a 3.742 g sample of a compound containing only carbon and hydrogen wasanalyzed by combustion and found to contain 3.140 g of carbon and 0.602 gof hydrogen. mass spectral analysis indicates that the molar mass for thiscompound is 100.2. what is the molecular formula for this compound?

Answers

Answer 1

Answer : The molecular formula for this compound is C7H14

To determine the molecular formula of the compound, we need to first calculate its empirical formula using the given mass percentages of carbon and hydrogen. The mass percent of carbon in the compound is: (3.140 g / 3.742 g) x 100% = 83.9%

The mass percent of hydrogen in the compound is: (0.602 g / 3.742 g) x 100% = 16.1%. Assuming a 100 g sample of the compound, we can calculate the masses of carbon and hydrogen in the sample: Mass of carbon = 83.9 g and Mass of hydrogen = 16.1 g

Next, we need to convert these masses to moles, using the atomic masses of carbon and hydrogen:1 mol C = 12.01 g, 1 mol H = 1.008 g. Moles of carbon = 83.9 g / 12.01 g/mol = 6.983 mol, Moles of hydrogen = 16.1 g / 1.008 g/mol = 15.95 mol. Dividing each mole value by the smallest mole value, we get the following mole ratio: C:H = 6.983 / 6.983 = 1.000 : 2.285

The empirical formula for the compound is therefore CH2. To determine the molecular formula, we need to find the molecular weight of the empirical formula, and then divide the given molar mass by this value to get the molecular formula multiplier. Molecular weight of CH2 = 12.01 + 2(1.008) = 14.026 g/mol, Molecular formula multiplier = 100.2 g/mol / 14.026 g/mol = 7.146. Multiplying the empirical formula by this multiplier, we get the molecular formula: C7H14

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Related Questions

ethers are fairly unreactive. which reagent can transform symmetrical ethers into two equivalents of the same alkyl halide? a. pbr3 b. hbr c. socl2/pyridine d. h2so4/h2o

Answers

Ethers are fairly unreactive. The reagent that can transform symmetrical ethers into two equivalents of the same alkyl halide is a. PBr3.

Symmetrical ethers have two identical alkyl or aryl groups attached to the oxygen atom in the ether. In addition, symmetrical ethers can be synthesized by the reaction between alkoxides and alkyl halides. Symmetrical ethers are stable and have a low reactivity with most nucleophiles and electrophiles. They are susceptible to acid-catalyzed cleavage of their C-O bond. Thus, in most cases, the breaking of the ether linkage requires strong acids, making the cleavage of ethers a slow reaction.

The reaction of symmetrical ethers with PBr3, a strong nucleophile, can transform symmetrical ethers into two equivalents of the same alkyl halide. PBr3 reacts with the ether oxygen atom to produce a bromide anion, which is then displaced by the alkyl group. A second equivalent of PBr3 can then react with the alkyl halide product to produce another alkyl bromide.

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Which statement best compares the energy and frequency of green waves to orange waves?

Green waves have a lower frequency and contain less energy than orange waves.
Green waves have a higher frequency and contain more energy than orange waves.
Orange waves have a higher frequency and contain less energy than green waves.
Orange waves have a lower frequency and contain more energy than green waves.

Answers

Orange waves have a lower frequency and contain less energy than green waves.

What is Wave?

A wave is a disturbance or oscillation that travels through space and time, accompanied by the transfer of energy without the transfer of matter. Waves can take many different forms, including sound waves, light waves, water waves, and seismic waves. They can be described in terms of their frequency, wavelength, amplitude, and velocity, among other properties. Waves play a fundamental role in many areas of science and technology, including communication, medicine, and engineering.

The energy of a wave is directly proportional to its frequency, which means that higher frequency waves contain more energy than lower frequency waves. The frequency of a wave refers to the number of complete cycles or oscillations that the wave undergoes per second, and is measured in units of Hertz (Hz).

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write the formula the following compounds and determine the shape of each of the following compounds:sodium tetrahydroxochromate iiipotassium hexachlorocobaltate iihexaaquairon iii chloride

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The formula of sodium tetrahydroxochromate III is Na₂Cr(OH)₄, and its shape is tetrahedral.

The formula of potassium hexachlorocobaltate II is K₃CoCl₆, and its shape is octahedral.

The formula of hexaaquairon III chloride is [Fe(H₂O)₆]Cl₃, and its shape is octahedral.

Tetrahedral Geometry: The tetrahedral geometry is characterized by four electron pairs that are distributed around a central atom. It has an angle of 109.5 degrees between adjacent hydrogen atoms.

Octahedral Geometry: An octahedron is a polygon with eight faces. This geometry has an angle of 90 degrees between adjacent molecules.

The six surrounding atoms are all positioned at the same distance from the central atom. The hexaaquairon III chloride compound has six water molecules that are coordinated to an iron center, giving it octahedral geometry.

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which of the following could be used to initiate the polymerization of isobutylene: a.) sulfuric acid, b.) boron trifluoride etherate, c.) water, or d.) butyllithium? (more than one answer can be selected)

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Option (a) and (b) is correct. Both sulfuric acid and boron trifluoride etherate could be used to initiate the polymerization of isobutylene.

Isobutylene is found as a colorless gas. It has a faint petroleum like odor. For transportation isobutylene may be stanched. It is shipped to other as a liquefied gas under its own vapor pressure. Polymerization is defined as any process in which relatively small molecules called monomers and combine chemically to produce a very large chainlike or network molecule called a polymer. The monomer molecules are called  be all alike or they may represent two, three, or more different compounds. Polymerization of Isobutylene occurs via carbo-cationic polymerization processes to form products with very broad molecular weight distributions and exhibiting consistencies which vary from liquids, for oligomers, to solids, for high-molecular-weight polymers.

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What are the spectator ions in Na+ + OH + H+ + Cl → H2O + Na+ + Cl?

O

A. Na+, OH, H+, and CM

OB. OH' and H+

O

O

C. Na+ and CH

D. OH", H+, and H20

E PREVIOUS

9A

Answers

The spectator ions in  [tex]Na^{+}[/tex] + [tex]OH^{-}[/tex] + [tex]H^{+}[/tex] + [tex]Cl^{-}[/tex]  → [tex]H_{2} O[/tex] +  [tex]Na^{+}[/tex] +  [tex]Cl^{-}[/tex] is sodium ions and chloride ions.

The spectator ion are defined as the ions which do not participate in chemical reactions and present the same on both sides of the reactions. If we write a net chemical reaction the spectator ions are cancelled from both sides of the equation.

          [tex]Na^{+}[/tex] + [tex]OH^{-}[/tex] + [tex]H^{+}[/tex] + [tex]Cl^{-}[/tex]  → [tex]H_{2} O[/tex] +  [tex]Na^{+}[/tex] +  [tex]Cl^{-}[/tex]

If we compare the chemical solutions before and after the reaction, sodium and chloride ions are present in both solutions but they do not undergo any chemical change at all. These ions present in the solution are called spectator ions since they don't participate in the chemical reaction at all.

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The correct question is,

What are the spectator ions in

 [tex]Na^{+}[/tex] + [tex]OH^{-}[/tex] + [tex]H^{+}[/tex] + [tex]Cl^{-}[/tex]  → [tex]H_{2} O[/tex] +  [tex]Na^{+}[/tex] +  [tex]Cl^{-}[/tex] ?

1. personal connections describe your reaction to the events in act i. why might you want to continue reading? explain.

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Readers may want to continue reading a work if they are intrigued by the characters, interested in the plot, or invested in the themes and messages presented.

Why will a reader continue reading?

In general, act sets the stage for the rest of the work, introducing key characters, establishing conflicts, and setting the tone and mood.

If a reader finds these elements compelling or engaging, they may be motivated to continue reading to see how the story unfolds and how the characters develop. Additionally, Act I may introduce questions or mysteries that pique the reader's curiosity and encourage them to keep reading to find the answers.

Thus, a reader may want to continue reading a work if they are in interested in the plot.

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plot a theoretical distillation curve of temperature (y-axis) vs. volume in ml (x-axis) for a 15 ml of a mixture containing 60% 1-propanol and 40% 2-propanol. are these two compounds easier to separate by distillation than cyclohexane and toluene? explain your answer. (6 pts)

Answers

To plot a theoretical distillation curve please follow the steps while we continue our discussion. Since their boiling point difference is higher it is easier to separate Cyclohexane and toluene by distillation than 1-propanol and 2-propanol.

How to separate two compounds by distillation?

Plot a theoretical distillation curve of temperature (y-axis) vs. volume in ml (x-axis) for a 15 ml mixture containing 60% 1-propanol and 40% 2-propanol, follow these steps:

1. Determine the boiling points of 1-propanol and 2-propanol. 1-propanol has a boiling point of 97°C, while 2-propanol has a boiling point of 82°C.

2. Calculate the volumes of each compound in the mixture. 60% of 15 ml is 9 ml (1-propanol) and 40% of 15 ml is 6 ml (2-propanol).

3. Plot the boiling points of each compound on the y-axis, and their respective volumes on the x-axis.

4. Draw a curve connecting the two points to represent the theoretical distillation curve.

To determine if 1-propanol and 2-propanol are easier to separate by distillation than cyclohexane and toluene, compare the boiling point differences between the compounds. The boiling point difference between 1-propanol and 2-propanol is 15°C (97°C - 82°C). The boiling point difference between cyclohexane and toluene is 34°C (110°C - 76°C).

Since the boiling point difference between cyclohexane and toluene is greater than that of 1-propanol and 2-propanol, it can be concluded that cyclohexane and toluene are easier to separate by distillation than 1-propanol and 2-propanol.

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the half life of 2n-71 is 2.4 minutes. if we started with 50g at the beginning, how many grams would be left after 12 minutes?

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The half life of 2n-71 is 2.4 minutes. if we started with 50g at the beginning, approximately 0.781 g grams would be left after 12 minutes.

Given that the half-life of N-71 is 2.4 minutes. Hence, T₁/₂=2.4 minutes.

Initial mass of N-71 is 50 g.

We need to find out the mass of N-71 left after 12 minutes. We know that half-life is the time required to reduce the initial quantity to half of its value.

Therefore, we can use the following formula: M(t) = Mo (1/2)^{(t/T1/2)}

Where, M(t) is the mass of the isotope at time 't'.

Mo is the initial mass of the isotope.

T₁/₂ is the half-life of the isotope.

t is the time at which the isotope mass is measured.

Substituting the given values in the above formula, we get:

M(12) = 50 (1/2)^{(12/2.4)}

= 50 (1/2)^{(5)}

= 50/32

= 1.5625 g.

Therefore, the number of grams left after 12 minutes would be approximately 0.781 g.

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halogenated hydrocarbons will eventually break into more harmful component parts if they are exposed to:

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Answer: Halogenated hydrocarbons will eventually break into more harmful component parts if they are exposed to ultraviolet radiation.

Halogenated hydrocarbons are organic compounds that contain one or more halogen atoms in the form of fluorine, chlorine, bromine, or iodine. When they react with other elements, they produce alkyl radicals and halogen atoms, both of which are reactive.

This reaction can be initiated by exposure to light or heat, which can cause the halogen-carbon bond to break and release halogen atoms.

Thus, halogenated hydrocarbons are a significant source of pollution, particularly in the atmosphere. They are also very durable and will linger in the environment for a long time. As a result, they have a significant effect on the environment and human health.

When exposed to ultraviolet radiation, halogenated hydrocarbons break down into more dangerous component parts that can be toxic to humans and animals.

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the molecule to the right is common in many reactions involving electron transfer. which of the circled components are most directly involved in the redox chemistry?

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Answer: The X+ and Y- ions are most directly involved in redox chemistry, as the transfer of electrons between them is the basis of the reaction.

The molecule to the right is a diatomic molecule composed of a positively charged cation, X+, and a negatively charged anion, Y-.

The circled components are the X+ and Y- ions. The redox chemistry involves the transfer of electrons between these two components. In a redox reaction, electrons are transferred from the X+ ion (oxidation) to the Y- ion (reduction). This transfer of electrons results in changes to the oxidation states of the ions, X+ and Y-.

The net effect is the conversion of energy, which can be used to drive various chemical reactions.

In summary, the X+ and Y- ions are most directly involved in redox chemistry, as the transfer of electrons between them is the basis of the reaction.


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a reaction in which simple compounds are assembled into more complex compounds is most accurately described as

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A reaction in which simple compounds are assembled into more complex compounds is most accurately described as a Anabolic reaction.The correct answer is "Anabolic Reaction."

An anabolic reaction is the process of creating more complicated molecules from simpler molecules or small substances. They usually require energy to take place, so anabolic reactions often occur in the body when the energy is released, such as when a person eats food. In biological organisms, this reaction process is important since it allows the organism to build more complicated structures necessary for life and growth.The following is a summary of the five types of chemical reactions:Oxidation-reduction reactionsAcid-base reactionsPrecipitation reactionsComplexation reactionsExchange reactionsAnabolic reactions belong to the family of oxidation-reduction reactions. Anabolic reactions need the input of energy to synthesize more complex molecules. Therefore, they are endergonic. Catabolic reactions, on the other hand, break down molecules into simpler forms and produce energy. They are exergonic as they release energy.A chemical reaction refers to a chemical transformation that involves the breaking and forming of bonds between atoms. Chemical reactions take place in the natural world, and they can be observed every day. The transformation of food into energy in our bodies, for example, is an example of a chemical reaction.

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What mass of carbon dioxide is lost when 2.5 g of magnesium carbonate is decomposed by heating?
What mass of potassium chloride is formed when 2.8 g of potassium hydroxide is completely neutralized by hydrochloric acid?

Answers

Thee mass of carbon dioxide lost when 2.5 g of magnesium carbonate is decomposed by heating is 1.30 g. The mass of potassium chloride formed is 3.72 g.

1- To find the mass of carbon dioxide lost when 2.5 g of magnesium carbonate is decomposed by heating, we need to know the chemical equation for the reaction:

MgCO₃(s) -> MgO(s) + CO₂(g)

From the equation, we can see that 1 mole of magnesium carbonate produces 1 mole of carbon dioxide. The molar mass of magnesium carbonate is 84.31 g/mol, so we can calculate the number of moles of magnesium carbonate as:

n = m/M = 2.5 g / 84.31 g/mol = 0.0296 mol

m = nM = 0.0296 mol * 44.01 g/mol = 1.30 g

2- To find the mass of potassium chloride formed when 2.8 g of potassium hydroxide is completely neutralized by hydrochloric acid, we need to know the balanced chemical equation for the reaction:

KOH(aq) + HCl(aq) -> KCl(aq) + H2O(l)

we can calculate the number of moles of potassium hydroxide as:

n = m/M = 2.8 g / 56.11 g/mol = 0.0499 mol

m = nM = 0.0499 mol * 74.55 g/mol = 3.72 g

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a student titrates a 25 ml of an unknown concentration of hcl with 35 ml of a 0.890 m solution of koh toreach the equivalence point. what is the ph of the unknown hcl solution?

Answers

In order to determine the pH of the unknown HCl solution, a titration calculation must be performed and the pH is 0.903.

The process of adding a standard solution to another solution with the aim of determining the concentration of the second solution is known as titration. HCl is a strong acid, while KOH is a strong base, which implies that when they react, their equivalence point is pH 7.  The pH scale is used to measure the acidity or basicity of a solution. pH is defined as the negative logarithm of the hydrogen ion concentration of a solution. pH is a measure of the acidity or basicity of a solution. It is a dimensionless value that ranges from 0 to 14.

1. Before the titration of the HCl solution with the KOH solution,

Let's calculate the number of moles of KOH using the formula given below:

Number of moles of KOH = concentration of KOH × volume of KOH solution

Number of moles of KOH = 0.890 M × 0.035 L

                                          = 0.03115 mol

We now convert moles of KOH to moles of HCl to find the concentration of HCl using the equation given below:

Moles of KOH = Moles of HCl

0.03115 mol KOH = Moles of HCl

25 mL of HCl = 0.025 L of HCl

Therefore, the concentration of HCl = 0.03115 mol / 0.025 L

                                                            = 1.246 M

We have now found the concentration of the HCl solution to be 1.246 M.

2. To find the pH of HCl, let's first recall that the concentration of H+ ions in a solution of a strong acid is equal to its concentration.

Since HCl is a strong acid, its pH can be found using the formula:

pH = -log[H+]

pH = -log[1.246]

pH = 0.903

Hence, the pH of the unknown HCl solution is 0.903.

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you have a 100 ml solution of 0.02 m sodium carbonate (na2 co3 ). you are given the following information:

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Student question: You have a 100 mL solution of 0.02 M sodium carbonate (Na2 CO3 ). You are given the following information:

Your answer: To work with this 0.02 M sodium carbonate (Na2CO3) solution, you can follow these steps:

Step 1: Calculate the moles of Na2CO3 in the solution.
To do this, use the formula:

Moles = Molarity × Volume (in L)
Moles = 0.02 M × 0.100 L (since 100 mL = 0.100 L)
Moles = 0.002 mol Na2CO3

Step 2: Utilize the information given in the problem.
As you haven't provided any additional information, you can now use the 0.002 moles of Na2CO3 in the 100 mL solution for your further calculations or reactions, depending on the context of your problem.

a solution made up of 40% alcohol by volume is mixed with 4 liters of solution that is 10% alcohol by volume. how much, in liters, of the 40% alcoholic solution is needed to make a mixture that is 25% alcohol by volume?

Answers

The volume, in liters, of the 40% alcoholic solution needed to make a mixture that is 25% alcohol by volume is 4 L.

To find the amount of 40% alcoholic solution needed to make a mixture that is 25% alcohol by volume, we need to use the following formula:

C₁V₁ + C₂V₂ = CfVf

where C₁ is the concentration of the first solution, V₁ is the volume of the first solution, C₂ is the concentration of the second solution, V₂ is the volume of the second solution, Cf is the desired concentration of the resulting mixture, and Vf is the volume of the resulting mixture.

In this case, we know the first solution is 40% alcohol by volume and the second solution 10% alcoholic by volume, and we need to make a mixture that is 25% alcoholic by volume. We need to know the volume of the first solution, V₁.

Plugging in the values, we get:

C₁V₁ + C₂V₂ = CfVf

0.40V₁ + (0.10)(4) = (0.25)(4 + V₁ )

Solving for the value of V₁, we get:

0.40V₁ + 0.40 = 1 + 0.25V₁

0.15V₁ = 0.60

V₁ = 4

Therefore, 4 liters of the first solution is needed.

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A certain liquid X has a normal freezing point of −0.10∘ C and a freezing point depression constant K f =2.85∘C⋅kg ′mol −1, A solution is prepared by dissolving some urea (CH4N2O) in 600.g of X. This solution freezes at −2.1∘C. Calculate the mass of CH4N 2O that was dissolved. Round your answer to 2 significant digits.

Answers

To calculate the mass of CH4N2O dissolved,

we need to follow these steps:

1. Determine the freezing point depression:
ΔTf = T(freezing point of pure X) - T(freezing point of solution) = -0.10°C - (-2.1°C) = 2°C

2. Calculate the molality (m) of the solution using the freezing point depression constant (Kf) and the freezing point depression (ΔTf):
ΔTf = Kf × m
m = ΔTf / Kf = 2°C / 2.85°C·kg/mol = 0.7018 mol/kg

3. Find the moles of CH4N2O in the solution:
moles of CH4N2O = molality × mass of solvent (in kg)
moles of CH4N2O = 0.7018 mol/kg × 0.600 kg = 0.4211 mol

4. Calculate the mass of CH4N2O using its molar mass (60.06 g/mol):
mass of CH4N2O = moles × molar mass = 0.4211 mol × 60.06 g/mol = 25.29 g

Rounded to 2 significant digits,

the mass of CH4N2O dissolved is 25 g.

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the typical concentration of acetic acid in commercial vinegar is 5.0% w/v. calculate the molarity of this solution

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The molarity of the commercial vinegar is 0.833 M.

To calculate the molarity of the commercial vinegar, we need to know the formula of acetic acid, which is CH3COOH. Then, we need to convert the percentage w/v to grams per liter (g/L) by assuming 100 mL of solution.

Finally, we can use the formula of molarity to calculate the concentration of acetic acid in moles per liter (mol/L). Here are the steps:

Step 1: Determine the formula of acetic acid (CH3COOH).

Step 2: Convert the percentage w/v to g/L by assuming 100 mL of solution.5.0% w/v = 5.0 g/100 mL = 50 g/L

Step 3: Calculate the molar mass of acetic acid. C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol.Molar mass = (2 x C) + (4 x H) + (2 x O) = 60.05 g/mol

Step 4: Calculate the number of moles of acetic acid in 1 L of solution.Number of moles = mass / molar massNumber of moles = 50 g / 60.05 g/mol = 0.8327 mol

Step 5:Calculate the molarity of the solution.Molarity = number of moles / volume Molarity = 0.8327 mol / 1 L = 0.833 M

Therefore, the molarity of the commercial vinegar is 0.833 M.

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write the equilibrium equation for the saturated barium chromate solution. barium chromate equilibrium:

Answers

The balanced chemical equation for the reaction between barium chromate and water in order to form a saturated barium chromate solution can be written as follows:

[tex]BaCrO_4 (s) \rightarrow Ba^{2+} (aq) + CrO_4^{2-} (aq)[/tex]

The formula of barium chromate is [tex]BaCrO_4[/tex].

The solution will be saturated once the amount of [tex]BaCrO_4[/tex] dissolved in water reaches its maximum solubility, after which no more [tex]BaCrO_4[/tex] can dissolve in water.

Thus, at the saturation point, the equilibrium equation can be written as follows:

[tex]BaCrO_4 (s) \rightarrow Ba^{2+} (aq) + CrO_4^{2-} (aq)[/tex]

The law of mass action, states that at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This allows us to write an expression for the equilibrium constant (K) based on the concentrations of the reactants and products at equilibrium. The equilibrium constant expression varies depending on the balanced chemical equation for the reaction.

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David added dilute hydrochloric acid to solid calcium carbonate in a beaker. When he weighed the beaker after the bubbling had stopped, he noticed a reduction in mass. Propose why his results did not appear to agree with the law of conservation of mass

Answers

David's results did not appear to agree with the law of conservation of mass due to the release of carbon dioxide gas, which caused a reduction in the total mass of the beaker and its contents.

The law of conservation of mass states that in a chemical reaction, the total mass of the reactants is equal to the total mass of the products. In this case, David added dilute hydrochloric acid to solid calcium carbonate, which is a classic example of an acid-base reaction. The reaction can be represented by the following equation:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

The reaction produces calcium chloride, carbon dioxide, and water. Carbon dioxide gas is released as bubbles, which can be seen as effervescence.

When David weighed the beaker after the bubbling had stopped, he noticed a reduction in mass. This apparent violation of the law of conservation of mass can be explained by the fact that some of the products of the reaction escaped from the beaker in the form of gas. Since carbon dioxide is a gas, it was released into the air, causing a reduction in the total mass of the beaker and its contents. This means that some of the products were not present in the beaker at the end of the reaction, leading to an apparent decrease in mass.

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PLEASE HELP THIS IS URGENT

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The equation for the production of sulfur trioxide gas from sulfur dioxide (57.50 g) and oxygen (20.0 L) using the ideal gas law indicates;

The volume of sulfur trioxide that will be formed at STP is 20.1 L

The volume of sulfur trioxide formed at 15.0°C and 98920 Pa is 21.7 L

What is the ideal gas law?

The ideal gas law is an equation of state that describes an ideal gas behavior. It relates the pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) of the gas and the universal gas constant. The equation is written as P·V = n·R·T

The balanced chemical equation for the reaction is: 2SO₂ (g) + O₂ (g) --> 2SO₃ (g)

First, we need to convert the given amounts of reactants to moles. We can do this by using the molar mass of SO₂ (64.07 g/mol) and the ideal gas law for O₂ (P·V = n·R·T). At STP (Standard Temperature and Pressure), the temperature is 0°C (273.15 K) and the pressure is 1 atm (101325 Pa). The gas constant R is 8.314 J/Kmol.

The number of moles of SO₂ is: 57.50 g/(64.07 g/mol) = 0.897 moles

The number of moles of O₂ is; (101325 Pa)·(20.0 L)/(8.314 J/K.mol)·(273.15 K) = 0.892 moles

Since the ratio of SO₂ to O₂ in the balanced equation is 2:1, SO₂ is the limiting reactant and will determine the amount of product formed.

The number of moles of SO₃ produced is; (0.897 mol SO₂)·(2 mol SO₃/2 mol SO₂) = 0.897 mol (Which is based on the number of moles of SO₂ in the reactant side of the equation)

At STP, one mole of any gas occupies a volume of 22.4 L, so the volume of SO₃ produced at STP is: (0.897 mol) × (22.4 L/mol) ≈ 20.1 L

To find the volume of SO₃ at 15°C and 98920 Pa, we can use the ideal gas law again; P·V = n·R·T

V = (n·R·T)/P = ((0.897 mol)·(8.314 J/K.mol)·(288.15 K))/(98920 Pa) ≈ 21.7 L

Therefore, the volume of sulfur trioxide formed at STP is 20.1 L and at 15°C and 98920 Pa is 21.7 L

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how many ml of alcohol and how many ml of water are needed to prepare a 35% alcohol solution containing 15.0 ml alcohol

Answers

To prepare a 35% alcohol solution containing 15.0 ml of alcohol, you will need 27.9 ml of water and 15 ml of alcohol.

To calculate this, you can use the equation C1V1 = C2V2, where C1 is the concentration of the alcohol (in this case, 35%), V1 is the volume of alcohol you need (15 ml), C2 is the desired concentration of the solution (35%), and V2 is the total volume of the solution (25 ml).

To prepare a 35% alcohol solution containing 15.0 ml alcohol, you will require 27.9 ml of water. The amount of alcohol and water required to prepare a 35% alcohol solution containing 15.0 ml alcohol is given below:

Given data:

Volume of alcohol = 15 ml% of alcohol = 35%

Let us find the amount of water required.

Volume of solution = Volume of alcohol + Volume of water

Using the above formula, Volume of solution = 15 + Volume of water

Let us find the percentage of water in the solution.

35% alcohol solution implies that the solution contains 35 ml of alcohol in 100 ml of solution. Therefore, the amount of solution that contains 1 ml of alcohol is:

1 ml of alcohol = (100 / 35) ml of solution = 20 / 7 ml of solution= 2.86 ml of solution.

Therefore, the amount of solution required to prepare 15 ml of alcohol is:

15 ml of alcohol = 15 × (2.86 ml of solution) = 42.9 ml of solution.

Using the formula for volume of solution, 42.9 ml of solution = 15 ml of alcohol + Volume of water.

Volume of water = 42.9 ml of solution - 15 ml of alcohol= 27.9 ml.

Therefore, you will require 15 ml of alcohol and 27.9 ml of water to prepare a 35% alcohol solution containing 15 ml of alcohol.

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Although ATP is the main energy currency in cells, other molecules, such as NAD, play a central role in some metabolic pathways by transferring electrons. The oxidized form of NAD is NAD+, and the reduced form is NADH. Identify the components of NAD+ and ATP. NH, O=P 0 NH, N OH OH O 0 NH 0 0 0 O=P-0 OH OH OH OH ATP NAD Answer Bank deoxyribose phosphate adenine nicotinamide ribose Select the components that are common to both ATP and NAD. ribose adenine deoxyribose phosphate nicotinamide

Answers

The components that are common to both ATP and NAD are: adenine and ribose. Adenine and ribose are both found in ATP and NAD molecules.

What are ATP and NAD?

ATP stands for Adenosine Triphosphate, which is the primary energy carrier in cells. ATP is an energy-rich molecule that stores energy that can be used by the cell. It is composed of three phosphate groups, an adenine base, and a ribose sugar.

NAD stands for Nicotinamide Adenine Dinucleotide, which is an electron carrier molecule that is involved in many cellular metabolic reactions. It is composed of two nucleotides (adenine and ribose) linked by two phosphate groups.

The oxidized form of NAD is NAD+ while the reduced form is NADH.

The components of ATP and NAD are: Adenine and ribose are the two components common to both ATP and NAD.

Other components are specific to each molecule, as follows: ATP components: Three phosphate groups An adenine base A ribose sugar NAD+ components: Nicotinamide (a type of vitamin B3)Adenine A ribose sugar Two phosphate groups.

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when methanol (ch3oh) is dissolved in water, the temperature of the mixture drops. what does this indicate?

Answers

This indicates that when methanol is dissolved in water, an endothermic reaction is taking place, as the heat energy of the mixture is used up to break the bonds of the methanol molecules and dissolve them in the water.


When methanol (CH3OH) is dissolved in water, the temperature of the mixture drops. This indicates that the dissolution of methanol in water is exothermic. During the dissolution process, energy is released, which is transferred from the mixture to the surroundings. This leads to a decrease in the temperature of the mixture.

The following equation represents the dissolution of methanol in water: CH3OH (l) + H2O (l) → CH3OH(aq). When the methanol and water molecules interact with each other, hydrogen bonds are formed between them. The hydrogen bonds lead to the release of energy, which is the cause of the temperature drop.

Therefore, when methanol is dissolved in water, the temperature of the mixture drops, indicating that the process is exothermic.

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what is the total mass in grams of precipitate that can be produced by mixing a solution made from 300g of solid barium chlorate dissolved in 760 ml of a soloution and 540ml of 0.67m lithium sulfate soloution

Answers

The total mass of precipitate (BaSO4) that can be produced is 175.6 grams.

What is total mass?

Total mass refers to the weight of the shell, its service and structural apparatus, and the largest cargo permitted to be carried

To determine the mass of precipitate that can be produced when solutions of barium chlorate and lithium sulfate are mixed, we need to first write and balance the chemical equation for the reaction:

Ba(ClO3)2 (aq) + Li2SO4 (aq) → BaSO4 (s) + 2LiClO3 (aq)

The balanced equation shows that for every one mole of barium chlorate that reacts, one mole of barium sulfate is produced. Therefore, we need to calculate the number of moles of barium chlorate in the solution to determine the maximum amount of barium sulfate that can be formed.

First, we need to calculate the number of moles of barium chlorate in the solution:

Mass of solid barium chlorate = 300 g

Molar mass of barium chlorate = 2 x atomic mass of Ba + 6 x atomic mass of Cl + 6 x atomic mass of O = 2(137.33 g/mol) + 6(35.45 g/mol) + 6(16.00 g/mol) = 398.22 g/mol

Number of moles of barium chlorate = mass / molar mass = 300 g / 398.22 g/mol = 0.753 mol

Next, we need to calculate the maximum amount of barium sulfate that can be formed from this amount of barium chlorate:

According to the balanced equation, 1 mole of Ba(ClO3)2 produces 1 mole of BaSO4

Therefore, the maximum number of moles of BaSO4 that can be formed is also 0.753 mol

Finally, we can calculate the mass of BaSO4 that can be formed using its molar mass:

Molar mass of BaSO4 = atomic mass of Ba + atomic mass of S + 4 x atomic mass of O = 137.33 g/mol + 32.06 g/mol + 4(16.00 g/mol) = 233.39 g/mol

Mass of BaSO4 = number of moles x molar mass = 0.753 mol x 233.39 g/mol = 175.6 g

Therefore, the total mass of precipitate (BaSO4) that can be produced is 175.6 grams.

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calculate the final molarity of h c l h c l the resulting solution when 5.56 ml of 2.896 m h c l 5.56 ml of 2.896 m h c l is added to 4.44 ml 4.44 ml of water.

Answers

The final molarity of HCl of the resulting solution when 5.56 ml of 2.896 m HCl is added to 4.44 ml of water is 1.61 m.

The final molarity of HCl in the resulting solution can be calculated using the formula:

M₁V₁ = M₂V₂

where M₁ and M₂ are the concentrations of the first HCl solution and the resulting solution, and V₁ and V₂ are the volumes of the first solution and the resulting solution.

For this particular question, M₁ is equal to 2.896 mol/L, V₁ is equal to 5.56 mL, and V₂ is equal to (5.56 + 4.44) = 10 mL.

Substituting in the values, we can get the final concentration in molarity of the resulting solution.

M₂ = M₁V₁ / V₂

M₂ = (2.896 mol/L)(5.56 mL) / 10 mL

M₂ = 1.61 mol/L

In summary, when 5.56 mL of 2.896 m HCl is added to 4.44 mL of water, the final molarity of HCl in the resulting solution is 1.61 mol/L.

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how many moles of potassium phosphate (k3po4) are produced from 8.0 8.0 mol of potassium hydroxide (koh)?

Answers

8.0 moles of potassium hydroxide (KOH) produces 2.67 moles of potassium phosphate (K₃PO₄).


The reaction is written as:
3KOH + H₃PO₄ -> K₃PO₄ + 3H₂O

This reaction can be described as a double replacement reaction, meaning that two reactants swap partners and form two new products. In this case, the two reactants are KOH and H₃PO₄ and the two products are K₃PO and H₂O.

According to the reaction, 3 moles of KOH will give 1 mole of K₃PO₄.


Therefore, since 8.0 moles of KOH are reacted with phosphoric acid, the number of moles of potassium phosphate (K₃PO₄)produced will be (8/3) = 2.67 moles.

The amount of potassium phosphate produced is directly proportional to the amount of potassium hydroxide used in the reaction.


Therefore, when  8.0 moles of potassium hydroxide (KOH) reacts with H₃PO₄, 2.67 moles of potassium phosphate(K₃PO₄) are produced.

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which of the following statements is true for real gases? choose all that apply. the volume occupied by the molecules can cause an increase in pressure compared to the ideal gas. as attractive forces between molecules increase, deviations from ideal behavior become more apparent at relatively high temperatures. attractive forces between molecules cause an increase in pressure compared to the ideal gas. as molecules increase in size, deviations from ideal behavior become more apparent at relatively high pressures.

Answers

True statements for real gases are:

Option a): The volume occupied by the molecules can cause an increase in pressure compared to the ideal gasOption b): As attractive forces between molecules increase, deviations from ideal behavior become more apparent at relatively high temperatures

Real gases are gases that do not behave perfectly like ideal gases at all conditions of temperature and pressure. They deviate from ideal behavior under certain conditions, especially at high pressures and low temperatures.

The assumptions of the Kinetic Theory of Gases that make gases to be called ideal gases are not valid under all conditions of temperature and pressure. However, ideal gases serve as a reference point for understanding the behavior of real gases. The molecules of a real gas do occupy some space and have some volume. Therefore, they will cause an increase in pressure compared to ideal gases.

The attractive forces between the molecules of a real gas cause a decrease in the volume of the gas compared to the ideal gas. This results in an increase in pressure.Therefore, statement a is true.

Attractive forces between the molecules of gas become more significant as the temperature is decreased. This will result in deviations from ideal behavior. The attractive forces between the molecules cause them to stay close to each other. Therefore, the size of the molecules is more apparent at high pressures. Thus, statement b is true.

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how many kj of energy will be released when 4.72 g of carbon react with excess oxygen gas to produce carbon dioxide

Answers


When 4.72 g of carbon react with excess oxygen gas to produce carbon dioxide, 609.6 kJ of energy will be released

This is because the reaction between carbon and oxygen is exothermic, meaning that energy is released when the reaction takes place.

For carbon, the energy released per mole is 717 kJ. For oxygen, the energy released per mole is 498 kJ.

The total energy released in the reaction, you need to multiply the energy released per mole by the number of moles of each element present in the reaction.

In this reaction, 4.72 g of carbon and excess oxygen are present. The atomic mass of carbon is 12.0107 g/mol, which means that 0.3948 moles of carbon are present in 4.72 g.

The atomic mass of oxygen is 15.999 g/mol, which means that 6.26 moles of oxygen are present.

Multiplying the energy released per mole of each element by the number of moles present in the reaction yields the total energy released.

This is equal to 717 kJ/mol x 0.3948 mol = 282.3 kJ, and 498 kJ/mol x 6.26 mol = 3127.48 kJ.

Adding these two values together gives the total energy released in the reaction,

which is equal to 3127.48 kJ + 282.3 kJ = 3409.78 kJ. Since 1 kJ = 1000 J, the total energy released in this reaction is 3409.78 kJ = 3.40978 x 10^6 J.

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how can the chemical potential energy in an endothermic reaction best be described?(1 point) responses product and reactant chemical potential energy varies in different environments. product and reactant chemical potential energy varies in different environments. products and reactants have the same chemical potential energy. products and reactants have the same chemical potential energy. products have higher chemical potential energy than reactants. products have higher chemical potential energy than reactants. reactants have higher chemical potential energy than products.

Answers

The chemical potential energy in an endothermic reaction can best be described as products having a higher chemical potential energy than reactants.

The chemical potential energy of endothermic reactions

The chemical potential energy in an endothermic reaction can best be described as:

"Products have higher chemical potential energy than reactants."

In an endothermic reaction, energy is absorbed by the system, and the products of the reaction have a higher potential energy than the reactants. This increase in potential energy is typically in the form of heat, which is absorbed from the environment.

Therefore, the correct option products have higher chemical potential energy than reactants.

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if you mix 312 gram(s) of a solute in water and bring it to a final volume of 378 ml, what will be the concentration of the resulting solutions in g/ml? answers cannot contain more than one decimal place.

Answers

Answer : If you mix 312 g of a solute in water and bring it to a final volume of 378 mL, the concentration of the resulting solution will be 0.82 g/mL.

To calculate the concentration, divide the mass of solute (312 grams) by the volume of the resulting solution (378 mL). Thus, 312 g/cc/378 mL = 0.82 g/mL. The concentration of a solution is the ratio of the amount of solute (in this case, 312 grams) to the total volume of the solution (in this case, 378 mL). The concentration can be expressed as either grams of solute per milliliter of solution (g/mL) or moles of solute per liter of solution (mol/L).

The concentration of a solution can be increased by either adding more solute or decreasing the volume of the solution. For example, if you mix 500 g of a solute in 500 mL of water, the concentration of the resulting solution will be 1 g/mL. If you then reduce the volume of the solution to 250 mL, the concentration will increase to 2 g/mL.

It is also important to note that the concentration of a solution cannot be greater than the solubility of the solute. This means that the solute must be completely dissolved before it can be added to the solution in order for the concentration to be increased.



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