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Tambaya 1 Rahoto
3H2(g) + N2 ⇔ 2NH3(g) ; H= -ve
In the reaction above, lowering of temperature will
Tambaya 3 Rahoto
To what temperature must a gas at 273k be heated in order to double both its volume and pressure?
Tambaya 4 Rahoto
During the electrolysis of copper II sulphate between platinum electrodes, if litmus solution is added to the anode compartment
Bayanin Amsa
During the electrolysis of copper II sulphate between platinum electrodes, if litmus solution is added to the anode compartment, the litmus will turn red and oxygen gas will be evolved. This is because during electrolysis, the positively charged copper ions (Cu2+) in the copper II sulphate solution are attracted to the negative cathode electrode, where they gain electrons and are reduced to form solid copper. At the same time, the negatively charged sulphate ions (SO42-) are attracted to the positive anode electrode, where they lose electrons and are oxidized to form oxygen gas and water. The litmus added to the anode compartment turns red because of the formation of oxygen gas, which is a highly reactive oxidizing agent that can react with the litmus to cause it to turn red. No hydrogen gas is evolved because hydrogen is produced at the cathode, which is in a separate compartment from the anode where the litmus is added.
Tambaya 5 Rahoto
Suitable reagents for the laboratory preparation nitrogen are
Tambaya 6 Rahoto
The type of bonding in [Cu(NH3 )4 ]2+ is
Bayanin Amsa
The type of bonding in [Cu(NH3)4]2+ is coordinate bonding. Coordinate bonding (also known as dative covalent bonding) is a type of covalent bonding where one atom (in this case, the nitrogen atom in NH3) donates a pair of electrons to another atom or ion (in this case, the copper ion Cu2+). The donating atom is called the ligand, and the receiving atom or ion is called the central metal ion. In [Cu(NH3)4]2+, each ammonia molecule (NH3) donates a lone pair of electrons on the nitrogen atom to the copper ion, forming four coordinate bonds between the ligands and the central copper ion. The presence of coordinate bonds is indicated by the use of square brackets around the coordination compound, and the charge on the compound is indicated by the superscript outside the brackets. Therefore, the answer is option A: coordinate.
Tambaya 7 Rahoto
2SO2 (g) + O2 (g) ↔ 2SO3 (g) ΔH = -395.7kJmol−1
In the equation, an increase in temperature will shift the equilibrium position to the
Bayanin Amsa
Tambaya 9 Rahoto
The refreshing and characteristic taste of soda water and other soft drinks is as a result of the presence of
Tambaya 10 Rahoto
Which of the following metals cannot replace hydrogen from water or steam?
Tambaya 11 Rahoto
The figure above shows the electrolysis of molten sodium chloride. Z is the
Bayanin Amsa
The figure shows the electrolysis of molten sodium chloride. During electrolysis, an electric current is passed through a molten or dissolved ionic compound to separate the ions. The positive ions move towards the negative electrode (cathode) and the negative ions move towards the positive electrode (anode). In the figure, the electrode connected to the positive terminal of the battery is the anode and the electrode connected to the negative terminal is the cathode. At the anode, the negatively charged chloride ions (Cl-) lose electrons and are oxidized to form chlorine gas (Cl2). At the cathode, the positively charged sodium ions (Na+) gain electrons and are reduced to form liquid sodium metal (Na). Therefore, the answer is (a) anode where the Cl- ions are oxidized. Z is the anode in the figure.
Tambaya 12 Rahoto
Which of the following separation techniques can be employed in obtaining solvent from its solution?
Bayanin Amsa
The separation technique that can be employed in obtaining a solvent from its solution is evaporation. Evaporation is a process that involves heating a solution to vaporize the solvent, leaving behind the solute. The vaporized solvent can then be condensed and collected as a pure liquid. This technique is commonly used in industry and laboratory settings to recover solvents from solutions, as it is a simple and effective way to purify liquids. Distillation can also be used to separate a solvent from a solution, but it is a more complex process that involves boiling the solution and then condensing the vapors in a separate apparatus. Filtration and precipitation are not suitable for separating a solvent from a solution, as they are primarily used to separate solid particles from a liquid mixture.
Tambaya 13 Rahoto
The collision theory explains reaction rates in terms of
Bayanin Amsa
The collision theory explains reaction rates in terms of the frequency of collision of the reactants. In other words, the theory suggests that for a chemical reaction to occur, the reactant particles must collide with sufficient energy and with the correct orientation. The frequency of these collisions is an important factor in determining the rate of the reaction. The more frequently the reactant particles collide, the more likely it is that they will react and form products. Therefore, increasing the frequency of collisions between reactant particles can increase the rate of a chemical reaction. The size of the reactants or the products does not play a significant role in the collision theory.
Tambaya 14 Rahoto
A correct electrochemical series can be obtained from Na, Ca, Al, Mg, Zn, Fe, Pb, H, Cu, Hg, Ag, Au by interchanging
Bayanin Amsa
Tambaya 15 Rahoto
The knowledge of half-life can be used to
Tambaya 16 Rahoto
When large hydrocarbon molecules are heated at high temperature in the presence of a catalyst to give smaller molecules, the process is known as
Bayanin Amsa
The process of breaking down large hydrocarbon molecules into smaller molecules by heating them at high temperatures in the presence of a catalyst is known as cracking. This process is used to convert heavy, high-molecular-weight hydrocarbon molecules into lighter, more valuable products such as gasoline and diesel fuel. The high temperatures cause the large molecules to break apart into smaller ones, and the catalyst helps speed up the reaction. This process is important in the petrochemical industry, as it allows for the production of a wider range of useful products from crude oil.
Tambaya 17 Rahoto
A quantity of electricity liberates 3.6g of Silver from its salt. What mass of aluminium Will be liberated from its salt by the same quantity of electricity? [Al = 27, Ag = 108].
Bayanin Amsa
The amount of substance liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the solution. This is known as Faraday's laws of electrolysis. The key to solving this problem is to recognize that the same quantity of electricity is used to liberate both silver and aluminum from their respective salts. We can use the ratio of their molar masses to determine the mass of aluminum liberated. The molar mass of silver (Ag) is 108 g/mol, while the molar mass of aluminum (Al) is 27 g/mol. This means that it takes four times as many moles of aluminum to make the same mass as one mole of silver. Since the same quantity of electricity liberates 3.6g of silver from its salt, it will liberate four times as many moles of aluminum. Therefore, the mass of aluminum liberated is: (4 moles of Al) x (27 g/mol) = 108 g So, the mass of aluminum liberated is 0.108 g, or 0.1 g to one significant figure. Therefore, the answer is option D: 0.3g.
Tambaya 18 Rahoto
The conductivity of an acid solution depends on the
Bayanin Amsa
The conductivity of an acid solution depends on the amount of ions present and their mobilities. When an acid dissolves in water, it forms ions that can carry an electric charge. These ions are what allows the solution to conduct electricity. The more ions there are in the solution, the better it can conduct electricity. However, not all ions have the same mobility or ability to move around in the solution. Ions with a higher mobility can move more easily through the solution, leading to a higher conductivity. Therefore, the conductivity of an acid solution is determined by both the amount of ions present and their mobilities. Other factors such as temperature can also affect conductivity, but the primary factors are the amount and mobility of ions.
Tambaya 19 Rahoto
In the upper atmosphere, the ultra-violet light breaks off a free chlorine atom from chlorofluorocarbon molecule. The effect of this is that the free chlorine atom will
Bayanin Amsa
The free chlorine atom that breaks off from a chlorofluorocarbon molecule will be very reactive and will attack ozone in the upper atmosphere. Ozone is a molecule made up of three oxygen atoms, and when the free chlorine atom reacts with ozone, it breaks the ozone molecule into two separate oxygen molecules. This reaction reduces the amount of ozone in the atmosphere, which is known as ozone depletion. Over time, this can lead to a thinning of the ozone layer, which protects life on Earth from harmful ultraviolet radiation from the sun.
Tambaya 20 Rahoto
At what temperature is the solubility of potassium trioxonitrate(V ) equal to that of sodium trioxonitrate (V)?
Bayanin Amsa
Tambaya 21 Rahoto
Sieving is a technique used to separate mixtures containing solid particles of
Bayanin Amsa
Sieving is a technique used to separate mixtures containing solid particles of different sizes. A sieve is a mesh or perforated screen that is used to separate particles based on their size. The mixture is poured onto the sieve, and the particles that are too large to pass through the holes are left on top, while the smaller particles fall through the holes and are collected below. This process allows for the separation of the different-sized particles, making it easier to purify or further process the mixture.
Tambaya 22 Rahoto
An element used in the production of matches is
Bayanin Amsa
The element used in the production of matches is sulphur. Matches are small sticks made of wood or cardboard with a chemical mixture at one end. This chemical mixture, called the match head, contains several compounds including sulphur. When the match is struck against a rough surface, the friction generates heat that ignites the sulphur in the match head, causing a flame. This flame then ignites the other compounds in the match head, which in turn ignites the wood or cardboard stick. Sulphur is an important component of the match head because it is highly flammable and burns easily. It also helps to ignite the other compounds in the match head. However, sulphur by itself is not a good fuel, which means that it cannot sustain a flame on its own. Therefore, it needs other combustible materials, such as potassium chlorate or phosphorus, to make the match head burn. Overall, sulphur plays a crucial role in the chemistry of matches and allows us to easily start fires for various purposes.
Tambaya 24 Rahoto
(I). 3CuO(s) + 2NH3(g) -----> 3Cu(s) + 3H2O(l) + N2(g)
(II). 2NH3(g) + 3Cl2(g) -----> 6HCl(g) + N2(g)
(III). 4NH3(g) + 3O2(g) -----> 6H2O(l) + N2(g)
The reactions represented by the equations above demonstrate the
Bayanin Amsa
Tambaya 25 Rahoto
How many atoms are present in 6.0g of magnesium? [Mg = 24, N.A = 6.02 x 10 23 mol]
Tambaya 26 Rahoto
The situation obtained when a perfect gas expands into a vacuum is
Tambaya 27 Rahoto
Which of the following is used to power steam engines?
Bayanin Amsa
Coal is the fuel that is typically used to power steam engines. Coal is burned in a furnace to heat water and produce steam, which is then used to power a steam engine. The steam engine converts the energy from the steam into mechanical energy, which can be used to power machines or generate electricity. Coal is a fossil fuel that has been used for centuries as a source of energy, and it played a significant role in the industrial revolution, powering steam engines that were used to drive machines in factories and transport goods and people by train. Today, steam engines are less common as other forms of energy have taken their place, but they remain an important part of our history and technological development.
Tambaya 28 Rahoto
Aluminium does not react with either dilute or concentrated trioxonitrate (V) acid because
Tambaya 29 Rahoto
The boiling of fat and aqueous caustic soda is referred to as
Bayanin Amsa
The boiling of fat and aqueous caustic soda is referred to as saponification. Saponification is the process of converting fat into soap through a reaction with an alkaline substance, such as caustic soda. The reaction results in the formation of soap (a salt of a fatty acid) and glycerol. This process is important in the manufacture of soap, as it allows the fat to be converted into a useful cleaning product.
Tambaya 30 Rahoto
A basic postulate of the kinetic theory of gases is that the molecules of a gas move in straight lines between collisions. This implies that
Tambaya 31 Rahoto
The presence of ammonia gas in a desiccator can exclusively be removed by
Bayanin Amsa
Tambaya 32 Rahoto
The Sulphide which is insoluble in dilute hydrochloric acid is
Bayanin Amsa
The sulphide which is insoluble in dilute hydrochloric acid is Copper Sulphide (CuS). When metal sulphides react with hydrochloric acid, they undergo an acid-base reaction to produce hydrogen sulphide gas and the corresponding metal chloride. For example, when Iron Sulphide (FeS) reacts with hydrochloric acid, it forms hydrogen sulphide gas (H2S) and iron chloride (FeCl2) as follows: FeS + 2HCl → H2S + FeCl2 However, Copper Sulphide (CuS) does not react with dilute hydrochloric acid, as it is insoluble in this acid. This is due to the fact that CuS is a much less reactive metal sulphide compared to FeS and ZnS, and therefore it does not undergo an acid-base reaction with dilute hydrochloric acid. In summary, CuS is the sulphide which is insoluble in dilute hydrochloric acid due to its low reactivity with acids.
Tambaya 33 Rahoto
Calculate the percentage composition of oxygen in calcium trioxocarbonate(IV) [Ca=40, C=12, O=16]
Bayanin Amsa
To calculate the percentage composition of oxygen in calcium trioxocarbonate(IV), we first need to determine the molar mass of the compound. The compound has one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O). So, the molar mass of calcium trioxocarbonate(IV) can be calculated as follows: Molar mass = (1 × atomic mass of Ca) + (1 × atomic mass of C) + (3 × atomic mass of O) = (1 × 40) + (1 × 12) + (3 × 16) = 40 + 12 + 48 = 100 g/mol Next, we need to determine the mass of oxygen in one mole of calcium trioxocarbonate(IV). The compound has three oxygen atoms, each with an atomic mass of 16 g/mol. Therefore, the total mass of oxygen in one mole of the compound is: Mass of oxygen = 3 × 16 = 48 g/mol Finally, to determine the percentage composition of oxygen in calcium trioxocarbonate(IV), we divide the mass of oxygen by the molar mass of the compound and multiply by 100. Percentage of oxygen = (Mass of oxygen / Molar mass of compound) × 100 = (48 / 100) × 100 = 48% Therefore, the correct answer is 48, which represents the percentage composition of oxygen in calcium trioxocarbonate(IV).
Tambaya 34 Rahoto
Which of these alloys contains copper?
Bayanin Amsa
Bronze is the alloy that contains copper. Bronze is a metal alloy composed of copper and typically other elements such as tin, aluminum, silicon, or nickel. It is known for its strength, durability, and corrosion resistance. In fact, bronze is one of the earliest alloys created by humans, and it has been used for thousands of years to make tools, weapons, and decorative objects. Solder is an alloy of lead, tin, and sometimes other metals that is used to join metals together by melting the solder and allowing it to flow into the joint. Steel is an alloy of iron and carbon, and sometimes other elements like chromium, nickel, or manganese, that is known for its strength and durability. Permallory is a nickel-iron alloy with high magnetic permeability and low coercive force, which makes it useful in the production of electrical and electronic equipment. None of these alloys contain copper.
Tambaya 35 Rahoto
The Consecutive members of an alkane homologous series differ by
Bayanin Amsa
The consecutive members of an alkane homologous series differ by a CH2 unit. This means that each successive member of the alkane series has one more CH2 unit than the previous member. For example, consider the simplest alkane, methane (CH4). The next member of the series is ethane (C2H6), which differs from methane by one CH2 unit. The next member after that is propane (C3H8), which differs from ethane by another CH2 unit. This pattern continues for all members of the alkane homologous series. The reason for this is that each carbon atom in the alkane chain must be bonded to four other atoms, which are usually hydrogen atoms. This means that each carbon atom in the chain can only bond to one other carbon atom. Therefore, the length of the alkane chain can only increase by adding CH2 units to the end of the chain. In summary, the consecutive members of an alkane homologous series differ by a CH2 unit because this is the only way to add length to the alkane chain while maintaining the required number of bonds for each carbon atom in the chain.
Tambaya 36 Rahoto
The radio isotope used in industrial radiography for the rapid checking of faults in welds and casting is?
Tambaya 37 Rahoto
According to Charles' law, the volume of a gas becomes zero at
Bayanin Amsa
Charles' law states that the volume of a gas is directly proportional to its temperature, provided that the pressure remains constant. This means that as the temperature of a gas increases, its volume also increases. However, it is important to note that this law only applies to ideal gases, which are theoretical gases that perfectly follow the laws of thermodynamics. According to Charles' law, the volume of a gas becomes zero at absolute zero, which is approximately -273°C. At this temperature, the gas particles would have no kinetic energy and would be in their lowest energy state. The volume of a real gas would not actually become zero at absolute zero because the gas particles would have some residual intermolecular interactions that would prevent them from completely collapsing to a single point.
Tambaya 38 Rahoto
2KClO3(g) MNO3? 2KCl(s) + 3O2(g)
The importance of the catalyst in the reaction above is that
Tambaya 39 Rahoto
Which of the following produces relatively few ions in solution?
Bayanin Amsa
The correct answer is AI(OH)3. When ionic compounds dissolve in water, they dissociate into their constituent ions, producing charged particles in solution. The more ions a compound produces, the more conductive it is in solution. AI(OH)3, also known as aluminum hydroxide, produces relatively few ions in solution because it is a weak base. When AI(OH)3 dissolves in water, it releases a small amount of Al3+ and OH- ions. In contrast, NaOH, KOH, and Ca(OH)2 are strong bases that dissociate more completely in water and produce more ions in solution. NaOH and KOH produce one hydroxide ion for every sodium or potassium ion, while Ca(OH)2 produces two hydroxide ions for every calcium ion. Therefore, of the options listed, AI(OH)3 produces relatively few ions in solution.
Tambaya 40 Rahoto
The choice of method for extracting a metal from its ores depends on the
Bayanin Amsa
The choice of method for extracting a metal from its ores depends on the position of the metal in the electrochemical series. The electrochemical series is a list of metals arranged in order of their ability to gain or lose electrons. The metals at the top of the series (such as sodium and potassium) are very reactive and will readily lose electrons, while those at the bottom (such as gold and platinum) are less reactive and less likely to lose electrons. The position of a metal in the electrochemical series determines the method of extraction that should be used. For example, metals at the top of the series are usually extracted by electrolysis, which involves passing an electric current through a molten compound of the metal. This process is necessary because the metals at the top of the series are very reactive and are strongly bonded to other elements in their ores. On the other hand, metals at the bottom of the series are usually extracted by reduction with carbon or hydrogen. This is because these metals are less reactive and can be separated from their ores by reacting them with a reducing agent that can take away the oxygen and other impurities. Therefore, the position of the metal in the electrochemical series is a crucial factor in determining the method of extraction that should be used to extract it from its ores.
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