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Tambaya 1 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 2 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 3 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 4 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 5 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 6 Rahoto
An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5? [ C =12, H = 1, N = 14, O = 16].
Bayanin Amsa
To find the molecular formula of the compound, follow these steps:
1. Determine the Empirical Formula:
Start by assuming you have 100 grams of the compound. This means you have:
Now, convert these masses to moles using their atomic masses (C = 12, H = 1, N = 14, O = 16):
Next, divide each by the smallest number of moles to get the simplest ratio:
This gives us the empirical formula: C5H7NO2.
2. Determine the Molecular Formula:
The molecular formula is a multiple of the empirical formula. To determine this multiple, we need to find the empirical formula mass and compare it with the molar mass derived from the given vapor density.
Calculate the empirical formula mass:
The molar mass can be calculated from the vapor density:
Now, find the ratio of the molar mass to the empirical formula mass:
This ratio is approximately 1, indicating the molecular formula is the same as the empirical formula. Since empirical formulas typically should perfectly match the atomic proportions we derive from experiments, our calculations regarding the assumptions on the vapour and empirical formula mass remains our best match.
Therefore, the molecular formula is C5H7NO2.
Tambaya 7 Rahoto
The compound of Copper used as a fungicide is
Bayanin Amsa
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).
Tambaya 8 Rahoto
The Van der waals forces of attraction operates between
Bayanin Amsa
The Van der Waals forces of attraction operate between molecules. These are weak forces of attraction that occur due to momentary changes in the electron distribution within molecules. Here's a simple explanation:
Therefore, the forces can affect the physical properties of molecular compounds, such as boiling and melting points, but do not generally involve charged particles like cations or anions.
Tambaya 9 Rahoto
Determine the empirical formula of an oxide of sulphur containing 60% of oxygen
[S = 32, O = 16 ]
Bayanin Amsa
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.
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
The composition of alloy permalloy is iron and
Bayanin Amsa
The alloy known as **permalloy** is composed primarily of **iron** and **nickel**. Permalloy is a well-known magnetic alloy that typically consists of about **80% nickel and 20% iron**. It is renowned for having high magnetic permeability, meaning it can become magnetized easily, which makes it extremely useful in a variety of electrical and magnetic applications, such as transformers, memory storage, and magnetic shielding. The nickel in permalloy enhances the magnetic properties of the iron, giving the alloy its unique characteristics.
Tambaya 12 Rahoto
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]
Bayanin Amsa
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.
Tambaya 13 Rahoto
The basicity of tetraoxophosphate(V) acid is
Bayanin Amsa
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.
Tambaya 14 Rahoto
Scandium is not regarded as a transition metal because its ion has
Bayanin Amsa
Scandium is not regarded as a transition metal because its ion has no electron in the d-orbital.
To understand this, let's first define a transition metal. A transition metal is defined as an element that has an incomplete d-subshell in either its elemental form or in any of its common oxidation states.
When Scandium (Sc) loses electrons to form its most common ion (Sc3+), it loses three electrons. These electrons are removed from the 4s and 3d orbitals. The electron configuration for Scandium is [Ar] 3d1 4s2. Upon losing three electrons to form Sc3+, the resulting electron configuration is [Ar], which means there are:
As a result, there are no electrons in the d-orbital of the Scandium ion, which does not meet the criteria for a transition metal.
Tambaya 15 Rahoto
The chemical formula for potassiumhexacyanoferrate(II) is
Bayanin Amsa
The chemical formula for potassiumhexacyanoferrate(II) is K4Fe(CN)6.
Let's break down the name to understand why:
1. Potassium (K): The compound includes potassium ions. In this case, four potassium ions are present, indicated by the subscript 4 in K4.
2. Hexacyano: The prefix "hexa" means six, which signifies there are six cyanide ions (CN-) in the complex. This is represented as (CN)6.
3. Ferrate (II): The word "ferrate" suggests the presence of iron (Fe). The Roman numeral (II) indicates that the iron is in the +2 oxidation state.
Overall, the complex ion is [Fe(CN)6] with a charge of 4-, so to balance the charge, four potassium ions (each with a charge of +1) are needed, resulting in the formula K4Fe(CN)6.
Tambaya 16 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 17 Rahoto
What accounts for the low melting and boiling points of covalent molecules?
Bayanin Amsa
The low melting and boiling points of covalent molecules are primarily due to the presence of weak intermolecular forces between the molecules. While covalent molecules consist of atoms bonded together by strong covalent bonds, the forces between separate molecules, known as van der Waals forces or London dispersion forces, are much weaker. These weak forces require significantly less energy to overcome, which explains why covalent molecules tend to have lower melting and boiling points.
Although covalent molecules have definite shapes and possess shared electron pairs, these characteristics have little influence on the melting and boiling points. The focus is instead on how much energy is needed to separate the molecules from one another.
Covalent molecules are not typically three-dimensional structures like ionic compounds or metals which form intricate lattices and require more energy to disrupt. Thus, the primary reason for their lower melting and boiling points is the presence of weak intermolecular forces that can be more easily overcome with minimal energy input.
Tambaya 18 Rahoto
Concentrated sodium chloride solution is electrolyzed using mercury cathode and graphite anode. The products at the anode and the cathode respectively are
Bayanin Amsa
When a concentrated sodium chloride solution is electrolyzed using a mercury cathode and graphite anode, the products are hydrogen gas at the cathode and chlorine gas at the anode
At the anode, 2Cl− → Cl2 + 2e−
At the cathode, 2H+ + 2e− → H2
During the electrolysis, hydrogen and chloride ions are removed from solution whereas sodium and hydroxide ions are left behind in solution. This means that sodium hydroxide is also formed during the electrolysis of sodium chloride solution.
Tambaya 19 Rahoto
When a specie undergoes oxidation, its
Bayanin Amsa
When a species undergoes oxidation, it experiences an increase in its oxidation number. Oxidation is a chemical process where a species loses electrons. In terms of oxidation number, electrons have a negative charge, so losing them results in an increase in charge. Thus, the oxidation number of the species becomes more positive or less negative.
To help understand, consider sodium (Na) reacting with chlorine (Cl2) to form sodium chloride (NaCl):
This change clearly shows that when sodium is oxidized, its oxidation number increases.
Therefore, the correct explanation is: a species undergoing oxidation will have its oxidation number increase.
Tambaya 20 Rahoto
If a salt weighs 2g and upon exposure to the atmosphere weighs 1.5g, this is as a result of
Bayanin Amsa
The observation that a salt initially weighs 2g, but reduces to 1.5g after exposure to the atmosphere is primarily due to the process called efflorescence.
Efflorescence occurs when a salt loses water molecules from its crystal structure when exposed to air, which is why the weight of the salt decreases over time. This loss of water is because some salts contain water of crystallization, and when such salts are exposed to the atmosphere, they can release this water, leading to a reduction in weight.
In this specific case, the salt has lost 0.5g of water, leading to the weight change from 2g to 1.5g. This process is different from hygroscopy, which involves absorbing moisture from the atmosphere, or deliquescence, where a substance absorbs moisture and eventually dissolves in it. It's also not related to effervescence, which is the escape of gas from an aqueous solution.
Tambaya 21 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 22 Rahoto
The constituents of Alnico are Aluminium, Nickel and
Bayanin Amsa
Alnico is a type of alloy that is known for its strong magnetic properties. The name "Alnico" comes from the elements it is primarily composed of: Aluminum (Al), Nickel (Ni), and Cobalt (Co). These elements are combined to form an alloy that retains its magnetism well and can operate at high temperatures, making it ideal for applications like electric motors, sensors, and various electronic devices.
While there are different variations of Alnico, the presence of Cobalt (Co) is essential for enhancing the magnetic properties of the alloy. The other elements listed, such as Magnesium (Mg), Manganese (Mn), and Copper (Cu), are not typical core constituents of Alnico. Although trace amounts of other elements like copper may sometimes be included in specific formulations, the primary and most significant component responsible for Alnico's powerful magnetic characteristics is Cobalt (Co).
Tambaya 23 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 24 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 25 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 26 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 27 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 28 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 29 Rahoto
At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that
Bayanin Amsa
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:
Tambaya 30 Rahoto
The ions responsible for permanent hardness in water are sulphates of
Bayanin Amsa
Permanent hardness in water is mainly caused by the presence of certain metal ions, specifically the **sulfates (SO₄²⁻)** and **chlorides (Cl⁻)** of calcium (Ca) and magnesium (Mg). These compounds do not precipitate out when the water is boiled, which means they remain dissolved and continue to contribute to the hardness of the water.
Among the options you provided, the ions responsible for permanent hardness in water are the **sulfates of calcium (Ca²⁺)** and **magnesium (Mg²⁺)**. The presence of calcium sulfate (CaSO₄) and magnesium sulfate (MgSO₄) in water keeps it hard.
When compared to temporary hardness, which can be removed by boiling the water to precipitate bicarbonates, **permanent hardness cannot be removed by boiling**. Instead, methods such as ion exchange or the use of water softeners are required to remove these ions from the water.
In summary, the ions causing permanent hardness in water are the **sulfates of calcium (Ca²⁺)** and **magnesium (Mg²⁺)**. These ions remain dissolved and continue to make the water hard, despite boiling.
Tambaya 31 Rahoto
When Calcium ethynide is decomposed by water, the gas produced is
Bayanin Amsa
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.
Tambaya 32 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 33 Rahoto
Hydrochloric acid is regarded as a strong acid because it
Bayanin Amsa
Hydrochloric acid (HCl) is regarded as a strong acid because it ionizes completely in water. This means that when HCl is dissolved in water, it breaks down entirely into hydrogen ions (H+) and chloride ions (Cl-). In a solution, there are no molecules of HCl left; only its ions are present.
This complete ionization results in a high concentration of hydrogen ions, which is a key characteristic of strong acids. Because there are more hydrogen ions available, hydrochloric acid can readily participate in chemical reactions, particularly those involving proton transfers, like neutralization reactions with bases.
In summary, the reason HCl is considered strong is due to its ability to consistently and completely ionize in an aqueous solution, not because of its physical state, source, or reactive nature with bases. Therefore, the property that defines it as a strong acid is that it ionizes completely.
Tambaya 34 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 35 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 36 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 37 Rahoto
An organic compound with general formula RCOR' is an
Bayanin Amsa
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.
Tambaya 38 Rahoto
The IUPAC nomenclature of the compound above is
Bayanin Amsa
The IUPAC nomenclature of the compound above is 2-methylpropan-2-ol.
Tambaya 39 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 40 Rahoto
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 ).
Bayanin Amsa
The problem requires calculating the **vapor density** of the gas. Vapor density is defined as the mass of a certain volume of a gas compared to the mass of an equal volume of hydrogen, where the hydrogen standard is 2 g/mol (as the molecular weight of hydrogen gas, H₂, is 2).
Here's a step-by-step explanation:
The calculated vapor density of the gas is 22.
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