OH) reacts with 3 moles of oxygen (O2) to produce 2 moles of carbon dioxide (CO2).
First, let's calculate how many moles of oxygen 6 g represents. The molecular weight of oxygen (O2) is approximately 32 g/mol. Therefore, the number of moles of oxygen is:
Number of moles of O2 = 6 g / 32 g/mol = 0.1875 moles
According to the balanced equation, 3 moles of O2 produce 2 moles of CO2. Hence, the relationship between moles of O2 and moles of CO2 is:
2 moles of CO2 / 3 moles of O2 = x moles of CO2 / 0.1875 moles of O2
Solving for x, we have:
x = (2/3) * 0.1875 = 0.125
Therefore, 0.125 moles of CO2 are produced when 6g of oxygen is used to burn ethanol.
Tambaya 31 Rahoto
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?
Bayanin Amsa
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
Tambaya 32 Rahoto
In a chemical reaction, surface area of reactants can affect
Bayanin Amsa
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.
Tambaya 33 Rahoto
Biuret test is a chemical test used for detecting the presence of
Bayanin Amsa
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.
Tambaya 34 Rahoto
The shape of the molecule of Carbon(IV) oxide is
Bayanin Amsa
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.
Tambaya 35 Rahoto
H2 S(g) + Cl2 (g) → 2HCl(g) + S(s)
What is the change in oxidation state of sulphur from reactant to product?
Bayanin Amsa
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**.
Tambaya 36 Rahoto
The volume in cm3 of a 0.12 moldm−3 HCl required to completely neutralize a 20cm3 of 0.20 moldm−3 of NaOH is
Bayanin Amsa
To find the volume of HCl that is required to completely neutralize the NaOH solution, we need to use the concept of a neutralization reaction. A neutralization reaction occurs when an acid and a base react to form water and a salt, thus neutralizing each other.
In this particular reaction, the balanced chemical equation is:
HCl + NaOH → NaCl + H2O
Here, the equation tells us that one mole of HCl reacts with one mole of NaOH. Therefore, the molar ratio of HCl to NaOH is 1:1.
First, let's determine the number of moles of NaOH present in 20 cm3 solution:
Number of moles of NaOH = Concentration (mol/dm3) × Volume (dm3)
= 0.20 mol/dm3 × 20 cm3 × (1 dm3 / 1000 cm3)
= 0.20 × 0.020
= 0.004 moles
Since the reaction is in a 1:1 ratio, the number of moles of HCl required is also 0.004 moles.
Now, let's determine the volume of HCl solution required:
Volume of HCl (dm3) = Number of moles / Concentration
= 0.004 moles / 0.12 mol/dm3
= 0.03333 dm3
Convert this volume from dm3 to cm3:
0.03333 dm3 × 1000 cm3 / dm3 = 33.33 cm3
Therefore, the volume of HCl required is 33.33 cm3.
Tambaya 37 Rahoto
The amount of water a substance chemically combined with is called water of
Bayanin Amsa
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.
Tambaya 38 Rahoto
Rust on the surface of a metal sheet contains
Bayanin Amsa
Rust on the surface of a metal, specifically on **iron**, is primarily composed of **hydrated iron(III) oxide**. The rusting process occurs when **iron** reacts with **oxygen** and **water** from the environment. This chemical reaction typically produces a compound called **iron(III) oxide**, which is then combined with water molecules, resulting in **hydrated iron(III) oxide**. This hydrated state gives rust its characteristic flaky and reddish-brown appearance.
Tambaya 39 Rahoto
A radioactive element of mass 1g has half-life of 2 minutes, what fraction of the substance would have disintegrated after 10 minutes?
Bayanin Amsa
Originalmass2n
= Residual mass
Where n = number of activity = exposuretimehalflife
Given:
Original mass = 1g, exposure time = 10 minutes , half life = 2 minutes, Residual mass = ?
Substituting all the given parameters appropriately, we have
n = 102
n = 5
Originalmass2n = Residual mass
125
5 = Residual mass
132 = Residual mass
Residual mass = 132
or 0.03125g
Tambaya 40 Rahoto
The group VIII elements are the inert gases because they
Bayanin Amsa
The group VIII elements, also known as the noble gases, are called inert gases primarily because they all have completely filled valence shells. In a very simplified explanation:
1. Complete Valence Shells: All the noble gases have their outermost shells completely filled with electrons. This configuration is considered very stable and requires no additional electrons to reach stability, unlike other elements that may gain, lose, or share electrons to achieve a full valence shell.
2. Highly Stable: Due to this completely filled valence shell, the noble gases do not readily react with other elements to form compounds. Their stability comes from the fact that they do not need to bond with other elements to achieve a more stable state.
3. Examples: For instance, Helium (He) has two electrons filling its first shell, Neon (Ne) has eight electrons in its second shell, and similarly, other noble gases also have fully occupied outer shells.
This property is why the noble gases are termed "inert," which means they are largely non-reactive.
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