Carbon and its Compounds - Class 10 Science - Chapter 4 - Notes, NCERT Solutions & Extra Questions
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Notes - Carbon and its Compounds | Class 10 NCERT | Science
Carbon and its compounds form the basis of life on Earth and a large variety of materials around us. In this detailed guide, we will explore the fascinating world of carbon, its unique properties, and its compounds.
Basics of Carbon
Elemental Carbon
Carbon is an essential element found in various forms in nature. It is a fundamental building block of life and a significant component of many materials we use daily, such as plastics, synthetic fibers, and fuels.
Occurrence of Carbon
Although the percentage of carbon in Earth's crust and atmosphere is relatively small, its impact is massive. The Earth's crust contains carbon in forms like carbonates, coal, and petroleum, while the atmosphere has carbon dioxide. Both these reservoirs of carbon play a crucial role in supporting life and human activities.
Bonding in Carbon: The Covalent Bond
Ionic vs. Covalent Compounds
While ionic compounds involve the transfer of electrons between atoms, covalent compounds form by sharing electrons. Carbon mainly forms covalent bonds, leading to the development of an extensive range of compounds.
Electron Dot Structure
To understand how carbon bonds with other atoms, we use electron dot structures. For example, methane (CH₄) can be depicted as follows:
Unique Properties of Carbon
Valence Electrons and Reactivity
Carbon has four valence electrons. Its ability to share these electrons with other atoms leads to the formation of a variety of stable compounds. The electron configuration of carbon encourages it to form four covalent bonds.
Catenation in Carbon Compounds
Carbon is unique due to its ability to form long chains, branched chains, and rings, known as catenation. For example, carbon atoms can link together in single, double, or triple bonds to form a myriad of compounds.
Saturated and Unsaturated Compounds
Carbon compounds are classified as saturated when they have single bonds, like ethane (C₂H₆), and as unsaturated when they have double or triple bonds, like ethene (C₂H₄).
Allotropes of Carbon
Diamond and Graphite
Carbon exists in different forms called allotropes. The most well-known allotropes are diamond and graphite. In a diamond, each carbon atom bonds to four other carbon atoms, forming a rigid 3D structure. In graphite, each carbon atom bonds to three others in flat hexagonal layers.
Buckminsterfullerene
Another interesting allotrope is Buckminsterfullerene (C60), which has a soccer ball-like structure made up of 60 carbon atoms.
Versatile Nature of Carbon
Formation of Millions of Compounds
Carbon can form millions of compounds due to its tetravalency (ability to form four bonds) and catenation. This property results in various compounds with different properties and uses.
Functional Groups in Carbon Compounds
Functional groups such as alcohols, ketones, and carboxylic acids confer specific properties to carbon compounds, regardless of their carbon chain lengths.
Homologous Series
A homologous series is a series of carbon compounds where each successive compound differs by a -CH₂- unit. For instance, methane (CH₄) and ethane (C₂H₆) differ by this unit, showcasing a gradation in physical properties.
Systematic Naming (Nomenclature)
Carbon compounds are named based on their carbon chain length and functional groups. For example, a three-carbon chain with an alcohol group is named propanol.
Chemical Properties of Carbon Compounds
Combustion Reactions
Carbon and its compounds typically undergo combustion to produce carbon dioxide, water, and energy. For example:
$ C + O₂ \rightarrow CO₂ + \text{heat and light} $
$ CH₄ + 2O₂ \rightarrow CO₂ + 2H₂O + \text{heat and light} $
Oxidation
Oxidation involves the addition of oxygen to a carbon compound, transforming an alcohol into a carboxylic acid. Potassium permanganate is a common oxidizing agent for this process.
Addition Reactions
Unsaturated hydrocarbons can undergo addition reactions, where elements add to the compound without breaking the carbon chain.
Substitution Reactions
In substitution reactions, an atom in a molecule is replaced by another atom or group, such as chlorine replacing a hydrogen atom in a hydrocarbon.
Important Carbon Compounds
Ethanol
Ethanol (C₂H₅OH) is a liquid used in alcoholic beverages, medicines, and industrial applications. It reacts with sodium to produce sodium ethoxide and hydrogen gas.
Ethanoic Acid
Ethanoic acid (CH₃COOH), also known as acetic acid, is used in vinegar. It forms esters with alcohols, a reaction used to produce fragrant compounds.
Soaps and Detergents
Mechanism of Action
Soaps and detergents clean by emulsifying oily dirt. Soap molecules form micelles that trap oil, allowing it to be washed away with water.
Effectiveness in Hard Water
While soap forms scum with hard water ions, detergents do not, making them effective in both soft and hard water.
Conclusion
Carbon is a versatile element essential for life and numerous materials. Understanding its properties and compounds gives us insights into the world around us, from the fuels we use to the detergents that clean our clothes. Studying carbon compounds in class 10 lays the foundation for more advanced chemistry concepts in the future.
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Extra Questions - Carbon and its Compounds | NCERT | Science | Class 10
The graph given below compares the price (in Rs) and weight of 6 bags (in kg) of sugar of different brands A, B, C, D, E, F.
Which brand(s) costs/cost more than Brand D? A E and F B E and C C F D E
The correct option is A: E and F.
When examining the graph, it's clear that the points representing brands E and F lie above the point representing brand D. This means that brands E and F have a higher cost compared to brand D.
In carbon dioxide, the ratio of carbon atoms to oxygen atoms is always:
A. True
B. False
C. 3:2
D. 3:1
The correct option is A: True
In carbon dioxide, the ratio of carbon atoms to oxygen atoms is always $1:2$. This means that one carbon atom combines with two oxygen atoms to form each molecule of $\mathrm{CO}_{2}$. Therefore, the ratio of carbon to oxygen atoms in carbon dioxide is consistently $1:2$.
Which one of the following represents 3-D structure?
A.
B.
C.
D. All of the above.
Correct Answer: C
A 3-D structure of a molecule is depicted using a combination of dash, wedge, and straight lines.
Dashed lines typically represent bonds that are going behind the plane of the page.
Wedge-shaped lines represent bonds that are coming out of the plane towards the viewer.
Straight lines represent bonds that lie in the plane of the page.
Therefore, option C correctly illustrates a 3-D structure.
Draw the structures of the following compounds:
(a) Prop-1-ene
(b) 2,3-Dimethylbutane
(c) 2-Methylpropane
(d) 3-Hexene
(e) Prop-1-yne
Compound Structures:
(a) Prop-1-ene
Prop-1-ene is a three-carbon alkene with a double bond between the first and second carbon atoms. Its structure is: $$ \text{CH}_2=\text{CH}-\text{CH}_3 $$
(b) 2,3-Dimethylbutane
2,3-Dimethylbutane is a branched alkane. It has four carbons in the main chain with methyl groups attached to the second and third carbon atoms: $$ \text{CH}_3-\text{C}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\text{CH}_3 $$
(c) 2-Methylpropane
2-Methylpropane, also known as isobutane, is an alkane with three carbon atoms in the main chain and a methyl group attached to the second carbon: $$ \text{CH}_3-\text{C}(\text{CH}_3)-\text{CH}_3 $$
(d) 3-Hexene
3-Hexene is an alkene with a six-carbon chain and a double bond between the third and fourth carbon atoms: $$ \text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_3 $$
(e) Prop-1-yne
Prop-1-yne is a three-carbon alkyne with a triple bond between the first and second carbon atoms: $$ \text{CH}\equiv \text{C}-\text{CH}_3 $$
These structural formulas clearly represent the molecular arrangement of atoms in each compound, ensuring correct identification and understanding of their chemical properties.
Ionic compounds exist in solid state at room temperature because they have strong electrostatic forces of attraction.
A. solid
B. liquid
C. electrostatic forces
D. van der Waals forces
The correct options are:
A: Solid
C: Electrostatic forces
Ionic compounds have strong electrostatic forces of attraction between the ions, which results in the formation of a lattice structure of ions. Consequently, they exist as solids at room temperature. To separate these ions and overcome the attractive forces, a high amount of energy is required. As a result, ionic compounds typically have high melting points.
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Ask Chatterbot AINCERT Solutions - Carbon and its Compounds | NCERT | Science | Class 10
Ethane, with the molecular formula $\mathrm{C}_{2} \mathrm{H}_{6}$ has
(a) 6 covalent bonds.
(b) 7 covalent bonds.
(c) 8 covalent bonds.
(d) 9 covalent bonds.
Ethane $\mathrm{C}_{2} \mathrm{H}_{6}$ consists of two carbon atoms and six hydrogen atoms. In ethane, each carbon atom forms four covalent bonds: three with hydrogen atoms and one with another carbon atom.
Let's count the covalent bonds:
Each carbon atom has 3 bonds with hydrogen atoms, totalling $3 \times 2 = 6$ carbon-hydrogen bonds.
There is 1 carbon-carbon bond.
Adding these together gives the total number of covalent bonds in ethane:
$$ 6 (\mathrm{C}-\mathrm{H}) + 1 (\mathrm{C}-\mathrm{C}) = 7 \text{ covalent bonds.} $$
So, the correct answer is: (b) 7 covalent bonds.
Butanone is a four-carbon compound with the functional group
(a) carboxylic acid.
(b) aldehyde.
(c) ketone.
(d) alcohol.
Butanone is a four-carbon compound with the functional group:
(c) ketone.
The name "butanone" indicates that it is a ketone with four carbon atoms in its carbon chain (from "butane"), and the "one" suffix denotes the ketone functional group, which is a carbonyl group attached to two carbon atoms within the carbon skeleton.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that
(a) the food is not cooked completely.
(b) the fuel is not burning completely.
(c) the fuel is wet.
(d) the fuel is burning completely.
When the bottom of a cooking vessel gets blackened on the outside, it indicates that the fuel is not burning completely. So the correct option is:
(b) the fuel is not burning completely.
Explain the nature of the covalent bond using the bond formation in $\mathrm{CH}_{3} \mathrm{Cl}$.
The covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These pairs of electrons are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms when they share electrons is known as covalent bonding.
In the case of $\mathrm{CH}_3\mathrm{Cl}$ (Methyl Chloride), the covalent bonding can be explained as follows:
Carbon ($C$) has four valence electrons and needs four more to complete its octet.
Each hydrogen ($H$) atom has one valence electron and needs one more to complete its duplet.
Chlorine ($Cl$) has seven valence electrons and needs one more to complete its octet.
The carbon atom forms three single covalent bonds with three hydrogen atoms by sharing one of its valence electrons with the valence electron of each hydrogen atom. This leaves carbon with one unpaired electron:
The '?' marks the unpaired electron on carbon that is available for bonding. Now, the carbon atom can form a single covalent bond with the chlorine atom by sharing its unpaired electron with one of the chlorine's unpaired electrons, thus giving both the carbon and the chlorine a complete octet:
Each line (-- or |) represents a pair of shared electrons, i.e., a single covalent bond. The nature of these covalent bonds in $\mathrm{CH}_3\mathrm{Cl}$ is that the electrons are shared between atoms to attain a stable electronic configuration. The difference in electronegativity between carbon and hydrogen is small, so the sharing is quite equal, making these bonds nearly nonpolar. However, because chlorine is more electronegative than carbon, the bond between carbon and chlorine is polar, with a partial negative charge ($δ^-$) on the chlorine and a partial positive charge ($δ^+$) on the carbon. This difference in electronegativity leads to an uneven distribution of electron density in the bond, with the shared electrons being pulled closer to the chlorine atom.
Draw the electron dot structures for
(a) ethanoic acid.
(b) $\mathrm{H}_{2} \mathrm{~S}$.
(c) propanone.
(d) $\mathrm{F}_{2}$.
Here are the electron dot structures for the compounds you've asked:
(a) Ethanoic acid:
(b) $ \mathrm{H_2S} $:
(c) Propanone:
(d) $ \mathrm{F_2} $:
What is an homologous series? Explain with an example.
A homologous series is a series of compounds in which the same functional group substitutes for hydrogen in a carbon chain. These compounds differ from each other by a constant unit, commonly a $-CH_2-$ group. For example, the homologous series of alkanes starts with methane $CH_4$, followed by ethane $C_2H_6$, and propane $C_3H_8$, with each successive compound differing by a $-CH_2-$ unit from the previous compound .
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Ethanol and ethanoic acid can be differentiated based on their physical and chemical properties as follows:
Physical Properties:
Ethanol: Ethanol is a liquid at room temperature and is commonly known as alcohol. It is the active ingredient of all alcoholic drinks and is a good solvent. It is used in various medicines, such as tincture iodine, cough syrups, and many tonics. Furthermore, ethanol is soluble in water in all proportions.
Ethanoic Acid: Ethanoic acid, commonly known as acetic acid, is also a carboxylic acid. A 5-8% solution of acetic acid in water is called vinegar, which is used as a preservative in pickles. The melting point of pure ethanoic acid is 290 K (16.85°C), and it often freezes during winter in cold climates, which is why it is sometimes referred to as glacial acetic acid.
Chemical Properties:
Ethanol: One of the chemical properties of ethanol can be demonstrated through its psychoactive effects when consumed in small quantities, leading to drunkenness.
Ethanoic Acid: This acid has the typical acidic properties of carboxylic acids, and it is a weak acid unlike strong mineral acids.
These differing properties make it possible to distinctively identify ethanol and ethanoic acid.
Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Micelles form when soap is added to water due to the amphipathic nature of soap molecules. Each soap molecule has a hydrophilic (water-attracting) 'head' and a hydrophobic (water-repelling) 'tail'. When soap is dissolved in water, the hydrophobic tails avoid water and the hydrophilic heads interact with it. This leads to the formation of micelles which are spherical structures where the hydrophobic tails are oriented towards the center, shielded from water by the hydrophilic heads.
In water, this structure minimizes the unfavorable interaction between the hydrophobic tails of the soap and the polar water molecules. The hydrophilic heads, on the other hand, are in a favorable interaction with water on the micelle's exterior.
The formation of micelles depends on the nature of the solvent. In the case of ethanol, which is a less polar solvent compared to water, the soap may behave differently. Since ethanol is not as polar as water, the hydrophobic effect, which drives the formation of micelles in water, is not as strong in ethanol. Thus, micelles may not readily form in ethanol or other nonpolar solvents, because the hydrophobic tails of the soap molecules will not be as incompatible with the surrounding solvent, reducing the driving force for micelle formation.
However, it is important to note that the propensity for micelle formation also depends on other factors, such as the concentration of soap, temperature, and the specific properties of the soap and the solvent in question. At certain concentrations and conditions, it's possible that micelles or similar structures could form in ethanol, but they may have different characteristics than those formed in water.
Why are carbon and its compounds used as fuels for most applications?
Carbon and its compounds are used as fuels for most applications because they are highly stable. The strong bonds that carbon forms with other elements, specifically because of its small size, contribute to this stability. This enables the nucleus to hold onto the shared pairs of electrons strongly. The carbon-carbon bond is very strong and hence stable, which makes these compounds useful as fuels that can release energy when burnt. Moreover, carbon can form covalent bonds with a range of other elements like hydrogen, oxygen, sulphur, nitrogen, and chlorine, leading to a variety of compounds with specific properties useful in fuel applications. Carbon compounds with hydrogen, such as hydrocarbons, release a large amount of heat and light upon combustion, which is essential for a substance to be a good fuel .
Explain the formation of scum when hard water is treated with soap.
The formation of scum when hard water is treated with soap occurs because the soap reacts with calcium and magnesium salts, which cause the hardness of water. These salts react with soap to form an insoluble substance called scum. This happens because soap molecules, which are sodium or potassium salts of long-chain carboxylic acids, have one end that is ionic and interacts with water, while the other end, the hydrocarbon 'tail', is hydrophobic and interacts with oil. In hard water, the ionic-end of the soap reacts with calcium and magnesium ions, leading to the formation of a white curdy precipitate, which is the scum. As a result, a larger amount of soap is needed to create foam and effectively clean in hard water .
What change will you observe if you test soap with litmus paper (red and blue)?
If you test soap with litmus paper, you will observe the following changes:
Blue litmus paper will remain blue, indicating that soap is basic in nature.
Red litmus paper will turn blue, confirming the basicity of soap.
Soap is a mild alkali and thus causes red litmus paper to turn blue, which is a characteristic test for bases. It does not affect blue litmus paper because blue litmus paper is used to test for acidic substances, and it stays blue when in contact with a base.
What is hydrogenation? What is its industrial application?
Hydrogenation is a chemical reaction that involves the addition of hydrogen to another compound. In the context of organic chemistry, hydrogenation commonly refers to the addition of hydrogen to unsaturated hydrocarbons in the presence of a catalyst, such as palladium or nickel, to give saturated hydrocarbons. This is a type of addition reaction where the unsaturated bonds (double or triple bonds) of hydrocarbons become saturated by attaching hydrogen atoms.
An important industrial application of hydrogenation is the hydrogenation of vegetable oils using a nickel catalyst. Vegetable oils have long unsaturated carbon chains, and by the process of hydrogenation, they are converted into saturated hydrocarbons. This process is used to convert liquid vegetable oils into solid or semi-solid fats, like margarine and shortening, which are used in baking and cooking .
Which of the following hydrocarbons undergo addition reactions: $\mathrm{C}_{2} \mathrm{H}_{6}, \mathrm{C}_{3} \mathrm{H}_{8}, \mathrm{C}_{3} \mathrm{H}_{6}, \mathrm{C}_{2} \mathrm{H}_{2}$ and $\mathrm{CH}_{4}$.
Hydrocarbons that undergo addition reactions typically possess multiple bonds, such as double or triple bonds, between carbon atoms. This is because addition reactions involve breaking these multiple bonds and adding new atoms or groups to the carbon atoms involved.
Let's examine each of the given hydrocarbons:
$\mathrm{C}_{2}\mathrm{H}_{6}$ (ethane) - This is an alkane with a single bond between the carbon atoms, so it does not typically undergo addition reactions but rather substitution reactions.
$\mathrm{C}_{3}\mathrm{H}_{8}$ (propane) - This is also an alkane with only single bonds between carbon atoms, so it, too, does not typically undergo addition reactions.
$\mathrm{C}_{3}\mathrm{H}_{6}$ (propene) - This is an alkene with a double bond between two of the carbon atoms, so it can undergo addition reactions.
$\mathrm{C}_{2}\mathrm{H}_{2}$ (acetylene or ethyne) - This is an alkyne with a triple bond between the carbon atoms, so it can also undergo addition reactions.
$\mathrm{CH}_{4}$ (methane) - This is an alkane with only single bonds between the carbon atom and hydrogen atoms, so it does not typically undergo addition reactions.
Therefore, the hydrocarbons that can undergo addition reactions from the list provided are $\mathrm{C}_{3}\mathrm{H}_{6}$ (propene) and $\mathrm{C}_{2}\mathrm{H}_{2}$ (acetylene or ethyne).
Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
To differentiate between saturated and unsaturated hydrocarbons, you can use the bromine water test. Here's how it works:
Saturated hydrocarbons like alkanes do not react with bromine water, so the bromine water retains its original brownish-orange color.
Unsaturated hydrocarbons like alkenes and alkynes will react with bromine water, decolorizing it from brownish-orange to colorless.
The underlying chemical reaction involves the addition of bromine across the double or triple bonds in unsaturated hydrocarbons, which does not occur in saturated hydrocarbons because they have only single bonds and are thus fully saturated with hydrogen.
Explain the mechanism of the cleaning action of soaps.
The cleaning action of soaps is based on their ability to act as emulsifying agents, helping to disperse oil and grease in water. Soap molecules have two distinct parts: a hydrophobic (water-hating) end and a hydrophilic (water-loving) end. The hydrophobic end, which is the long carbon chain, is attracted to oils and grease, while the hydrophilic ionic end is attracted to water.
When soap is added to water containing dirt or grease, soap molecules surround the oil droplets. The hydrophobic tails of the soap molecules embed themselves in the grease, while the hydrophilic heads remain in the water. This creates a spherical structure known as a micelle, with the hydrophobic tails of the soap molecules tucked into the center away from the water, thereby trapping the grease and dirt.
As a result, the grease becomes emulsified in the water, forming smaller droplets surrounded by soap molecules. These droplets are then able to be washed away with the water, effectively removing the dirt or grease from the surface being cleaned. The formation of micelles also ensures that the removed dirt gets suspended in the water and does not redeposit onto the surface, allowing for efficient cleaning .
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Ask Chatterbot AIExtra Questions and Answers - Carbon and its Compounds | NCERT | Science | Class 10
How many bonds are there in $CO_2$ molecule?
A) $1$
B) $2$
C) $3$
D) $4$
The molecule carbon dioxide ($CO_2$) consists of one carbon atom double bonded to two oxygen atoms. Here is how the molecule is typically structured:
Carbon forms two double bonds, one with each of the two oxygen atoms.
Therefore, there are two double bonds in a $CO_2$ molecule.
Answer: B) $2$.
The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus, and 39% oxygen. Find the empirical formula of the compound."
To find the empirical formula of sodium phosphate, we start by determining the number of moles of each element in a 100g sample and then express these quantities as simple ratios.
Step-by-Step Analysis:
Break down the percentage composition:
Sodium (Na): 42.1%
Phosphorus (P): 18.9%
Oxygen (O): 39.0%
Divide each percentage by the atomic weight of the respective element to find the number of moles:
Atomic weights: Na = 23, P = 31, O = 16
$\text{Mo}(\text{Na}) = \frac{42.1}{23} = 1.83$
$\text{Mo}(\text{P}) = \frac{18.9}{31} = 0.61$
$\text{Mo}(\text{O}) = \frac{39.0}{16} = 2.44$
Obtain the simplest whole number ratio by dividing each result by the smallest moles value obtained, which in this case is 0.61 for Phosphorus (P):
$\text{Ratio}_{\text{Na}} = \frac{1.83}{0.61} = 3$
$\text{Ratio}_{\text{P}} = \frac{0.61}{0.61} = 1$
$\text{Ratio}_{\text{O}} = \frac{2.44}{0.61} = 4$
Combine the ratios to determine the empirical formula:
The simplest whole number ratios are 3 (Na), 1 (P), and 4 (O).
Conclusion:
The compound's empirical formula is deduced as $\textbf{Na}_3\textbf{PO}_4$ based on the provided analysis.
The electronic configuration of Phosphorus (atomic number 15) will be:
(A) $2,8,7$ (B) $2,8,6$ (C) $2,8,5$ (D) $2,8,4$
The correct option is (C):
$$ 2, 8, 5 $$
Phosphorus (P) with an atomic number of 15 can be arranged into its shells based on the maximum capacity of each shell. Typically, the maximum number of electrons that can occupy the:
$K$ shell (first shell) is 2,
$L$ shell (second shell) is 8, and
$M$ shell (third shell) begins with capacity for 18 electrons.
Given 15 electrons:
The $K$ shell is filled with 2 electrons.
The $L$ shell is occupied by 8 electrons.
The remainder, which is 5 electrons, will occupy the $M$ shell.
Thus, the electronic configuration of Phosphorus is represented by the arrangement 2, 8, 5.
Statement 1: Coal and coke are examples of crystalline forms of carbon.
Statement 2: Graphite and diamonds are examples of non-crystalline forms of carbon.
A) Statement 1 is true and statement 2 is false.
B) Statement 1 is false and statement 2 is true.
C) Both statements 1 and 2 are true.
D) Both statements 1 and 2 are false.
The correct option is D) Both statements 1 and 2 are false.
Carbon can be found in two primary forms based on its atomic arrangement:
Crystalline form: In this form, carbon atoms are arranged in a well-defined geometric pattern. Examples include diamond, graphite, and fullerenes.
Amorphous form: Here, the carbon atoms lack a well-defined geometric arrangement. Examples include coal, coke, and charcoal.
Thus, both given statements are incorrect because:
Statement 1 erroneously classifies coal and coke as crystalline forms of carbon.
Statement 2 mistakenly lists graphite and diamonds as non-crystalline forms of carbon.
Which of the following molecular formulas corresponds to ethyl butanoate ester? (a) $\mathrm{C}{5} \mathrm{H}{10} \mathrm{O}{2}$ (b) $\mathrm{C}{6} \mathrm{H}{12} \mathrm{O}{2}$ (c) $\mathrm{C}{7} \mathrm{H}{14} \mathrm{O}{2}$ (d) $\mathrm{C}{8} \mathrm{H}{18} \mathrm{O}{2}$
The correct molecular formula for ethyl butanoate ester is:
(b) $ \mathrm{C}6\mathrm{H}{12}\mathrm{O}_2 $
Ethyl butanoate is an ester formed from ethyl alcohol (ethyl group, $\mathrm{C}_2\mathrm{H}_5$) and butanoic acid (butyric acid, $\mathrm{C}_3\mathrm{H}_7\mathrm{COO}$). Combining these, the compound's molecular formula totals to:
Carbon atoms: $ 2 (\text{from } \mathrm{C}_2\mathrm{H}_5) + 3 (\text{from } \mathrm{C}_3\mathrm{H}_7\mathrm{COO}) + 1 (\text{from carbonyl CO}) = 6 $
Hydrogen atoms: $5 (\text{from } \mathrm{C}_2\mathrm{H}_5) + 7 (\text{from } \mathrm{C}_3\mathrm{H}_7\mathrm{COO}) = 12$
Oxygen atoms: $2 (\text{from the ester group } \mathrm{COO})$
Thus, the total molecular formula is $\mathrm{C}6\mathrm{H}{12}\mathrm{O}_2$.
The ____ is at a higher energy level in an alpha decay.
A) daughter nucleus
B) parent nucleus
C) alpha particle
The correct answer is B) parent nucleus.
In alpha decay, the reaction is spontaneous and one-way, indicating that the reactants (parent nucleus) must have more energy than the products (daughter nucleus and alpha particle). Therefore, the parent nucleus is at a higher energy level, enabling it to release energy and particles (like the alpha particle) during the decay process. This ensures irreversible flow of the decay process, confirming that the parent nucleus starts at a higher energy state.
The main source of carbon monoxide in the atmosphere is:
A) Vehicles
B) Industries
C) Bio wastes
D) Photochemical reactions in the troposphere.
The correct answer is Option D: Photochemical reactions in the troposphere.
Photochemical reactions in the troposphere are the principal source of carbon monoxide in the atmosphere. These reactions produce approximately $ 5 \times 10^{12} $ kilograms of carbon monoxide each year. Natural events like volcanoes and forest fires also contribute to atmospheric carbon monoxide levels.
Carbon monoxide is formed through the partial oxidation of carbon-containing compounds when there is insufficient oxygen to produce carbon dioxide. This often occurs during the incomplete combustion of fuels such as gasoline, natural gas, oil, coal, and wood. Among human activities, vehicle emissions represent the largest man-made source of carbon monoxide.
Which of the following is responsible for acid rain?
A Nitrogen oxide
B Sulphur dioxide
C Iron oxide
D Both Nitrogen oxide and Sulphur dioxide
The correct answer is D: Both Nitrogen oxide and Sulphur dioxide.
Fossil fuels often contain sulphur and nitrogen. When these fuels are burned, sulphur dioxide (SO2) and nitrogen oxides (NOx) are released into the atmosphere. These gases can then dissolve in rainwater to form sulphuric acid (H2SO4) and nitric acid (HNO3), respectively, which contribute to the formation of acid rain. Consequently, both nitrogen oxide and sulphur dioxide are responsible for acid rain.
Which gas in the atmosphere is primarily responsible for global warming?
A. Ozone
B. Sulphur dioxide
C. Carbon dioxide
D. Nitrous oxide
The correct answer is C. Carbon dioxide.
Global warming refers to the gradual increase in Earth's average surface temperature. The primary culprit behind this phenomenon is Carbon dioxide (CO$_2$). This gas is predominantly released through the burning of fossil fuels such as coal, oil, and natural gas. CO$_2$ is particularly effective at trapping heat within the atmosphere, thereby enhancing the greenhouse effect and contributing significantly to global warming.
How many electrons does a carbon atom share with another carbon atom in the formation of an acetylene molecule?
A Three
B Two
C One
D Four
The correct answer is D: Four.
Acetylene, also known as ethyne, is classified chemically as an alkyne. Alkynes are characterized by having a carbon-carbon triple bond. This triple bond is made up of one sigma bond and two pi bonds.
Acetylene is the simplest alkyne and contains two carbon atoms.
Each carbon in the molecule shares four electrons with the other carbon, comprising three electrons for the triple bond and one more through a sigma bond.
Additionally, each carbon atom shares its remaining electrons with a hydrogen atom, thus completing their valency.
Hence, each carbon shares four electrons with the other carbon in the formation of an acetylene molecule. This exemplifies a typical triple bond configuration in chemistry, where a stronger and shorter bond is formed through the sharing of multiple electrons between two atoms.
In a compound $\mathrm{C}, \mathrm{H}$, & $\mathrm{N}$ are present in the ratio of $18: 2: 7$ by weight. The empirical formula of the compound is:
(A) $\mathrm{C}_{2}\mathrm{H}_{6}\mathrm{N}_{2}$
(B) $\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{N}$
(C) $\mathrm{C}_{2}\mathrm{H}_{8}\mathrm{N}_{2}$
(D) $\mathrm{C}_{9}\mathrm{H}_{12}\mathrm{N}_{3}$
The correct option is (B) $\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{N}$
Given the ratio of weights of elements in the compound as $\textbf{C} : \textbf{H} : \textbf{N} = 18:2:7$. We then convert this weight ratio to a molar ratio by dividing each weight by the respective atomic masses of carbon (C, 12 g/mol), hydrogen (H, 1 g/mol), and nitrogen (N, 14 g/mol):
$$ \text{Molar ratio of } C:H:N = \frac{18}{12}:\frac{2}{1}:\frac{7}{14} = \frac{3}{2}:2:\frac{1}{2} $$
To simplify these fractions into whole numbers, multiply each term by 2:
$$ 3:4:1 $$
Therefore, the empirical formula based on the simplest whole number ratio of moles is
$$ \mathrm{C}_{3}\mathrm{H}_{4}\mathrm{N} $$
Thus, option (B) $\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{N}$ is the correct answer.
On heating anhydrous $\mathrm{Na}_{2} \mathrm{CO}_{3}$, _________ is evolved.
A) $\mathrm{CO}_{2}$
B) Water vapour
C) $\mathrm{CO}$
D) No gas
The correct answer is D) No gas.
Anhydrous $\mathrm{Na}_2\mathrm{CO}_3$, also known as soda ash, does not decompose upon heating. Since it is anhydrous, it contains no water molecules, and its stable chemical structure prevents it from releasing any gases like $\mathrm{CO}_2$ or $\mathrm{CO}$ at normal temperatures used for heating.
Which of the following is not a greenhouse gas?
a) Carbon dioxide
b) Sulphur dioxide
c) Methane
d) Nitrogen
Nitrogen is not a greenhouse gas. However, Carbon dioxide, Sulfur dioxide, and Methane are recognized as greenhouse gases.
Acid rain is due to an increase in atmospheric concentration of:
A) ozone and dust
B) $\mathrm{CO}_{2}$ and $\mathrm{CO}$
C) $\mathrm{SO}_{2}$ and $\mathrm{CO}$
D) $\mathrm{SO}_{2}$ and $\mathrm{NO}_{2}$
The correct answer is D) $\mathrm{SO}_{2}$ and $\mathrm{NO}_{2}$.
Sulfur dioxide ($\mathrm{SO}_{2}$) and nitrogen dioxide ($\mathrm{NO}_{2}$) are primary pollutants contributing to the formation of acid rain. When present in the atmosphere in high concentrations, these gases can dissolve in water vapor. They subsequently form sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$) and nitric acid ($\mathrm{HNO}_3$) as they combine with water. These acidic compounds are then carried by rainwater, leading to what is commonly known as acid rain. This phenomenon can cause significant damage to ecosystems, including forests and vegetation.
Which amongst the following is not an allotrope of carbon?
A) Carbon dioxide
B) Fullerene
C) Diamond
D) Graphite
The correct answer is A) Carbon dioxide.
Allotropes refer to different physical forms in which an element can exist in the same physical state. Well-known allotropes of carbon include diamond, graphite, and fullerene.
However, carbon dioxide $\left(\mathrm{CO}_{2}\right)$ is not an allotrope, but a chemical compound consisting of carbon and oxygen. Thus, option A is the one that is not an allotrope of carbon.
Why is graphite used to make electrodes?
Graphite is employed in the creation of electrodes primarily due to its ability to conduct electricity. This electrical conductivity arises because of significant electron delocalization within its carbon layers, which is referred to as aromaticity. The valence electrons in graphite are not bound to any specific atom but are free to move across the structure. This freedom allows graphite to conduct electricity efficiently, making it an ideal choice for electrodes.
Venus is the hottest planet because of the presence of carbon dioxide and clouds.
A. Methane
B. Helium
C. Carbon dioxide
D. Sulphuric acid
The correct option is D. Sulphuric acid
Venus is not the closest planet to the sun but is still the hottest.
This exceptional heat is due to Venus's thick atmosphere, which efficiently traps sunlight.
The atmospheric thickness is attributed to both clouds of carbon dioxide and sulphuric acid.
$10 , \mathrm{dm}^{3}$ of $\mathrm{N}_{2}$ gas and $10 , \mathrm{dm}^{3}$ of gas $X$ at the same temperature contain the same number of molecules and have the same mass. The gas $X$ is:
A $\mathrm{CO}_{2}$
B $\mathrm{CO}$
C $\mathrm{H}_{2}$
D $\mathrm{N}$
The correct answer is Option B: $\mathrm{CO}$.
Given that both $10 , \mathrm{dm}^{3}$ of $\mathrm{N}_{2}$ and $10 , \mathrm{dm}^{3}$ of gas $X$ at the same temperature contain the same number of molecules, and also weigh the same, we can conclude that both have identical molar masses.
Key Information Given:
Equal volumes under the same conditions implies equal number of molecules (due to Avogadro's Law).
Equal mass implies equal total molar mass.
Key Calculation:The molar mass of $\mathrm{N}_2$ is given by: $$ \text{Molecular weight of } \mathrm{N}_2 = 28 , \mathrm{amu} $$
Since the number of molecules and masses are equal, gas $X$ must also have the same molar mass: $$ \text{Molar mass of gas } X = 28 , \mathrm{amu} $$
Comparing this with the molar mass of various gas choices:
$\mathrm{CO}_2$: approximately 44 amu,
$\mathrm{CO}$: exactly 28 amu,
$\mathrm{H}_2$: approximately 2 amu,
$\mathrm{N}$ (atomic nitrogen, unusual in diatomic form would be $\mathrm{N}_2$): 14 amu (but typically exists as $\mathrm{N}_2$).
Conclusion:The gas $X$ that matches both the molecular weight and other given conditions is $\mathrm{CO}$. Hence, the answer is Option B: $\mathrm{CO}$.
Carbon and hydrogen are detected by heating the organic compound with:
A. Magnesium oxide
B. Cupric oxide
C. Zinc oxide
D. Sulphur dioxide
The correct answer is B. Cupric oxide.
Cupric oxide (CuO) plays a pivotal role in detecting carbon and hydrogen in organic compounds, a process fundamental to qualitative analysis. When an organic compound is heated strongly with cupric oxide, it aids in the oxidation of carbon to carbon dioxide (CO2) and hydrogen to water (H2O).
For Carbon: The reaction can be summarized as: $$ \mathrm{C} + 2 \mathrm{CuO} \xrightarrow{\Delta} \mathrm{CO}_2 + 2 \mathrm{Cu} $$ The produced CO2 can further be confirmed by passing it through lime water (calcium hydroxide solution), where a formation of calcium carbonate precipitate indicates the presence of carbon dioxide: $$ \mathrm{CO}_2 + \mathrm{Ca(OH)}_2 \rightarrow \mathrm{CaCO}_3 + \mathrm{H}_2O $$
For Hydrogen: The detection involves the reaction: $$ 2 \mathrm{H} + \mathrm{CuO} \xrightarrow{\Delta} \mathrm{H}_2O + \mathrm{Cu} $$ The water formed can subsequently be tested using anhydrous copper sulfate, which turns from white to blue in the presence of water: $$ \mathrm{CuSO}_4 + 5 \mathrm{H}_2O \rightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2O $$
Thus, through these chemical transformations and subsequent confirmatory tests, both carbon and hydrogen are efficiently detected in the given organic compound with the aid of cupric oxide.
Which of the following is most electronegative?
A) Carbon
B) Silicon
C) Lead
D) Tin
The correct answer is A) Carbon.
Electronegativity refers to the ability of an atom to attract shared electrons in a chemical bond. Among the elements listed, carbon is the most electronegative. This is because electronegativity typically decreases as you move down a group in the periodic table. Silicon, tin, and lead are situated below carbon in Group 14, thus exhibiting progressively lower electronegativities compared to carbon.
Which element in the modern periodic table is similar to eka-silicon?
A) Phosphorus
B) Carbon
C) Germanium
D) Aluminium
The correct answer is C) Germanium.
Dmitri Mendeleev, while developing his periodic table, predicted the existence of several elements that had not yet been discovered. He left placeholders for these elements, referring to them by prefixing with "eka" (meaning one beyond in Sanskrit) the name of the element above it in the group.
One such element he predicted was eka-silicon. Mendeleev described properties for eka-silicon that matched fairly closely with the element Germanium when it was discovered later. Therefore, in the modern periodic table, Germanium is the element that corresponds to Mendeleev's eka-silicon.
Write down the names of compounds represented by the following formulae: $\mathrm{CaCl}_{2}$
In the chemical formula $\mathrm{CaCl}_{2}$, $\mathrm{Ca}$ represents calcium and $\mathrm{Cl}$ represents the chloride ion. Since calcium is the cation and chloride is the anion in this formula, and remembering that the name of the cation precedes the anion in naming, the compound is thus named Calcium chloride.
The correct order of ionization energy for comparing carbon, nitrogen, and oxygen atoms is
A) $\quad \mathrm{C} > \mathrm{N} > \mathrm{O}$
B) $\mathrm{C} > \mathrm{N} < \mathrm{O}$
C) $\mathrm{C} < \mathrm{N} > \mathrm{O}$
D) $\mathrm{C} < \mathrm{N} < \mathrm{O}$
The correct option is C: $$ \mathrm{C} < \mathrm{N} > \mathrm{O} $$
Ionization energy typically increases as you move from left to right across a period in the periodic table. This trend occurs because the atomic radius generally decreases, which means the outer electrons are closer to the nucleus and more tightly bound to it. Therefore, more energy is required to remove an electron. As per the trend, one would expect: $$ \mathrm{C} < \mathrm{N} < \mathrm{O} $$
However, nitrogen ($\mathrm{N}$) possesses a half-filled $p$ orbital configuration ($2p^3$), which is notably more stable due to symmetry and exchange energy considerations. This half-filled structure makes electrons in nitrogen less willing to be removed compared to the usual trend, leading to an unusually high ionization energy for nitrogen.
Thus, due to this additional stability in nitrogen:
Carbon ($\mathrm{C}$), having a $2p^2$ configuration, is less stable compared to the half-filled configuration of nitrogen.
Oxygen ($\mathrm{O}$), having a $2p^4$ configuration, experiences inter-electron repulsion that slightly reduces its ionization energy compared to nitrogen.
As a result, the exception in the general trend due to the half-filled stability of nitrogen leads to the order: $$ \mathrm{C} < \mathrm{N} > \mathrm{O} $$ Thus, confirming that Option C is correct.
Name an allotrope of carbon which contains both single and double bonds between carbon atoms.
A) Graphite
B) Diamond
C) Fullerene
D) No allotrope meets such a condition.
The correct choices are B: Graphite and C: Fullerene.
In diamond, each carbon atom has single bonds with four other carbon atoms, forming a tetrahedral lattice. This structure requires significant energy to disrupt due to the strong covalent bonding.
On the other hand, in graphite, each carbon atom is covalently bonded to three other atoms in planar layers. This differing bond structure includes both single and double carbon bonds, which facilitates electrical conductivity along the planes.
Fullerenes are another intricate carbon allotrope, characterized by rings containing both five and six carbon atoms. These molecules exhibit a mesh-like structure composed of single and double bond arrangements, forming closed or partially closed spheres or tubes.
Thus, both graphite and fullerenes display the presence of single and double carbon bonds, making options B and C correct.
In Duma's method for estimation of nitrogen, $0.25~\mathrm{g}$ of an organic compound gave 40mL of nitrogen collected at 300K temperature and 725mm pressure. If the aqueous tension at 300K is 25 mm, the percentage of nitrogen in the compound is :
A) 15.76
B) 17.36
C) $\mathbf{18.20}$
D) 16.76
The correct option is C) $18.20$
Given:
Mass of the organic compound = $0.25 , \text{g}$
Volume of nitrogen collected = $40 , \text{mL}$
Temperature = $300 , \text{K}$
Total pressure = $725 , \text{mm}$ of mercury
Aqueous tension at $300 , \text{K}$ = $25 , \text{mm}$ of mercury
First, correct the total pressure for the aqueous tension: $$ P_1 = 725 , \text{mm} - 25 , \text{mm} = 700 , \text{mm} $$
We use the ideal gas equation to convert the volume of nitrogen collected to the conditions at Standard Temperature and Pressure (STP - $273 , \text{K}$ and $760 , \text{mm}$ of mercury): $$ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} $$ Where:
$P_1 = 700 , \text{mm}$ (corrected pressure)
$V_1 = 40 , \text{mL}$ (volume of nitrogen)
$T_1 = 300 , \text{K}$ (temperature of nitrogen)
$P_2 = 760 , \text{mm}$ (pressure at STP)
$T_2 = 273 , \text{K}$ (temperature at STP)
Solving for $V_2$ (volume at STP): $$ V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{700 \times 40 \times 273}{300 \times 760} = 33.52 , \text{mL} $$
We calculate the mass of nitrogen by using its proportionate weight at STP where $22,400 , \text{mL}$ of nitrogen weighs $28 , \text{g}$: $$ \text{Weight of } N_2 = \frac{28 \times 33.52}{22400} = 0.0419 , \text{g} $$
Finally, we calculate the percentage of nitrogen in the organic compound: $$ \text{Percentage of nitrogen} = \left(\frac{0.0419}{0.25}\right) \times 100 = 16.76% $$
Therefore, the correct answer is D) $16.76$.
The product obtained on passing excess carbon dioxide through lime water is
A) $\mathrm{CaCO}_{3}$
B) $\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$
C) $\mathrm{CaHCO}_{3}$
D) $\mathrm{Ca}_{2}\mathrm{CO}_{3}$
The correct answer is B) $ \mathrm{Ca}(\mathrm{HCO}_3)_2 $.
Lime water is essentially an aqueous solution of calcium hydroxide, denoted by $ \mathrm{Ca(OH)}_2 $. Initially, when carbon dioxide ($ \mathrm{CO}_2 $) is passed through lime water, it reacts to form calcium carbonate ($ \mathrm{CaCO}_3 $), which causes the solution to turn milky due to the formation of a precipitate.
However, upon passing excess $ \mathrm{CO}_2 $, the previously formed precipitate of $ \mathrm{CaCO}_3 $ begins to dissolve, leading to the formation of calcium bicarbonate, represented by $ \mathrm{Ca}(\mathrm{HCO}_3)_2 $. This chemical reaction can be represented as:
$$ \mathrm{Ca(OH)_2} + \mathrm{H}_2\mathrm{O} + 2\mathrm{CO}_2 \rightarrow \mathrm{Ca}(\mathrm{HCO}_3)_2 $$
This transformation explains why option B) $ \mathrm{Ca}(\mathrm{HCO}_3)_2 $ is the correct product obtained when an excess of carbon dioxide is passed through lime water.
Decay of organic matter releases the gas into the air:
A. Nitrogen B. Ammonia C. Carbon dioxide D. None
C) Carbon dioxide
Decomposition of organic matter includes the process where organic material is progressively broken down, eventually releasing several gases. Among the gases released during the decomposition of organic matter, carbon dioxide (CO₂) is a primary component. This process is influenced by various factors and occurs through multiple key steps:
1. Photo-oxidation: This involves the breakdown due to exposure to light, along with leaching, comminution (breakdown into smaller pieces), and mineralization. Organisms such as arthropods play a crucial role in comminution and mineralization.
2. Decomposition: This is commonly assessed by measuring the respiration rate, which involves the conversion of organic matter back into carbon dioxide. The decomposition rate varies with environmental conditions, generally being higher in moist (mesic) environments compared to dry (arid) ones. Decomposition can be quantitatively described using models that depict it as a function declining exponentially over time.
3. Comminution: This step increases the surface area of detrital materials, enhancing microbial colonization and subsequent decay, though not all organic material is converted directly to CO₂ during the decomposition.
Thus, among the options, carbon dioxide is indeed the gas prominently released during the decay of organic matter.
Which decomposes on heating?
(A) $\mathrm{K}_{2} \mathrm{CO}_{3}$
(B) $\mathrm{Rb}_{2} \mathrm{CO}_{3}$
(C) $\mathrm{Na}_{2} \mathrm{CO}_{3}$
(D) $\mathrm{MgCO}_{3}$
The correct option is (D) $\mathbf{MgCO}_{3}$.
Magnesium carbonate ($\mathrm{MgCO}_{3}$) decomposes upon heating. This decomposition is due to the high polarizing power of the $\mathrm{Mg}^{2+}$ ion, which destabilizes the carbonate ion ($\mathrm{CO}_{3}^{2-}$), leading to the release of carbon dioxide ($\mathrm{CO}_{2}$) and the formation of magnesium oxide ($\mathrm{MgO}$). The reaction can be represented as:
$$ \mathrm{MgCO}{3} \rightarrow \mathrm{MgO} + \mathrm{CO}{2} $$
The compound that has the least heat of decomposition among the following is
A) $\text{BeCO}_{3}$
B) $\text{MgCO}_{3}$
C) $\text{CaCO}_{3}$
D) $\text{SrCO}_{3}$
The correct answer is Option A: $\text{BeCO}_3$.
Beryllium carbonate ($\text{BeCO}_3$) is the least stable among the given carbonates. This lack of stability is primarily due to the polarization effects of the large carbonate ion ($\text{CO}_3^{2-}$) by the small $\text{Be}^{2+}$ ion. The small cation can polarize the larger anion, leading to a higher covalent character in the compound. As a result, $\text{BeCO}_3$ can decompose more easily compared to other listed carbonates, implying it has the least heat of decomposition.
If the diameter of a carbon atom is $0.15 \mathrm{~nm}$, calculate the number of carbon atoms that can be placed side by side in a straight line across a scale with a length of 20 cm.
Conversion Factors: The conversion between meters and centimeters is: $$ 1 \text{ m} = 100 \text{ cm} \quad \text{and} \quad 1 \text{ cm} = 10^{-2} \text{ m} $$
Length of the Scale: The provided scale length is $20 \text{ cm}$. Converting this to meters: $$ 20 \text{ cm} = 20 \times 10^{-2} \text{ m} = 0.2 \text{ m} $$
Diameter of a Carbon Atom: The given diameter is $0.15 \text{ nm}$. Converting nanometers to meters: $$ 0.15 \text{ nm} = 0.15 \times 10^{-9} \text{ m} $$
Each carbon atom takes $0.15 \times 10^{-9} \text{ m}$ of space.
Calculating the Number of Carbon Atoms: To find out how many atoms can fit in a straight line across the scale, we divide the total length of the scale by the diameter of one carbon atom. Therefore: $$ \text{Number of carbon atoms} = \frac{0.2 \text{ m}}{0.15 \times 10^{-9} \text{ m}} $$ This simplifies to: $$ \text{Number of carbon atoms} = \frac{0.2}{0.15 \times 10^{-9}} = 1.33 \times 10^{9} $$
Conclusion: Approximately $1.33 \times 10^{9}$ carbon atoms can be placed side by side in a straight line across a scale that is 20 cm long.
Two or more elements that are chemically bound together in a fixed ratio is known as a:
Compound
Homogeneous Mixture
Mixture
The correct option is A) Compound.
When two or more elements chemically combine in a fixed ratio by mass, the resultant product is known as a compound.
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