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Tambaya 1 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 2 Rahoto
The situation obtained when a perfect gas expands into a vacuum is
Tambaya 3 Rahoto
Which of the following metals cannot replace hydrogen from water or steam?
Tambaya 4 Rahoto
H2 S(g) + Cl2(g) → 2HCl(g) + S(g) In the reaction above, the substance that is reduced is
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Tambaya 5 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 6 Rahoto
The end products of burning a candle in the atmosphere are water and
Tambaya 7 Rahoto
In the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralise 10cm3 of 1.25 molar sulphuric acid?
Tambaya 8 Rahoto
The general formula of alkanones is
Tambaya 9 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
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Tambaya 10 Rahoto
ME + nF -----> pG + qH
In the equation shown, the equilibrium constant is given by?
Bayanin Amsa
The equilibrium constant for a chemical reaction is a measure of the balance between the reactants and products of a reaction at a particular temperature. The equilibrium constant is given by the ratio of the product of the concentration of the products raised to their stoichiometric coefficients, to the product of the concentration of the reactants raised to their stoichiometric coefficients. In the equation ME + nF -> pG + qH, the correct expression for the equilibrium constant is [G]^p * [H]^q / [E]^m * [F]^n, represented by.
Tambaya 11 Rahoto
The hydrogen ion concentration of a sample of orange juice is 2.0 X 10−11 moldm−3 . What is its pOH ? [log102 = 0.3010]
Tambaya 12 Rahoto
Which of the compounds is composed of Al, Si, O and H?
Bayanin Amsa
The compound composed of Al, Si, O and H is clay. Clay is a type of sedimentary rock that is made up of very small mineral particles, including hydrated aluminum silicates and other minerals such as quartz and feldspar. These minerals are rich in aluminum, silicon, oxygen, and hydrogen, which gives clay its unique chemical composition. Clay is formed through a process of weathering and erosion of rocks containing these minerals over a long period of time. As water and other natural forces break down the rocks, the mineral particles become suspended in water and are eventually deposited in sedimentary layers. Over time, these layers become compacted and cemented together, forming the solid clay deposits we see today. Therefore, the answer is option C: Clay.
Tambaya 13 Rahoto
The ionic radii of metals are usually
Bayanin Amsa
The ionic radii of metals are usually smaller than their atomic radii. The size of an atom is determined by the distance between the nucleus and the outermost electrons, which is known as the atomic radius. When a metal atom loses one or more electrons to form a positive ion (or cation), the resulting ion has a smaller size than the original atom. This is because the positive charge of the ion attracts the remaining electrons closer to the nucleus, making the ion smaller in size. So, when a metal forms a cation, its ionic radius is typically smaller than its atomic radius. This is a general trend in the periodic table, although there are some exceptions.
Tambaya 14 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 15 Rahoto
What volume of oxygen will remain after reacting 8cm of hydrogen gas with 20cm of oxygen gas
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Tambaya 16 Rahoto
To what temperature must a gas at 273k be heated in order to double both its volume and pressure?
Tambaya 17 Rahoto
Elements P, Q, R, S have 6, 11, 15, 17 electrons respectively, therefore,
Bayanin Amsa
Elements form bonds with other elements in order to attain a stable electron configuration, like the one found in noble gases. There are two types of bonds: covalent and ionic (also called electrovalent). In covalent bonds, two elements share electrons to attain a stable electron configuration. This type of bond is formed between two non-metal elements. In ionic bonds, one element donates electrons to another element, creating ions. This type of bond is formed between a metal and a non-metal element. Based on the information given, we can deduce the following: - P is a metal, as it has only 6 electrons. - Q is a non-Metal, as it has 11 electrons. - R is a metal, as it has 15 electrons. - S is a non-Metal, as it has 17 electrons. So, from this information, we can conclude that: - P will form an ionic bond with R, as P is a metal and R is a metal. - Q will form a covalent bond with S, as Q is a non-Metal and S is a non-Metal. Therefore, the correct answer is "Q will form a covalent bond with S."
Tambaya 18 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 19 Rahoto
2KClO3(g) MNO3? 2KCl(s) + 3O2(g)
The importance of the catalyst in the reaction above is that
Tambaya 20 Rahoto
A balanced chemical equation obeys the law of
Bayanin Amsa
A balanced chemical equation obeys the law of conservation of mass. This means that in a chemical reaction, the total mass of the reactants must be equal to the total mass of the products. In other words, atoms cannot be created or destroyed during a chemical reaction, only rearranged. For example, if we burn a piece of wood, the mass of the ashes and the gases released will be equal to the mass of the original wood. This is because the atoms in the wood (carbon, hydrogen, oxygen, etc.) are rearranged during the burning process to form new molecules, but the total number of atoms remains the same. By balancing a chemical equation, we ensure that the same number and type of atoms are present on both sides of the equation, which satisfies the law of conservation of mass.
Tambaya 21 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 22 Rahoto
The periodic classification is an arrangement of the elements
Bayanin Amsa
The periodic classification is an arrangement of the elements based on their atomic numbers. The periodic table is a chart that lists all the known chemical elements in order of increasing atomic number, arranged in rows and columns according to their electronic structure and chemical properties. The atomic number of an element is the number of protons in the nucleus of an atom of that element. Each element has a unique atomic number, which determines its position in the periodic table. The elements are arranged in rows called periods, and in columns called groups or families. Elements in the same group have similar properties because they have the same number of valence electrons, which are the electrons in the outermost shell of the atom. The periodic table is an incredibly useful tool for chemists because it allows them to predict the properties of elements based on their position in the table. For example, elements in the same group tend to form similar compounds, so if you know the properties of one element in a group, you can often predict the properties of the other elements in that group. In summary, the periodic classification is an arrangement of the elements based on their atomic numbers. The periodic table is a chart that organizes the elements into rows and columns based on their electronic structure and chemical properties, allowing scientists to make predictions about the behavior of the elements based on their position in the table.
Tambaya 24 Rahoto
What mass of Cu would be produced by the cathodic reduction of Cu2+ when 1.60A of current passes through a solution of CuSO4 for 1 hour. (F=96500Cmol−1 , Cu=64)
Bayanin Amsa
The reduction reaction that occurs at the cathode during the electrolysis of CuSO4" tabindex="0" class="mjx-chtml MathJax_CHTML" id="MathJax-Element-1-Frame">4, is: Cu2+" tabindex="0" class="mjx-chtml MathJax_CHTML" id="MathJax-Element-2-Frame">2+ + 2e- -> Cu(s) From this, we can see that each Cu2+ ion requires two electrons to be reduced to copper metal. Given the current (I = 1.60 A), time (t = 1 hour = 3600 s), and Faraday's constant (F = 96500 C/mol), we can calculate the total amount of charge that passes through the solution: Q = I*t = 1.60 A * 3600 s = 5760 C Using Faraday's law, we can relate the amount of charge that passes through the solution to the number of moles of electrons transferred during the reduction reaction: n = Q/F = 5760 C / 96500 C/mol = 0.0597 mol e- Since each Cu2+ ion requires 2 electrons to be reduced to copper metal, the number of moles of copper produced is half the number of moles of electrons transferred: mol Cu = 0.0597 mol e- / 2 = 0.0299 mol Cu Finally, we can convert the moles of copper produced to grams using the molar mass of copper: mass Cu = 0.0299 mol Cu * 64 g/mol = 1.91 g Therefore, the answer is 1.91 g of Cu produced. is correct.
Tambaya 25 Rahoto
According to the Kinetic Theory an increase in temperature causes the kinetic energy of particles to
Bayanin Amsa
The kinetic energy of particles increases with an increase in temperature. In the Kinetic Theory, temperature is related to the average kinetic energy of the particles in a substance. The higher the temperature, the faster the particles move, and the more energy they have. Think of it like this: if you throw a ball, it will have more energy and travel farther if you throw it harder. Similarly, if you heat up a substance, its particles will move faster and have more energy. So, the answer is that an increase in temperature causes the kinetic energy of particles to increase.
Tambaya 26 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 27 Rahoto
Which of the following pairs of substances will react further with oxygen to form a higher oxide?
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Tambaya 29 Rahoto
In the preparation of oxygen by heating KCIO, in the presence of MnO2 only moderate heat is needed because the catalyst acts by 2
Bayanin Amsa
The presence of MnO2 acts as a catalyst in the reaction of KCIO2 to produce oxygen. A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the reaction itself. MnO2 acts by lowering the energy barrier of the reaction, which means it reduces the amount of energy required for the reaction to take place. This makes it easier for the reaction to occur, and thus the reaction proceeds at a faster rate. As a result, only moderate heat is needed to provide the initial energy required for the reaction to start. Therefore, the correct answer is: lowering the energy barrier of the reaction.
Tambaya 30 Rahoto
A correct electrochemical series can be obtained from Na, Ca, Al, Mg, Zn, Fe, Pb, H, Cu, Hg, Ag, Au by interchanging
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Tambaya 31 Rahoto
Which of the following statements is correct about the periodic table?
Tambaya 32 Rahoto
A sample of hard water contains some calcium sulphate and calcium hydrogen carbonate. The total hardness may therefore be removed by
Tambaya 33 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 34 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 35 Rahoto
3H2(g) + N2 ⇔ 2NH3(g) ; H= -ve
In the reaction above, lowering of temperature will
Tambaya 36 Rahoto
What is the PH of 0.00 1 moldm3 solution of the sodium hydroxide
Tambaya 37 Rahoto
Aluminium does not react with either dilute or concentrated trioxonitrate (V) acid because
Tambaya 38 Rahoto
The salt that reacts with dilute hydrochloric acid to produce a pungent smelling gas which decolourizes acidified purple potassium tetraoxomanganate (VII) solution is
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Tambaya 39 Rahoto
If the molecular mass of tetraoxosulphate (VI) acid is 98, calculate its vapour density
Tambaya 40 Rahoto
An element X forms the following compounds with chlorine; XCl4 , XCl3 , XCl2 . This illustrates the
Bayanin Amsa
The element X forming different compounds with chlorine (XCl4, XCl3, and XCl2) illustrates the law of multiple proportions. This law states that when two elements combine to form more than one compound, the ratio of the masses of one element that combine with a fixed mass of the other element is always a whole number ratio. In this case, the ratio of chlorine to X in the different compounds (XCl4, XCl3, and XCl2) is 4:1, 3:1, and 2:1, respectively, which are all whole number ratios.
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