Ana loda....
|
Latsa & Riƙe don Ja Shi Gabaɗaya |
|||
|
Danna nan don rufewa |
|||
Tambaya 1 Rahoto
The IUPAC nomenclature of the compound above is
Bayanin Amsa
The IUPAC nomenclature of the compound above is 2-methylpropan-2-ol.
Tambaya 2 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 3 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 4 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 5 Rahoto
The IUPAC Nomenclature of CH3 CH2 C(CH3 )=C(CH3 )2 for the compound is
Bayanin Amsa
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.
Tambaya 6 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 7 Rahoto
When n = 3, the quantum number of an element is
Bayanin Amsa
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.
Tambaya 8 Rahoto
Fog is a colloid in which
Bayanin Amsa
**Fog** is a type of colloid, which is a mixture where very small particles of one substance are evenly distributed throughout another substance. In the case of fog, it consists of tiny **liquid droplets** that are dispersed in a **gas**. Specifically, these are tiny droplets of water suspended in the air. When you walk through fog, you are essentially walking through air that contains these minute water droplets.
Thus, the correct description of fog as a colloid is that it consists of **liquid particles dispersed in a gas medium**. The liquid here is water, and the gas is air.
Tambaya 9 Rahoto
Boyle's law can be expressed mathematically as
Bayanin Amsa
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.
Tambaya 10 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 11 Rahoto
Sulphur(IV)oxide can be used as a
Bayanin Amsa
Sulphur(IV) oxide has many uses including food preservation, refrigeration, laboratory reagent and solvent, sulphuric acid production, fumigant etc.Sulphur(IV) oxide is a good refrigerant because it has a high heat of evaporation and can be easily condensed.
Tambaya 12 Rahoto
The table above shows the formulae of some ions. In which of these compounds is the formula not correct?
Bayanin Amsa
To assess the correctness of the chemical formulae for the given compounds, let's break down each compound:
Aluminium Tetraoxosulphate(VI), Al2(SO4)3:
Aluminium ion is denoted as Al3+, and the sulphate ion is SO42-. To balance the charges between the positive and negative ions:
2 x (+3) from aluminium ions = +6
3 x (-2) from sulphate ions = -6
Thus, the charges balance out, making the formula correct.
Calcium Trioxonitrate(V), Ca(NO3)2:
Calcium ion is Ca2+, and the nitrate ion is NO3-. To balance the charges:
1 x (+2) from calcium ion = +2
2 x (-1) from nitrate ions = -2
The charges balance out, therefore, this formula is also correct.
Iron(III) Bromide, Fe3Br:
Iron(III) ion is Fe3+, and bromide ion is Br-. Each iron ion would pair with three bromide ions to balance the charges:
FeBr3, where:
1 x (+3) from iron = +3
3 x (-1) from bromide = -3
The charges balance out in the correct formula which should be FeBr3, making the given formula Fe3Br incorrect.
Potassium Sulphide, K2S:
Potassium ion is K+, and sulphide ion is S2-. To balance the charges:
2 x (+1) from potassium ions = +2
1 x (-2) from sulphide ion = -2
The charges balance out, making this formula correct.
Therefore, the compound with the incorrect formula is Iron(III) Bromide where the proper chemical formula should be FeBr3, not Fe3Br.
Tambaya 13 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 14 Rahoto
Cx Hy O + 5O2 → 4CO2 + 4H2 O
Cx Hy O in the equation is
Bayanin Amsa
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
Tambaya 15 Rahoto
Bayanin Amsa
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.
Tambaya 16 Rahoto
23892 U + 10 n → 23992 U
The process above produces
Bayanin Amsa
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.
Tambaya 17 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 18 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 19 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 20 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 21 Rahoto
The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is
Bayanin Amsa
The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is water.
Here's why:
Hence, the correct answer is water, as it is the substance that both reacts with sodium to form an alkali and changes the color of anhydrous copper(II) tetraoxosulphate (VI) to blue.
Tambaya 22 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 23 Rahoto
How many isomers has the organic compound represented by the formula C3 H8 O ?
Bayanin Amsa
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.
Tambaya 24 Rahoto
An example of a physical change is
Bayanin Amsa
An example of a physical change is the boiling of water. Let me explain why this is considered a physical change:
A physical change is a change where the substances involved do not change their chemical composition, meaning they remain the same substance, just in a different form or appearance. In the case of boiling water, when water is heated to its boiling point, it changes from a liquid to a gas (steam), but it is still comprised of water molecules (H2O). The change is reversible, so the gas can condense back into liquid water without any new substance being formed.
On the other hand:
Thus, boiling water is an excellent example of a physical change as it involves only the change in the state of matter without altering the substance's identity.
Tambaya 25 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 26 Rahoto
What method is suitable for the separation of gases present in air?
Bayanin Amsa
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.
Tambaya 27 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 28 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 29 Rahoto
The electronic configuration of an atom of Nitrogen is 1s2 2s2 2p1x 2p1y 2p1z because the atom is
Bayanin Amsa
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.
Tambaya 30 Rahoto
The molecular formular of a hydrocarbon with an empirical formula of CH3 and a molar mass of 30 is
Bayanin Amsa
To find the molecular formula of a hydrocarbon given its empirical formula and molar mass, you need to compare the empirical formula mass with the given molar mass.
The empirical formula given is CH3. The molar mass of the empirical formula is calculated as follows:
Total empirical formula mass = 12 + 3 = 15 g/mol
The provided molar mass of the compound is 30 g/mol. To determine how many empirical units are in the molecular formula, divide the molecular mass (given) by the empirical formula mass:
Number of empirical units = 30 g/mol / 15 g/mol = 2
Therefore, the molecular formula is twice the empirical formula:
Empirical formula: CH3
Molecular formula: (CH3)2 = C2H6
The correct molecular formula is C2H6.
Tambaya 31 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 32 Rahoto
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
Bayanin Amsa
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**.
Tambaya 33 Rahoto
CuOs + H2 (g ) ⇌ Cus + H2 O(g )
In the equation above, the effect of increased pressure on the equilibrium position is that
Bayanin Amsa
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.
Tambaya 34 Rahoto
The number of molecules of helium gas contained in 11.5g of the gas is
Bayanin Amsa
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.
Tambaya 35 Rahoto
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
Bayanin Amsa
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.
Tambaya 36 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 37 Rahoto
Hydrogen chloride gas and ammonia can be used to demonstrate the fountain experiment because they are
Bayanin Amsa
In the fountain experiment, hydrogen chloride gas (HCl) and ammonia (NH₃) are used to demonstrate the creation of a visible 'fountain' due to their high solubility in water. Here's a simple explanation:
When hydrogen chloride gas and ammonia gas come into contact with water, they dissolve very quickly and react vigorously. This is because both gases are very soluble in water. As they dissolve, a vacuum-like pressure is created inside the container where the gases are held, pulling water up into it, creating the 'fountain' effect.
Moreover, when HCl and NH₃ gases react with each other, they form a white, solid product known as ammonium chloride (NH₄Cl), which is a demonstration of how both gases can effectively dissolve and react with not just water, but also with each other.
Thus, the ability of these gases to create a fountain effect is primarily because they are very soluble in water, which allows them to dissolve rapidly and create the pressure differential necessary for the water to be pulled into the container dynamically.
Tambaya 38 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 39 Rahoto
A gas that turns lime water milky is likely to be from
Bayanin Amsa
The gas that turns lime water milky is **Carbon Dioxide**. This is because carbon dioxide reacts with calcium hydroxide, which is the main component of lime water, to form calcium carbonate. This chemical reaction can be represented by the equation:
Ca(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)
In this equation, calcium hydroxide ({Ca(OH)2}) in the lime water reacts with carbon dioxide ({CO2}) to produce calcium carbonate ({CaCO3}) and water ({H2O}).
The result is a milky or cloudy appearance due to the formation of insoluble calcium carbonate precipitate in the lime water. This reaction is a common test for the presence of carbon dioxide gas.
Among the options given, **Trioxocarbonate(IV)** is another name for the Carbonate group involving the gas carbon dioxide ({CO2}). Hence, the gas related to Trioxocarbonate(IV) is the one that turns lime water milky.
Tambaya 40 Rahoto
The principle which states that no two electrons in the same orbitals of an atom have same value for all four quantum numbers is the
Bayanin Amsa
The principle that states that no two electrons in the same orbitals of an atom can have the same value for all four quantum numbers is the Pauli Exclusion Principle.
To understand this principle, it's important to know a bit about the structure of an atom and what quantum numbers are:
Quantum Numbers:
1. **Principal Quantum Number (n):** This describes the energy level or shell of the electron.
2. **Angular Momentum Quantum Number (l):** This describes the subshell or shape of the orbital (s, p, d, f...).
3. **Magnetic Quantum Number (ml):** This describes the specific orbital within a subshell where the electron is located.
4. **Spin Quantum Number (ms):** This describes the spin direction of the electron, which can be either +1/2 or -1/2.
The Pauli Exclusion Principle asserts that each electron in an atom has a unique set of these four quantum numbers. While electrons can share the first three quantum numbers if they are in the same orbital (meaning they share the same energy level, the same subshell, and the same specific orbital within that subshell), they must have different Spin Quantum Numbers. This means that in any given orbital, one electron can have a spin of +1/2 and the other must have a spin of -1/2. This principle is fundamental in explaining the electronic structure of atoms and, consequently, the behavior and properties of elements.
Za ka so ka ci gaba da wannan aikin?