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Vraag 1 Verslag
The IUPAC nomenclature of the structure above is
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CH2CH2 - CCH3CH2
Start the numbering from the terminal carbon.
Vraag 2 Verslag
The gas that can be collected by downward displacement of air is
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Upward delivery works well for hydrogen and ammonia, which are both less densed than air. Sometimes, they are collected over water.
Vraag 3 Verslag
Water for town supply is chlorinate to make it free from
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Water for town supply is chlorinated to make it free from bacteria. Chlorine is added to water in small amounts to kill harmful bacteria and other microorganisms that may be present in the water. These microorganisms can cause illnesses such as typhoid, cholera, and dysentery if ingested. Chlorination is an effective and inexpensive method to disinfect water and protect public health. Chlorine works by breaking down the cell walls of microorganisms, which leads to their death. Chlorine also helps to control the growth of algae and other microorganisms in the water supply, which can cause taste and odor problems. Chlorination does not remove other impurities from the water such as bad color, temporary hardness, or permanent hardness. These impurities may require additional treatment methods such as filtration or water softening to be removed.
Vraag 4 Verslag
Tin is unaffected by air at ordinary temperature due to its
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Tin is unaffected by air at ordinary temperature because of its high reactivity, which means it doesn't easily react or corrode with other elements such as oxygen, the main component of air. Tin has a high reactivity, which helps it resist corrosion and oxidation. This makes tin a great material for products that need to be resistant to corrosion, such as food containers and metal pipes.
Vraag 5 Verslag
The acid anhydride that will produce weak acid in water is
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H2CO3 is an example of a weak acid while H2SO4 and HNO3 are examples of a strong acid.
CO2 combines with water to give a weak trioxocarbonate (IV) acid.
CO2 + H2O ? H2CO3
Vraag 6 Verslag
The shape of ammonia molecules is
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The shape of ammonia (NH3) molecules is tetrahedral. The ammonia molecule has one nitrogen atom and three hydrogen atoms. The nitrogen atom is at the center of the molecule, with the three hydrogen atoms arranged around it. The shape of the molecule is determined by the arrangement of the electrons in the outermost energy level of the nitrogen atom, which is called the valence shell. In the valence shell of the nitrogen atom, there are four pairs of electrons, one of which is a lone pair, while the other three are bonded to hydrogen atoms. These four pairs of electrons repel each other, and try to get as far away from each other as possible. This results in a tetrahedral arrangement of the atoms around the nitrogen atom. A tetrahedron is a three-dimensional shape with four triangular faces, and it looks like a pyramid with a triangular base. In the case of ammonia, the three hydrogen atoms form a triangular base, and the lone pair of electrons sits at the top of the pyramid, creating a tetrahedral shape. Therefore, the shape of ammonia molecules is tetrahedral.
Vraag 7 Verslag
The constituent common to duralumin and alnico is
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The constituent common to duralumin and alnico is aluminum (Al). Duralumin is a type of aluminum alloy that typically contains around 4% copper and small amounts of other metals such as magnesium and manganese. It is known for its high strength-to-weight ratio and was commonly used in aircraft construction in the early 20th century. Alnico, on the other hand, is a type of permanent magnet alloy that contains aluminum, nickel, and cobalt (hence the name "alnico"). It also often contains small amounts of other metals such as copper and iron. Alnico magnets are known for their high magnetic strength and resistance to demagnetization. Although duralumin and alnico are very different in their properties and applications, they both contain aluminum as a common constituent. Aluminum is a versatile metal that is widely used in many different alloys due to its light weight, strength, and corrosion resistance.
Vraag 8 Verslag
A given mass of gas occupies 2dm3 at 300k. At what temperature will its volume be doubled, keeping the pressure constant?
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To solve this problem, we can use the combined gas law, which states that the product of pressure and volume divided by temperature is constant, as long as the amount of gas and the pressure remain constant. In this case, the pressure is constant and the initial conditions of the gas are: - Volume (V1) = 2 dm3 - Temperature (T1) = 300 K Let's call the final temperature T2, and the final volume V2 = 2V1 = 4 dm3, since we want the volume to be doubled. Using the combined gas law, we can set up the following equation: P * V1 / T1 = P * V2 / T2 Simplifying the equation, we get: V1 / T1 = V2 / T2 Substituting the values we know, we get: 2 / 300 = 4 / T2 Cross-multiplying and solving for T2, we get: T2 = 4 * 300 / 2 = 600 K Therefore, the temperature at which the gas will double its volume, keeping the pressure constant, is 600 K. Option (d) is the correct answer.
Vraag 9 Verslag
The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500(mol-1)
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To calculate the mass of silver deposited, we can use Faraday's laws of electrolysis, which states that the amount of substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The formula to calculate the mass of substance produced is: Mass (g) = (Current × Time × Atomic mass) / (Faraday's constant × Valency) Here, we have a current of 10A passing through a solution of silver salt for 4830s, and we know that the atomic mass of silver is 108 g/mol and the valency is 1 (since silver has a charge of +1). The value of Faraday's constant is 96500 C/mol. Substituting the values in the formula, we get: Mass (g) = (10A × 4830s × 108 g/mol) / (96500 C/mol × 1) Mass (g) = 54.0g Therefore, the mass of silver deposited is 54.0g, which is option (A) in the given options.
Vraag 10 Verslag
In the laboratory preparation of ethyl ethanoate, the water present in the mixture is removed using a solution of
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This is used in the preparation of ethyl ethanoate to remove the water present since it serves as a dehydrated agent.
Vraag 11 Verslag
An isomer of C5H12 is
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C5H12 is the molecular formula for pentane, which has five carbon atoms and 12 hydrogen atoms. An isomer is a molecule that has the same molecular formula as another molecule, but with a different arrangement of atoms. Therefore, we need to look for molecules with the molecular formula C5H12 that have a different arrangement of atoms than pentane. Let's consider the options: - Butane (C4H10) has a different molecular formula than pentane (C5H12), so it cannot be an isomer of pentane. - 2-Methyl propane (C4H10) also has a different molecular formula than pentane, so it cannot be an isomer of pentane. - 2-Methyl butane (C5H12) has the same molecular formula as pentane, but with a different arrangement of atoms. Specifically, the carbon atoms in 2-methyl butane are arranged in a branched chain, while the carbon atoms in pentane are arranged in a straight chain. Therefore, 2-methyl butane is an isomer of pentane. - 2-Ethyl butane (C6H14) has a different molecular formula than pentane, so it cannot be an isomer of pentane. Therefore, the isomer of C5H12 is 2-methyl butane.
Vraag 12 Verslag
An oxide XO2 has a vapour density of 32. What is the atomic mass of X?
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Molecular mass = vapour density X2
Mm of XO2 = x + 16(2) = x + 32
vapour density = 32
∴ x + 32 = 32 × 2
x + 32 = 64
x = 64 - 32
x = 32
∴ the relative molecular mass of X is 32
Relative molecular mass = vapour density × 2
Vraag 13 Verslag
In the laboratory preparation of trioxonitrate (V) acid the nitrogen(iv) oxide formed as a by-product is removed by
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In the laboratory preparation of trioxonitrate (V) acid, nitrogen(iv) oxide is formed as a by-product. Nitrogen(iv) oxide is a harmful gas that can cause respiratory problems, so it needs to be removed from the reaction mixture. One way to remove nitrogen(iv) oxide is by bubbling air through the acid solution. When air is bubbled through the solution, the nitrogen(iv) oxide reacts with the oxygen in the air to form nitrogen dioxide, which is a brown gas. Nitrogen dioxide is not as harmful as nitrogen(iv) oxide and can be safely vented out of the laboratory. Therefore, bubbling air through the acid solution is the method used to remove nitrogen(iv) oxide formed as a by-product in the laboratory preparation of trioxonitrate (V) acid.
Vraag 14 Verslag
22688Ra → x86Rn + alpha particle
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The given equation represents the alpha decay of ^226Ra into an unknown nuclide ^86Rn and an alpha particle. During alpha decay, an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons bound together, resulting in a decrease of two in the atomic number and a decrease of four in the mass number. To determine the atomic number and mass number of the unknown nuclide ^86Rn, we can subtract the atomic and mass numbers of the alpha particle from those of ^226Ra: Atomic number: 88 (Ra) - 2 (alpha particle) = 86 (Rn) Mass number: 226 (Ra) - 4 (alpha particle) = 222 (Rn) Therefore, the unknown nuclide produced in the alpha decay of ^226Ra is ^222Rn, and the correct option is (D) 222.
Vraag 15 Verslag
In the diagram above. X is
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The setup represents the production of sulfur dioxide. And the cylinder marked X is SO2
Vraag 16 Verslag
Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]
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For an ideal gas PV = nRT
Amount in moles = n
Volume v = 10.5dm3
Pressure P = 6atm
Temperature T = 30°C + 273 = 303k
R, Gas constant = 0.082 atmdm3k-1 mol
Recall from ideal gas equation
pv = nRT
n = RVRT
n = 6×1050.082×303
n= 2.536mol
Vraag 17 Verslag
The alkyl group is represented by the general formula
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The alkyl group is represented by the general formula CnH2n+1, where "n" represents the number of carbon atoms in the group. An alkyl group is a type of organic molecule that consists of a chain of carbon atoms, each of which is bonded to two hydrogen atoms. The formula CnH2n+1 represents this structure, where "n" is the number of carbon atoms in the chain. The number of hydrogen atoms in the chain is always two more than the number of carbon atoms, hence the formula CnH2n+1.
Vraag 18 Verslag
The reaction of halogens in the presence of sunlight is an example of
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The reaction of halogens in the presence of sunlight is an example of a substitution reaction. In this type of chemical reaction, one atom or group of atoms in a molecule is replaced by another atom or group of atoms. In the case of halogens, such as chlorine or bromine, they can react with certain organic compounds by breaking the carbon-hydrogen bond and replacing the hydrogen atom with a halogen atom. This reaction is usually initiated by the energy from sunlight, which provides the necessary activation energy to break the bond. For example, in the presence of sunlight, chlorine gas can react with methane gas to form chloromethane and hydrogen chloride: CH4 + Cl2 -> CH3Cl + HCl In this reaction, one of the hydrogen atoms in methane is replaced by a chlorine atom, forming chloromethane, while hydrogen chloride is also produced as a byproduct. Overall, the reaction of halogens in the presence of sunlight is an example of a substitution reaction, where one atom or group of atoms is substituted for another.
Vraag 19 Verslag
Temporary hard water is formed when rain water containing dissolved carbon(IV) oxide flows over deposits of
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Temporary hard water is formed when rainwater, which is naturally soft, comes into contact with deposits of calcium carbonate (CaCO3) in the ground. As the rainwater seeps through the soil, it dissolves carbon dioxide (CO2) to form carbonic acid (H2CO3). This weak acid reacts with the calcium carbonate to form calcium bicarbonate (Ca(HCO3)2), which is soluble in water. As a result, the water that flows over these deposits contains dissolved calcium bicarbonate, which gives it temporary hardness. This temporary hardness can be removed by boiling the water, which causes the calcium bicarbonate to decompose back into calcium carbonate and carbon dioxide. The calcium carbonate then forms a white precipitate, which can be removed by filtration or settling. So, the correct answer to the given question is CaCO3.
Vraag 20 Verslag
CH2S(g) + O2(g) → 2Cn + SO2(g)
What is the change in the oxidation number of copper in the reaction?
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In the reactant;
Cu2S
2 Cu - 2(1) = 0
2 Cu = 2
Cu = 22
Cu = +1
In the product, Cu
Cu = O
The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively
Vraag 21 Verslag
When ΔH is negative, a reaction is said to be
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When ΔH is negative, a reaction is exothermic. In simpler terms, exothermic reactions release energy into their surroundings, which is often felt as heat. When ΔH is negative, it means that the enthalpy change of the reaction is negative, indicating that energy is being released. This can be seen as a decrease in temperature in the surrounding environment or a release of heat or light. Examples of exothermic reactions include combustion reactions, where a fuel combines with oxygen to release heat and light, or the reaction between an acid and a base to form a salt and water. In both cases, energy is released into the surrounding environment, and the enthalpy change is negative (ΔH < 0).
Vraag 22 Verslag
The densities of two gases, X and Y are 0.5gdm-3 and 2.0gdm-3 respectively. What is the rate of diffusion of X relative to Y?
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The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is Graham's Law of diffusion of gas.
R ∝ 1√Mm or R ∝ 1√D
Dx = 0.5gdm-3, Dy = 2gdm-3
R= K√D
R√D = k
R1√D1 = R1√D2
Rx√Dx = Ry√Dy
RxRy = √Dy√Dx
= √2√0.5
= 2.0
Vraag 23 Verslag
Aluminium hydroxide is used in the dyeing industry as a
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Aluminium hydroxide is used in the dyeing industry as a mordant. It combines with a dye and thereby fixes it in a material.
Vraag 24 Verslag
Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 Cmol-1]
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The amount of silver deposited can be calculated using Faraday's law of electrolysis, which states that the amount of a substance deposited at an electrode during electrolysis is proportional to the number of moles of electric charge passed through the solution. We can start by using the formula: moles of substance = charge passed / Faraday's constant Where Faraday's constant is equal to 96500 Cmol^-1. Plugging in the given values: moles of silver = 9650 C / 96500 Cmol^-1 moles of silver = 0.1 So, the amount of silver deposited is 0.1 moles.
Vraag 25 Verslag
Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O2(l) + X. In the reaction above X is
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In the given chemical equation, the reactants are solid calcium hydroxide (Ca(OH)2) and gaseous ammonium chloride (NH4Cl). When they react, they form solid calcium chloride (CaCl2), liquid water (H2O), and X, an unknown product. To determine the identity of X, we need to balance the chemical equation first. The balanced chemical equation is: Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O(l) + 2NH3(g) As per the balanced chemical equation, the product X is 2 molecules of ammonia (NH3). Therefore, the answer is (b) NH3.
Vraag 26 Verslag
The salt formed from a weak acid and a strong base hydrolyzes in water to form
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This is a solution formed from the hydrolyses of an alkali in water.
Vraag 27 Verslag
The oxidation number of iodine in KIO3 is
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The oxidation number of an atom is a measure of the number of electrons that it has gained or lost in a chemical reaction. To determine the oxidation number of iodine in KIO3, we need to use the fact that the sum of the oxidation numbers of all the atoms in a compound is equal to the overall charge of the compound. In KIO3, potassium (K) has an oxidation number of +1 because it belongs to the alkali metal group and it always has a +1 oxidation state. Oxygen (O) usually has an oxidation number of -2, unless it is combined with a more electronegative atom or a peroxide, which is not the case in KIO3. Let's assume that iodine (I) has an oxidation number of x. Since KIO3 is a neutral compound, the sum of the oxidation numbers of all its atoms must be zero. Therefore, we can write the following equation: (+1) + x + 3(-2) = 0 Simplifying the equation, we get: x = +5 Therefore, the oxidation number of iodine in KIO3 is +5.
Vraag 28 Verslag
The tincture of iodine means iodine dissolved in
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It is also called weak iodine solution. Tincture solutions are characterized by the presence of alcohol.
Vraag 29 Verslag
The shape of the S-orbital is
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The shape of the S-orbital is spherical. This means that it has a symmetrical, 3-dimensional shape that resembles a ball or a globe. It is not flat or elongated, but rather is evenly rounded in all directions. This is the most basic type of orbital and is used to describe the electrons in the outermost energy level of an atom.
Vraag 30 Verslag
The constituent of air necessary in the rusting process are
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The constituent of air includes O2, CO2, N2 and Noble gases. While for a rusting process to take place, the presence of O2, H2O and CO2 is important.
The constituent of air includes O2, CO2, N2 and Noble gases and for rusting process to take place.
The presence of O2 and CO2 as a constituent of air is indispensable
Vraag 31 Verslag
The ideal gas laws and equations are true for all gases at
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The ideal gas laws and equations are most accurate for gases under conditions of low pressure and high temperature. At low pressure, the gas molecules are far apart and experience fewer intermolecular forces, allowing them to behave more ideally. At high temperature, the gas molecules have more kinetic energy, which helps to overcome any intermolecular forces that may be present. Under these conditions, the volume of a gas is directly proportional to its temperature and inversely proportional to its pressure, as described by the ideal gas law. However, as pressure and/or temperature increase, the behavior of real gases becomes more complex and may deviate from ideal gas behavior.
Vraag 32 Verslag
The general formula of alkanones is
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The general formula of alkanones is R2CO, which represents a functional group that consists of a carbon atom double-bonded to an oxygen atom (C=O) and attached to two alkyl (or aryl) groups represented by the "R" symbol. In simpler terms, alkanones are a type of organic compound that contain a carbon atom double-bonded to an oxygen atom and two additional carbon-containing groups. The "R" groups can be any combination of alkyl or aryl groups, which are made up of carbon and hydrogen atoms arranged in a specific way. The general formula R2CO applies to all alkanones, regardless of the specific alkyl or aryl groups attached to the central carbon atom. By knowing this formula, we can easily recognize an alkanone when we see one and predict its chemical properties based on the functional group it contains.
Vraag 33 Verslag
The furring of kettles is caused by the presence in water of
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Furring of kettles is caused by the temporary hardness in water. Temporary hardness in water is caused by calcium and magnesium trioxocarbonate (IV)
CaCO3 causes the furring of kettles
Vraag 34 Verslag
The enzyme used in the hydrolysis of starch to dextrin and maltose is
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The enzyme used in the hydrolysis of starch to dextrin and maltose is called amylase. Starch is a complex carbohydrate found in many foods, including grains, potatoes, and vegetables. It consists of long chains of glucose molecules that are linked together. To digest starch, our bodies use the enzyme amylase, which breaks down the long chains into smaller molecules of dextrin and maltose. These smaller molecules are more easily absorbed into the bloodstream and used as energy by the body. Amylase is produced in the salivary glands and pancreas and is released into the digestive system when we eat. In the mouth, amylase begins breaking down starch as we chew our food, and then continues its work in the small intestine where it is released by the pancreas. In summary, amylase is the enzyme responsible for breaking down starch into smaller molecules of dextrin and maltose during digestion.
Vraag 35 Verslag
The enzyme used in the hydrolysis of starch to dextrin and maltose is
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This is any amylase or a mixture of amylases that converts starch to dextrin and maltose.
Vraag 36 Verslag
In the diagram above, Y is
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In the preparation of sulphur dioxide by the action of dilute acids on sulphates and bisulphites. conc H2SO4 helps to release SO2 from the mixture.
The setup represents the production of sulphur dioxide
Vraag 37 Verslag
Incomplete oxidation of ethanol yields
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C2H5OH → CH3CHO → CH3COOH
Ethanol.....oxidation Ethanal........Ethanoic Acid
Primary alcohol oxidises to aldehyde and later to carboxylic acid.
Secondary alcohol oxidises to ketones.
Ethanol is an example of primary alcohol and primary alcohol can be oxidised to aldehyde and carboxylic acid. Wherein, incomplete oxidation of primary alcohol yields aldehyde also known as alkaline while complete oxidation of primary alcohol yields carboxylic acid.
Vraag 38 Verslag
X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be
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Enthalpy Change [?H]Entropy Change [?S]Gibbs free Energy[?G]PositivePositivedepends on T, may be + or -NegativePositivealways negativeNegativeNegativedepends on T, may be + or -PositiveNegativealways positive
?G= ?H ? T?S.
To determine whether ?G will be positive or negative, the value of ?H(change in enthalpy) and ?S (change in entropy) must be given. Likewise the temperature.
Vraag 39 Verslag
Due to the high reactivity of sodium, it is usually stored under
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Na is kept under kerosene (paraffin) to avoid reactivity with air.
Paraffin is also known as Kerosene.
Na(sodium) is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide.
Vraag 40 Verslag
When few drops of concentrated trioxonitrate(V) acid is added to an unknown sample and wanned an intense yellow colouration is observed. The likely functional group present in the sample is
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Xanthopreitic test for the presence of protein, when conc nitric acid is added to the drop, an intense yellow colouration is observed.
It contains all the functional group of protein which includes the amino, alkanol and the carboxylic group. Adding few drops of conc HNO3 to a protein, gives an intense yellow colouration.
It is called Xanthopreitic test.
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