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Ajụjụ 1 Ripọtì
The situation obtained when a perfect gas expands into a vacuum is
Ajụjụ 2 Ripọtì
A substance that is used as a ripening agent for fruits is
Akọwa Nkọwa
The substance that is commonly used as a ripening agent for fruits is ethene. Ethene, also known as ethylene, is a natural plant hormone that is produced by fruits, especially during the ripening process. It is a colorless gas that can be easily synthesized and used as a ripening agent for fruits. When fruits are exposed to ethene, it triggers a series of biochemical reactions that accelerate the natural ripening process. This can help fruits to ripen faster and more uniformly, which is important for commercial purposes where fruits need to be sold quickly. The use of ethene as a ripening agent is regulated by food safety agencies, as excessive exposure to ethene can cause over-ripening and spoilage of fruits. However, when used in appropriate concentrations, ethene is a safe and effective way to promote the ripening of fruits.
Ajụjụ 3 Ripọtì
The type of bonding in [Cu(NH3 )4 ]2+ is
Akọwa Nkọwa
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.
Ajụjụ 4 Ripọtì
To what temperature must a gas at 273k be heated in order to double both its volume and pressure?
Ajụjụ 5 Ripọtì
Elements P, Q, R, S have 6, 11, 15, 17 electrons respectively, therefore,
Akọwa Nkọwa
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."
Ajụjụ 6 Ripọtì
3H2(g) + N2 ⇔ 2NH3(g) ; H= -ve
In the reaction above, lowering of temperature will
Ajụjụ 7 Ripọtì
Which of these alloys contains copper?
Akọwa Nkọwa
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.
Ajụjụ 8 Ripọtì
In the shown experiment (Fig. 1) the litmus paper will initially
Ajụjụ 9 Ripọtì
The knowledge of half-life can be used to
Ajụjụ 10 Ripọtì
Diamond is a bad conductor of electricity because its bonding electrons are used in
Akọwa Nkọwa
Diamond is a bad conductor of electricity because of its unique structure and bonding. The carbon atoms in diamond form a covalent network, where each carbon atom is bonded to four other carbon atoms. These bonds are strong and hold the atoms in a rigid three-dimensional structure called a crystal lattice. In a covalent bond, atoms share electrons to form a stable compound. In diamond, each carbon atom shares its valence electrons with four neighboring carbon atoms, forming a very strong covalent bond. All the valence electrons in the crystal lattice are used in covalent bond formation, which means there are no free or mobile electrons to carry an electric current. In other words, the electrons are tightly held in the covalent bonds, making it difficult for them to move around the crystal lattice and conduct electricity. In contrast, metals conduct electricity well because they have delocalized or free electrons that can move through the lattice of positively charged ions. So, diamond, being a covalent network solid, does not have free electrons that can carry an electric current, which is why it is a bad conductor of electricity.
Ajụjụ 11 Ripọtì
If 1 litre of 2.2M sulphuric acid is poured into a bucket containing 10 litres of water and the resulting solution mixed thoroughly, the resulting sulphuric acid concentration will be
Akọwa Nkọwa
When 1 liter of 2.2M sulphuric acid is added to 10 liters of water, the total volume of the resulting solution is 11 liters. To find the resulting concentration of sulphuric acid, we need to use the equation: M1V1 = M2V2 where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume. We can plug in the values we know: M1 = 2.2M (the initial concentration of the sulphuric acid) V1 = 1L (the initial volume of the sulphuric acid) M2 = ? (the final concentration we're trying to find) V2 = 11L (the final volume of the resulting solution) Solving for M2, we get: M2 = (M1 x V1) / V2 M2 = (2.2M x 1L) / 11L M2 = 0.2M Therefore, the resulting sulphuric acid concentration is 0.2M or 0.2 moles per liter. In summary, when 1 liter of 2.2M sulphuric acid is mixed with 10 liters of water, the resulting sulphuric acid concentration is diluted to 0.2M. This is because the total volume of the resulting solution is greater than the initial volume of the sulphuric acid, which leads to a decrease in concentration.
Ajụjụ 12 Ripọtì
The Consecutive members of an alkane homologous series differ by
Akọwa Nkọwa
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.
Ajụjụ 13 Ripọtì
The end products of burning a candle in the atmosphere are water and
Ajụjụ 14 Ripọtì
The radio isotope used in industrial radiography for the rapid checking of faults in welds and casting is?
Ajụjụ 15 Ripọtì
The general formula of alkanones is
Ajụjụ 16 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 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
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 20 Ripọtì
The presence of ammonia gas in a desiccator can exclusively be removed by
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
Aluminium does not react with either dilute or concentrated trioxonitrate (V) acid because
Ajụjụ 22 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 23 Ripọtì
What volume of oxygen will remain after reacting 8cm of hydrogen gas with 20cm of oxygen gas
Akọwa Nkọwa
Ajụjụ 25 Ripọtì
Which of the following pairs of substances will react further with oxygen to form a higher oxide?
Akọwa Nkọwa
Ajụjụ 26 Ripọtì
If one of the following oxides is heated with hydrogen or carbon using a bunsen burner. it is not reduced to the metal, Which one is it?
Akọwa Nkọwa
The oxide that cannot be reduced to the metal when heated with hydrogen or carbon using a Bunsen burner is magnesium oxide. Magnesium oxide is an ionic compound made up of positively charged magnesium ions and negatively charged oxygen ions. When heated with hydrogen or carbon, the oxygen ions are not easily removed from the compound. This is because the ionic bond between the magnesium and oxygen ions is very strong and requires a lot of energy to break. On the other hand, lead oxide, copper oxide, and tin oxide are all metal oxides and can be reduced to the metal by heating with hydrogen or carbon. This is because they have a weaker bond between the metal and oxygen ions, allowing the oxygen to be removed more easily when heated. In conclusion, magnesium oxide is the oxide that cannot be reduced to the metal when heated with hydrogen or carbon using a Bunsen burner.
Ajụjụ 27 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 28 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 29 Ripọtì
How many atoms are present in 6.0g of magnesium? [Mg = 24, N.A = 6.02 x 10 23 mol]
Ajụjụ 30 Ripọtì
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].
Akọwa Nkọwa
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.
Ajụjụ 31 Ripọtì
What is the concentration of a solution containing 2g of NaOH in 100cm3 of solution? [Na = 23, O =16, H = 1]
Akọwa Nkọwa
The concentration of a solution containing 2g of NaOH in 100cm3 of solution is 0.40 moldm-3. This can be calculated by using the formula: molarity (M) = number of moles of solute / volume of solution (in liters) First, we need to calculate the number of moles of NaOH in the solution. The molar mass of NaOH is (23 + 16 + 1) = 40 g/mol. So, 2g of NaOH is equal to 2/40 = 0.05 moles. Next, we need to convert the volume of the solution from cm3 to liters. 1 cm3 = 0.001 liters, so 100 cm3 = 0.1 liters. Finally, we can calculate the molarity as follows: M = 0.05 moles / 0.1 liters = 0.5 mol/L = 0.50 moldm-3 So, the concentration of the solution is 0.50 moldm-3.
Ajụjụ 32 Ripọtì
The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as;
Akọwa Nkọwa
The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as esterification. Esterification is the process of forming an ester, which is a type of organic compound, from an alcohol and an acid. The acid catalyst is used to speed up the reaction by providing a proton to the reaction mixture, which helps to form the ester. Esterification results in the loss of a water molecule from the reaction mixture, which makes the reaction a type of dehydration reaction. However, it is important to note that esterification is a specific type of dehydration reaction where the products are an ester and an alcohol. So, the answer is esterification.
Ajụjụ 33 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 34 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 35 Ripọtì
The number of electrons in the valence shell of an element of atomic number 14 is?
Akọwa Nkọwa
The number of electrons in the valence shell of an element can be determined by using the periodic table and the electron configuration of the element. The valence shell is the outermost shell that contains electrons that are involved in chemical reactions. For an element with atomic number 14, which is silicon, the electron configuration is 1s2 2s2 2p6 3s2 3p2. The valence shell of silicon is the third shell, which contains 3s2 and 3p2 electrons. Therefore, the number of electrons in the valence shell of silicon is 4 electrons.
Ajụjụ 36 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 37 Ripọtì
2KClO3(g) MNO3? 2KCl(s) + 3O2(g)
The importance of the catalyst in the reaction above is that
Ajụjụ 38 Ripọtì
An element used in the production of matches is
Akọwa Nkọwa
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.
Ajụjụ 39 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 40 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
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