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Ajụjụ 1 Ripọtì
The end products of burning a candle in the atmosphere are water and
Ajụjụ 2 Ripọtì
Calculate the pH of 0.05 moldm?3 H2 SO4
Akọwa Nkọwa
To solve this problem, we need to use the formula for calculating the pH of a solution, which is: pH = -log[H+] where [H+] is the concentration of hydrogen ions in moles per liter. The given chemical equation is: H2SO4 + 2H2O → 2H3O+ + SO42- From this equation, we can see that one molecule of sulfuric acid (H2SO4) can donate two hydrogen ions (H+) to the solution, which means that the concentration of hydrogen ions is twice the concentration of sulfuric acid. Therefore, the concentration of hydrogen ions in this solution is: [H+] = 2 x 0.05 moldm^-3 = 0.1 moldm^-3 Now we can use the formula for pH: pH = -log[H+] pH = -log(0.1) pH = 1.00 Therefore, the pH of the solution is 1.00.
Ajụjụ 3 Ripọtì
To what volume must 300cm3 of 0.60M sodium hydroxide solution be diluted to give a 0.40M solution?
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
H2 S(g) + Cl2(g) → 2HCl(g) + S(g) In the reaction above, the substance that is reduced is
Ajụjụ 5 Ripọtì
What volume of oxygen will remain after reacting 8cm of hydrogen gas with 20cm of oxygen gas
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
2SO2 (g) + O2 (g) ↔ 2SO3 (g) ΔH = -395.7kJmol−1
In the equation, an increase in temperature will shift the equilibrium position to the
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
If the molecular mass of tetraoxosulphate (VI) acid is 98, calculate its vapour density
Ajụjụ 8 Ripọtì
The choice of method for extracting a metal from its ores depends on the
Akọwa Nkọwa
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.
Ajụjụ 9 Ripọtì
How many atoms are present in 6.0g of magnesium? [Mg = 24, N.A = 6.02 x 10 23 mol]
Ajụjụ 10 Ripọtì
Which of the following produces relatively few ions in solution?
Akọwa Nkọwa
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.
Ajụjụ 11 Ripọtì
Which of the following separation techniques can be employed in obtaining solvent from its solution?
Akọwa Nkọwa
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.
Ajụjụ 12 Ripọtì
When air which contains the gases Oxygen, nitrogen, carbondioxide, water vapour and the rare gases, is passed through alkaline pyrogallol and then over quicklime, the only gases left are;
Akọwa Nkọwa
Ajụjụ 13 Ripọtì
A given amount of gas occupies 10.0dm5 at 4atm and 273°C. The number of moles of the gas present is [Molar volume of gas at s.t.p = 22.4dm3
]
Akọwa Nkọwa
The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. We can use this equation to solve for the number of moles of gas present. First, we need to convert the volume from dm5 to dm3, which is the same as liters (L). So, 10.0 dm5 is equal to 10.0/1000 = 0.01 dm3 or 0.01 L. Next, we need to convert the temperature from Celsius to Kelvin by adding 273 to get 546 K. Now we can plug in the values we have into the ideal gas law: 4 atm x 0.01 L = n x 0.0821 L·atm/K·mol x 546 K Simplifying, we get: 0.04 = n x 44.8 Solving for n, we get: n = 0.04/44.8 = 0.00089 mol Finally, we can compare this value to the molar volume of a gas at standard temperature and pressure (STP), which is 22.4 L/mol. To do this, we need to convert the volume of gas we have to STP conditions. Since the temperature is already at STP (273 K), we just need to adjust the pressure. Using the ideal gas law, we can solve for the volume at STP: 1 atm x V = 0.00089 mol x 0.0821 L·atm/K·mol x 273 K Simplifying, we get: V = 0.0224 L or 22.4 dm3 Therefore, the amount of gas present is equal to 0.00089 mol, which is less than 1 mol. So the answer is 0.89 mol.
Ajụjụ 14 Ripọtì
2KClO3(g) MNO3? 2KCl(s) + 3O2(g)
The importance of the catalyst in the reaction above is that
Ajụjụ 15 Ripọtì
(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
Akọwa Nkọwa
Ajụjụ 16 Ripọtì
Which of the following is an example of a chemical change?
Ajụjụ 17 Ripọtì
Which of the compounds is composed of Al, Si, O and H?
Akọwa Nkọwa
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.
Ajụjụ 18 Ripọtì
During the electrolysis of copper II sulphate between platinum electrodes, if litmus solution is added to the anode compartment
Akọwa Nkọwa
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.
Ajụjụ 19 Ripọtì
In the preparation of oxygen by heating KCIO, in the presence of MnO2 only moderate heat is needed because the catalyst acts by 2
Akọwa Nkọwa
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.
Ajụjụ 20 Ripọtì
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]
Ajụjụ 21 Ripọtì
A sample of hard water contains some calcium sulphate and calcium hydrogen carbonate. The total hardness may therefore be removed by
Ajụjụ 22 Ripọtì
The alkanoic acid found in human sweat is
Akọwa Nkọwa
The alkanoic acid found in human sweat is CH3CH2COOH, also known as propionic acid. Sweat is composed of various substances such as water, electrolytes, and waste products. One of these waste products is an oily substance called sebum, which is secreted by the sebaceous glands in the skin. When sebum breaks down, it forms various fatty acids, including propionic acid. Propionic acid has a slightly pungent odor, which is why sweat can sometimes smell sour or cheesy. However, the presence of propionic acid in sweat is actually beneficial, as it has antimicrobial properties that help to prevent the growth of harmful bacteria on the skin. In summary, the alkanoic acid found in human sweat is propionic acid, which is a fatty acid produced when sebum breaks down. Its antimicrobial properties help to keep the skin healthy.
Ajụjụ 23 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 24 Ripọtì
The conductivity of an acid solution depends on the
Akọwa Nkọwa
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.
Ajụjụ 26 Ripọtì
The ionic radii of metals are usually
Akọwa Nkọwa
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.
Ajụjụ 27 Ripọtì
The solubility of the solids that dissolves in a given solvent with the liberation of heat will
Akọwa Nkọwa
The solubility of solids in a given solvent is the amount of solid that can dissolve in the solvent to form a solution. When a solid dissolves in a solvent, it releases heat. The solubility of the solid in the solvent can be affected by changes in temperature. Generally, when the temperature of a solution increases, the solubility of the solid in the solvent increases as well. This is because the increased heat energy makes it easier for the solid particles to separate and dissolve in the solvent. As a result, the solubility of the solid in the solvent will increase with an increase in temperature. On the other hand, if the temperature decreases, the solubility of the solid in the solvent decreases. This is because the decreased heat energy makes it harder for the solid particles to separate and dissolve in the solvent. As a result, the solubility of the solid in the solvent will decrease with a decrease in temperature. In summary, the solubility of solids in a given solvent will generally increase with an increase in temperature and decrease with a decrease in temperature.
Ajụjụ 28 Ripọtì
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
Ajụjụ 29 Ripọtì
The collision theory explains reaction rates in terms of
Akọwa Nkọwa
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.
Ajụjụ 30 Ripọtì
Which of the following pairs of substances will react further with oxygen to form a higher oxide?
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
Suitable reagents for the laboratory preparation nitrogen are
Ajụjụ 32 Ripọtì
On which of the following is the solubility of a gaseous substance dependent?
I. Nature of solvent
II. Nature of solute
III. Temperature
IV. Pressure
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
The general formula of alkanones is
Ajụjụ 34 Ripọtì
The constituent common to duralumin and alnico is
Akọwa Nkọwa
The common constituent found in both duralumin and alnico is aluminum (Al). Duralumin is an alloy made up of aluminum, copper, manganese, and magnesium. It is known for its high strength and light weight, making it useful in various applications such as aerospace and construction. Alnico, on the other hand, is an alloy made of aluminum, nickel, cobalt, iron, and small amounts of other elements. It is used in the production of strong permanent magnets for various applications such as in motors, generators, and loudspeakers. So, even though duralumin and alnico have different properties and uses, they both contain the element aluminum.
Ajụjụ 35 Ripọtì
According to the Kinetic Theory an increase in temperature causes the kinetic energy of particles to
Akọwa Nkọwa
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.
Ajụjụ 37 Ripọtì
An element X forms the following compounds with chlorine; XCl4 , XCl3 , XCl2 . This illustrates the
Akọwa Nkọwa
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.
Ajụjụ 38 Ripọtì
The radio isotope used in industrial radiography for the rapid checking of faults in welds and casting is?
Ajụjụ 39 Ripọtì
To what temperature must a gas at 273k be heated in order to double both its volume and pressure?
Ajụjụ 40 Ripọtì
What is the PH of 0.00 1 moldm3 solution of the sodium hydroxide
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