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Vraag 1 Verslag
Determine the empirical formula of an oxide of sulphur containing 60% of oxygen
[S = 32, O = 16 ]
Antwoorddetails
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.
Vraag 2 Verslag
Hydrochloric acid is not suitable in the preparation of ethanoic acid because it
Antwoorddetails
Hydrochloric acid is not suitable for preparing ethanoic acid because it is too volatile.Being too volatile, means it has a low boiling point and is easily evaporated. Thus, HCl is not suitable because it cannot carry out the oxidation process required to convert alcohols into acids like ethanoic acid.
Ethanoic acid, also known as acetic acid, is a weak acid that doesn't fully dissociate in water, while hydrochloric acid is a strong acid that dissociates almost completely.
Vraag 3 Verslag
The reaction between alkanoic acids and alkanols in the presence of an acid catalyst is known as
Antwoorddetails
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.
Vraag 4 Verslag
The electronic configuration of an atom of Nitrogen is 1s2 2s2 2p1x 2p1y 2p1z because the atom is
Antwoorddetails
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.
Vraag 5 Verslag
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]
Antwoorddetails
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.
Vraag 6 Verslag
What accounts for the low melting and boiling points of covalent molecules?
Antwoorddetails
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.
Vraag 7 Verslag
H2 S(g) + Cl2 (g) → 2HCl(g) + S(s)
What is the change in oxidation state of sulphur from reactant to product?
Antwoorddetails
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**.
Vraag 8 Verslag
H2 SO4
C2 H5 OH → C2 H4
1700 C
The reaction above illustrates
Antwoorddetails
This reaction illustrates dehydration. In chemistry, dehydration refers to the process of removing water (H2O) from a compound. Let's break down the given reaction to understand this better.
The provided chemical equation is:
C2H5OH → C2H4 + H2O
This equation indicates that ethanol (C2H5OH) is being transformed into ethylene (C2H4) with the production of water (H2O).
The process involves the breaking of bonds in ethanol and the removal of a water molecule, as follows:
This reaction is typically carried out under certain conditions, in this case at a high temperature of 1700°C, to facilitate the dehydration process.
Therefore, this is indeed a dehydration reaction as it involves converting ethanol into ethylene by removing water.
Vraag 9 Verslag
If a salt weighs 2g and upon exposure to the atmosphere weighs 1.5g, this is as a result of
Antwoorddetails
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.
Vraag 10 Verslag
How many moles of CO2 are produced when ethanol is burnt with 6g of oxygen
Antwoorddetails
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.
Vraag 11 Verslag
Antwoorddetails
When a strong acid reacts with a strong base, the result is the formation of a neutral salt. This reaction is a part of a chemical process known as neutralization.
Let's break it down further:
During a neutralization reaction, the hydrogen ions (H⁺) from the acid combine with the hydroxide ions (OH⁻) from the base to form water (H₂O). Meanwhile, the remaining ions (for example, Na⁺ from NaOH and Cl⁻ from HCl) come together to form a compound known as a salt. This salt does not affect the acidity or basicity of the solution, hence it is considered neutral.
Therefore, the salt formed in such a reaction is a neutral salt, which is what is referred to as a normal salt in the options provided.
Vraag 12 Verslag
Strong acids can be distinguished from weak acids by any of the following methods, EXCEPT
Antwoorddetails
To distinguish between strong acids and weak acids, we can employ several methods based on their chemical properties:
Conductivity Measurement: Strong acids dissociate completely in water, releasing more ions. Because ion concentration is directly related to electrical conductivity, strong acids exhibit higher conductivity than weak acids, which only partially dissociate.
Litmus Paper: This method helps determine if a solution is acidic or basic but does not provide detailed information about the strength (strong or weak) of an acid. Both strong and weak acids turn blue litmus red. Therefore, **litmus paper cannot effectively distinguish between a strong and a weak acid.**
Measurement of pH: Strong acids have a lower pH because they fully dissociate to release more hydrogen ions (H+), whereas weak acids have a relatively higher pH as they do not dissociate completely. Thus, pH measurement can distinguish the extent of acidity.
Measurement of Heat of Reaction: The heat of reaction can give insights into the strength of an acid because it involves the degree of ionization and the energetics associated with it. A strong acid will exhibit a different calorimetric response compared to a weak acid.
In summary, **litmus paper is not suitable for distinguishing between a strong and a weak acid**, as it only indicates acidity but does not reveal the strength of the acid.
Vraag 13 Verslag
Aqueous solution of sodium hydroxide can be used to test for the presence of : I. Ca2+ , II. Zn2+ , III. Cu2+
Antwoorddetails
Aqueous solution of sodium hydroxide (NaOH) is a versatile reagent in chemistry, often used to test for the presence of metal ions. When sodium hydroxide is added to solutions containing certain metal ions, it forms precipitates that are characteristic of those ions. Here's how it interacts with each of the mentioned ions:
Calcium ions (Ca2+): When NaOH is added to a solution containing calcium ions, a white precipitate of calcium hydroxide (Ca(OH)2) can form. However, the precipitate is only slightly soluble in water, and this reaction is not the most definitive test for calcium ions.
Zinc ions (Zn2+): When sodium hydroxide is added to a solution containing zinc ions, a white gelatinous precipitate of zinc hydroxide (Zn(OH)2) forms. This precipitate is soluble in excess NaOH, leading to a clear, colorless solution. This reaction is used to test for zinc ions.
Copper ions (Cu2+): When NaOH is added to a solution containing copper ions, a pale blue precipitate of copper(II) hydroxide (Cu(OH)2) forms. This precipitate is insoluble even in excess NaOH, and the formation of this blue precipitate is a common test for copper ions.
Therefore, an aqueous solution of sodium hydroxide can be used to test for the presence of all three ions: calcium (Ca2+), zinc (Zn2+), and copper (Cu2+). The reaction and precipitate formation with each ion serve as indicators of their presence. Thus, the correct answer is:
I, II and III.
Vraag 14 Verslag
The difference in molecular mass between an alkene and alkyne with six carbon per mole is
Antwoorddetails
To determine the difference in molecular mass between an alkene and an alkyne, let's first take a look at their general formulas.
Alkene: An alkene is a hydrocarbon with at least one double bond between carbon atoms. For an alkene with six carbon atoms, the general formula is CnH2n. Therefore, for 6 carbon atoms, the molecular formula is C6H12.
Alkyne: An alkyne is a hydrocarbon with at least one triple bond between carbon atoms. For an alkyne with six carbon atoms, the general formula is CnH2n-2. Therefore, for 6 carbon atoms, the molecular formula is C6H10.
Now let's calculate the molecular masses:
Molecular mass of alkene (C6H12):
Molecular mass of alkyne (C6H10):
The **difference** in molecular mass between the alkene and alkyne is **84 g/mol - 82 g/mol** = 2 g/mol.
Vraag 15 Verslag
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?
Antwoorddetails
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.
Vraag 16 Verslag
The shape of ammonia molecule is
Antwoorddetails
The shape of the ammonia molecule (NH3) is trigonal pyramidal. To understand why, let's explore the electron and molecular geometry using a simple explanation:
Ammonia consists of one nitrogen (N) atom bonded to three hydrogen (H) atoms. The nitrogen atom has five valence electrons requiring three more electrons to complete its octet. These are acquired by forming covalent bonds with three hydrogen atoms. In addition to the three bonding pairs, there is one lone pair of electrons on the nitrogen atom.
According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, electron pairs, including bonding pairs and lone pairs, repel each other and arrange themselves as far apart as possible to minimize repulsion. In ammonia:
The presence of the lone pair on nitrogen creates a slight distortion, causing the molecule's shape to be trigonal pyramidal rather than perfectly tetrahedral. The lone pair occupies more space and pushes the hydrogen atoms slightly closer together. This results in a pyramidal shape, with nitrogen at the apex, and the three hydrogen atoms forming the base of the pyramid.
The trigonal pyramidal shape of ammonia is a result of this molecular geometry, not to be confused with any of the other options like V-shaped, tetrahedral, or co-planar.
Vraag 17 Verslag
The IUPAC name of the compound above is
Antwoorddetails
To determine the IUPAC name of a compound, follow these steps:
Hence, by following these steps, if the bromo and methyl groups are both attached to the second carbon (lowest numbering possible), the IUPAC name of the compound is "2-bromo, 2-methyl butane."
Vraag 18 Verslag
What is the vapour density of 560cm3 of a gas that weighs 0.4g at s.t.p?
[Molar Volume of gas at s.t.p = 22.4 dm3 ]
Antwoorddetails
To find the vapour density of a gas, you can use the formula:
Vapour density = (Molar mass of gas) / 2
However, first, we need to determine the molar mass of the gas. One can find the molar mass using the given data:
We know that at standard temperature and pressure (s.t.p.), 1 mole of any gas occupies 22.4 dm3. We need to convert the volume from cm3 to dm3 because the molar volume is given in dm3:
560 cm3 = 0.560 dm3
Now, let's find the number of moles in 0.560 dm3:
The number of moles (n) = Volume of gas (dm3) / Molar volume at s.t.p. (dm3/mol)
n = 0.560 dm3 / 22.4 dm3/mol
n = 0.025 moles
Given that the mass of the gas is 0.4 grams, we can find the molar mass by using the relation:
Molar Mass = Mass / Number of Moles
Molar Mass = 0.4 g / 0.025 moles
Molar Mass = 16 g/mol
Now that we have the molar mass, we can find the vapour density:
Vapour density = Molar mass / 2
Vapour density = 16 g/mol / 2
Vapour density = 8.0
Hence, the vapour density of the gas is 8.0.
Vraag 19 Verslag
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]
Antwoorddetails
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.
Vraag 20 Verslag
When n = 3, the quantum number of an element is
Antwoorddetails
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.
Vraag 21 Verslag
CuOs + H2 (g ) ⇌ Cus + H2 O(g )
In the equation above, the effect of increased pressure on the equilibrium position is that
Antwoorddetails
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.
Vraag 22 Verslag
Hydrogen chloride gas and ammonia can be used to demonstrate the fountain experiment because they are
Antwoorddetails
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.
Vraag 23 Verslag
The term that is not associated with petroleum industry is ?
Antwoorddetails
Cracking, saponification and polymerization are all terminologies associated with the petroleum industry but fermentation is associated with the brewery industry.
Cracking is a chemical process that breaks down heavy hydrocarbon molecules into lighter, more useful ones.
Saponification is a chemical reaction that converts fats and oils into soap. It's not directly involved in petroleum, but it can be used to analyze petroleum products.
Polymerization is a process in the petroleum industry that converts light olefin gases into higher molecular weight hydrocarbons.
Fermentation is the process in which a substance breaks down into a simpler substance. Microorganisms like yeast and bacteria usually play a role in the fermentation process, creating beer, wine, bread,yogurt and other foods.
Vraag 24 Verslag
What method is suitable for the separation of gases present in air?
Antwoorddetails
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.
Vraag 25 Verslag
The law which states that a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is
Antwoorddetails
The law that states a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is the law of definite proportion.
To explain this simply, let's consider water as an example. Water is made up of hydrogen and oxygen. According to the law of definite proportion, a sample of pure water taken from anywhere in the world will always contain the same ratio of hydrogen to oxygen by mass. Specifically, water will always have approximately 88.8% oxygen and 11.2% hydrogen by mass.
This is because a chemical compound has a fixed composition, regardless of the process used to create it or the source from which it is derived. The law of definite proportion, also known as the law of constant composition, is fundamental in chemistry because it supports the idea that chemical compounds are composed of elements in specific and fixed ratios. This does not change regardless of how the compound is prepared or where it is found.
Vraag 26 Verslag
Antwoorddetails
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.
Vraag 27 Verslag
The stability of atomic nucleus is determined by ratio of
Antwoorddetails
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.
Vraag 28 Verslag
If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are
Antwoorddetails
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.
Vraag 29 Verslag
The molecular formular of a hydrocarbon with an empirical formula of CH3 and a molar mass of 30 is
Antwoorddetails
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.
Vraag 30 Verslag
The shape of the molecule of Carbon(IV) oxide is
Antwoorddetails
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.
Vraag 31 Verslag
The reaction of hydrogen and chlorine to produce hydrogen chloride gas is explosive in
Antwoorddetails
The reaction between hydrogen and chlorine to produce hydrogen chloride gas is explosive in sunlight. This is because sunlight contains a broad range of electromagnetic radiation, including ultraviolet (UV) light, which is energetic enough to initiate the reaction.
Here is a simplified explanation:
In contrast, other forms of light like diffused light, infrared light, and Raman light do not provide enough energy to initiate this explosive reaction because they lack the necessary UV component found in sunlight.
Vraag 32 Verslag
The main constituent of water-glass is
Antwoorddetails
The main constituent of water-glass is sodium trioxosilicate(IV). Water-glass, also known as liquid glass, is common terminology for a mixture of sodium silicate and water. The primary chemical component in water-glass is sodium silicate, which includes sodium ions (Na+) bonded with silicate ions (SiO44-).
Essentially, when sodium silicate is dissolved in water, it results in a viscous liquid that can be utilized in various applications such as in cements, passive fire protection, textile and lumber processing, and as a sealant. Sodium trioxosilicate(IV) forms a significant part of this mixture as it reacts with other compounds to create a hardened, glass-like structure when it dries. Therefore, when water-glass is mentioned, it is mostly referring to solutions that have sodium trioxosilicate(IV) as their principal compound.
Vraag 33 Verslag
A major effect of oil pollution in coastal water is
Antwoorddetails
One of the major effects of oil pollution in coastal water is the destruction of aquatic life.
When oil spills into a water body, it forms a thin layer called a sheen on the surface of the water. This oil layer blocks sunlight from reaching aquatic plants and phytoplankton, inhibiting their ability to perform photosynthesis. As a result, these plants and microorganisms suffer, impacting the entire food chain.
Moreover, oil can coat the feathers of birds and the fur of marine mammals, which affects their insulation and buoyancy, leading to hypothermia, drowning, or inability to fly. Additionally, the toxic components in oil are harmful if ingested, causing internal damage to fish and other marine organisms. These combined effects can lead to significant mortality in aquatic ecosystems, threatening biodiversity and the natural balance of coastal waters.
Therefore, oil pollution can severely affect the health and survival of aquatic life, creating disruptions that can persist for many years.
Vraag 34 Verslag
Nitrogen obtained from air is not absolutely pure because it contains the following except
Antwoorddetails
Nitrogen obtained from air is not absolutely pure because it contains other gases, including:
Vraag 35 Verslag
The element which can combine with oxygen to form an acid anhydride of the form XO2 is
Antwoorddetails
An Acid anhydride can be defined as a non-metal oxide which forms an acidic solution when reacted with water.
Sulphur is the element that can combine with oxygen to form an acid anhydride of the form XO2 .
An acid oxide is a compound that forms an acid when it reacts with water. Non-metals in groups 4–7 form acidic oxides.
Vraag 36 Verslag
The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is
Antwoorddetails
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.
Vraag 37 Verslag
Benzene formed nitrobenzene at temperature of 600 C when it reacts with mixture of concentrated trioxonitrate(V) acid and concentrated
Antwoorddetails
The reaction described is the nitration of benzene to form nitrobenzene. This is an example of an electrophilic aromatic substitution reaction. **Nitration** involves replacing a hydrogen atom on a benzene ring with a nitro group (NO2). This reaction requires a nitrating mixture composed of concentrated nitric acid (trioxonitrate(V) acid) and concentrated sulfuric acid (tetraoxosulphate(VI) acid). Let me explain why:
Nitration is typically carried out using a mixture of **concentrated nitric acid and concentrated sulfuric acid** at a temperature of around **60°C**. The role of sulfuric acid in this mixture is to act as a catalyst and a dehydrating agent. It helps generate the nitronium ion (NO2+), which is the active electrophile that attacks the benzene ring.
Here's a simplified mechanism for this reaction:
None of the other options listed (hydrochloric acid, phosphoric acid, and hydrogen iodide) contain the necessary combination of properties to generate the nitronium ion and facilitate the nitration of benzene.
Therefore, the correct mixture to carry out the nitration of benzene, forming nitrobenzene at a temperature of 60°C, is a combination of **concentrated nitric acid and concentrated sulfuric acid (tetraoxosulphate(VI) acid)**.
Vraag 38 Verslag
The indicator used in a titration between strong acid and weak base is
Antwoorddetails
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.
Vraag 39 Verslag
At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that
Antwoorddetails
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:
Vraag 40 Verslag
During the fractional distillation of crude oil, the fraction that distills at 200 - 2500 C is
Antwoorddetails
The petroleum fractions that distill at 200–250°C are naphtha and kerosene,
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