Explain what an ion is. ​

The student expected to get exactly 3.5 grammes of Chromium (III) oxide from adding 1.75 grammes of Chromium trinitrate and 1.75 grammes of Sodium oxide. To create the needed amount of Chromium oxide.

To get the mass per mole, divide the reactant's mass by its molecular weight. As an alternative, we might multiply the millilitres of the reactant solution by the grammes per millilitre of the liquid solution. Next, divide the outcome by the reactant's molar mass.

This is because the chemical equation for the reaction between Chromium trinitrate and Sodium oxide is not balanced to produce Chromium(III) oxide.

2Cr(NO₃)₃ + 3Na₂O → Cr₂O₃ + 6NaNO₃

From this balanced equation, we can see that for every 2 moles of Chromium trinitrate, we need 3 moles of Sodium oxide to produce 1 mole of Chromium(III) oxide. This means that the amounts of Chromium trinitrate and Sodium oxide required to produce 3.5 grams of Chromium(III) oxide are:

Amount of Chromium trinitrate = (2/3) x (3.5 grams / molar mass of Chromium trinitrate) = 1.53 grams

Amount of Sodium oxide = (3/2) x (3.5 grams / molar mass of Sodium oxide) = 2.26 grams

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The forest's ability to store carbon would be lost to the atmosphere if all the trees were removed and burned.

As they expand, trees and other plants take in carbon dioxide from the atmosphere. This is changed into carbon and stored in the soil, the plant's roots, branches, leaves, and trunks. When forests are cut down or burned, carbon dioxide, which is primarily the form of stored carbon, is released into the atmosphere.

Once forested areas would become drier and more vulnerable to severe droughts without trees.

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The mass of Butane required to completely react to produce 8.70 mol of Carbon dioxide is 126.51 g.

Similarly, 1 mole of carbon dioxide is 44 gm as Carbon dioxide (12 gm + (2 x 16gm) = 44 gm. So, from the above two observations, it is clear that 16 gm of methane when burns completely in excess of air it forms 44 gm of carbon dioxide. In light of this, 44/16 x 8 = 22 gm of carbon dioxide will be produced from 8 gm of methane.

2 Butane(g) + 13 Oxygen(g) ---> 8 Carbon dioxide(g) + 10 Water(g)

2 mol Butane/ 8 mol Carbon dioxide = x mol Butane/ 8.70 mol Carbon dioxide

x mol Butane= (2 mol Butane/ 8 mol Carbon dioxide) x (8.70 mol Carbon dioxide) = 2.175 mol Butane

Convert the number of moles of Butane to grams:

To convert from moles to grams, we need to use the molar mass of Butane, which is 58.12 g/mol:

Number of moles times molar mass = mass of butane

Mass of Butane= 2.175 mol x 58.12 g/mol

Mass of Butane= 126.51 g

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Answer:

Explanation:

1

Locate the parentheses after the chemical formula, whether or not it is within the context of an equation.

2

Identify the parentheses as (s) for solid, (l) for liquid, (g) for gas, or (aq) for aqueous solution. An aqueous solution is a substance dissolved in water.

3

If no parentheses follow the chemical formula, look for key words or phrases. These are especially helpful in the context of chemical reactions. For example, a precipitate is a solid, a combustion reaction produces water and carbon dioxide in their gaseous forms, and in solubility experiments, ions are in aqueous solution.

Methane, also known as [tex]CH_{4}[/tex], is a substance that has 27% carbon by mass. [tex]C_{3}H_{8}[/tex] is a different molecule that has 27% mass-based carbon.

Carbon has a molar mass of 12.01 g/mol. Finding compounds with molar masses similar to 27 g/mol will help us figure out which chemical has about 27% carbon by mass, as 27% of 27 g/mol is roughly 7.3 g/mol (because 27% is comparable to 0.27).

The chemical [tex]CH_{4}[/tex], which has a molar mass of roughly 16 g/mol, is one that fulfils this definition. We can use the following formula to get the percentage of carbon in [tex]CH_{4}[/tex]:

(Mass of Element / Molar Mass of Compound) x 100% Equals % composition

The % content of carbon in methane is as follows:

Mass of carbon divided by the molar mass of methane yields the percentage of carbon.

% of carbon is equal to (12.01 g/mol / 16.04 g/mol) multiplied by 100.

Carbon content as a percentage: 74.97%

As a result, methane does not contain 27% of its mass in carbon.

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Answer:d 0.648 J/g*C

Explanation:

-4602/(42.0g x -169)

Formula Q/m^t=C

Answer:I think it’s c hopefully okay.

Explanation:

Boyle's law states that, at a certain temperature, a gas's pressure and volume are inversely related, meaning that as the volume falls, the pressure rises and vice versa.

According to Boyle's law, the volume of the container has an inverse relationship to the gas's pressure. In other words, a big volume container will have low pressure, and a low volume container will have high pressure. Breathing is a biological example of how this law operates.

According to Boyle's law, a gas's pressure and volume are inversely proportional. When the temperature stays the same, pressure rises as volume rises and vice versa.

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Answer:

The molarity of H2C2O4 in the given solution is 0.222 M

Explanation:

To find the molarity of H2C2O4, we need to know the number of moles of H2C2O4 present in a given volume of its solution. We can use the following formula to calculate the molarity:

Molarity (M) = Number of moles of solute ÷ Volume of solution in liters

To calculate the number of moles of H2C2O4, we need to know the mass of H2C2O4 present and its molar mass. The molar mass of H2C2O4 is:

Molar mass of H2C2O4 = (2 x Atomic mass of H) + (2 x Atomic mass of C) + (4 x Atomic mass of O) = (2 x 1.01 g/mol) + (2 x 12.01 g/mol) + (4 x 16.00 g/mol) = 90.04 g/mol

Suppose we have 5.0 g of H2C2O4 in 250 mL of its solution. First, we need to convert the volume of the solution from milliliters to liters:

Volume of solution = 250 mL = 0.250 L

Next, we can use the following formula to calculate the number of moles of H2C2O4 in the solution:

Number of moles of H2C2O4 = Mass of H2C2O4 ÷ Molar mass of H2C2O4

Number of moles of H2C2O4 = 5.0 g ÷ 90.04 g/mol = 0.05553 mol

Now, we can use the formula for molarity to calculate the molarity of H2C2O4:

Molarity (M) = Number of moles of solute ÷ Volume of solution in liters

Molarity (M) = 0.05553 mol ÷ 0.250 L = 0.222 M

MARK ME BRAINLIEST

The theoretical yield of [tex]Pl_{3}[/tex] from the given reaction is 73.2 grams.

What is theoretical yield ?

To calculate the theoretical yield of [tex]Pl_{3}[/tex], we first need to determine which reactant is limiting. We can do this by calculating the amount of product that would be formed from each reactant, and comparing the values.

From the balanced chemical equation, we can see that 2 moles of P reacts with 3 moles of [tex]I_{2}[/tex] to produce 2 moles of [tex]Pl_{3}[/tex]. So, the number of moles of P and [tex]I_{2}[/tex] present in the reaction can be calculated as:

moles of P = mass of P / molar mass of P

moles of P = 27.0 g / 30.97 g/mol

moles of P = 0.871 mol

moles of [tex]I_{2}[/tex] = mass of [tex]I_{2}[/tex] / molar mass of [tex]I_{2}[/tex]

moles of [tex]I_{2}[/tex] = 68.0 g / 253.81 g/mol

moles of [tex]I_{2}[/tex] = 0.268 mol

Now, we can use the stoichiometry of the balanced chemical equation to calculate the number of moles of [tex]Pl_{3}[/tex] that can be produced from each reactant:

mol of [tex]Pl_{3}[/tex] produced from P = (0.871 mol P) / (2 mol P per 2 mol [tex]Pl_{3}[/tex]) * (2 mol [tex]Pl_{3}[/tex] / 2 mol P)

mol of [tex]Pl_{3}[/tex] produced from P = 0.871 mol [tex]Pl_{3}[/tex]

mol of [tex]Pl_{3}[/tex] produced from [tex]I_{2}[/tex] = (0.268 mol [tex]I_{2}[/tex]) / (3 mol [tex]I_{2}[/tex] per 2 mol [tex]Pl_{3}[/tex]) * (2 mol [tex]Pl_{3}[/tex] / 1 mol [tex]I_{2}[/tex])

mol of [tex]Pl_{3}[/tex] produced from [tex]I_{2}[/tex] = 0.178 mol [tex]Pl_{3}[/tex]

Since the amount of [tex]Pl_{3}[/tex] produced is limited by the amount of [tex]I_{2}[/tex], [tex]I_{2}[/tex] is the limiting reactant. Therefore, the theoretical yield of [tex]Pl_{3}[/tex] is the amount of [tex]Pl_{3}[/tex] that would be produced if all of the limiting reactant ([tex]I_{2}[/tex]) were completely consumed:

mass of [tex]Pl_{3}[/tex] = moles of [tex]Pl_{3}[/tex] * molar mass of [tex]Pl_{3}[/tex]

mass of [tex]Pl_{3}[/tex] = 0.178 mol * 411.81 g/mol

mass of [tex]Pl_{3}[/tex] = 73.2 g

Therefore, the theoretical yield of [tex]Pl_{3}[/tex] from the given reaction is 73.2 grams.

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An ICE (Initial, Change, Equilibrium) table is a tool used to organize the information and calculations involved in solving equilibrium problems.

A) KC = 4.00

The balanced equation for the reaction is:

2 A (aq) ⇌ B(aq)

The equilibrium expression is:

KC = [B]¹ / [A]²

We are given that KC = 4.00 and the initial concentration of A is 4.00 M. Let x be the change in concentration of A and B at equilibrium.

Since two moles of A react to form one mole of B, the change in concentration of B is half the change in concentration of A.

Using the equilibrium expression, we can write:

4.00 = x² / (4.00 - 2x)²

Solving for x:

x = 0.62 M

Therefore, the equilibrium concentrations are [A] = 2.76 M and [B] = 0.62 M.

B) KC = 200

The equilibrium expression is:

KC = [B]¹ / [A]²

We are given that KC = 200. Substituting this value into the equilibrium expression and solving for x, we get:

x = 0.20 M

Therefore, the equilibrium concentrations are [A] = 3.60 M and [B] = 0.20 M.

C) We start with an initial concentration of A = 4.00 M.

Let x be the change in concentration for A and B. At equilibrium, the concentration of A will be 4.00 - 2x, and the concentration of B will be x.

Using the expression for KC, we can write:

KC = [B] / [A]²

Substituting the equilibrium concentrations:

8.00 x 10³= x / (4.00 - 2x)²

Simplifying:

8.00 x 10⁻³ = x / (16 - 16x + 4x²)

32 - 32x + 8x² = 1/x

8x³ - 32x³ + 32x - 1 = 0

This is a cubic equation that can be solved using numerical methods or approximated by trial and error. One possible solution is x = 0.033 M.

Therefore, at equilibrium:

[A] = 4.00 - 2x = 3.94 M

[B] = x = 0.033 M

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Answer:

Answer:Ester and anhydride would be expected to have a similar boiling point to a ketone of similar size.

The volume (in mL) of 0.100 M Na₂CO₃ needed to produce 1.00 g of CaCO₃ is 100 mL

First, we shall determine the mole in 1.00 g of CaCO₃. Details below:

Mole = mass / molar mass

Mole of CaCO₃ = 1 / 100.09

Mole of CaCO₃ = 0.01 mole

Next, we shall obtain the mole of Na₂CO₃. Details below:

Na₂CO₃ + CaCl₂ -> 2NaCl + CaCO₃

From the balanced equation above,

1 mole of CaCO₃ were obtained from 1 mole of Na₂CO₃

Therefore,

0.01 moles of CaCO₃ will also be obtain from 0.01 mole of Na₂CO₃

Finally, we shall determine the volume of Na₂CO₃ needed. Details below:

Volume = mole / molarity

Volume of Na₂CO₃ = 0.01 / 0.1

Volume of Na₂CO₃ = 0.1 L

Multiply by 1000 to express in mL

Volume of Na₂CO₃ = 0.1  1000 =

Volume of Na₂CO₃ = 100 mL

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The reaction is neutralization reaction and the pH after addition of 10.0 mL of NaOH is 4.87.

By neutralising a process in water, surplus hydrogen or hydroxide ions are removed from the solution.

Neutralization reactions are quite useful in daily life. The following applications are significant ones:It is imperative to treat wastewater before it harms the environment, and this can only be done by using the right chemical reagents to neutralise the wastewater's strong base.In the form of antacid tablets or gels, which provide relief from acidity brought on by gastric acid in the stomach, they are frequently used in daily life.

[tex]n(H3O+) = 1.8 * 10^-^5 mol/L × 0.030 L - 0.001 mol = 5.4 * 10^-^7 mol[/tex]

The total volume of the solution is now 30.0 mL + 10.0 mL = 40.0 mL = 0.040 L. Therefore, the concentration of H3O+ is:

[tex][H3O+] = n(H3O+) / V(total) = 5.4 * 10^-^7 mol / 0.040 L = 1.35 * 10^-^5 mol/L[/tex]

Taking the negative logarithm of the concentration gives the pH:

[tex]pH = -log[H3O^+] = -log(1.35 * 10^-^5) = 4.87[/tex]

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Answer:

HI is the Arrhenius acid

CsOH is the Arrhenius base

Explanation:

Acid-Base reactions are reactions that involve the transfer of protons.

Arrenhius Acids

Arrenhius acids are molecules that have an H⁺ ion. In an acid-base reaction, the acid will donate the H⁺ ion to the base. The molecule HI (called hydroiodic acid) is the acid because it has an H⁺ ion. Note that an H⁺ ion is simply a proton because it has no electrons or protons.

H₂O can also be considered an Arrhenius acid because it does have an H⁺ ion, but it is a very weak acid.

Arrenhius Bases

Arrhenius bases are molecules that have an OH⁻ ion. In an acid-base reaction, the base will accept an H⁺ ion. The molecule CsOH (called cesium hydroxide) is the base because it has an OH⁻ ion. The OH⁻ ion is one of the most common polyatomic ions and is called hydroxide.

H₂O can also be considered an Arrhenius base because it has an OH⁻ ion. However, it is a weak base. Still, water is very important to acid-base reactions because it can be either an acid or a base.

In the reaction HI + CsOH + H2O, HI is the Arrhenius acid and CsOH is the Arrhenius base.

In the given reaction HI + CsOH + H2O, HI is the Arrhenius acid and CsOH is the Arrhenius base. In the Arrhenius theory of acids and bases, an acid is a substance that produces hydrogen ions (H+) in aqueous solution, while a base is a substance that produces hydroxide ions (OH-) in aqueous solution. HI can dissociate in water to produce H+ ions, making it an acid. CsOH can dissociate in water to produce OH- ions, making it a base.

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In 375 grammes of the metal Jurupium, there are 3 moles of the element.

On 1 November 1952, the first thermonuclear explosion, which occurred on a Pacific atoll, left behind debris that contained the element Einsteinium. A neighbouring atoll's fallout material was shipped to Berkeley, California, for analysis. It belongs to the group of elements with atomic number 63 that are numerically squeezed between barium and hafnium. A lanthanide, or europium, is one of those mysterious elements outside the periodic table's core.

Divide the mass of Jurupium by its molar mass to see how many moles there are in 375 grammes of Jurupium metal:

375 g / 125 g/mol = 3 moles

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Although sodium and phosphorus are both found in the same period of the periodic table, the nature of their oxides is different. Although P2O5 is acidic, Na2O is basic.

Metal oxides are basic, whereas non-metal oxides are acidic. The metallic character of the elements lessens as you move from left to right over time, while the non-metallic character grows.

Since the oxides in an element's higher oxidation state are more acidic than those in its lower oxidation state, dinitrogen pentoxide is the most acidic oxide compared to nitrogen dioxide and nitrogen tetraoxide.

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Answer:

To find the percentage by volume of alcohol, we need to divide the volume of alcohol by the total volume of the solution, then multiply by 100%:

Volume of alcohol = 19 mL

Total volume of solution = 19 mL + 31 mL = 50 mL

Percentage by volume of alcohol = (19/50) x 100% = 38%

Therefore, the percentage by volume of alcohol in the solution is 38%.

Answer:

20% volume percentage I think

Answer: Oh, just found the question, the answer to your question is B.

Explanation:

According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in temperature, pressure or concentration, the system will adjust its position of equilibrium to counteract the change.

In the given exothermic reaction, the forward reaction (2SO2(g) + O2(g) --> 2SO3(g)) releases heat, meaning that it is favored at lower temperatures. Therefore, increasing the temperature will shift the equilibrium position to the left (towards the reactants), in order to counteract the increase in temperature.

So, if the temperature of the system is increased, the concentration of sulfur dioxide will decrease.

The answer is B: The amount of sulfur dioxide decreases.

Molecules are in constant motion due to their thermal energy, which is related to their temperature.

As the molecules move faster, they are more likely to overcome the intermolecular forces that hold them together, causing them to break apart and become less cohesive. This can cause a solid to melt into a liquid, or a liquid to evaporate into a gas.

In summary, heating a substance increases the kinetic energy of its molecules, causing them to move faster, collide with one another with greater force, and spread apart from each other, resulting in an increase in volume and thermal expansion.

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Because the speed limit sign is in kilometres per hour and the driver is accustomed to miles per hour, no other cars are passing too quickly. The permitted speed is roughly 62 mph.

A regulatory sign is the speed limit sign. The purpose of speed limit signs is to inform drivers of the legal maximum and minimum speeds that are required of them. Drivers are not permitted to go over the limit stated by the sign.

When approaching or passing a troop, police, or military procession, or when driving past construction workers fixing roads, a motor vehicle's driver must travel at a pace of no more than 25 kph.

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There are 3.44 x 10^{22} particles in 0.057 moles of lithium bromide.

The lithium bromide formula also known as the lithium monobromide formula or Bromo lithium formula is explored. It is a counterion bromide-based salt of lithium.

we have to use Avogadro's constant,

Avogadro's constant, is approximately equal to 6.022 x 10^{22} particles per mole.

we can use the following formula:

number of particles = moles x Avogadro's constant

Substitute the values,

number of particles = 0.057 moles x 6.022 x 10^{23} particles/mol

Simplifying the equation

number of particles = 3.44 x 10^{22} particles

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Answer:

(i) The molar mass of sucrose (C12H22O11) can be calculated by adding the molar masses of the individual atoms:

molar mass of C12H22O11 = 12 x 12.01 + 22 x 1.01 + 11 x 16.00 = 342.3 g/mol

To calculate the number of moles of sucrose in the sample, we divide the mass of the sample by the molar mass:

moles of sucrose = mass of sample / molar mass

moles of sucrose = 5.8 / 342.3

moles of sucrose = 0.0169 mol

Therefore, there are 0.0169 moles of sucrose in the sample.

(ii) To calculate the number of moles of carbon in the sample, we need to determine the number of moles of C atoms present in each molecule of sucrose. Sucrose contains 12 carbon atoms per molecule.

moles of carbon = moles of sucrose x number of carbon atoms per molecule

moles of carbon = 0.0169 x 12

moles of carbon = 0.203 mol

Therefore, there are 0.203 moles of carbon in the sample.

(iii) To calculate the total number of carbon atoms in the sample, we multiply the number of moles of carbon by Avogadro's constant:

number of carbon atoms = moles of carbon x Avogadro's constant

number of carbon atoms = 0.203 x 6.02 x 10^23

number of carbon atoms = 1.22 x 10^23

Therefore, there are approximately 1.22 x 10^23 carbon atoms in the sample of sucrose.

Explanation:

The right response is an atom with 8 valence electrons, or (c). Because of its stable electron configuration, an atom with an entire octet (8 valence electrons) is not reactive.

Valence electrons range in number from 1 to 8 depending on the element. Bonding: Atoms with 8 valence electrons tend to be more stable. Because they have 8 valence electrons, neon, argon, krypton, and xenon atoms are nonreactive.

The least reactive elements are noble gases. They have eight valence electrons, which fill their outer energy level, and this explains why. Noble gases rarely interact with other elements to form compounds since this is the arrangement of electrons that is the most stable.

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Answer:

The formula for the dissociation constant for a base (Kb) is given by the expression: Kb = [BH+][OH-]/[B], where [BH+] is the concentration of the conjugate acid, [OH-] is the concentration of hydroxide ions and [B] is the concentration of the base. In this case, the base is pyridine (C5H5N), so the formula for its dissociation constant would be: Kb = [C5H5NH+][OH-]/[C5H5N].

Explanation:

The dissociation constant for pyridine, a weak base, is represented by Kb and is calculated using the formula Kb = [C5H5NH+][OH-] / [C5H5N].

The dissociation constant of a weak base, such as pyridine, is represented by Kb. The formula for Kb is given by:

Kb = [C5H5NH+][OH-] / [C5H5N]

Where [C5H5NH+] represents the concentration of the conjugate acid (pyridinium ion), [OH-] represents the concentration of hydroxide ions, and [C5H5N] represents the concentration of pyridine.

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The value of the equilibrium constant Kc or Kp for ((ab)²)/((a₂)(b²))

The equilibrium constant expression for the given chemical reaction is:

Kc = [(ab)²] / [(a₂)(b²)]

where Kc is the equilibrium constant.

Let's assume that the gases have partial pressures of p(ab), p(a2), and p(b). Then, the equilibrium constant expression can be written as:

Kp = [(p(ab))²] / [(p(a₂))(p(b²))]

Now, we can use the ideal gas law to relate the partial pressures to the number of moles of each gas:

p(ab) = (n(ab) * RT) / V

p(a2) = (n(a2) * RT) / V

p(b) = (n(b) * RT) / V

where n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and V is the volume.

Substituting these expressions into the equilibrium constant expression, we get:

Kp = [(n(ab) * RT / V)²] / [(n(a₂) * RT / V) * (n(b) * RT / V)²]

Simplifying, we get:

Kp = (n(ab)² / n(a₂) * n(b)²) * (RT / V)²

We can rewrite the expression in terms of the number of moles of the reactants and products, as:

Kp = ([ab]² / [a₂] * [b]²) * (RT / P)²

where [ab], [a₂], and [b] are the molar concentrations of the reactants and products, and P is the total pressure of the gases.

Therefore, the value of the equilibrium constant Kc or Kp for ((ab)²)/((a₂)(b²)) can be calculated using the above expression. However, the specific numerical value of Kc or Kp will depend on the temperature and pressure conditions of the reaction.

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Answer:

C

Explanation:

A displacement reaction occurs when a more reactive element replaces a less reactive element in a compound.

The general reaction can be represented as:

A + BC → AC + B

Therefore, the reactant that is more reactive than the element in the compound will undergo a displacement reaction.Out of the given options, the reactant that will not undergo a displacement reaction is C. Fluorine + sodium chloride. This is because fluorine is the most electronegative element and it cannot be displaced by any other element. In other words, fluorine is the most reactive element and there is no other element that is more reactive than it. Therefore, the reaction between fluorine and sodium chloride will not undergo a displacement reaction.

Option A, Magnesium + sodium fluoride, will undergo a displacement reaction because magnesium is more reactive than sodium, and can displace sodium from sodium fluoride.

Option B, Chlorine + sodium chloride, will undergo a displacement reaction because chlorine is more reactive than sodium, and can displace sodium from sodium chloride.

Option D, Zinc + copper (ll) chloride, will undergo a displacement reaction because zinc is more reactive than copper, and can displace copper from copper (ll) chloride.

Thus, the correct answer is C. Fluorine + sodium chloride.

Bronsted-Lowry acid-base reaction, Lewis acid base reaction, Lewis acid-base reaction, and Arrhenius acid-base reaction and Bronsted-Lowry acid-base reaction are some examples of acid-base reactions.

Lewis acidity and basicity are based on the sharing of an electron pair, unlike the Brnsted-Lowry and Arrhenius classifications, which are based on the transfer of protons. Lewis bases can give away an electron pair, whereas Lewis acids can absorb one.

Nitric acid, hydrobromic acid, and sulfuric acid (H2SO4) are further Arrhenius acids. (HNO3). Sodium hydroxide (NaOH) and potassium hydroxide are examples of Arrhenius bases. (KOH).

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Answer:

Explanation:

To find the empirical formula of the hydrocarbon, we need to determine the relative amounts of carbon and hydrogen in the compound. We can do this by using the masses of CO2 and H2O produced during combustion.

First, let's calculate the moles of CO2 and H2O produced:

moles of CO2 = 4.40 g / 44.01 g/mol = 0.100 mol

moles of H2O = 2.70 g / 18.02 g/mol = 0.150 mol

Next, let's determine the number of moles of carbon and hydrogen in the hydrocarbon. During combustion, the hydrocarbon reacts with oxygen to produce CO2 and H2O. The balanced chemical equation for the combustion of a hydrocarbon can be written as:

CnHm + (n + m/4)O2 → nCO2 + m/2H2O

where n and m are the coefficients that balance the equation.

From the equation, we can see that for every n moles of CO2 produced, there must be n moles of carbon in the original hydrocarbon. Similarly, for every m/2 moles of H2O produced, there must be m/2 moles of hydrogen in the original hydrocarbon.

Using this information, we can calculate the number of moles of carbon and hydrogen in the hydrocarbon:

moles of carbon = 0.100 mol

moles of hydrogen = 0.150 mol * 2 = 0.300 mol

Now, we need to find the empirical formula of the hydrocarbon by dividing the number of moles of each element by the smallest number of moles:

empirical formula CnHm = C(0.100 mol / 0.100 mol)H(0.300 mol / 0.100 mol) = CH3

Therefore, the empirical formula of the hydrocarbon is CH3.