A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. The empirical formula of the compound was found to be C3H6O2. Show how this was calculated.

What does the empirical formula tell you about the compound?

The molar mass of the actual compound was found to be 222.27g/mol. Find the molecular formula of this compound. What does the molecular formula tell you about the compound?

Can you see what type of functional group this compound could have?

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  initial concentration is  [tex]C_f = 0.0022 \ M[/tex]

Explanation:

From the question we are told that

    The volume of solution A is  [tex]V_i = 10.0 mL[/tex]

    The concentration of A is [tex]C_i = 0.0200 \ M[/tex]

    The volume of solution B  is  [tex]V_B = 10.0mL[/tex]

    The volume of water is  [tex]V_{w } = 70.0 mL[/tex]

Generally the law of dilution is mathematically represented as

             [tex]C_i * V_i = C_f * V_f[/tex]

Where  [tex]C_f[/tex] is the concentration of  the mixture

            [tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as

            [tex]V_f = 10 + 10 + 70[/tex]

           [tex]V_f = 90mL[/tex]

So  

      [tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]

substituting values

       [tex]C_f = \frac{0.0200 * 10 }{90}[/tex]

       [tex]C_f = 0.0022 \ M[/tex]

Note the mixture obtained is  [tex]KIO_3[/tex]

Answer:

- N2 does not exist as a liquid at pressures below 0.127 atm.

- N2 is a solid at 16.7 atm and 56.5 K.

- N2 is a liquid at 1.00 atm and 73.9 K

- N2 is a gas at 0.127 atm and 84.0 K.

Explanation:

Hello,

At first, we organize the information:

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).

- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

Best regards.

Answer:

The answer to your question is given below

Explanation:

The balanced equation for the reaction is given below:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

1. Writing an expression for the equilibrium constant, K.

The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, we can write the equilibrium constant, K for the reaction as follow:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²

2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.

Answer:

Explanation: M(PCL5)= 31 + 5(35.5)

=208.5g/mol

M(H20)= 18g/mol

n(PCL5) = 75.5÷208.5

= 0.362mol

n(HCl)/n(PCL5)= 5/1

n(HCl)= 5×0.362

=1.81mol of HCl

Answer:

The base is involved in the rate determining step of an E2 reaction mechanism

Explanation:

Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;

Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.

For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;

Rate = k[alkyl halide] [base]

The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.

Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.

Answer:

1.1 × 10⁻³ g

Explanation:

Step 1: Given data

Henry's Law constant for methane (k): 1.4 × 10⁻³ M/atm

Volume of water (=volume of solution): 75 mL

Partial pressure of methane (P): 0.68 atm

Step 2: Calculate the concentration of methane in water (C)

We will use Henry's law.

[tex]C = k \times P = 1.4 \times 10^{-3}M/atm \times 0.68atm = 9.5 \times 10^{-4}M[/tex]

Step 3: Calculate the moles of methane in 75 mL of water

[tex]\frac{9.5 \times 10^{-4}mol}{L} \times 0.075 L = 7.1 \times 10^{-5}mol[/tex]

Step 4: Calculate the mass corresponding to 7.1 × 10⁻⁵ mol of methane

The molar mass of methane is 16.04 g/mol.

[tex]7.1 \times 10^{-5}mol \times \frac{16.04g}{mol} = 1.1 \times 10^{-3} g[/tex]

Answer:

The calculator add the CO2 released from the use of electricity, released from driving and the CO2 from the waste that we disposed.

Explanation:

The carbon dioxide, CO2 is what the human body does not need, therefore, we breathe it out, hence taking in oxygen(respiration process). The plants need oxygen for the production of their own food.

The carbon calculator estimate the amount of CO2 that each individual releases into the atmosphere through the consideration of several factors such as the kind of food that we eat.

Therefore, if we are to use the carbon calculator to determine the amount of CO2 that each individual releases into the atmosphere we will have:

The amount of CO2 that each individual releases into the atmosphere =( CO2 released from the use of electricity) + (CO2 released from driving) + (the CO2 from the waste that we disposed).

Answer

Naphthalene is a non electrolyte

If the unknown compound is an electrolyte it gives 2 or more ions in solution

( NaCl >> Na+ + Cl- => 2 ions

Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)

the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )

For naphthalene

delta T = 1.86 x m

for a salt that gives 2 ions

delta T = 1.86 x m x 2

hence the lowering in freezion point of unkown is greater then napthalene

Answer:

Oxidation by FAD  

Explanation:

1. Oxidation by NAD⁺

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;                  E°´ =  -0.031 V  

NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺;                      E°´ = -0.320 V

Succinate + NAD⁺ ⇌ Fumarate  + NADH + H⁺; E°' =  -0.351 V

2. Oxidation by FAD

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻;        E°´ = -0.031 V  

FAD + 2H⁺ + 2e⁻ ⇌ FADH₂;                    E°´  = -0.219 V

Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V

Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.

FAD is the stronger oxidizing agent.

The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.  

Answer:

Gaseous

Explanation:

Gasses can move freely and do not form the shape of their containers

Liquids are more free than solids, but they conform to the shape of their container

Solids are not free

Answer:

The new solution is 1.4% m/V

Explanation:

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:

[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]

We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

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Answer:

1. stability factor

2. steric hindrance factor

Explanation:

stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.

ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive

Answer:

[tex]V=27992L=28.00m^3[/tex]

Explanation:

Hello,

In this case, the combustion of methane is shown below:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]

Best regards.

Answer and Explanation:

The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.

If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.

Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol

Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol

moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-

According to this, the following are the answers to the sentences:

a. The number of moles of CH₃COOH will remain the same ⇒ FALSE

The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions

b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE

Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)

c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE

The equilibrium concentration of OH⁻ is increased

d. The pH will decrease⇒ FALSE

pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:

[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]

pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15

Thus, the pH will increase.

Answer:

a. boiling

Explanation:

Answer:

here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.

hope it helps...

The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.

Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.

There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.

The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.

Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

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Answer:

the density if vinegar will also be needed

Explanation:

Because this is an experiment of volumetric analysis

Answer:

i think option a is correct answer because when there is low temperature then the kinetic enegry will be very less and the atoms moves very slowly.

Answer:

A. very, very slowly

Explanation:

A is the answer because atoms will move faster in hot gas than in cold gas.

Answer:

1 mole of CaC₂ will produce 26g of C₂H₂ or 64.1g of CaC₂ will produce 26g of C₂H₂

Explanation:

Hello,

To solve this question, we'll require a balanced chemical equation of reaction between calcium carbide and water.

Equation of reaction

CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

Molar mass of calcium carbide (CaC₂) = 64.1g/mol

Molar mass of water (H₂O) = 18g/mol

Molar mass of calcium hydroxide (Ca(OH)₂) = 74g/mol

Molar mass of ethyne (C₂H₂) = 26g/mol

From the equation of reaction, 1 mole of CaC₂ will produce 1 mole of C₂H₂

1 mole of CaC₂ = mass / molar mass

Mass = 1 × 64.1

Mass = 64.1g

1 mole of C₂H₂ = mass / molar mass

Mass = 1 × 26

Mass = 26g

Therefore, 1 mole of CaC₂ will produce 26g of C₂H₂

Note: this is a hypothetical calculation since we were not given the initial mass of CaC₂ that starts the reaction

Answer:

143pm is the radius of an Al atom

Explanation:

In a face centered cubic structure, FCC, there are 4 atoms per unit cell.

First, you need to obtain the mass of an unit cell using molar mass of Aluminium  and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.

Mass of an unit cell

As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:

4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g

Edge length

As density of aluminium is 2.71g/cm³, the volume of an unit cell is:

1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³

And the length of an edge of the cell is:

∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m

Radius:

As in FCC structure, Edge = √8 R, radius of an atom of Al is:

4.044x10⁻¹⁰m = √8 R

1.430x10⁻¹⁰m = R.

In pm:

1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =

The radius of the atom of Al in the FCC structure has been 143 pm.

The FCC lattice has been contributed with atoms at the edge of the cubic structure.

The FCC has consisted of 4 atoms in a lattice.

[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole

4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles

The mass of 1 mole Al has been 26.98 g/mol.

The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g

The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.

Density = [tex]\rm \dfrac{mass}{volume}[/tex]

Volume = Density × Mass

The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g

The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]

Volume = [tex]\rm edge\;length^3[/tex]

6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]

Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm

Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm

Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.

Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]

4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]

Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.

1 m = [tex]\rm 10^1^2[/tex] pm

1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.

The radius of the atom of Al in the FCC structure has been 143 pm.

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Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

[tex][A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M[/tex]

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

[tex]\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%[/tex]

The question is incomplete as the options are missing, however, the correct complete question is attached.

Answer:

The correct answer is option A. ( check image)

Explanation:

The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.

Thus, the correct answer is option A. ( check image)

Answer:

2ErF3 + 3Mg → 2Er + 3MgF2

Explanation:

Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.

Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2

The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6

Answer:

Compounds whose real bond angle are the same as ideal bond angle;

SF6, BF3, CH4, PCI5

Compounds whose real bond angles differ from ideal bond angles;

H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5

Explanation:

According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.

The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.

For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.

Considering the definition of bond and the different type of bonds, valence is not one of  the types of bonds.

A chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.

In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.

The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.

An ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.

Metallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.

In summary, valence is not one of  the types of bonds. The types of bonds are covalent, ionic and metallic.

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Answer:

A. None of these.  

Explanation:

A crystal structure is an arrangement of atoms or ions in a repeating three-dimensional array.

B. is wrong. Metal atoms, such as gold, arrange themselves into a crystal structure.

C. is wrong. Ionic solids, such as sodium chloride, arrange themselves into a crystal structure.

D. is wrong. Macromolecules (network solids), such as diamond, arrange themselves into a crystal structure.

The correct answer is None of these.  

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Answer: 10 moles/kg.

Explanation:

Given, Mass of solute = 24.2 g

Molar mass of solute = 24.2 g/mol

[tex]\text{Moles of solute =}\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}\\\\=\dfrac{24.2}{24.2}=1[/tex]

Mass of solvent = 100.0g = 0.1 kg  [1 g=0.001 kg]

[tex]\text{Molality}=\dfrac{\text{Moles of solute}}{\text{kilograms of Solvent}}\\\\=\dfrac{1}{0.1}\\\\=10\ moles/kg[/tex]

Hence, the molality of the resulting solution is 10 moles/kg.

Answer:

Answer:

C. An electron at this electrode has a higher potential energy than it has at a standard hydrogen electrode.

Explanation:

The standard hydrogen electrode (SHE) is used to measure the electrode potential of substances. The standard hydrogen electrode is arbitrarily assigned an electrode potential of zero. Recall that electrode potentials are always measured as reduction potentials in electrochemical systems.

For an electrode that has a negative electrode potential, electrons at this electrode have a higher potential energy compared to electrons at the standard hydrogen electrode. Electrons flow from this electrode to the hydrogen electrode.

On the other hand, a positive electrode potential implies that an electron at this electrode has a lower potential energy than it has at a standard hydrogen electrode. Hence electrons will flow from the standard hydrogen electrode to this electrode.

Answer: The empirical formula is [tex]PbO_2[/tex]

Explanation:

Mass of Pb =  4.33 g

Mass of O = (5.00-4.33) g = 0.67 g

Step 1 : convert given masses into moles

Moles of Pb =[tex]\frac{\text{ given mass of Pb}}{\text{ molar mass of Pb}}= \frac{4.33g}{207g/mole}=0.021moles[/tex]

Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.67g}{16g/mole}=0.042moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Pb = [tex]\frac{0.021}{0.021}=1[/tex]

For O = [tex]\frac{0.042}{0.021}=2[/tex]

The ratio of Pb O=  1: 2

Hence the empirical formula is [tex]PbO_2[/tex]