15) Of the following, __________ is a weak acid. A) HF B) HCl C) HBr D) HNO3
E) HCIO4

At a temperature of approximately 353.54 K, the reaction will have a ΔG of 0, indicating that it is at equilibrium.

For a reaction to have a ΔG (Gibbs free energy change) of 0, it must be at equilibrium. To determine the temperature at which this occurs, we can use the following equation:
ΔG = ΔH - TΔS
In this case, ΔH = -35 kJ and ΔS = -99 J/K. First, let's convert ΔH to J by multiplying by 1000:
ΔH = -35,000 J
Now, we can rewrite the equation with the given values:
0 = -35,000 J - T(-99 J/K)
To solve for the temperature (T), first isolate T by adding 35,000 J to both sides of the equation:
35,000 J = 99 J/K * T
Now, divide both sides by 99 J/K to find the temperature:
T ≈ 35,000 J / 99 J/K ≈ 353.54 K
At a temperature of approximately 353.54 K, the reaction will have a ΔG of 0, indicating that it is at equilibrium.

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From the choices below, the major force controlling tertiary protein structure are hydrogen bonding, disulfide bonds and ion pairs hydrophobic effect.

Protein tertiary structure is the three dimensional shape of a protein. The tertiary structure will have a single polypeptide chain "backbone" with one or more protein secondary structures, the protein domains. Amino acid side chains may interact and bond in a number of ways.

Important to tertiary structure are hydrophobic interactions, in which amino acids with nonpolar, hydrophobic R groups cluster together on the inside of the protein, leaving hydrophilic amino acids on the outside to interact with surrounding water molecules.

Therefore, From the choices below, the major force controlling tertiary protein structure are hydrogen bonding, disulfide bonds and ion pairs hydrophobic effect.

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Yes, a substance can have multiple routes of exposure. The route of exposure can affect the way in which a substance affects the body.

Routes of exposure refer to the ways in which a substance can enter the body. These can include:

1. Inhalation: breathing in the substance through the lungs

2. Ingestion: swallowing the substance

3. Dermal contact: skin contact with the substance

4. Injection: injection of the substance into the body through a needle or other means

Some substances can have multiple routes of exposure. For example, chemicals used in industrial processes can be inhaled as a gas or a vapor, ingested through contaminated food or water, or absorbed through the skin.

For example, inhalation of certain substances may cause respiratory problems, while ingestion of the same substance may affect the digestive system.

Therefore, it is important to consider all possible routes of exposure when evaluating the potential health effects of a substance.

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The structure of acetanilide is:

H H

| |

C=O-NH-C6H5

Acetanilide is an organic compound that contains an amide functional group (-CONH-) and a phenyl group (C6H5) attached to the nitrogen atom of the amide group.

Here is the method of Recrystallization

The first step is to drawing the structure of acetanilide is to draw the amide functional group, which consists of a carbonyl group (C=O) and an amine group (NH) attached to a central carbon atom.

The second step is to attach the phenyl group to the nitrogen atom of the amide group, replacing one of the hydrogen atoms.

This results in the final structure of acetanilide, where the phenyl group is attached to the nitrogen atom of the amide group via a single bond.

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When using sugar substitutes in baking, it should be substituted by weight instead of by volume.

When substituting sugar with a sugar substitute, it's important to keep in mind that sugar substitutes are often much sweeter than sugar, so a little goes a long way. Therefore, it's best to measure them by weight instead of volume to ensure accuracy. This is especially important when baking, as the amount of sugar can affect the texture, rise, and overall outcome of the baked goods. Measuring by weight also helps to avoid any inconsistencies that may arise from measuring by volume, such as settling, air pockets, or variations in the scoop size. To determine the correct amount of sugar substitute to use, consult the product's packaging for conversion information, or use a conversion chart to find the equivalent weight of sugar for the amount called for in the recipe.

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The 10. 00 g sample contains 7. 494 g C and 1. 260 g H. 1.25g is the mass of oxygen in gram of oxygen are in the carbohydrate sample.

A body's mass is an inherent quality. Prior to the discoveries of the atom or particle physics, it was widely considered to be tied to the amount of matter within a physical body. It was discovered that, despite having the same quantity of matter in theory, different atoms and elementary particles have varied masses. There are various conceptions of mass in contemporary physics that are theoretically different but physically equivalent.

Mass of sample = mass of carbon + mass of Hydrogen + mass of oxygen

10. 00=  7. 494 +  1. 260 + mass of oxygen

mass of oxygen= 1.25g

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One of the compounds present in carnauba wax was isolated, purified, and then treated with aqueous sodium hydroxide to yield an alcohol with 30 carbon atoms and a carboxylate ion with 20 carbon atoms. The likely structure of the compound is a fatty acid ester.

Carnauba wax is a mixture of esters derived from fatty acids and fatty alcohols. By isolating and purifying one of the compounds present in carnauba wax, it is possible to obtain a single ester molecule.

When this ester is treated with aqueous sodium hydroxide, it undergoes saponification, which involves breaking down the ester into its corresponding carboxylic acid and alcohol components.
The alcohol component has 30 carbon atoms, indicating that it is a long-chain fatty alcohol. The carboxylate ion component has 20 carbon atoms, indicating that it is a long-chain fatty acid. The original ester must have had 30 carbon atoms in its alcohol component and 20 carbon atoms in its acid component.
Therefore, the likely structure of the compound is a fatty acid ester with a 30-carbon alcohol component and a 20-carbon acid component.
The given information suggests that the compound in question is a fatty acid ester, with a 30-carbon alcohol component and a 20-carbon acid component, derived from carnauba wax.

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Answer: The molecules in hot water move faster and are slightly further apart.

Explanation:

There's more space between the molecules, the volume of hot water has fewer molecules in it and weighs a little bit less than the same volume of cold water. So hot water is less dense than cold water.

To determine the molar concentration of H3O+ ions in cola with a pH of 4.240, we can use the relationship between pH and H3O+ concentration.

pH is defined as the negative logarithm (base 10) of the H3O+ concentration:

pH = -log[H3O+]

Rearranging the equation, we can express the H3O+ concentration in terms of pH:

[H3O+] = 10^(-pH)

Substituting the given pH value:

[H3O+] = 10^(-4.240)

Using a calculator, we can evaluate this expression:

[H3O+] ≈ 4.08 × 10^(-5) mol/L

Therefore, the molar concentration of H3O+ ions in the cola is approximately 4.08 × 10^(-5) mol/L.

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True. When hydrogen is combined with a metal, its oxidation number is generally +1.

When hydrogen is combined with a metal, it usually forms an ionic compound in which hydrogen has an oxidation number of +1. This is because hydrogen is less electronegative than most metals, and therefore tends to lose its electron to become a positively charged ion (H+).

The metal, on the other hand, tends to gain the electron from hydrogen to become a negatively charged ion. For example, in the compound sodium hydride (NaH), hydrogen has an oxidation number of +1, while sodium has an oxidation number of -1.

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The Keq for this reaction at 298 K is 3.3.

To calculate the Keq for this reaction, we use the equation:

Keq = ([NO]^2[O3])/[NO2]^2

We can use the initial and equilibrium concentrations of NO and O3 to find the concentration of NO2 at equilibrium:

2NO(g) + O3(g) -> 2NO2(g)

Initially, [NO2] = 0 M. At equilibrium, we can use the stoichiometry of the reaction to find that:

[NO2] = (0.70 M - 0.28 M)/2 = 0.21 M

Substituting the concentrations into the Keq equation, we get:

Keq = ([0.28 M]^2[1])/[0.21 M]^2 = 3.3

Therefore, the Keq for this reaction at 298 K is 3.3.

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Waves travel quickly in a solid because the molecules are closely packed and physically bonded together.

This allows for efficient energy transfer, resulting in faster wave propagation. In liquids and gases, waves travel through a medium by causing the molecules to vibrate or oscillate back and forth. As the wave moves through the medium, it transfers energy to neighboring molecules, which in turn transfer the energy to their neighboring molecules, and so on. This transfer of energy from molecule to molecule creates a wave that propagates through the medium. so, Waves travel quickly in a solid because the molecules are closely packed and physically bonded together. Waves travel the quickest in solids because the molecules in solids are closely packed and physically bonded together. This allows the wave to transfer energy quickly through the material.

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The equilibrium constant has a definite value for every reversible reaction at a particular temperature. However, it varies with change in temperature. The equilibrium constant is independent of the initial concentration of reactants. Here the equilibrium constant is 77. The correct option is C.

Equilibrium constant can be expressed in terms of the partial pressures of the reactants and products. When the partial pressures are used, then the equilibrium constant is Kp.

Kp for the reverse reaction is:

Kp' = (pBr2) × (pNO)² /p(NOBr)²

Kp' = 1/Kp = 1/0.013 =  76.9 ≈ 77

Thus the correct option is C.

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The biggest risk of mercury if it spills is its toxicity. To avoid you should: option a) option all windows and doors to the outside and option c) it can't be cleaned from certain surfaces.

Mercury can cause harm to the nervous system, kidneys, and other organs. If a spill occurs, it is important to immediately open all windows and doors to the outside to increase ventilation and decrease exposure. It is crucial to avoid allowing children to help clean up the spill as they may not understand the potential danger and could be exposed to the mercury. It is also important to note that mercury can not be cleaned up easily from certain surfaces such as carpet, fabrics, and porous materials. In these cases, it may be necessary to seek professional help in cleaning up the spill.
The biggest risk of mercury spills is mercury vapor exposure, which can be harmful to health. If mercury spills, you should:

a. Open all windows and doors to the outside: This helps to ventilate the area and reduce the concentration of mercury vapor in the air.

c. Mercury cannot be cleaned up easily from certain surfaces: It's important to know that mercury can be challenging to clean up from porous surfaces, such as carpet, fabric, or wood. In such cases, it may be necessary to remove and dispose of the contaminated materials properly.

Please note that option b (allowing children to help you clean up the spill) is incorrect, as it's crucial to keep children and pets away from the spill area to prevent exposure.

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NH3 is a Bronsted-Lowry base because it can accept a proton (H+) to form NH4+.

HClO is a Bronsted-Lowry acid because it can donate a proton (H+) to form ClO-.

NH4+ is a Bronsted-Lowry acid because it can donate a proton (H+) to form NH3.

ClO- is a Bronsted-Lowry base because it can accept a proton (H+) to form HClO.

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The balanced equation for the neutralization reaction between oxalic acid (H₂C₂O₄) and potassium hydroxide is: H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O

In this reaction, the acid (H₂C₂O₄) reacts with the base (KOH) to form a salt (K₂C₂O₄) and water (H₂O). The coefficients in the balanced equation indicate that one mole of oxalic acid reacts with two moles of potassium hydroxide to form one mole of potassium oxalate and two moles of water.

The steps to write the balanced equation of oxalic acid (H₂C₂O₄) and potassium hydroxide are

1. Write the unbalanced equation: H₂C₂O₄ + KOH → K₂C₂O₄ + H₂O.

2. Balance the potassium atoms by adding a coefficient of 2 in front of KOH: H₂C₂O₄ + 2KOH → K₂C₂O₄ + H₂O.

3. Check that all other atoms are balanced: 2 hydrogen atoms from oxalic acid and 2 hydrogen atoms from the two potassium hydroxide molecules combine to form 2 water molecules, and the carbon and oxygen atoms are also balanced. So the balanced equation for the neutralization reaction between oxalic acid (H₂C₂O₄) and potassium hydroxide is H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O.

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The carbon skeletons of amino acids Processed in Catabolism by degradation.

Amino acids are grouped according to their major degradative end product. Some amino acids are listed more than once because different parts of their carbon skeletons are degraded to different end products. The figure shows the most important catabolic pathways in vertebrates, but there are minor variations among vertebrate species.

Threonine, for instance, is degraded via at least two different pathways and the importance of a given pathway can vary with the organism and its metabolic conditions.

The glucogenic and ketogenic amino acids are also delineated in the figure, by color shading.  The amino acids degraded to pyruvate are also potentially ketogenic. Only two amino acids, leucine and lysine, are exclusively ketogenic.

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Growing rice results in the release of methane (CH4) into the atmosphere. Option A is correct.

Rice cultivation is a major source of methane emissions globally. Methane is a potent greenhouse gas, with a global warming potential more than 25 times greater than carbon dioxide over a 100-year time horizon.

Methane is produced during the anaerobic decomposition of organic matter in flooded rice paddies, where oxygen is limited. Rice plants also release methane from their roots through a process called methanogenesis, which is facilitated by certain types of bacteria that live in the soil.

In addition to rice cultivation, methane is also produced by livestock, natural gas and oil production, and landfills. Reducing methane emissions is an important strategy for mitigating climate change, as methane has a significant impact on the Earth's radiative balance and contributes to the warming of the planet.

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True.  Phosphorylation can be uncoupled from electron flow by agents that dissipate the proton gradient

Phosphorylation is the process by which ATP is synthesized from ADP and inorganic phosphate. This process occurs in the mitochondria of eukaryotic cells and is coupled to electron flow through the electron transport chain (ETC). As electrons are passed along the ETC, protons are pumped across the inner mitochondrial membrane, creating a proton gradient. This proton gradient is essential for the production of ATP by ATP synthase, as it drives the rotation of the enzyme's rotor, which in turn drives the synthesis of ATP from ADP and inorganic phosphate.

However, agents that dissipate the proton gradient can uncouple phosphorylation from electron flow. One such agent is the chemical dinitrophenol (DNP), which can shuttle protons across the inner mitochondrial membrane, effectively short-circuiting the proton gradient. As a result, ATP synthase can no longer use the proton gradient to drive ATP synthesis, and phosphorylation is uncoupled from electron flow. Instead, the energy released by electron flow is dissipated as heat.

Overall, phosphorylation is tightly coupled to electron flow through the ETC, but this coupling can be disrupted by agents that dissipate the proton gradient.

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There are four stereoisomers of 4-chloro-2-methylpentane that exist. Stereoisomers are molecules that have the same molecular formula, but a different arrangement of atoms in space due to the presence of one or more chiral centers. The correct option is d.

In this case, there is one chiral center in the molecule, which is the carbon atom bonded to the chlorine atom.
The four possible stereoisomers can be identified by assigning priority to the four different groups attached to the chiral center based on atomic number.

The lowest priority group, in this case, is the hydrogen atom. The remaining three groups are the methyl group, the ethyl group, and the chlorine atom.
Starting with the highest priority group, the methyl group, and tracing a path through the other two groups, we can determine the configuration of the chiral center. If the path is clockwise, the configuration is labeled R. If the path is counterclockwise, the configuration is labeled S.
Using this method, we can determine that there are two possible R stereoisomers and two possible S stereoisomers of 4-chloro-2-methylpentane. Therefore, the total number of stereoisomers is four (2R,3S-4-chloro-2-methylpentane, 3R,2S-4-chloro-2-methylpentane, 2S,3R-4-chloro-2-methylpentane, and 3S,2R-4-chloro-2-methylpentane).

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The pH of a buffer prepared by combining 50.0 mL of 1.00 M ammonia and 50.0 mL of 1.00 M ammonium nitrate is 4.74 . Option (c).

To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa (or pKb) of the weak acid (or base) and the ratio of the concentrations of the weak acid (or base) and its conjugate base (or acid):

pH = pKb + log([conjugate acid]/[weak base])

We can first calculate the pKb of ammonia using its Kb:

Kb = [NH4+][OH-]/[NH3] = 1.8 x 10^-5

pKb = -log(Kb) = 4.74

Next, we can calculate the concentrations of ammonia and ammonium ion in the buffer solution:

[ammonia] = (1.00 M)(50.0 mL)/(100.0 mL) = 0.50 M

[ammonium ion] = (1.00 M)(50.0 mL)/(100.0 mL) = 0.50 M

The ratio of [ammonium ion] to [ammonia] is 1:1, so we can substitute these values into the Henderson-Hasselbalch equation:

pH = pKb + log([conjugate acid]/[weak base])

pH = 4.74 + log(0.50/0.50)

pH = 4.74

Therefore, the pH of the buffer solution is 4.74, which corresponds to answer choice C).

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The molar solubility of AgBr in the 3.0 x [tex]10^{-2[/tex] M [tex]AgNO_3[/tex] solution is approximately 1.67 x 10^-11 M.

To calculate the molar solubility of AgBr in a 3.0 x [tex]10^{-2[/tex] M [tex]AgNO_3[/tex] solution, we can use the solubility product constant (Ksp) and the concept of an ion product (Q). The solubility product constant is a measure of how soluble a substance is in a given solvent, and the ion product is the product of the concentrations of ions in a solution.
For AgBr, the dissolution reaction can be represented as:
AgBr(s) ⇌ [tex]Ag^+[/tex](aq) +[tex]Br^-[/tex](aq)
The Ksp of AgBr is 5.0 x [tex]10^{-14[/tex].
In the presence of 3.0 x [tex]10^{-2[/tex] M [tex]AgNO_3[/tex], the concentration of Ag+ ions is increased. The solubility equilibrium will shift to minimize this change according to Le Chatelier's principle. Let's represent the molar solubility of AgBr as "s." Since the [tex]AgNO_3[/tex] solution already has a concentration of 3.0 x [tex]10^{-2[/tex] M, the Ag+ concentration is 3.0 x [tex]10^{-2[/tex] + s. The Br- concentration remains as "s."
Now, we can set up an equation using Ksp and Q:
Ksp = [Ag+][Br-]
5.0 x [tex]10^{-13[/tex] = (3.0 x [tex]10^{-2[/tex] + s)(s)
Since s is very small compared to 3.0 x [tex]10^{-2[/tex], we can approximate the equation as:
5.0 x [tex]10^{-13[/tex] ≈ (3.0 x [tex]10^{-2[/tex])(s)
Next, we can solve for "s":
s ≈ (5.0 x [tex]10^{-13[/tex]) / (3.0 x [tex]10^{-2[/tex])
s ≈ 1.67 x [tex]10^{-11[/tex] M
Thus, the molar solubility of AgBr in the 3.0 x [tex]10^{-2[/tex] M [tex]AgNO_3[/tex] solution is approximately 1.67 x [tex]10^{-11[/tex] M.

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The chemist should mix 11.25 gallons of water and 18.75 gallons of the 40% alcohol solution to obtain 30 gallons of a 25% alcohol solution.

Let's assume the chemist needs to mix x gallons of water with y gallons of the 40% alcohol solution to obtain 30 gallons of a 25% alcohol solution.

To find the solution, we need to consider the amount of alcohol in each component before and after mixing.

Amount of alcohol in the water: 0% (water contains no alcohol)

Amount of alcohol in the 40% alcohol solution: 40% of y gallons = 0.4y gallons

After mixing the solutions, the total amount of alcohol in the mixture is 25% of 30 gallons = 0.25 * 30 = 7.5 gallons.

So, we can set up the following equation to represent the alcohol balance:

0.4y + 0 = 7.5

Solving for y:

0.4y = 7.5

y = 7.5 / 0.4

y = 18.75

The chemist needs to mix 18.75 gallons of the 40% alcohol solution.

To find the amount of water needed, we subtract the amount of the 40% alcohol solution from the total volume:

x = 30 - y

x = 30 - 18.75

x = 11.25

The chemist needs to mix 11.25 gallons of water.

Therefore, the chemist should mix 11.25 gallons of water and 18.75 gallons of the 40% alcohol solution to obtain 30 gallons of a 25% alcohol solution.

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We can conclude that the reaction will shift to the right in order to reach equilibrium due to equilibrium constant being [tex]4.63 * 10^(20)[/tex] which is much greater than 1.

In order to determine which way the reaction would shift to reach equilibrium, we need to look at the reaction quotient (Qc) and compare it to the equilibrium constant (Kc).

At standard state conditions, the concentrations of all species are assumed to be 1 M. Therefore, the initial Qc for this reaction would be:

[tex]Qc = [SO3]^2 / [SO2]^2[O2] = 1^2 / 1^2(1) = 1[/tex]

If the reaction shifts to the right, the concentration of SO3 will increase, while the concentrations of SO2 and O2 will decrease. This would result in a decrease in Qc, as the denominator would become smaller.

If the reaction shifts to the left, the concentration of SO3 will decrease, while the concentrations of SO2 and O2 will increase. This would result in an increase in Qc, as the numerator would become smaller.

Since the equilibrium constant for this reaction is [tex]Kc = [SO3]^2 / [SO2]^2[O2] = 4.63 * 10^(20)[/tex], which is much greater than 1, we can conclude that the reaction will shift to the right in order to reach equilibrium. This means that the concentration of SO3 will increase, while the concentrations of SO2 and O2 will decrease, until the reaction reaches equilibrium.

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To find the rate constant for a first-order reaction, we can use the equation:

ln([A]0/[A]t) = kt

where [A]0 is the initial concentration of A, [A]t is the concentration of A at time t, k is the rate constant, and ln is the natural logarithm.

Using the data given, we can calculate the rate constant as follows:

ln(1.60/0.40) = k(10.0 s)
ln(0.40/0.10) = k(20.0 s)

Simplifying these equations, we get:

k = (ln(1.60/0.40))/10.0 s
k = (ln(0.40/0.10))/20.0 s

Calculating these values, we get:

k = 0.231 s^-1 (rounded to three significant figures)

Therefore, the rate constant for this reaction is 0.231 s^-1.

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At the first halfway point of the titration of H2CO3 with NaOH, the primary species in solution would be HCO3-.

This is because at this point, half of the H2CO3 has been neutralized by the NaOH, forming HCO3-. H2CO3 is a weak acid, and as NaOH is added, it reacts with the H+ ions to form water, which increases the pH of the solution. As the pH increases, the H2CO3 molecule loses a proton to form HCO3-. At the halfway point, the pH of the solution is around 8.3, which is close to the pKa of H2CO3 (6.35), indicating that about half of the H2CO3 has been converted to HCO3-. Therefore, at the first halfway point, the primary species in solution is HCO3-.

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Hydrocyanic acid is a weak acid, Hydrofluroic acid is a weak acid and phenol is a weak acid and the correct option is option A.

Acid strength is the measure of the ability of the acid to lose its H+ ion

The dissociation of a strong acid in solution is finely complete, omitting in its most concentrated solutions.

A weak acid partially dissociates with both the undissociated acid and its dissociation products in the solution, in equilibrium to each other.

Acid strength depends on the strength of the H and A bond. The weaker the bond, the lesser the energy that will be required to break it. Thus, the acid is strong.

Thus, the ideal selection is option A.

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The amount (grams) of oxygen present in 10g of the carbohydrate sample is 1.256grams.

A carbohydrate or carbs are the sugars, starches, and dietary fiber that occur in plant foods and dairy products.

Carbohydrates contain only carbon, hydrogen and oxygen atoms; prior to any oxidation or reduction, most have the empirical formula Cm(H2O)n.

According to this question, 10.00g sample contains 7.484g C and 1.260g H. This means that the amount of oxygen can be calculated as;

Mass of O = 10 - (7.484 + 1.260)

Mass of O = 10 - 8.744 = 1.256grams.

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