Ana loda....
|
Latsa & Riƙe don Ja Shi Gabaɗaya |
|||
|
Danna nan don rufewa |
|||
Tambaya 1 Rahoto
Na2 X ⇌ 2Na+ + X2−
The bond between Na and X is likely to be
Bayanin Amsa
The bond between Na and X is most likely to be ionic. Let's break this down simply:
In the equation provided:
Na2X ⇌ 2Na+ + X2−
The sodium (Na) atoms become positively charged ions (Na+), while X becomes a negatively charged ion (X2−). This change in charge occurs because sodium atoms donate electrons to the X atom. The donation of electrons by sodium to X indicates a transfer of electrons, which is a hallmark of an ionic bond.
In an ionic bond, electrons are transferred from one atom to another, resulting in a positively charged ion and a negatively charged ion. These oppositely charged ions attract each other, forming a strong ionic bond.
In summary, since sodium (Na) donates electrons to X forming ions, the bond between Na and X is most likely to be ionic.
Tambaya 2 Rahoto
An example of a compound that is acidic in solution is
Bayanin Amsa
Phosphoric acid is a weak acid that can donate three hydrogen ions in water. Phosphoric acid partially ionizes when dissolved in an aqueous solution.
Tambaya 3 Rahoto
The molecular formular of a hydrocarbon with an empirical formula of CH3 and a molar mass of 30 is
Bayanin Amsa
To find the molecular formula of a hydrocarbon given its empirical formula and molar mass, you need to compare the empirical formula mass with the given molar mass.
The empirical formula given is CH3. The molar mass of the empirical formula is calculated as follows:
Total empirical formula mass = 12 + 3 = 15 g/mol
The provided molar mass of the compound is 30 g/mol. To determine how many empirical units are in the molecular formula, divide the molecular mass (given) by the empirical formula mass:
Number of empirical units = 30 g/mol / 15 g/mol = 2
Therefore, the molecular formula is twice the empirical formula:
Empirical formula: CH3
Molecular formula: (CH3)2 = C2H6
The correct molecular formula is C2H6.
Tambaya 4 Rahoto
A factor that does not affect the rate of a chemical reaction is
Bayanin Amsa
In evaluating the factors that affect the rate of a chemical reaction, we can look at each of the possible influences: surface area, temperature, volume, and catalyst.
Surface Area: When you increase the surface area of reactants, it allows more particles to collide with each other per unit of time, which in turn increases the rate of reaction. Imagine smaller particles like powders reacting faster than larger chunks because they have a greater surface exposed to the other reactants.
Temperature: Increasing the temperature usually increases the rate of reaction. Higher temperatures cause particles to move faster, increasing the energy of collisions, and therefore increasing the chance of successful reactions.
Catalyst: A catalyst is a substance that increases the rate of a chemical reaction without being consumed by it. It lowers the activation energy needed for the reaction to occur, thus allowing it to proceed faster.
Volume: The volume of the container or the amount of space in which a reaction occurs generally does not directly affect the rate of the reaction. While changing the volume can alter pressure or concentration in gaseous reactions, which in turn affects the rate, the volume itself is not a direct factor affecting reaction rate.
Therefore, the factor that does not directly affect the rate of a chemical reaction is volume. It indirectly affects reaction rates by altering concentration or pressure in certain reaction conditions, but it is not a direct influencing factor on its own.
Tambaya 5 Rahoto
Kerosene is used as solvent for
Bayanin Amsa
Kerosene is commonly used as a solvent for paints. Let me explain why in a simple way:
Kerosene is a type of fuel that is composed of hydrocarbons, which are molecules made up of hydrogen and carbon atoms. These hydrocarbons give kerosene the ability to dissolve other similar substances.
Paints often contain oils and other hydrocarbon-based compounds. Since kerosene is also hydrocarbon-based, it can effectively dissolve and thin these compounds. This makes it suitable for use as a solvent in paints, allowing the paint to be thinned or cleaned up after use. This property makes kerosene a good choice for cleaning brushes and other painting tools or for dissolving dried paint.
On the other hand, sulphur, gums, and fats are typically not dissolved effectively by kerosene because of their different chemical properties. Therefore, kerosene as a solvent is primarily useful in the context of working with paints and similar hydrocarbon-based materials.
Tambaya 6 Rahoto
In the graph above, y represents
Bayanin Amsa
To understand what y represents in the graph, we need to think about what graphs in chemistry, specifically regarding energy changes in reactions, generally show.
Chemical reaction energy diagrams often depict a reaction's energy change as a curve from the reactants to the products, showing different energy levels throughout the process. The energy required to start a reaction or to transform the reactants into an activated complex (also known as the transition state) is crucial.
The height of this energy barrier is called the activation energy. This is the minimum amount of energy required to start a chemical reaction. The activation energy is represented by the peak in the energy graph between the reactant energy level and the top of the curve.
Therefore, in this context, y represents the activation energy needed for the reaction to proceed. Understanding activation energy is vital as it determines how quickly a reaction will occur. Reactions with a high activation energy tend to happen more slowly because it is less probable that the necessary energy for the reaction to occur spontaneously will be present.
Tambaya 7 Rahoto
Determine the half-life of a first order reaction with constant 4.5 x 10−3 sec−1 .
Bayanin Amsa
To determine the half-life of a first-order reaction, you can use the formula:
Half-life (\(t_{1/2}\)) = \(\frac{0.693}{k}\)
where \(k\) is the rate constant of the reaction. For the given problem, the rate constant (\(k\)) is 4.5 x 10-3 s-1.
Substituting the value of \(k\) into the formula, we have:
\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}}\)
Perform the division:
\(t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}} \approx 154\) s
Therefore, the half-life of the reaction is 154 seconds.
Tambaya 8 Rahoto
What is the vapour density of 560cm3 of a gas that weighs 0.4g at s.t.p?
[Molar Volume of gas at s.t.p = 22.4 dm3 ]
Bayanin Amsa
To find the vapour density of a gas, you can use the formula:
Vapour density = (Molar mass of gas) / 2
However, first, we need to determine the molar mass of the gas. One can find the molar mass using the given data:
We know that at standard temperature and pressure (s.t.p.), 1 mole of any gas occupies 22.4 dm3. We need to convert the volume from cm3 to dm3 because the molar volume is given in dm3:
560 cm3 = 0.560 dm3
Now, let's find the number of moles in 0.560 dm3:
The number of moles (n) = Volume of gas (dm3) / Molar volume at s.t.p. (dm3/mol)
n = 0.560 dm3 / 22.4 dm3/mol
n = 0.025 moles
Given that the mass of the gas is 0.4 grams, we can find the molar mass by using the relation:
Molar Mass = Mass / Number of Moles
Molar Mass = 0.4 g / 0.025 moles
Molar Mass = 16 g/mol
Now that we have the molar mass, we can find the vapour density:
Vapour density = Molar mass / 2
Vapour density = 16 g/mol / 2
Vapour density = 8.0
Hence, the vapour density of the gas is 8.0.
Tambaya 9 Rahoto
The law which states that a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is
Bayanin Amsa
The law that states a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is the law of definite proportion.
To explain this simply, let's consider water as an example. Water is made up of hydrogen and oxygen. According to the law of definite proportion, a sample of pure water taken from anywhere in the world will always contain the same ratio of hydrogen to oxygen by mass. Specifically, water will always have approximately 88.8% oxygen and 11.2% hydrogen by mass.
This is because a chemical compound has a fixed composition, regardless of the process used to create it or the source from which it is derived. The law of definite proportion, also known as the law of constant composition, is fundamental in chemistry because it supports the idea that chemical compounds are composed of elements in specific and fixed ratios. This does not change regardless of how the compound is prepared or where it is found.
Tambaya 10 Rahoto
The Van der waals forces of attraction operates between
Bayanin Amsa
The Van der Waals forces of attraction operate between molecules. These are weak forces of attraction that occur due to momentary changes in the electron distribution within molecules. Here's a simple explanation:
Therefore, the forces can affect the physical properties of molecular compounds, such as boiling and melting points, but do not generally involve charged particles like cations or anions.
Tambaya 11 Rahoto
| COMPOUND | S | T | U | V | W |
| FORMULA | ROR' | RCOOH' | RCOR' | ROH' | RCOOR' |
From the table above, which of these two compounds can form functional group isomers?
Bayanin Amsa
ROH' and ROR' can form functional group isomers because they are the functional groups of alcohols and ethers, respectively.
Ethers have a pair of alkyl or aromatic groups attached to a linking oxygen atom. ROH is the functional group of alcohols, which are derivatives of water with one hydrogen atom replaced by an alkyl group.
Alcohols (ROH) and ethers (ROR') can form functional group isomers because they have the same chemical formula but different functional groups. E.g CH3 CH2 OH and CH3 OCH3
Tambaya 12 Rahoto
The reaction between alkanoic acids and alkanols in the presence of an acid catalyst is known as
Bayanin Amsa
The reaction between alkanoic acids and alkanols in the presence of an acid catalyst is known as esterification.
An alkanoic acid, also known as a carboxylic acid, is a type of organic acid that contains a carboxyl group (-COOH). An alkanol, commonly referred to as an alcohol, contains a hydroxyl group (-OH).
When an alkanoic acid reacts with an alkanol in the presence of an acid catalyst (commonly sulfuric acid), they combine to form an ester and water. This particular reaction is termed esterification. The acid catalyst speeds up the reaction by donating protons, which helps in breaking and forming new bonds.
Here's a simplified view of the reaction:
1. Alkanoic Acid (R-COOH) + Alkanol (R'-OH) -> Ester (R-COOR') + Water (H2O)
The key characteristics of esterification are:
Therefore, in summary, the process described is esterification.
Tambaya 13 Rahoto
An example of a physical change is
Bayanin Amsa
A physical change involves a change in the physical properties of a substance, without a change in its chemical composition. This means that the substance remains the same at the molecular level, despite how it might appear differently.
An example of a physical change from the given options is the liquefaction of liquids. In this process, a substance transitions from a solid or gas to a liquid state. This change is purely physical because the molecular structure of the substance does not change; only its state or form does. Importantly, such a change is usually reversible, meaning the substance can return to its original state. For instance, water can change into ice (frozen) or steam (vapor), and can still revert back to liquid water.
On the other hand, the other options involve chemical changes, where the original substances undergo chemical reactions to form new substances with different properties, thus altering the molecular structure depending on the option.
Tambaya 14 Rahoto
The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is
Bayanin Amsa
The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is water.
Here's why:
Hence, the correct answer is water, as it is the substance that both reacts with sodium to form an alkali and changes the color of anhydrous copper(II) tetraoxosulphate (VI) to blue.
Tambaya 15 Rahoto
In a chemical reaction, surface area of reactants can affect
Bayanin Amsa
The surface area of reactants affects the rate of a reaction between limestone and hydrochloric acid because it increases the number of collisions between the particles of the reactants. For example, if you have a large marble chip of calcium carbonate and hydrochloric acid, the acid can't reach all the calcium carbonate in the middle of the chip. If you break the marble chip into smaller pieces, you'll have a larger surface area for the acid to react with, and the reaction will happen faster.
Tambaya 16 Rahoto
Bayanin Amsa
In the Contact Process, the catalyst used for the conversion of sulphur(IV) oxide (SO2) to sulphur(VI) oxide (SO3) is vanadium(V) oxide, also chemically represented as V2O5. This catalyst is preferred because it is more cost-effective and significantly more durable under reaction conditions than other catalysts such as platinum. Moreover, while platinum is also an effective catalyst, it is prone to poisoning by impurities that may be present in the reaction mixture. Vanadium(V) oxide, on the other hand, offers a better balance of efficiency, cost, and durability, making it the catalyst of choice in industrial applications of the Contact Process.
Tambaya 17 Rahoto
An oxide of nitrogen that can rekindle a glowing splint is
Bayanin Amsa
The ability to rekindle a glowing splint is an indicator of the presence of an oxidizing agent, typically oxygen or a substance that releases oxygen. Among oxides of nitrogen, only a few are capable of doing this.
Nitrogen(I) oxide, commonly known as nitrous oxide (N2O), is not a strong enough oxidizer to rekindle a glowing splint.
Nitrogen(II) oxide, known as nitric oxide (NO), is not stable in the presence of oxygen and does not have the ability to rekindle a glowing splint because it does not actively release oxygen.
Nitrogen(IV) oxide or nitrogen dioxide (NO2), can support combustion by releasing oxygen as it decomposes. It is a brown gas and an effective oxidizer.
Dinitrogen tetraoxide (N2O4) is in equilibrium with nitrogen dioxide (NO2). However, at standard conditions, it is not as effective an oxidizer for rekindling a glowing splint as pure NO2.
In conclusion, the oxide of nitrogen that can rekindle a glowing splint is nitrogen(IV) oxide or nitrogen dioxide (NO2) due to its ability to release oxygen and support combustion.
Tambaya 18 Rahoto
The general molecular formula Cn H2n?2 represents that of an
Bayanin Amsa
The molecular formula CnH2n-2 represents an alkyne.
To understand this, let's take a look at the characteristics of hydrocarbons, which are compounds made up of hydrogen and carbon:
The formula CnH2n-2 indicates the presence of two fewer hydrogen atoms than in an alkene. This deficiency of hydrogen atoms is characteristic of a triple bond, which is a key feature of alkynes. Therefore, hydrocarbons with this formula must contain at least one triple carbon-carbon bond.
Tambaya 19 Rahoto
The basicity of tetraoxophosphate(V) acid is
Bayanin Amsa
The term basicity of an acid refers to the number of hydrogen ions (H⁺) that an acid can donate when it dissociates in water. In simpler terms, it's the number of replaceable hydrogen ions in one molecule of the acid.
Tetraoxophosphate(V) acid is another name for phosphoric acid, which has the chemical formula H₃PO₄. In this molecule, there are three hydrogen (H) atoms bonded to the phosphate group (PO₄).
When H₃PO₄ dissolves in water, it donates hydrogen ions in three steps:
Therefore, phosphoric acid, or tetraoxophosphate(V) acid, can donate a total of three hydrogen ions. Hence, the basicity of tetraoxophosphate(V) acid is 3.
Tambaya 20 Rahoto
A typical chemical reaction will be spontaneous if
Bayanin Amsa
In thermodynamics, a chemical reaction is considered spontaneous when it occurs naturally under a given set of conditions without needing to be driven by an external force. The spontaneity of a reaction is best determined by the Gibbs Free Energy change, denoted as ΔG.
The criteria for spontaneity is as follows:
Now, let's relate this to the given options:
Thus, a chemical reaction is spontaneous when the Gibbs Free Energy change (ΔG) is negative.
Tambaya 21 Rahoto
For chemical reaction to be spontaneous, ∆G must be
Bayanin Amsa
In the context of chemical reactions, the spontaneity of a reaction is determined by the Gibbs Free Energy change, represented by the symbol ΔG. A chemical reaction is considered to be spontaneous if it proceeds on its own without needing continuous external input of energy.
For a reaction to be spontaneous, the value of ∆G must be negative. This is based on the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
Where:
A negative value for ΔG indicates that the process releases energy and will proceed spontaneously. This means the system is moving towards a lower energy and more stable state, naturally favoring the products over the reactants.
In contrast, a positive ΔG indicates that the reaction is non-spontaneous and requires energy input. If ΔG is zero, the system is at equilibrium, meaning there is no net change taking place, but this doesn't indicate spontaneity.
Therefore, in summary, for a reaction to be spontaneous, ∆G must be negative.
Tambaya 22 Rahoto
In the treatment of water for municipal supply, chlorine is used to
Bayanin Amsa
In the treatment of water for municipal supply, chlorine is used to kill germs. This process is known as chlorination. Chlorine is a very effective disinfectant and is used to eliminate harmful microorganisms such as bacteria, viruses, and protozoans that may be present in the water. By doing so, chlorine helps to ensure that the water is safe for human consumption and protects public health by preventing waterborne diseases. It is important to note that **chlorine is not used to prevent tooth decay, prevent goitre, or to remove colour or odour** in water treatment for municipal supply.
Tambaya 23 Rahoto
A major effect of oil pollution in coastal water is
Bayanin Amsa
One of the major effects of oil pollution in coastal water is the destruction of aquatic life.
When oil spills into a water body, it forms a thin layer called a sheen on the surface of the water. This oil layer blocks sunlight from reaching aquatic plants and phytoplankton, inhibiting their ability to perform photosynthesis. As a result, these plants and microorganisms suffer, impacting the entire food chain.
Moreover, oil can coat the feathers of birds and the fur of marine mammals, which affects their insulation and buoyancy, leading to hypothermia, drowning, or inability to fly. Additionally, the toxic components in oil are harmful if ingested, causing internal damage to fish and other marine organisms. These combined effects can lead to significant mortality in aquatic ecosystems, threatening biodiversity and the natural balance of coastal waters.
Therefore, oil pollution can severely affect the health and survival of aquatic life, creating disruptions that can persist for many years.
Tambaya 24 Rahoto
The shape of the molecule of Carbon(IV) oxide is
Bayanin Amsa
The shape of the molecule of Carbon(IV) oxide, also known as carbon dioxide (CO2), is linear. This is because of the following reasons:
Due to this arrangement, carbon dioxide has a symmetric shape, making it non-polar despite having polar covalent bonds. The pulling forces of the two oxygen atoms on either side of the carbon atom cancel each other out, reinforcing its linear configuration.
Tambaya 25 Rahoto
Concentrated sodium chloride solution is electrolyzed using mercury cathode and graphite anode. The products at the anode and the cathode respectively are
Bayanin Amsa
When a concentrated sodium chloride solution is electrolyzed using a mercury cathode and graphite anode, the products are hydrogen gas at the cathode and chlorine gas at the anode
At the anode, 2Cl− → Cl2 + 2e−
At the cathode, 2H+ + 2e− → H2
During the electrolysis, hydrogen and chloride ions are removed from solution whereas sodium and hydroxide ions are left behind in solution. This means that sodium hydroxide is also formed during the electrolysis of sodium chloride solution.
Tambaya 26 Rahoto
Aqueous solution of sodium hydroxide can be used to test for the presence of : I. Ca2+ , II. Zn2+ , III. Cu2+
Bayanin Amsa
Aqueous solution of sodium hydroxide (NaOH) is a versatile reagent in chemistry, often used to test for the presence of metal ions. When sodium hydroxide is added to solutions containing certain metal ions, it forms precipitates that are characteristic of those ions. Here's how it interacts with each of the mentioned ions:
Calcium ions (Ca2+): When NaOH is added to a solution containing calcium ions, a white precipitate of calcium hydroxide (Ca(OH)2) can form. However, the precipitate is only slightly soluble in water, and this reaction is not the most definitive test for calcium ions.
Zinc ions (Zn2+): When sodium hydroxide is added to a solution containing zinc ions, a white gelatinous precipitate of zinc hydroxide (Zn(OH)2) forms. This precipitate is soluble in excess NaOH, leading to a clear, colorless solution. This reaction is used to test for zinc ions.
Copper ions (Cu2+): When NaOH is added to a solution containing copper ions, a pale blue precipitate of copper(II) hydroxide (Cu(OH)2) forms. This precipitate is insoluble even in excess NaOH, and the formation of this blue precipitate is a common test for copper ions.
Therefore, an aqueous solution of sodium hydroxide can be used to test for the presence of all three ions: calcium (Ca2+), zinc (Zn2+), and copper (Cu2+). The reaction and precipitate formation with each ion serve as indicators of their presence. Thus, the correct answer is:
I, II and III.
Tambaya 27 Rahoto
CH3 -CH2 -OH and CH3 -O-CH3
The relationship between the two compounds above, is that they are
Bayanin Amsa
The relationship between the two compounds is that they are isomers.
To understand why these compounds are isomers, let's break down their structures and definitions:
1. Structures of the Compounds:
2. Definitions:
Both compounds have the same molecular formula: C2H6O. However, they have different arrangements of their atoms. Ethanol has a hydroxyl group (-OH) attached to an ethyl group (CH3-CH2-), while dimethyl ether involves two methyl groups (CH3-) bonded to an oxygen atom (O). This difference in structure leads to different chemical and physical properties, despite having the same molecular formula. Hence, these two compounds are classified as isomers.
Tambaya 28 Rahoto
One of the following is not a water pollutant?
Bayanin Amsa
Water pollutants are substances that, when introduced into the water, cause harm to ecosystems, human health, and the overall quality of the water. Each of the options provided has the potential to be considered a water pollutant, except for one. Let's explain them:
1. Inorganic fertilizers: These are substances mainly composed of synthetic chemicals, including nitrates and phosphates. When these fertilizers enter water bodies, they can lead to nutrient pollution, which causes excessive growth of algae (eutrophication), leading to a decrease in oxygen levels in the water, harming aquatic life.
2. Warm water affluent: This refers to the discharge of heated water into natural water bodies. This heat contamination can change the temperature of the water, affecting the metabolism of aquatic life and leading to thermal pollution.
3. Oxygen gas: Oxygen gas is a fundamental component of the Earth's atmosphere and is not considered a water pollutant. In fact, dissolved oxygen is crucial for the survival of aquatic organisms. Rather than causing any harm, adequate levels of dissolved oxygen in water bodies are essential for maintaining healthy aquatic ecosystems.
4. Biodegradable waste: These are organic materials that decompose in the environment. When introduced in large quantities into water bodies, they can consume a significant amount of dissolved oxygen as they decompose, which can lead to depletion of oxygen levels and cause harm to aquatic life, making them pollutants in aquatic ecosystems.
Given the explanations above, oxygen gas is the option that is not a water pollutant. It is vital for the health of aquatic ecosystems, unlike the other options, which can all lead to some form of pollution in water bodies.
Tambaya 29 Rahoto
How many isomers has the organic compound represented by the formula C3 H8 O ?
Bayanin Amsa
The molecular formula C3H8O represents organic compounds that contain 3 carbon atoms, 8 hydrogen atoms, and 1 oxygen atom. Let's elucidate the possible isomers, which are molecules with the same molecular formula but different structural arrangements.
1. Alcohols: One class of compounds that can form isomers for this formula are alcohols, which include a functional group -OH.
a. Propan-1-ol: This is a straight-chain alcohol where the -OH group is on the first carbon. The structure is as follows:
CH3-CH2-CH2-OH
b. Propan-2-ol: This is another alcohol where the -OH group is on the second carbon, giving it a different structure and properties:
CH3-CH(OH)-CH3
2. Ethers: This is another class of possible isomers, where the oxygen atom is bonded to two alkyl groups.
c. Methoxyethane: Also known as ethyl methyl ether, it has a structure where the oxygen is in a bridge position between a methyl group and an ethyl group:
CH3-O-CH2-CH3
These are the possible structural isomers for this molecular formula. Therefore, the compound C3H8O has three isomers overall:
Thus, the answer is three distinct isomers.
Tambaya 30 Rahoto
The IUPAC Nomenclature of CH3 CH2 C(CH3 )=C(CH3 )2 for the compound is
Bayanin Amsa
The compound in question is written as CH₃₃CH₂₂C(CH₃₃)=C(CH₃₃)₂₂, which seems to be intended as (CH3)3CH2CH=C(CH3)3. The IUPAC nomenclature of organic compounds follows specific rules to name the compound uniquely such that it is understood universally. Here is a comprehensive breakdown:
1. Select the longest carbon chain that includes the highest-order functional group, which, in this case, is the alkene group (double bond).
2. The longest chain consists of 5 carbons, which gives us the root name "pentene". We choose the carbon chain such that the double bond gets the lowest possible number, starting from the end of the chain closest to the double bond.
3. Number the carbon atoms in the chain from the end closest to the double bond. The numbering direction will determine the position of the double bond and substituents. The double bond starts on carbon 2.
4. Identify and name the substituents attached to the carbon chain. In this case, there are two methyl groups on carbon 3. This means it is dimethyl as there are two of them.
Thus, the complete name of the compound is 2,3-dimethylpent-2-ene. Here, "2,3-dimethyl" indicates the position and quantity of methyl groups, "pent" indicates the longest chain with 5 carbons, and "-2-ene" indicates a double bond starting at the second carbon.
Tambaya 31 Rahoto
What accounts for the low melting and boiling points of covalent molecules?
Bayanin Amsa
The low melting and boiling points of covalent molecules are primarily due to the presence of weak intermolecular forces between the molecules. While covalent molecules consist of atoms bonded together by strong covalent bonds, the forces between separate molecules, known as van der Waals forces or London dispersion forces, are much weaker. These weak forces require significantly less energy to overcome, which explains why covalent molecules tend to have lower melting and boiling points.
Although covalent molecules have definite shapes and possess shared electron pairs, these characteristics have little influence on the melting and boiling points. The focus is instead on how much energy is needed to separate the molecules from one another.
Covalent molecules are not typically three-dimensional structures like ionic compounds or metals which form intricate lattices and require more energy to disrupt. Thus, the primary reason for their lower melting and boiling points is the presence of weak intermolecular forces that can be more easily overcome with minimal energy input.
Tambaya 32 Rahoto
An organic compound with general formula RCOR' is an
Bayanin Amsa
The general formula RCOR' represents a class of organic compounds known as ketones. In this formula, R and R' are alkyl groups, which are chains of carbon and hydrogen atoms. The CO in the middle is a carbonyl group, which consists of a carbon atom double-bonded to an oxygen atom. Therefore, with the presence of two alkyl groups on either side of the carbonyl group, the compound is categorized as a ketone, scientifically referred to as an alkanone.
Here is a simple breakdown of the terms:
Hence, by looking at the general formula RCOR', the organic compound in question is undoubtedly an alkanone.
Tambaya 33 Rahoto
The number of molecules of helium gas contained in 11.5g of the gas is
Bayanin Amsa
To find the number of molecules of helium gas in a given mass, we can use Avogadro's number and the molar mass of helium.
Step 1: Determine the molar mass of helium.
Helium is a noble gas with an atomic mass of approximately 4 grams per mole (g/mol).
Step 2: Calculate the number of moles in 11.5 grams of helium.
The formula to find the number of moles is:
Number of moles = Mass (g) / Molar Mass (g/mol)
So for helium:
Number of moles = 11.5 g / 4 g/mol = 2.875 moles
Step 3: Use Avogadro's number to find the number of molecules.
Avogadro's number is 6.022 x 1023 molecules per mole.
The formula to find the number of molecules is:
Number of molecules = Number of moles x Avogadro's Number
Number of molecules = 2.875 moles x 6.022 x 1023 molecules/mole
Number of molecules ≈ 1.73 x 1024 molecules
Therefore, the number of molecules of helium gas in 11.5g of helium is approximately 1.73 x 1024.
Tambaya 34 Rahoto
If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are
Bayanin Amsa
To answer this question, let's break it down step by step:
Mass Number: The mass number is the total number of protons and neutrons in an atom's nucleus. In this case, the mass number is given as 31.
Stable Neutral Atom: A stable neutral atom has no overall electrical charge, meaning the number of protons (positively charged) must equal the number of electrons (negatively charged).
If we symbolize the number of protons by the atomic number (Z), we can say:
1. **Protons = Electrons** in a neutral atom.
2. **Mass Number (A) = Protons + Neutrons**.
Given that the mass number is 31, we have the equation:
A = Protons + Neutrons = 31.
Assuming a commonly known stable element like Phosphorus, which has an atomic number (Z) of 15, it means:
1. **Protons = 15**.
2. **Electrons = 15** (because it's a neutral atom).
3. To find Neutrons: Neutrons = Mass Number - Protons = 31 - 15 = 16.
So, in this scenario, the number of electrons is 15 and the number of neutrons is 16. This combination is found in the first option given.
Tambaya 35 Rahoto
The hybridization scheme in ethyne is
Bayanin Amsa
Ethyne, also known as acetylene, is a simple alkyne with the chemical formula C2H2. In ethyne, each carbon atom is bonded to two other atoms: one hydrogen atom and the other carbon atom. The molecular structure of ethyne is linear, with a triple bond between the two carbon atoms.
To determine the hybridization scheme in ethyne, we need to examine the arrangement of the electron pairs around each carbon atom. In ethyne, each carbon atom is forming two sigma (σ) bonds and two pi (π) bonds. Let's explain:
When we consider the hybridization of the carbon atoms, we focus on the formation of sigma bonds and lone pairs. In ethyne, each carbon atom utilizes two orbitals to form sigma bonds: one with the hydrogen atom and one with the other carbon atom. This implies that each carbon atom in ethyne must use two hybrid orbitals.
The two hybrid orbitals formed by each carbon atom in ethyne are a result of mixing one s orbital with one p orbital. This hybridization is referred to as sp hybridization, characterized by a linear electron geometry. The remaining two unhybridized p orbitals on each carbon atom are responsible for forming the two pi bonds in the triple bond.
In conclusion, the hybridization scheme in ethyne is sp.
Tambaya 36 Rahoto
The IUPAC nomenclature of the complex K4 Fe(CN)6 is
Bayanin Amsa
The compound in question is K4[Fe(CN)6]. To name this complex using IUPAC nomenclature, let's break it down into parts:
Next, consider the oxidation state of Fe:
Finally, we consider the oxidation state of the iron. Since calculations show that it is +2, the complex ion is named based on its oxidation state.
Hence, the IUPAC name of this compound is potassium hexacyanoferrate(II).
Tambaya 37 Rahoto
The reaction of hydrogen and chlorine to produce hydrogen chloride gas is explosive in
Bayanin Amsa
The reaction between hydrogen and chlorine to produce hydrogen chloride gas is explosive in sunlight. This is because sunlight contains a broad range of electromagnetic radiation, including ultraviolet (UV) light, which is energetic enough to initiate the reaction.
Here is a simplified explanation:
In contrast, other forms of light like diffused light, infrared light, and Raman light do not provide enough energy to initiate this explosive reaction because they lack the necessary UV component found in sunlight.
Tambaya 38 Rahoto
A radioactive element of mass 1g has half-life of 2 minutes, what fraction of the substance would have disintegrated after 10 minutes?
Bayanin Amsa
Originalmass2n
= Residual mass
Where n = number of activity = exposuretimehalflife
Given:
Original mass = 1g, exposure time = 10 minutes , half life = 2 minutes, Residual mass = ?
Substituting all the given parameters appropriately, we have
n = 102
n = 5
Originalmass2n = Residual mass
125
5 = Residual mass
132 = Residual mass
Residual mass = 132
or 0.03125g
Tambaya 39 Rahoto
The amount of water a substance chemically combined with is called water of
Bayanin Amsa
The amount of water that is chemically combined with a substance is referred to as water of crystallization. This is the water present in the crystalline form of a compound, necessary to maintain the structure of the crystals.
When certain substances crystallize from an aqueous solution, they incorporate a specific amount of water molecules into their crystal lattice structure. These water molecules are an integral part of the crystal and often affect its color, stability, and solubility. The water is combined in stoichiometric amounts, which means it is present in a fixed ratio relative to the rest of the molecule.
An example of this is copper(II) sulfate pentahydrate, which consists of copper(II) sulfate combined with five molecules of water per formula unit, represented as CuSO4·5H2O.
Tambaya 40 Rahoto
The amount of Faraday required to discharge 4.5 moles of Al3+ is
Bayanin Amsa
To determine the amount of Faraday required to discharge 4.5 moles of Al3+ ions, it is essential to understand Faraday's laws of electrolysis and the concept of moles in chemistry.
When discharging Al3+ ions to form aluminum metal (Al), the reduction half-reaction involved is:
Al3+ + 3e- → Al
From this equation, it can be seen that 3 moles of electrons (e-) are required to discharge 1 mole of Al3+ ions to form 1 mole of aluminum metal.
A Faraday is the amount of electric charge carried by one mole of electrons. Therefore, 1 Faraday corresponds to the charge needed to discharge 1 mole of electrons.
Now, to discharge 4.5 moles of Al3+, we need:
4.5 moles of Al3+ × 3 moles of electrons (e-)/mole of Al3+ = 13.5 moles of electrons
Since each Faraday discharges 1 mole of electrons, 13.5 moles of electrons correspond to 13.5 Faradays of charge.
Hence, the amount of Faraday required to discharge 4.5 moles of Al3+ ions is 13.5 Faradays.
Za ka so ka ci gaba da wannan aikin?