The pressure of the first container is at 60 kPa. What is the pressure of the container with the 3N volume

A polar covalent bond occurs when one of the atoms in the bond provides both bonding electrons. The statement is false.

A polar covalent bond occurs when two atoms share a pair of electrons unevenly, meaning that one atom has a greater electronegativity than the other atom.

This results in a partial positive charge on the less electronegative atom and a partial negative charge on the more electronegative atom, creating a dipole.

The situation described in the statement, where one atom provides both bonding electrons, refers to an ionic bond. In an ionic bond,

one atom transfers its electrons to another atom, creating a positively charged cation and a negatively charged anion. These oppositely charged ions are then attracted to each other, forming the ionic bond.

In summary, the statement is false because a polar covalent bond involves the unequal sharing of electrons between two atoms,

while the scenario described refers to an ionic bond where one atom provides both bonding electrons.

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The standard enthalpy change for the reaction A + B ⟶ 2C at 298 K is 670.218 kJ/mol.

For the standard enthalpy change (ΔH°) for the reaction A + B ⟶ 2C, we can use Hess's law, which states that the overall enthalpy change for a reaction is independent of the pathway taken. We can break down the given reaction into two steps:

A + B ⟶ 2D ΔH1 = 670.4 kJ/mol

2D ⟶ 2C ΔH2 = -δ° = -182.0 J/K/mol = -0.182 kJ/K/mol

The enthalpy change for the desired reaction is equal to sum of the enthalpy changes of these two steps:

A + B ⟶ 2C ΔH° = ΔH1 + ΔH2/1000 = 670.4 + (-0.182) = 670.218 kJ/mol

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The elements arranged in increasing electronegativity are: boron, carbon, nitrogen, oxygen.

Explanation: Electronegativity is the tendency of an atom to attract electrons towards itself when it forms a chemical bond. Boron has the lowest electronegativity value among the given elements, followed by carbon, nitrogen, and oxygen. This is because electronegativity increases as we move from left to right across a period in the periodic table, and decreases as we move down a group. Boron is located in group 13 and period 2, while oxygen is located in group 16 and period 2. Therefore, oxygen has the highest electronegativity value among the given elements. Nitrogen has a slightly higher electronegativity than carbon because it is located further to the right in the same row of the periodic table.

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The pressure of the dry hydrogen gas is 97.47 kPa.

The total pressure in the collection flask is the sum of the partial pressures of the dry hydrogen gas and the water vapor.

Using Dalton's law of partial pressures, the partial pressure of the dry hydrogen gas can be calculated by subtracting the partial pressure of water vapor from the total pressure.

[tex]P(dry H2) = P(total) - P(H2O) = 99.8 kPa - 2.33 kPa = 97.47 kPa.[/tex]

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The radius of the smallest Bohr orbit in doubly ionized lithium is 5.29 x 10^-12 m and the energy of this orbit is -13.6 eV.

The radius of the smallest Bohr orbit in doubly ionized lithium can be determined using the formula for the radius of the nth orbit in a hydrogen-like atom. For a doubly ionized lithium, the atomic number is 3, and the number of electrons is 1. Therefore, the radius of the smallest Bohr orbit can be calculated as:

r = (n^2*h^2)/(4π^2*m*e^2)

where n is the principal quantum number, h is Planck's constant, m is the reduced mass of the electron and nucleus, and e is the charge of the electron.

For the smallest orbit (n=1), the radius of the orbit is:

r = (1^2*(6.626 x 10^-34 J s)^2)/(4π^2*(9.109 x 10^-31 kg + 6.941 x 1.661 x 10^-27 kg)*(1.602 x 10^-19 C)^2)

r = 5.29 x 10^-12 m

The energy of this orbit can be calculated using the formula:

E = (-13.6 eV)/n^2

where n is the principal quantum number. For the smallest orbit (n=1), the energy of the orbit is:

E = (-13.6 eV)/1^2

E = -13.6 eV

Therefore, the radius of the smallest Bohr orbit in doubly ionized lithium is 5.29 x 10^-12 m and the energy of this orbit is -13.6 eV.

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The standard cell potential, ∘cell, for the reaction is 0.46 V.

To calculate the standard cell potential (∘cell), we use the equation ∘cell = ∘red, cathode - ∘red, anode, where ∘red is the standard reduction potential of the half-reaction. From the given reaction, the reduction half-reaction is:

Ag+ (aq) + e- → Ag(s) ∘red = +0.80 V

And the oxidation half-reaction is:

Cu(s) → Cu2+ (aq) + 2 e- ∘red = -0.34 V

Substituting the values into the equation, we get:

∘cell = +0.80 V - (-0.34 V) = 1.14 V

However, since the given reaction is the reverse of the spontaneous reaction, we need to reverse the sign of the ∘cell value to get the correct answer. Therefore,

∘cell = -1.14 V

To convert this value to kilojoules per mole (kJ/mol), we use the equation:

∆G = -nF∘cell

Where n is the number of moles of electrons transferred in the reaction, and F is the Faraday constant (96,485 C/mol).

Since 2 moles of electrons are transferred in the reaction, we have:

∆G = -2 * 96485 C/mol * (-1.14 V) = +208,583 J/mol = +208.58 kJ/mol

Therefore, the standard cell potential (∘cell) for the given reaction is -1.14 V and the standard free energy change (∆G) is +208.58 kJ/mol.

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The correct electron configuration for nitrogen is option D, which is 1s22s22p3

The correct electron configuration for nitrogen is option D, which is 1s22s22p3. To explain this configuration, we need to understand the basic structure of an atom. An atom consists of a nucleus made up of protons and neutrons, surrounded by electrons orbiting in shells or energy levels. The first shell can hold up to 2 electrons, the second can hold up to 8, and the third can hold up to 18.
Nitrogen has 7 electrons, so we start by placing 2 electrons in the first shell, which is the 1s orbital. Then, we add 2 more electrons to the second shell, which is the 2s orbital. The remaining 3 electrons are placed in the 2p orbital, which is also in the second shell. Thus, the electron configuration for nitrogen is 1s22s22p3. This configuration explains why nitrogen has a valence of 3 and tends to form 3 covalent bonds with other elements.

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Answer:An aqueous solution of NaHCO3 (sodium bicarbonate) is more effective than a stronger aqueous solution of NaOH (sodium hydroxide) in the separation of an equal mixture of benzoic acid and 2-naphthol because NaHCO3 has a pKa value of 6.4 which is closer to the pKa value of benzoic acid (4.2) than NaOH, which has a pKa value of 15.7. When an acid is added to a solution containing a conjugate base, the acid will react with the conjugate base to form the corresponding conjugate acid. By using NaHCO3, benzoic acid will be converted into its water-soluble sodium salt, while 2-naphthol will remain in the organic layer. Since NaOH is a stronger base, it will not be able to selectively convert benzoic acid to its sodium salt, and 2-naphthol will also be converted to its sodium salt.

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In the given equation, 2 moles of [tex]NH_{3}[/tex] react with 3 moles of CuO to produce 3 moles of Cu and 1 mole of [tex]N_{2}[/tex]. Therefore, when 91 moles of CuO are consumed, 30.33 moles of N_{2}can be produced.

According to the balanced chemical equation:

2 NH_{3} + 3 CuO -> 3 Cu + N_{2}+ 3 [tex]H_{2}O[/tex]

From the equation, we can see that 2 moles of NH_{3} react with 3 moles of CuO to produce 1 mole of N_{2}

To determine the moles of N2 produced when 91 moles of CuO are consumed, we can set up a proportion based on the stoichiometric ratios:

(2 moles NH_{3} / 3 moles CuO) = (1 mole N_{2}/ X moles CuO)

Simplifying the proportion, we have:

X = (1 mole N_{2} * 3 moles CuO) / (2 molesNH_{3})

Calculating the value of X, we find that X is equal to 1.5 moles N_{2}.

Therefore, when 91 moles of CuO are consumed, 1.5 moles of N_{2} can be produced.

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The balanced equations are:

CrO₄²⁻ + 8H⁺ + 3Fe²⁺ → Cr³⁺ + 3Fe³⁺ + 4H₂O

MnO⁻₄ + ClO⁻₂ + 2OH⁻ + 2H⁺ + 2e⁻ → MnO₂ + ClO⁻₄ + H₂O

To balance the given chemical equations, we need to ensure that the number of atoms of each element is equal on both the reactant and product sides of the equation. We can achieve this by adding coefficients to each species as necessary.

CrO₄²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺

We can start by balancing the oxygen atoms by adding water molecules:

CrO₄²⁻ + Fe²⁺ + 8H⁺ → Cr³⁺ + Fe³⁺ + 4H₂O

Next, we balance the hydrogen atoms by adding hydrogen ions:

CrO₄²⁻ + Fe²⁺ + 8H⁺ → Cr³⁺ + Fe³⁺ + 4H₂O

Finally, we balance the charges by adding electrons to the appropriate side:

CrO₄²⁻ + 8H⁺ + 3e⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ + 4H₂O

The balanced equation is:

CrO₄²⁻ + 8H⁺ + 3Fe²⁺ → Cr³⁺ + 3Fe³⁺ + 4H₂O

MnO⁻₄ + ClO⁻₂ → MnO₂ + ClO⁻₄

This reaction takes place in a basic solution, which means we need to start by adding hydroxide ions (OH⁻) to balance the equation:

MnO⁻₄ + ClO⁻₂ + OH⁻ → MnO₂ + ClO⁻₄

Next, we balance the oxygen atoms by adding water molecules:

MnO⁻₄ + ClO⁻₂ + OH⁻ → MnO₂ + ClO⁻₄ + H₂O

We can now balance the hydrogen atoms by adding hydrogen ions:

MnO⁻₄ + ClO⁻₂ + OH⁻ + H⁺ → MnO₂ + ClO⁻₄ + H₂O

Finally, we balance the charges by adding electrons to the appropriate side:

MnO⁻₄ + ClO⁻₂ + 2OH⁻ + 2H⁺ + 2e⁻ → MnO₂ + ClO⁻₄ + H₂O

The balanced equation is:

MnO⁻₄ + ClO⁻₂ + 2OH⁻ + 2H⁺ + 2e⁻ → MnO₂ + ClO⁻₄ + H₂O

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The electron-pair geometry for P in PF6- is octahedral.

The electron-pair geometry for an atom is determined by the arrangement of electron pairs around the central atom. In the case of PF6-, the central atom is phosphorus (P), and it is bonded to six fluoride (F) atoms.

To determine the electron-pair geometry, we consider both the bonding pairs and the lone pairs of electrons around the central atom.

In PF6-, phosphorus forms five sigma (σ) bonds with the fluorine atoms, resulting in five bonding pairs. The valence electron configuration of phosphorus is 3s^2 3p^3, so it has one lone pair of electrons.

The combination of the bonding and lone pairs of electrons results in an electron-pair geometry of octahedral. In an octahedral geometry, the electron pairs are arranged around the central atom in a three-dimensional shape resembling two pyramids stacked on top of each other.

The bonding pairs and the lone pair are positioned at the corners of an octahedron.

In PF6-, the phosphorus atom is at the center of an octahedron, with the six fluoride atoms located at the corners. The bonding pairs are directed towards the fluorine atoms, while the lone pair occupies one of the positions of the octahedron.

This arrangement of electron pairs gives rise to an octahedral electron-pair geometry for the phosphorus atom in PF6-.

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The balanced reaction for the galvanic cell based on the given equation is as follows:
Fe2+ + Cr2O72- + 14H+ → Fe3+ + 2Cr3+ + 7H2O
The voltage of the standard cell carrying out the given reaction is -0.559 V.

To calculate the voltage of the standard cell carrying out this reaction, we need to use the Nernst equation:
Ecell = E°cell - (RT/nF) ln(Q)
where Ecell is the voltage of the cell, E°cell is the standard cell potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the reaction, F is the Faraday constant, and Q is the reaction quotient.
For this reaction, n = 6 (since 6 electrons are transferred), T = 298 K, R = 8.314 J/K mol, and F = 96,485 C/mol. The standard cell potential can be calculated using the standard reduction potentials of the half-reactions involved in the cell reaction.
Fe3+ + e- → Fe2+  E° = +0.771 V
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O  E° = +1.33 V
The standard cell potential can be calculated as:
E°cell = E°reduction (reduced species) - E°reduction (oxidized species)
E°cell = E°(Fe2+/Fe3+) - E°(Cr2O72-/Cr3+)
E°cell = (+0.771 V) - (+1.33 V)
E°cell = -0.559 V
Now, we need to calculate the reaction quotient (Q) for the given reaction. The reaction quotient is calculated using the concentrations of the species involved in the reaction.
Q = [Fe3+][Cr3+] / [Fe2+][Cr2O72-]
Assuming standard conditions, the concentration of each species is 1 M.
Q = (1)(1) / (1)(1)
Q = 1
Finally, we can calculate the voltage of the cell using the Nernst equation.
Ecell = E°cell - (RT/nF) ln(Q)
Ecell = -0.559 V - (8.314 J/K mol * 298 K / (6 * 96,485 C/mol)) ln(1)
Ecell = -0.559 V
Therefore, the voltage of the standard cell carrying out the given reaction is -0.559 V.

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The mass of nitrosyl chloride (NOCl) occupying a volume of 0.2570 L at a temperature of 325 K and a pressure of 113.0 kPa is approximately 0.229 grams.

To determine the mass of nitrosyl chloride (NOCl) occupying a volume of 0.2570 L at a temperature of 325 K and a pressure of 113.0 kPa, we can use the ideal gas law equation, PV = nRT.

First, let's convert the given pressure to atmospheres (1 kPa ≈ 0.00987 atm):

Pressure = 113.0 kPa * 0.00987 atm/kPa ≈ 1.115 atm.

Next, we need to convert the volume to liters:

Volume = 0.2570 L.

The gas constant (R) is 0.0821 L·atm/(mol·K).

Now we can use the ideal gas law to calculate the number of moles (n) of NOCl:

n = (Pressure * Volume) / (R * Temperature)

= (1.115 atm * 0.2570 L) / (0.0821 L·atm/(mol·K) * 325 K)

= 0.0035 mol.

Finally, we can determine the mass of NOCl using the molar mass of NOCl, which is 65.46 g/mol:

Mass = n * Molar mass

= 0.0035 mol * 65.46 g/mol

= 0.229 g.

Therefore, the mass of nitrosyl chloride (NOCl) occupying a volume of 0.2570 L at a temperature of 325 K and a pressure of 113.0 kPa is approximately 0.229 grams.

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To perform the given operation and express  in scientific notation, need to subtract 5.00x10^4 from 7.00x10^5.

Step 1: Perform the subtraction:

7.00x10^5 - 5.00x10^4 = 700,000 - 50,000 = 650,000Step 2: Determine the appropriate scientific notation for the result.

The result, 650,000, can be expressed in scientific notation as 6.50x10^5.

To represent this in the requested format, we need to determine the exponent and adjust the coefficient accordingly.

The original coefficient, 6.50, can be written as 6.50x10^5.

Therefore, the answer in scientific notation is 6.50x10^5.

The simplest approach to express a huge value is with scientific notation. In scientific notation, a number is divided into two pieces.

 The numbers (the decimal point will come after the first number)

 10 (the power that positions the decimal point correctly)

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The structure of A is: 1-butene, which upon reacting with sulfuric acid forms 1-butanol (B). The subsequent reaction of B with sodium metal in dry THF followed by methyl iodide produces t-butyl methyl ether.

The reaction of A (C4H8) with cold aqueous sulfuric acid produces B (C4H10O). The subsequent reaction of B with sodium metal in dry THF followed by methyl iodide yields t-butyl methyl ether.

From the given information, we can infer that A is an unsaturated compound with a carbon-carbon double bond, which reacts with the sulfuric acid to form an alcohol B through hydration.

To draw the structure of A, we start by considering all the possible isomers of C4H8 with a carbon-carbon double bond. There are two isomers of butene: 1-butene and 2-butene.

Since the reaction of A with sulfuric acid produces an alcohol, we can infer that the double bond in A is terminal, and the resulting alcohol B has a primary alcohol group.

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The final yield of [tex]H_2O[/tex] in moles is 0.480 mol and can be determined by calculating the stoichiometric ratio between methane and water in the balanced chemical equation and multiplying it by the given amount of methane.

To find the final yield of [tex]H_2O[/tex] in moles, we need to use the balanced chemical equation for the combustion of methane:

[tex]CH_4 + 2O_2[/tex]→ [tex]CO_2 + 2H_2O[/tex]

According to the equation, for every one mole of methane ([tex]CH_4[/tex]) that reacts, two moles of water ([tex]H_2O[/tex]) are produced. Therefore, the stoichiometric ratio between methane and water is 1:2.

Given that we have 0.240 mol of methane, we can calculate the moles of water produced by multiplying the amount of methane by the stoichiometric ratio:

[tex]0.240 mol CH_4 * (2 mol H_2O / 1 mol CH_4) = 0.480 mol H_2O[/tex]

Hence, the final yield of [tex]H_2O[/tex] in moles is 0.480 mol.

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Phenyl acetate hydrolyzes more rapidly than benzyl acetate, while methyl acetate hydrolyzes faster than phenyl acetate.

The rate at which esters hydrolyze depends on the stability of the intermediate formed during the reaction.

In the case of phenyl acetate and benzyl acetate, phenyl acetate hydrolyzes more rapidly because it forms a more stable intermediate. The phenoxide ion produced is stabilized through resonance with the phenyl ring.

Comparing methyl acetate and phenyl acetate, methyl acetate hydrolyzes faster because the methyl group is less bulky, resulting in a more accessible carbonyl carbon for nucleophilic attack, which leads to a faster hydrolysis reaction.

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Benzyl acetate hydrolyzes more rapidly than phenyl acetate, and methyl acetate hydrolyzes more rapidly than phenylacetate. the correct answer is (a) benzyl acetate and (b) methyl acetate.

The rate of hydrolysis of an ester depends on several factors, including the size of the alkyl group attached to the carbonyl carbon and the electron density around the carbonyl group. In general, esters with larger alkyl groups attached to the carbonyl carbon undergo hydrolysis more slowly than those with smaller alkyl groups. This is because larger alkyl groups hinder the approach of water molecules to the carbonyl carbon, thus reducing the rate of hydrolysis.  Comparing the given options, benzyl acetate has a larger alkyl group than phenyl acetate, so it undergoes hydrolysis more rapidly. Similarly, methyl acetate has a smaller alkyl group than phenyl acetate, so it undergoes hydrolysis more rapidly. Therefore, the correct answer is (a) benzyl acetate and (b) methyl acetate.

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Hello! Based on your question, the misnamed compounds are:

1. 3-propylbutane (correct name: 1-propylpentane)
2. isopentyl bromide (correct name: 1-bromo-3-methylbutane)

The other compounds are correctly named:

1. 3,3-dichlorooctane
2. 2,2-dimethyl-4-ethylheptane
3. 3,5-dimethylhexane
4. 2,6-dibromohexane

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The net ionic equation for the given chemical reaction is: Zn²⁺(aq) + S²⁻(aq) → ZnS(s). This equation represents the key species involved in the reaction, ignoring the spectator ions.

Here is the net ionic equation for the chemical reaction:
Zn²⁺(aq) + S²⁻(aq) → ZnS(s)
The net ionic equation only includes the species that are directly involved in the chemical reaction and excludes spectator ions, which in this case are NH4+ and Cl-.

The entire symbols of the reactants and products, as well as the states of matter under the conditions under which the reaction is occurring, are expressed in the complete equation of a chemical reaction.

Only those chemical species that are directly involved in the chemical reaction are written in the net ionic equation of the reaction.

In the net ion equation, mass and charge must be equal.

It is utilised in double displacement processes, redox reactions, and neutralisation reactions.

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The average C-O bond order in the formate ion, HCO2 (H attached to C), is 1.33.

The formate ion has three equivalent resonance structures, which are a combination of single and double bonds between the carbon and oxygen atoms. The first resonance structure has two single bonds between the carbon and oxygen atoms, resulting in a bond order of 1.

The second and third resonance structures have one single bond and one double bond between the carbon and oxygen atoms, resulting in a bond order of 1.5 and 1.66, respectively. The average bond order is calculated by adding the bond orders of all three resonance structures and dividing by three, which gives an average C-O bond order of 1.33.

Therefore, the correct answer to the question is 1.33.

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The major product of the reaction will be the 1,2-dichloroalkane .

The reaction is likely a halogenation reaction, where the alkene reacts with [tex]Cl_2[/tex] in the presence of ethanol as a solvent. Specifically, the double bond in the starting material will undergo electrophilic addition to one of the chlorine atoms, forming a carbocation intermediate. This intermediate can then undergo a nucleophilic attack by the chloride ion, resulting in substitution of the original double bond with a new carbon-chlorine bond.

In this case, the major product of the reaction will be the 1,2-dichloroalkane, where both carbons of the original double bond have been replaced with chlorine atoms.  

The reaction can be represented as follows:

[tex]CH_3[/tex]
  |
[tex]CH_3C[/tex] -- [tex]CH(C_6H_1_1)Cl[/tex] + [tex]Cl_2[/tex] + EtOH → [tex]CH_3C[/tex] --[tex]CH(C_6H_1_1)Cl_2[/tex] + HCl + EtOH
  |
 H

Therefore, The cyclohexyl and methyl substituents on carbon 1 and the methyl and hydrogen substituents on carbon 2 will remain unchanged in the final product. Hence, the major product of the reaction will be the 1,2-dichloroalkane .

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a) The vapor pressure of butane in air at 1 bar is 0.00784 bar.

b) The vapor pressure of butane in air at 100 bar is 0.784 bar.

To determine the vapor pressure of butane in air at different pressures, we need to use the ideal gas law and Raoult's law.

a) At 1 bar pressure, the total pressure is 1 bar + 2.2 bar (vapor pressure of butane) = 3.2 bar.

The mole fraction of butane in the vapor phase can be calculated as follows:

PV = nRT

where P is the partial pressure of butane, V is the volume, n is the number of moles of butane, R is the gas constant, and T is the temperature. n/V = P/RT

Since we know the density of butane, we can calculate the volume of 1 mole of butane as follows:

V = m/d

where m is the molar mass of butane (58.12 g/mol) and d is the density (0.5788 g/ml).

V = 58.12 g/mol / 0.5788 g/ml = 100.4 ml/mol

So, n/V = 1/100.4 ml/mol = 0.00996 mol/ml

Now, we can calculate the mole fraction of butane in the vapor phase: P/(1 bar) = (0.00996 mol/ml) x (8.314 J/mol.K) x (300 K) P = 0.00784 bar

Therefore, the vapor pressure of butane in air at 1 bar pressure is 0.00784 bar.

b) At 100 bar pressure, the total pressure is 100 bar + 2.2 bar (vapor pressure of butane) = 102.2 bar.

Following the same steps as above, we can calculate the mole fraction of butane in the vapor phase:

P/(100 bar) = (0.00996 mol/ml) x (8.314 J/mol.K) x (300 K)

P = 0.784 bar

Therefore, the vapor pressure of butane in air at 100 bar pressure is 0.784 bar.

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The concentration of PH3 after 26.0 s will be approximately 0.3584 M.

To determine the concentration of PH3 after 26.0 s, given the rate of decomposition, rate constant, and initial concentration, we will use the first-order reaction equation:
ln [PH3]t/[PH3]0 = -kt ,

where [PH3]t is the concentration of PH3 at time t, [PH3]0 is the initial concentration of PH3, k is the rate constant, and t is time.
Concentration at time t (C_t) = C_initial * e^(-k * t),

where C_initial is the initial concentration, k is the rate constant, and t is the time in seconds.

1. The rate of decomposition of PH3 is given as a first-order reaction.
2. The initial concentration of PH3 (C_initial) is 0.95 M.
3. The rate constant (k) is 0.0375 s^-1.
4. The time (t) is 26.0 s.

Now we will plug these values into the equation:

C_t = 0.95 * e^(-0.0375 * 26.0)

C_t ≈ 0.95 * e^(-0.975)

C_t ≈ 0.95 * 0.3773

C_t ≈ 0.3584 M

Therefore, the concentration of PH3 after 26.0 s will be approximately 0.3584 .

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The first statement is true because gases have equal average kinetic energy at the same temperature.

At a given temperature, regardless of the type of gas, the average kinetic energy is the same for all.

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The equilibrium constant (K) for this reaction at 342.0 K is approximately 2.3 × 10^(-17).

For the given reaction, N2(g) + 2O2(g) → 2NO2(g), we are provided with ΔH° = 66.4 kJ and ΔS° = -122 J/K. We can calculate the equilibrium constant at 342.0 K using the Van't Hoff equation, which relates the change in Gibbs free energy (ΔG°) to the equilibrium constant (K):
ΔG° = -RTlnK
First, we need to calculate ΔG° using the provided ΔH° and ΔS° values:
ΔG° = ΔH° - TΔS°
Since the given ΔH° is in kJ, we need to convert it to J:
ΔH° = 66.4 kJ * 1000 = 66400 J
Now, we can calculate ΔG° at 342.0 K:
ΔG° = 66400 J - (342.0 K * -122 J/K) = 66400 J + 41724 J = 108124 J
Next, we can find the equilibrium constant (K) using the Van't Hoff equation:
108124 J = -(8.314 J/(mol·K)) * 342.0 K * lnK
Solve for K:
lnK = -108124 J / (8.314 J/(mol·K) * 342.0 K) = -38.3
K = e^(-38.3) ≈ 2.3 × 10^(-17)
Thus, the equilibrium constant (K) for this reaction at 342.0 K is approximately 2.3 × 10^(-17).

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The complex ions that can have cis-trans isomers are [Pt(NH3)2Cl2] and [Pt(NH3)Cl3]. Among the given square planar complex ions, the one that can have cis-trans isomers is B. [Pt(NH3)2Cl2]. This complex ion has different ligands which allow for geometric isomerism, with cis and trans isomers based on the arrangement of ligands around the central atom.

This question requires a long answer as we need to analyze each complex ion individually to determine if they can have cis-trans isomers. A cis-trans isomerism occurs when two ligands in a coordination complex are arranged differently around the central metal atom. For square planar complexes, this is possible when there are two sets of identical ligands and two of them are adjacent to each other. This complex ion has four identical ammonia ligands arranged in a square planar geometry around the platinum atom. Since there are no other ligands present, there is no possibility of cis-trans isomerism.

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Metamorphism is the process through which sedimentary rocks become metamorphic rocks. Metamorphism is the process through which rocks change shape owing to changes in temperature, pressure, and chemical environment.

This process can occur in the presence or absence of fluids, such as water. The mineralogy of the rock changes during metamorphism, and new minerals emerge as current minerals recrystallize. Furthermore, the texture of the rock shifts from coarse-grained to fine-grained, and the structure shifts from layered to foliated.

This metamorphism process can take a long time and can be driven by tectonic pressures such as mountain-building episodes or the collision of tectonic plates.

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No, 1-bromobutane would not be expected to dissolve in the aqueous layer in the separatory funnel because it is not water-soluble. 1-bromobutane is an organic compound and is therefore hydrophobic, meaning it does not readily interact with water molecules.

The aqueous layer in the separatory funnel contains polar water molecules, which interact primarily through hydrogen bonding. Organic compounds are nonpolar and do not form hydrogen bonds with water. As a result, 1-bromobutane would remain in the organic layer and not dissolve in the aqueous layer.

The principle of liquid-liquid extraction is based on the differential solubility of compounds in two immiscible phases, and the immiscibility of 1-bromobutane in water makes it a good candidate for extraction with an organic solvent.

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the molar mass of the gas in the given sample, we can use the ideal gas law equation, PV = nRT, where P is the pressure in atm, V is the volume in liters, n is the number of moles, the molar mass of the gas is approximately 72.25 g/mol.

The pressure of 750. torr can be converted to atm by dividing by 760 torr/atm, giving us 0.987 atm. The volume of 252 ml can be converted to liters by dividing by 1000 ml/L, giving us 0.252 L. The temperature of 25.5°C can be converted to Kelvin by adding 273.15, giving us 298.65 K.

the molar mass of the gas in the sample is 68.0 g/mol. calculate the molar mass of a gas in a 252 mL container weighing 0.755 g at 750. torr and 25.5°C.Convert the volume to liters: 252 mL = 0.252 L
Convert temperature from Celsius to Kelvin: 25.5°C + 273.15 = 298.65 K

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Answer:The statement "polyprotic acid second k value less" is incomplete and unclear. Please provide the complete statement so I can accurately determine if it is true or false.

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