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Swali 1 Ripoti
A factor that does not affect the rate of a chemical reaction is
Maelezo ya Majibu
In evaluating the factors that affect the rate of a chemical reaction, we can look at each of the possible influences: surface area, temperature, volume, and catalyst.
Surface Area: When you increase the surface area of reactants, it allows more particles to collide with each other per unit of time, which in turn increases the rate of reaction. Imagine smaller particles like powders reacting faster than larger chunks because they have a greater surface exposed to the other reactants.
Temperature: Increasing the temperature usually increases the rate of reaction. Higher temperatures cause particles to move faster, increasing the energy of collisions, and therefore increasing the chance of successful reactions.
Catalyst: A catalyst is a substance that increases the rate of a chemical reaction without being consumed by it. It lowers the activation energy needed for the reaction to occur, thus allowing it to proceed faster.
Volume: The volume of the container or the amount of space in which a reaction occurs generally does not directly affect the rate of the reaction. While changing the volume can alter pressure or concentration in gaseous reactions, which in turn affects the rate, the volume itself is not a direct factor affecting reaction rate.
Therefore, the factor that does not directly affect the rate of a chemical reaction is volume. It indirectly affects reaction rates by altering concentration or pressure in certain reaction conditions, but it is not a direct influencing factor on its own.
Swali 2 Ripoti
A type of isomerism that ClCH=CHCl can exhibit is
Maelezo ya Majibu
ClCH=CHCl can exhibit geometrical isomerism and positional isomerism. ClCH=CHCl can exhibit positional isomerism because the positions of the functional groups or substituent atoms are different. Positional isomerism occurs when compounds with the same molecular formula have different properties due to the difference in the position of a functional group, multiple bond, or branched chain.
Swali 3 Ripoti
Rust on the surface of a metal sheet contains
Maelezo ya Majibu
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.
Swali 4 Ripoti
Water gas obtained from the gasification of coke is made up of
Maelezo ya Majibu
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.
Swali 5 Ripoti
One of the following is not a water pollutant?
Maelezo ya Majibu
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.
Swali 6 Ripoti
What method is suitable for the separation of gases present in air?
Maelezo ya Majibu
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.
Swali 7 Ripoti
147 N + X → 146 C + 11 P
In the reaction above, X is
Maelezo ya Majibu
To determine what particle X is, we need to understand the reaction given:
N + X → \146\\ C + \11\ \P
The notation in nuclear reactions is important. The numbers on top (superscripts) are the mass numbers, which represent the total number of protons and neutrons. The numbers on the bottom (subscripts) are the atomic numbers, which represent the number of protons.
Here's what we have:
Let's consider the conservation of mass and charge:
1. **Conservation of Mass Number:** The mass number of the reactants should equal the mass number of the products. If N has a mass number 'a' and X has a mass number 'b', then:
a + b = 146 + 11 = 157
2. **Conservation of Atomic Number:** The total number of protons should also be conserved. If N has an atomic number 'c' and X has an atomic number 'd', then:
c + d = 6 + 1 = 7
To satisfy these rules:
- Option X could be a **neutron**, as neutrons have a mass number of 1 and an atomic number of 0, which means they do not affect the atomic number but contribute to the mass number.
Let's verify:
- Assume X is a neutron with a mass number of 1 and an atomic number of 0, which fits the requirement for conservation of atomic mass:
Therefore, X is a neutron because it helps conserve both the mass number and the atomic number in the given nuclear reaction.
Swali 8 Ripoti
How many isomers has the organic compound represented by the formula C3 H8 O ?
Maelezo ya Majibu
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.
Swali 9 Ripoti
The shape of the molecule of Carbon(IV) oxide is
Maelezo ya Majibu
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.
Swali 10 Ripoti
The constituent of petroleum fraction used in surfacing road is
Maelezo ya Majibu
Among the options listed, the constituent of petroleum used in surfacing roads is bitumen. Bitumen, also known as asphalt, is a sticky, black, and highly viscous liquid or semi-solid form of petroleum. It is the last fraction obtained when crude oil is distilled and is often left over after the lighter components are extracted.
Reasons why bitumen is used for road surfacing:
Due to these properties, bitumen is extensively used in road construction and surfacing, ensuring roads are durable, smooth, and safe for travel.
Swali 11 Ripoti
A gas that turns lime water milky is likely to be from
Maelezo ya Majibu
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.
Swali 12 Ripoti
The molecular formular of a hydrocarbon with an empirical formula of CH3 and a molar mass of 30 is
Maelezo ya Majibu
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.
Swali 13 Ripoti
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
Maelezo ya Majibu
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**.
Swali 14 Ripoti
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 ).
Maelezo ya Majibu
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.
Swali 15 Ripoti
The scientist that performed the experiment on discharged tubes that led to the discovery of the cathode rays as a sub-atomic particle is
Maelezo ya Majibu
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.
Swali 16 Ripoti
H2 S(g) + Cl2 (g) → 2HCl(g) + S(s)
What is the change in oxidation state of sulphur from reactant to product?
Maelezo ya Majibu
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**.
Swali 17 Ripoti
CH3 -CH2 -OH and CH3 -O-CH3
The relationship between the two compounds above, is that they are
Maelezo ya Majibu
The relationship between the two compounds is that they are isomers.
To understand why these compounds are isomers, let's break down their structures and definitions:
1. Structures of the Compounds:
2. Definitions:
Both compounds have the same molecular formula: C2H6O. However, they have different arrangements of their atoms. Ethanol has a hydroxyl group (-OH) attached to an ethyl group (CH3-CH2-), while dimethyl ether involves two methyl groups (CH3-) bonded to an oxygen atom (O). This difference in structure leads to different chemical and physical properties, despite having the same molecular formula. Hence, these two compounds are classified as isomers.
Swali 18 Ripoti
The ions responsible for permanent hardness in water are sulphates of
Maelezo ya Majibu
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.
Swali 19 Ripoti
If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are
Maelezo ya Majibu
To answer this question, let's break it down step by step:
Mass Number: The mass number is the total number of protons and neutrons in an atom's nucleus. In this case, the mass number is given as 31.
Stable Neutral Atom: A stable neutral atom has no overall electrical charge, meaning the number of protons (positively charged) must equal the number of electrons (negatively charged).
If we symbolize the number of protons by the atomic number (Z), we can say:
1. **Protons = Electrons** in a neutral atom.
2. **Mass Number (A) = Protons + Neutrons**.
Given that the mass number is 31, we have the equation:
A = Protons + Neutrons = 31.
Assuming a commonly known stable element like Phosphorus, which has an atomic number (Z) of 15, it means:
1. **Protons = 15**.
2. **Electrons = 15** (because it's a neutral atom).
3. To find Neutrons: Neutrons = Mass Number - Protons = 31 - 15 = 16.
So, in this scenario, the number of electrons is 15 and the number of neutrons is 16. This combination is found in the first option given.
Swali 20 Ripoti
23892 U + 10 n → 23992 U
The process above produces
Maelezo ya Majibu
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.
Swali 21 Ripoti
The IUPAC nomenclature of the complex K4 Fe(CN)6 is
Maelezo ya Majibu
The compound in question is K4[Fe(CN)6]. To name this complex using IUPAC nomenclature, let's break it down into parts:
Next, consider the oxidation state of Fe:
Finally, we consider the oxidation state of the iron. Since calculations show that it is +2, the complex ion is named based on its oxidation state.
Hence, the IUPAC name of this compound is potassium hexacyanoferrate(II).
Swali 22 Ripoti
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].
Maelezo ya Majibu
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.
Swali 23 Ripoti
CuOs + H2 (g ) ⇌ Cus + H2 O(g )
In the equation above, the effect of increased pressure on the equilibrium position is that
Maelezo ya Majibu
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.
Swali 24 Ripoti
An example of a compound that is acidic in solution is
Maelezo ya Majibu
Phosphoric acid is a weak acid that can donate three hydrogen ions in water. Phosphoric acid partially ionizes when dissolved in an aqueous solution.
Swali 25 Ripoti
The stability of atomic nucleus is determined by ratio of
Maelezo ya Majibu
The stability of an atomic nucleus is primarily determined by the neutron/proton ratio. This refers to the number of neutrons in relation to the number of protons within the nucleus. Let's break down why this ratio is crucial for nuclear stability:
The right balance between the number of neutrons and protons helps in achieving nuclear stability.
An imbalance in this ratio often results in an unstable nucleus, leading to radioactive decay as the nucleus attempts to reach a more stable form. This is why the neutron/proton ratio is a fundamental factor in the stability of the atomic nucleus.
Swali 26 Ripoti
Kerosene is used as solvent for
Maelezo ya Majibu
Kerosene is commonly used as a solvent for paints. Let me explain why in a simple way:
Kerosene is a type of fuel that is composed of hydrocarbons, which are molecules made up of hydrogen and carbon atoms. These hydrocarbons give kerosene the ability to dissolve other similar substances.
Paints often contain oils and other hydrocarbon-based compounds. Since kerosene is also hydrocarbon-based, it can effectively dissolve and thin these compounds. This makes it suitable for use as a solvent in paints, allowing the paint to be thinned or cleaned up after use. This property makes kerosene a good choice for cleaning brushes and other painting tools or for dissolving dried paint.
On the other hand, sulphur, gums, and fats are typically not dissolved effectively by kerosene because of their different chemical properties. Therefore, kerosene as a solvent is primarily useful in the context of working with paints and similar hydrocarbon-based materials.
Swali 27 Ripoti
The basicity of tetraoxophosphate(V) acid is
Maelezo ya Majibu
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.
Swali 28 Ripoti
In the treatment of water for municipal supply, chlorine is used to
Maelezo ya Majibu
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.
Swali 29 Ripoti
When n = 3, the quantum number of an element is
Maelezo ya Majibu
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.
Swali 30 Ripoti
A gas when mixed with oxygen, it produces a very hot and early controllable flame. What is the name of the flame and where is it used?
Maelezo ya Majibu
The Oxy-ethylene flame is a type of flame produced when oxygen is mixed with a gas called ethylene. This mixture results in a flame that is extremely hot and can be easily controlled. Such a flame is often used in industrial applications related to cutting and welding metals. The heat generated by an oxy-ethylene flame is sufficient to melt metals, allowing them to be welded together or cut apart efficiently.
Swali 31 Ripoti
Hydrochloric acid is regarded as a strong acid because it
Maelezo ya Majibu
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.
Swali 32 Ripoti
Which of the following represents an order of increasing reactivity?
Maelezo ya Majibu
To determine the order of increasing reactivity of the elements listed, it's important to understand the general trends in metal reactivity. Metals react by losing electrons, and their reactivity is often influenced by their ability to lose these electrons easily. In many cases, generally, alkali metals are the most reactive, and noble metals are the least reactive. Here's a basic description of the reactivity of the given metals:
With these considerations in mind, the order of increasing reactivity from the given options would be:
Gold (Au) < Copper (Cu) < Tin (Sn) < Iron (Fe) < Calcium (Ca)
This is the order where the least reactive element is first (gold), and the most reactive element is last (calcium). Hence, the correct option represents the order: Au < Cu < Sn < Fe < Ca.
Swali 33 Ripoti
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
Maelezo ya Majibu
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.
Swali 34 Ripoti
A liquid hydrocarbon obtained from fractional distillation of coal tar that is used in the pharmaceutical industry is
Maelezo ya Majibu
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.
Swali 35 Ripoti
If the solubility of KNO3 at 300 C is 3.10 mol/dm3 a solution containing 303g/dm3 KNO3 is likely to be
Maelezo ya Majibu
To determine the condition of the solution containing KNO3 at 300C, let's start by calculating the molarity of the given solution.
The molecular weight of KNO3 (Potassium Nitrate) is approximately:
Thus, KNO3 = 39 + 14 + (16 * 3) = 101 g/mol.
Now, to determine the molarity of the given solution:
Compare with the solubility at 300C:
If we compare the values:
Hence, the solution is unsaturated because it can still dissolve more KNO3 until it reaches the solubility limit of 3.10 mol/dm3.
Swali 36 Ripoti
The term strong and weak acids is used to indicate the
Maelezo ya Majibu
The terms strong and weak acids are used to indicate the extent of ionization of an acid. This means how completely an acid dissociates into its ions in water.
Strong acids completely dissociate in water. This means that nearly all the acid molecules break down into positive hydrogen ions (H+) and their respective anions. Examples include hydrochloric acid (HCl), sulfuric acid (H2SO4), and nitric acid (HNO3).
Weak acids, on the other hand, only partially dissociate in water. This means that only a small fraction of the acid molecules break down into ions. Most of the acid remains in its molecular form. An example of a weak acid is acetic acid (CH3COOH), which is found in vinegar.
Therefore, the strength of an acid in terms of its classification as strong or weak is about how fully it dissociates into ions in an aqueous solution, not about the number of H+ ions or the strength of its action on substances.
Swali 37 Ripoti
When a few drops of Millon reagents is added to egg-white solution in a test tube, the white precipitate changes to
Maelezo ya Majibu
When a few drops of Millon's reagent is added to an egg-white solution in a test tube and the solution is boiled, the white precipitate turns brick red. This indicates the presence of proteins.
Swali 38 Ripoti
At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that
Maelezo ya Majibu
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:
Swali 39 Ripoti
Determine the empirical formula of an oxide of sulphur containing 60% of oxygen
[S = 32, O = 16 ]
Maelezo ya Majibu
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.
Swali 40 Ripoti
An organic compound with general formula RCOR' is an
Maelezo ya Majibu
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.
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