The density of an aqueous solution containing 10.0 percent of ethanol (C2H5OH) by mass is 0.984 g/mL. (a) Calculate the molality of this solution. (b) Calculate its molarity. (c) What volume of the solution would contain 0.125 mole of ethanol

Answer:

B) Iron(lll) phosphate, FePO4

Option.B-Iron(lll) phosphate, FePO4  is the correct answer.

A monoclinic crystal powder with the chemical formula FePO4, ferric phosphate is sometimes referred to as ferric phosphate and ferric orthophosphate.

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Answer : The value of change in entropy for vaporization of water is [tex]1.09\times 10^2J/mol[/tex]

Explanation :

Formula used :

[tex]\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}[/tex]

where,

[tex]\Delta S^o[/tex] = change in entropy  of vaporization = ?

[tex]\Delta H^o_{vap}[/tex] = change in enthalpy of vaporization = 40.7 kJ/mol

[tex]T_b[/tex] = boiling point temperature of water = [tex]100^oC=273+100=373K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}[/tex]

[tex]\Delta S^o=\frac{40.7kJ/mol}{373K}[/tex]

[tex]\Delta S^o=1.09\times 10^2J/mol[/tex]

Therefore, the value of change in entropy for vaporization of water is [tex]1.09\times 10^2J/mol[/tex]

The entropy change ∆S° for the vaporization of 1.00 mol of water at its boiling point is calculated to be 109.09 J/K·mol using the given enthalpy of vaporization.

To calculate ∆S° for the vaporization of 1.00 mol water at 100.°C and 1.00 atm pressure when ∆H°vap = 40.7 kJ/mol, we can use the thermodynamic relation ∆G° = ∆H° - T∆S° and the fact that at the boiling point, ∆G° for the phase change is zero. This would mean the ∆S° can be calculated as ∆S° = ∆H°vap/T. Plugging in the known values,

∆S° = 40.7 kJ/mol / 373.15 K

∆S° = 0.10909 kJ/K·mol

Since 1 kJ = 1000 J, we convert this to J:

∆S° = 109.09 J/K·mol

Therefore, the entropy change ∆S° for the vaporization of 1.00 mol of water at its boiling point is 109.09 J/K·mol.

Answer:

See explanation below

Explanation:

3. Each atom has unique electron energy level when compared to other, when electron relaxes from an excited energy level, it emits energy in the form of electromagnetic waves with energy equals to the difference between energy of ground and excited state. As electromagnetic wave wavelength obeys planck relation

[tex]c = \lambda f[/tex]

And the energy of each photon in electromangetic wave of specific frequency is

[tex]E=hf[/tex]

Where h is planck's constant

We can see that the wavelength, which determines color and position of bright line in spectroscope corresponds directly to the energy of electromagnetic wave, and thus, the characteristic energy of atom.

4. From the answer above, you now know that electronic transition can emit electron with specific wavelength. In strontium, emission of colored light occurs by the relaxation of electrons in excited state such that the difference between excited energy level and ground level falls in the energy of visible light spectral range.

5. In Na, ground state configuration is 2-8-1. 2-7-1-1 indicates that one of electron in ground state got excited by external energy to excited state, and thus, indicates that Na atom is in excited state

Final answer:

Bright-line spectra viewed through a spectroscope can be used to identify the metal ions in salts used in flame tests.

Explanation:

Bright-line spectra viewed through a spectroscope can be used to identify the metal ions in the salts used in flame tests. Each metal ion has a unique set of energy levels, and when these ions are heated in a flame, their electrons get excited and jump to higher energy levels. As the electrons return to their ground states, they release energy in the form of light. This emitted light can be observed through a spectroscope, which separates the different wavelengths of light, creating a bright-line spectrum. By comparing the bright-line spectrum of an unknown salt to the known spectra of different metal ions, the metal ions present in the unknown salt can be identified.

Answer:

elemental form,  readily form compounds

Explanation:

Copper, silver, and gold have all been know since ancient times because they appear in nature in_elemental form___ and were thus discovered thousands of years ago. But the majority of elements__readily form compounds_and, consequently, are hard to find in nature.

Answer:

68325.8 g/mol

Explanation:

Given that:

Pressure = 743 torr

Temperature = 37 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (37 + 273.15) K = 310.15 K  

T = 310.15 K  

Volume = 2.30 mL = 0.00230 L ( 1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.364 L.torr/K.mol

Applying the equation as:

743 torr × 0.00230 L = n × 62.364 L.torr/K.mol  × 310.15 K  

⇒n = 8.84 × 10⁻⁵ moles

Since, given that:-

4 moles of oxygen gas is transferred by 1 mole of haemoglobin.

8.84 × 10⁻⁵ moles  of oxygen gas is transferred by 8.84 × 10⁻⁵/4 moles of haemoglobin.

Moles of haemoglobin = 2.21 × 10⁻⁵ moles

Mass = 1.51 g

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]2.21\times 10^{-5}\ mol= \frac{1.51\ g}{Molar\ mass}[/tex]

[tex]Molar\ mass= 68325.8\ g/mol[/tex]

The question is asking for the molar mass of hemoglobin, which can be calculated using the Ideal Gas Law and the information provided. Hemoglobin, a protein in red blood cells, combines with four oxygen molecules, the fact that is fundamental to this calculation.

The molar mass of hemoglobin can be calculated using the Ideal Gas Law (PV=nRT). In this case, the volume of O2 (V) is 2.30 mL or 0.0023 L, the pressure (P) is 743 torr (or approximately 0.977 atm), the gas constant (R) is 0.0821 L⋅atm/mol⋅K, and the temperature (T) is 37 °C or 310 K. We can solve this equation to find the amount of moles of O2 (n). As each molecule of hemoglobin combines with four molecules of O2, we can then find out the amount of moles of hemoglobin, and knowing its mass (1.51 g), finally calculate its molar mass (grams per mole).

To confirm the correctness of this procedure, remember that hemoglobin is a protein found in red blood cells which combines with what? Oxygen molecules - four of them, in fact! This is crucial for the transport of oxygen throughout the body, and it also directly pertains to and informs our calculation of hemoglobin's molar mass.

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Answer: [tex]\Delta H^0=+0.3kJ/mol[/tex].

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

[tex]S_{rhombic}+O_2(g)\rightarrow SO_2(g)[/tex]    [tex]\Delta H^0_1=-296.06kJ[/tex]   (1)

[tex]S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ[/tex]  (2)

The final reaction is:  

[tex]S_{rhombic}\rightarrow S_{monoclinic}[/tex]  [tex]\Delta H^0_3=?[/tex]   (3)

By subtracting (1) and (2)

[tex]\Delta H^0_3=\Delta H^0_1-\Delta H^0_2=-296.06kJ-(-296.36kJ)=0.3kJ[/tex]

Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ

The enthalpy change for the transformation from rhombic sulfur to monoclinic sulfur is -0.30 kJ/mol.

The enthalpy change for the transformation S(rhombic) → S(monoclinic), we will use the enthalpy changes provided for the reactions involving sulfur and oxygen:

Given these values, the enthalpy change for the transformation from rhombic sulfur to monoclinic sulfur can be found as follows:

ΔH (transformation) = ΔHo(S(monoclinic) → SO₂) - ΔHo(S(rhombic) → SO₂)

Using the provided enthalpy changes:

ΔH (transformation) = -296.36 kJ/mol - (-296.06 kJ/mol)

ΔH (transformation) = -296.36 kJ/mol + 296.06 kJ/mol

ΔH (transformation) = -0.30 kJ/mol

The enthalpy change for the transformation S(rhombic) → S(monoclinic) is -0.30 kJ/mol.

Answer : The correct option is, (E) 14

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

The given chemical reaction is,

[tex]S^{2-}+Cr_2O_7^{2-}\rightarrow S+Cr^{3+}[/tex]

The oxidation-reduction half reaction will be :

Oxidation : [tex]S^{2-}\rightarrow S[/tex]

Reduction : [tex]Cr_2O_7^{2-}\rightarrow Cr^{3+}[/tex]

First balance the main element in the reaction.

Oxidation : [tex]S^{2-}\rightarrow S[/tex]

Reduction : [tex]Cr_2O_7^{2-}\rightarrow 2Cr^{3+}[/tex]

Now balance oxygen atom on both side.

Oxidation : [tex]S^{2-}\rightarrow S[/tex]

Reduction : [tex]Cr_2O_7^{2-}\rightarrow 2Cr^{3+}+7H_2O[/tex]

Now balance hydrogen atom on both side.

Oxidation : [tex]S^{2-}\rightarrow S[/tex]

Reduction : [tex]Cr_2O_7^{2-}+14H^+\rightarrow 2Cr^{3+}+7H_2O[/tex]

Now balance the charge.

Oxidation : [tex]S^{2-}\rightarrow S+2e^-[/tex]

Reduction : [tex]Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O[/tex]

In order to balance the charge, we multiply oxidation reaction by 3. Thus, added both equation, we get the balanced redox reaction.

The balanced chemical equation in acidic medium will be,

[tex]3S^{2-}+Cr_2O_7^{2-}+14H^+\rightarrow 3S+2Cr^{3+}+7H_2O[/tex]

From the balanced chemical equation we conclude that, the number of moles of [tex]H^+[/tex] consumed from every mole of [tex]Cr_2O_7^{2-}[/tex] are 14 moles.

Hence. the correct option is, (E) 14

Answer:

The answer to your question is 21.45 g of KBr

Explanation:

Chemical reaction

                               2K + Br₂   ⇒   2KBr

                                       14.4            ?

Process

1.- Calculate the molecular mass of bromine and potassium bromide

Bromine = 2 x 79.9 = 159.8g

Potassium bromide = 2(79.9 + 39.1) = 238 g

2.- Solve it using proportions

              159.8 g of Bromine ------------ 238 g of potassium bromide                    

                14.4 g of Bromine  ------------  x

                        x = (14.4 x 238) / 159.8

                        x = 3427.2 / 159.8

                        x = 21.45g of KBr

Answer:

Internal energy

Explanation:

Particles have two main energies that scientists acknowledge.

When these two types of energies are considered collectively, we call that internal energy

Answer:

[tex]Age=2.52*10^9 years[/tex]

Explanation:

To determine the age, we need to know how much U have become Pb, so first we have to calculate the Pb moles present in a sample:

[tex]n_{Pb}=\frac{0.337g}{206g/mol}=0.00164 mol[/tex]

[tex]n_{U}=\frac{1 g}{238g/mol}=0.0042 mol[/tex]

The percentage of U degradation:

[tex]P=\frac{n_{Pb}}{n_{Pb}+n_{U}}[/tex]

[tex]P=\frac{0.00164}{0.00164+0.0042}=0.28[/tex]

Assuming that the life time is linear:

[tex]Age=\frac{4.5*10^9 years}{0.5 life time}*0.28 life time[/tex]

[tex]Age=2.52*10^9 years[/tex]

Answer:

7.1 x 10^9 years

Explanation:

This is the answer according to Mastering Chemistry.

Answer:

The process is called Nitrogen fixation

Explanation:

The nitrogen fixation is a process carried out by some prokaryotic microorganisms (bacteria), specifically those have the presence of the nitrogenase enzyme. The bacteria absorb the atmospheric nitrogen (N2) from the roots of plants, and the nitrogenase enzyme, with the help of two proteins that act as electron donors and acceptors (nitrogenase complex) reduce the nitrogen to ammonia (NH3), then the ammonia is ionized to NH4+ (ammonium). Followed by that, the ammonia is oxidated to nitrates and nitrites, which are finally absorbed again by plants.

Answer:

P = 3,63 atm

Explanation:

Using PV=nRT and knowing the temperature in the problem remains constant it is possible to rewrite this formula, thus:

PV = kn, where k is RT (Constant), P is pressure, V is volume and n are moles.

The first bulb contains:

4,87L×4,38atm = 21,3k moles

And the second bulb:

4,87L×2,90atm = 14,1k moles

When gases are mixed, the total moles are:

21,3k moles + 14,1k moles = 35,4k moles

As total volume is 4,87L + 4,87L = 9,74L

Replacing:

P = kn/V

P = 35,4k moles / 9,74L

P = 3,63 atm

I hope it helps!

Answer:

When the two gases are mixed, the ammonium chloride precipitates in the tube walls.

Explanation:

This is the reaction:

HCl (g)  +  NH₃(g)  →  NH₄Cl (s) ↓

As the product formed is solid at room temperature, a suspension is first formed in the internal air of the tube that appears as a cloud. Afterwards it finally precipitates into the walls forming a white layer

Explanation:

As the given reaction equation is as follows.

         [tex]N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)[/tex]

Hence, expression for [tex]K_{p}[/tex] of this reaction is as follows.

        [tex]K_{p} = \frac{P^{2}_{NO}}{P_{N_{2}} \times P_{O_{2}}}[/tex]

Now, putting the given values into the above expression as follows.

         [tex]K_{p} = \frac{P^{2}_{NO}}{P_{N_{2}} \times P_{O_{2}}}[/tex]

                 = [tex]\frac{(0.050)^{2} atm}{(0.15 atm) \times (0.33 atm)}[/tex]  

                 = [tex]5.05 \times 10^{-2}[/tex]

Thus, we can conclude that the value of [tex]K_{p}[/tex] is [tex]5.05 \times 10^{-2}[/tex].

Final answer:

The equilibrium constant, Kp, for the reaction N2(g) + O2(g) = 2 NO(g) at the given conditions is approximately 0.1010 when calculated using the given equilibrium partial pressures.

Explanation:

The question is asking us to calculate the equilibrium constant, KP, for the reaction given the equilibrium partial pressures of the reactants and products at a certain temperature. Using the reaction N2(g) + O2(g) = 2 NO(g), we can express the equilibrium constant KP in terms of the partial pressures of the gases:

KP = (PNO)2 / (PN2 * PO2)

Substituting in the given partial pressures:

KP = (0.050 ATM)2 / (0.15 ATM * 0.33 ATM) = 0.0050 / 0.0495 = 0.1010

So, the equilibrium constant KP at 2200°C for this reaction is approximately 0.1010.

Answer:

c) No, because Celsius is not an absolute temperature scale

Explanation:

converting  5 oC to kelvin which is the absolute temperature scale gives = 273 + 5 = 278 K

and converting 20 oC to kelvin = 20 + 273 = 293 K

the ratio = 278 / 293 = 0.94 approx 1 not 4

Ionic solids are more likely to dissolve in water and can become better conductors of electricity via chemical substitution.

Both covalent-network solids and ionic solids have different properties. Ionic solids dissolve readily in water and can conduct electricity when melted or dissolved because their ions are free to move. Covalent-network solids, on the other hand, are insoluble in water and are poor conductors of electricity in any state because their particles are electrically neutral.

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Answer:

-87.4 kJ/mol

Explanation:

We can rewrite Arrhenius' equation as:

ln(k₂/k₁) = -Ea/R * (1/T₂ - 1/T₁)

Where k₂ and k₁ are the rates of the reaction at temperatures T₂ and T₁, Ea is the activation energy and R is the universal gas constant.

For this problem:

k₂ = 5

k₁ = 17

T₂ = 15°C = 288.16 K

T₁ = 25°C = 298.16 K

R = 8.314 J/mol·K

We put the data in the equation and solve for Ea:

ln(5/17) = -Ea/8.314 J/mol·K * (1/288.16K - 1/298.16K)

-1.224 = -Ea/8.314 J/mol·K * 1.1639x10⁻⁴K⁻¹

Ea = -87416.78 J/mol ≅ -87.4 kJ/mol

Answer:

Enthalpy of vaporization = 30.8 kj/mol

Explanation:

Given data:

Mass of benzene = 95.0 g

Heat evolved = 37.5 KJ

Enthalpy of vaporization = ?

Solution:

Molar mass of benzene = 78 g/mol

Number of moles = mass/ molar mass

Number of moles = 95 g/ 78 g/mol

Number of moles = 1.218 mol

Enthalpy of vaporization =  37.5 KJ/1.218 mol

Enthalpy of vaporization = 30.8 kj/mol

Final answer:

The enthalpy of vaporization of benzene at 25°C is calculated by dividing the heat energy supplied by the number of moles vaporized, resulting in 30.8 kJ/mol.

Explanation:

The enthalpy of vaporization of benzene at 25°C can be calculated using the amount of heat supplied and the mass of benzene vaporized. Here's a step-by-step explanation:

First, convert the mass of benzene to moles using its molar mass (78.11 g/mol for C6H6).

Divide the given heat energy by the number of moles to find the enthalpy of vaporization per mole.

The calculation is as follows:

Moles of benzene = 95.0 g / 78.11 g/mol = 1.2169 mol

Enthalpy of vaporization (ΔHvap) = 37.5 kJ / 1.2169 mol = 30.8 kJ/mol

The correct answer is 30.8 kJ/mol, which is option 1.

Answer:

Newton's 2nd Law says the acceleration of an object depends on its mass and the amount of net force acting on it.

Explanation:

Definition of acceleration:

The acceleration is rate of change of velocity of an object with respect to time.

Formula:

a = Δv/Δt

a = acceleration

Δv = change in velocity

Δt = change in time

Units:

The unit of acceleration is m.s⁻².

Acceleration can also be determine through following formula,

F = m × a

a = F/m

This is the newton's second law:

"The acceleration of an object depends on its mass and the amount of net force acting on it"

The acceleration is depend directly on the force while inversely on the mass.

Answer:

Mass and Force

Explanation:

Answer:

B

Explanation:

Explanation:

Adsorption is defined as a process in which gas molecules tend to adsorb on the surface of a solid solvent.

Therefore, the statement solids are not able to act as a solvent if a gas is the solute, is false.

On the other hand, gases can also act as a solvent if a solid is the solute.

For example, air present in the surrounding acts as a solvent and solid substances present in the surrounding acts as a solute.

Hence, the statement gases are not able to act as a solvent if a solid is the solute, is false.

Thus, we can conclude that following are the statements which are incorrect.

Answer:

1.40 atm

Explanation:

To answer this question we can use Gay-Lussac's law, which states:

When volume and number of moles remain constant.

We put the known data in the equation and solve for P₂:

Using the ideal gas law, the pressure inside an aerosol can increases proportionately with temperature. If the can's temperature is raised to the boiling point of water (100°C or 373 K), the pressure increases from 1.11 atm to 1.40 atm.

The pressure inside an aerosol can, like this one which has a residual propellant, can be explained by the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature when the volume and amount of gas are held constant. If the can is heated to the boiling point of water (100°C or 373 K), we can expect the pressure to increase proportionately.

To find out the new pressure, we need to convert the initial temperature to Kelvin (23°C = 296 K), and then use the formula P1/T1 = P2/T2 where P1 is the initial pressure, T1 is the initial temperature, P2 is the pressure at elevated temperatures, and T2 is the temperature at the boiling point of water.

So, P2 = P1×T2/T1 = 1.11 atm × 373 K / 296 K = 1.40 atm. Therefore, the pressure inside the can at the boiling point of water is 1.40 atm.

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Answer:

True

Explanation:

All the chemical elements have their abbreviations as one or two letters

The first letter in the abbreviation is a capital letter and the second letter is always a small letter

Majority of abbreviations are based on their english names

Final answer:

The statement regarding a chemical symbol being a one- or two-letter abbreviation for a chemical element is true. These symbols are a shorthand for elements, such as 'O' for oxygen or 'Fe' for iron, and are crucial for anyone studying chemistry.

Explanation:

The statement that a chemical symbol is a one- or two-letter abbreviation for a chemical element is true. Chemical symbols provide a concise way to represent elements in scientific writings and discussions. For instance, the symbol 'O' stands for oxygen, while 'Zn' represents zinc. The first letter of a chemical symbol is always capitalized, and if there is a second letter, it is lowercase.

Elements that have been known for a long time often have symbols derived from their Latin names. For example, the symbol for iron is 'Fe', which comes from its Latin name 'ferrum'. It is important for those studying chemistry to become familiar with the chemical symbols of the elements, most of which you can find on the periodic table.

Explanation:

-101 mRNA codons

In the genetic code, an amino acid is encoded by 3 nucleotides, while there are just 4 bases . Each amino acid is specifically encoded by a codon- a triplet sequence of nucleotides...

Thus 99 amino acids= 99 codon sequences.

...along with a start and stop codon, this would require 101 mRNA codons

Further Explanation:

The nucleic acids are comprised of smaller units called nucleotides and function as storage for the body’s genetic information. These monomers include ribonucleic acid (RNA) or deoxyribonucleic acid (DNA). They differ from other macromolecules since they don’t provide the body with energy. They exist solely to encode and protein synthesis.

Basic makeup: C, H, O, P; they contain phosphate group 5 carbon sugar does nitrogen bases which may contain single to double bond ring.

Codons are three nucleotide bases encoding an amino acid or signal at the beginning or end of protein synthesis.

RNA codons determine certain amino acids so the order in which the bases occur within in the codon sequence designates which amino acid is to be made bus with the four RNA nucleotides (Adenine, Cysteine and Uracil) Up to 64 codons (with 3 as stop codons) determine amino acid synthesis. The stop codons ( UAG UGA UAA) terminate amino acid/ protein synthesis while the start codon AUG begins protein synthesis.

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Answer:

0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl

Explanation:

This is the reaction:

2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)

To make 3 moles of H₂, we need 2 moles of Al.

By conditions given, we will find out how many moles of H₂ do we have.

Let's use the Ideal Gas Law

P. V = n . R . T

1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K

(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n

0.182 mol = n

So the rule of three will be:

If 3 moles of H₂ came from 2 moles of Al

0.182 moles of H₂ will come from x

(0.182 .2) / 3 = 0.121 moles

Answer:

1.[tex]S^{-}(g)+e^{-} \rightarrow S^{2-}(g)[/tex]

2.[tex]Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}[/tex]

3.The electron affinity of  [tex]Mg^{2+}[/tex] is zero.

4.[tex]O^{2-}(g) \rightarrow O^{-}(g)+e^{-}[/tex]

Explanation:

1.

Electron affinity:

It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.

The electron affinity of [tex]S^{-}[/tex] is as follows.

[tex]S^{-}(g)+e^{-} \rightarrow S^{2-}(g)[/tex]

2.

Ionization energy:

Amount of energy required to removal of an electron from an isolated gaseous atom.

The third ionization energy of Titanium is as follows.

[tex]Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}[/tex]

3.

The electronic configuration of Mg: [tex]1s^{2}2s^{2}2p^{6}3s^{2}[/tex]

By the removal of two electrons from a magnesium element we get [tex]Mg^{2+}[/tex] ion.

[tex]Mg^{2+}[/tex] has inert gas configuration i.e,[tex]1s^{2}2s^{2}2p^{6}[/tex]

Hence, it does not require more electrons to get stability.

Therefore,the electron affinity of  [tex]Mg^{2+}[/tex] is zero.

4.

The ionization energy of [tex]O^{2-}[/tex] is follows.

[tex]O^{2-}(g) \rightarrow O^{-}(g)+e^{-}[/tex]

The electron affinity equation for S⁻ is S⁻(g) + e⁻ → S²⁻(g). The third ionization energy equation for titanium is Ti²⁺(g) → Ti³⁺(g) + e⁻. The equation representing the electron affinity of Mg²⁺ is Mg²⁺(g) + e⁻ → Mg³⁺(g). The ionization energy equation for O²⁻ is O²⁻(g) → O³⁻(g) + e⁻.

1. The electron affinity of S⁻: The electron affinity represents the energy change that occurs when an electron is added to a neutral atom or ion. For sulfur (S⁻), the equation representing this process is: S⁻(g) + e⁻ → S²⁻(g). Here, the physical state of S⁻ and S²⁻ is gas.

2. The third ionization energy of titanium: Ionization energy is the energy required to remove an electron from an atom or ion. For titanium, the equation representing the third ionization energy is: Ti²⁺(g) → Ti³⁺(g) + e⁻. The physical state of Ti²⁺ and Ti³⁺ is gas.

3. The electron affinity of Mg²⁺: The electron affinity of a cation, such as Mg²⁺, is the energy change when an electron is added to the cation. Here, the equation representing this process is: Mg²⁺(g) + e⁻ → Mg³⁺(g). The physical state of Mg²⁺ and Mg³⁺ is gas.

4. The ionization energy of O²⁻: Ionization energy is the energy required to remove an electron from an atom or ion. For oxygen, the equation representing the ionization energy of O²⁻ is: O²⁻(g) → O³⁻(g) + e⁻. The physical state of O²⁻ and O³⁻ is gas.

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Answer:they have one valence electron

Explanation:halogens are in group seven of the periodi table and they have seven electrons in their outermost shell thus need one electron to complete their octet condition. Thus goes into combination to accept one electrons from another element and that is their combining power which is their valency.

The halogens have seven valence electrons.  

• In the periodic table, the elements present in group 7A are known as halogens.  

• These are fluorine, chlorine, bromine, iodine, and astatine.  

• The halogen term means salt former.  

• The major characteristic of halogens is that they have seven valence electrons in their higher energy orbitals.  

• They are one electron behind in forming a full octet, so these elements would seem to produce anions exhibiting -1 charges and are known as halides.

Thus, halogens having seven valence electrons is the right statement.

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Answer:

Potassium

Explanation:

The description here fits potassium. Potassium is not found alone in nature as it is very reactive. Hence, it tends to occur in relation with other elements because of its high reactivity.

Because it is stated that it gives up its single valence electron, this shows that it belongs to group one of the periodic table. We might be tempted to think hydrogen is the correct answer. But it is interesting to note that hydrogen is not even supposed to be placed in group one at all. It is only placed there for convenience

Answer:

a. 5 0 0 +1/2

5 0 0 -1/2

Valid

b. 4 1 -1 +1/2

4 1 0 +1/2

4 1 +1 +1/2

Valid

c. 3 2 -1 +1/2

3 2 0 +1/2

3 2 +1 +1/2

3 2 0 +1/2

3 2 +2 +1/2

Pauli violation

d. 3 1 -1 +1/2

3 1 0 +1/2

3 3 +1 +1/2

other violation

Explanation:

The four quantum number and possible values are:

n = 1,2,3.....

l = 0 , (n-1), (n-2).....

m = +l , 0 , -l

s = [tex]+\frac{1}{2}[/tex] or [tex]-\frac{1}{2}[/tex]

Pauli's exclusion principle: No two electrons in an atom can have all the four quantum numbers same.

Let us check each case:

a. 5 0 0 +1/2

5 0 0 -1/2

Valid

b. 4 1 -1 +1/2

4 1 0 +1/2

4 1 +1 +1/2

Valid

c. 3 2 -1 +1/2

3 2 0 +1/2

3 2 +1 +1/2

3 2 0 +1/2

3 2 +2 +1/2

Pauli violation

The two electrons have same four quantum numbers

3 2 0 +1/2

3 2 0 +1/2

d. 3 1 -1 +1/2

3 1 0 +1/2

3 3 +1 +1/2

other violation

As mentioned above in the condition the value of "l" can be only less than "n"

So for 3 3 +1 +1/2 : n = 3 and l= 3, which is not valid.

The arrangement of electrons in orbitals are showed by four sets of quantum numbers.

According to the Pauli exclusion theory, no two electrons in an atom should have all the four quantum numbers as the same. According to this principle, the spin quantum number of electrons in an atom must differ even if they are in the same orbital.

For the first set;

5 0 0 +1/2

5 0 0 -1/2

This set, correctly corresponds to the 5s  orbital so it is valid.

For the second set;

4  1  -1  +1/2

4  1  0  +1/2

4  1  +1  +1/2

This should have corresponding to the 4p orbital so it is valid.

For the third set;

3  2 -1  +1/2

3  2  0  +1/2

3  2  +1  +1/2

3  2  0  +1/2

3  2 +2  +1/2

This set should correspond to a 3d orbital but we can see that are two electrons in the set that has exactly the same quantum numbers of 3 2 0 +1/2. This violates the Pauli exclusion theory so we should mark "Pauli violation".

For the fourth set:

3  1  -1  +1/2

3   1   0  +1/2

3  3  +1  +1/2

This set should have corresponded to a 3p orbital but remember that the values of l only range from 0 to (n - 1). This means that we can not have n =3, l=3 so the arrangement 3  3  +1  +1/2 is not possible. We should mark  "other violation".

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Answer: The concentration of [tex]Au^{3+}[/tex] is [tex]1.096\times 10^{-6}[/tex]

Explanation:

The given chemical cell follows:

[tex]Pt(s)|H_2(g,1atm)|H^+(aq,1.0M)||Au^{3+}(aq,?M)|Au(s)[/tex]

Oxidation half reaction: [tex]H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V[/tex]       ( × 3)

Reduction half reaction: [tex]Au^{3+}(aq,?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V[/tex]       ( × 2)

Net cell reaction: [tex]3H_2(g,1atm)+2Au^{3+}(aq,?M)\rightarrow 6H^{+}(aq,1.0M)+2Au(s)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Putting values in above equation, we get:

[tex]E^o_{cell}=1.50-(0.0)=1.50V[/tex]

To calculate the EMF of the cell, we use the Nernst equation, which is:

[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}[/tex]

where,

[tex]E_{cell}[/tex] = electrode potential of the cell = 1.23 V

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[tex][H^{+}]=1.0M[/tex]

[tex][Au^{3+}]=?M[/tex]

Putting values in above equation, we get:

[tex]1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})[/tex]

[tex][Au^{3+}]=-1.0906\times 10^{-6},1.096\times 10^{-6}[/tex]

Neglecting the negative value because concentration cannot be negative.

Hence, the concentration of [tex]Au^{3+}[/tex] is [tex]1.096\times 10^{-6}[/tex]

The percentage of chlorine by mass in the unknown chlorofluorocarbon is 56.8%

We'll begin by calculating the number of mole of Cl₂ produced using the ideal gas equation as illustrated below:

PV = nRT

0.989 × 0.564 = n × 0.0821 × 298

0.557796 = n × 24.4658

Divide both side by 24.4658

n = 0.557796 / 24.4658

n = 0.0228 mole

Next, we shall determine the mass of 0.0228 mole of Cl₂

Mass = mole × molar mass

Mass of Cl₂ = 0.0228 × 71

Mass of Cl₂ = 1.62 g

Finally, we shall determine the percentage of chlorine in the unknown chlorofluorocarbon.

Percentage = (mass / total mass) × 100

Percentage of chlorine =(1.62 / 2.85) × 100

Percentage of chlorine = 56.8%

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The percent chlorine (by mass) in the unknown chlorofluorocarbon is 0%.

In order to determine the percent chlorine (by mass) in the unknown chlorofluorocarbon, we need to calculate the mass of chlorine gas produced from the 2.85-g sample. From the equation 2.5g H and 7.5g C make up a 10.0g sample, we can calculate the percent composition of hydrogen and carbon to be 25% and 75%, respectively.

Knowing the percent composition of hydrogen and carbon, we can assume that the unknown chlorofluorocarbon consists of only hydrogen, carbon, and chlorine. Since chlorine gas is produced when the unknown chlorofluorocarbon is decomposed, the percent chlorine (by mass) can be calculated by subtracting the percent composition of hydrogen and carbon from 100%. Therefore, the percent chlorine is 100% - (25% + 75%) = 0%.

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