The four sets of lines in the hydrogen emission spectrum are known as Balmer, Brackett, Paschen, and Lyman. For each series, assign the energy (infrared, ultraviolet, or visible), the nr value, and all possible n, values up to 7 Series Balmer Brackett Paschen Lyman Energy nf ni visible) infrared ultraviolet 1 2 3 6 7

Answer:

Kc for this reaction is 0.43

Explanation:

This is the equilibrium:

N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)

And we have all the concentration at equilibrium:

N₂: 0.25M

H₂ : 1.3M

NO: 0.33M

H₂: 1.2M

They are ok, because they are in MOLARITY. (mol/L)

Let's make the expression for Kc

Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)

Kc = (0.33² . 1.2²) / (0.25 . 1.2²)

Kc = 0.4356

In two significant digits. 0.43

Final answer:

To calculate the equilibrium constant (Kc) for the reaction between nitrogen and water to form nitrogen monoxide and hydrogen, the concentration of nitrogen at equilibrium is required, which was not provided in the question.

Explanation:

The chemical equation for the reaction between nitrogen and water to form nitrogen monoxide and hydrogen is given by:

N2(g) + 2H2O(g) -> 2NO(g) + 2H2(g)

The equilibrium constant (Kc) for this reaction can be calculated using the concentrations of the reactants and products at equilibrium:

Kc = [NO]2 * [H2]2 / ([N2] * [H2O]2)

Substituting the given equilibrium concentrations into the equation:

Kc = (1.3 M)2 * (0.33 M)2 / ([N2] * (0.25 M)2)

However, we do not have the concentration of N2 at equilibrium provided in the question. The student needs to provide this concentration to calculate the value of Kc.

Answer : The specific heat of tungsten is, [tex]0.139J/g^oC[/tex]

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]c_1[/tex] = specific heat of tungsten = ?

[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of tungsten = 19.5 g

[tex]m_2[/tex] = mass of water = 78.5 g

[tex]T_f[/tex] = final temperature = [tex]23.20^oC[/tex]

[tex]T_1[/tex] = initial temperature of tungsten = [tex]97.80^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]22.58^oC[/tex]

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

[tex]19.5g\times c_1\times (23.20-97.80)^oC=-78.5g\times 4.18J/g^oC\times (23.20-22.58)^oC[/tex]

[tex]c_1=0.139J/g^oC[/tex]

Therefore, the specific heat of tungsten is, [tex]0.139J/g^oC[/tex]

The rate constant (k) is 16.2 dm³/mol·h. The half-life of reactant A is 5.1 × 10³ s and the half-life of reactant B is 2.1 × 10³ s.

To calculate the rate constant (k) of a second-order reaction, we can use the integrated rate law:

1/[A] - 1/[A]₀ = kt

Where [A] is the concentration of A at time t, [A]₀ is the initial concentration of A, k is the rate constant, and t is the time. Rearranging this equation, we get:

k = (1/[A] - 1/[A]₀) / t

Substituting the given values, k = (1/0.020 - 1/0.075) / 1.0 = 16.2 dm³/mol·h.

To find the half-life of each reactant, we can use the equation:

t½ = 1 / (k * [A]₀)

For reactant A, t½ = 1 / (16.2 * 0.075) = 5.1 × 10³ s.

For reactant B, t½ = 1 / (16.2 * 0.050) = 2.1 × 10³ s.

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The rate constant for this second-order reaction is 16.2 dm^3/mol*h, and the half-lives of reactants A and B are 5.1 × 10^3 s and 2.1 × 10^3 s respectively.

The rate constant for a second-order reaction can be calculated using the formula: k = 1 / ((1 / [A]2 - 1 / [A]1) * t). Here, [A]1 is the initial concentration of A (0.075 mol dm^-3), [A]2 is the final concentration of A (0.020 mol dm^-3), and t is the time elapsed (1.0 h). Plugging these values into the formula gives a rate constant, k = 1 / ((1 / 0.020 - 1 / 0.075) * 1.0)= 16.2 dm^3/mol*h.

For a second-order reaction, the half-life can be determined by the formula: t1/2 = 1 / (k * [A]0). Therefore, the half lives for reactants A and B are as follows: t1/2A = 1 / (16.2 * 0.075)≈ 0.820 h or 5.1 × 10^3 s and t1/2B = 1 / (16.2 * 0.050) ≈ 1.235 h or 2.1 × 10^3 s. Hence, the rate constant and half-lives of the reactants have been calculated using the provided concentrations and the second-order reaction rate law.

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Explanation:

-Filter help — delete some big unreacted, undesirable species (norit is probably from what you are sorting through, its only carbon which cleans up things)

— extract with DCM because you are probably in an aqueous phase, and some butanoate is in it

- Anhydrous sodium absorbs excess of  water (dries the material)

-evaporation in the hood to clear the DCM and crystallize the product.

Answer:

The percentage of amine is protonated is 63,07%

Explanation:

The reaction of an amine RNH₃ (weak base) with water is:

RNH₃ + H₂O ⇄ RNH₄⁺ + OH⁻

The kb is defined as:

[tex]kb = \frac{[RNH_{4}^+][OH^-]}{[NH_{3}]}[/tex]

As kb = 4,004x10⁻⁵ and [OH⁻] is [tex]10^{-(14-9,370)}=2,344x10^{-5}[/tex]:

[tex]4,004x10^{-5} = \frac{[RNH_{4}^+][2,34x10^{-5}]}{[NH_{3}]}[/tex]

1,708 = [RNH₄⁺] / [RNH₃] (1)

As the total amine is a 100%:

[RNH₄⁺] + [RNH₃] = 100% (2)

Replacing (1) in (2):

1,708 [RNH₃]+ [RNH₃] = 100%

2,708 [RNH₃] = 100%

[RNH₃] = 36,93%

Thus,

[RNH₄⁺] = 63,07%

The percentage of amine protonated is 63,07%

I hope it helps!

The percentage of the amine protonated at a pH of 9.370, with Kb of 4.004 × 10⁻⁵, is approximately 63%.

Use the relationship pKa + pKb = 14 to find the pKa of the amine's conjugate acid:

pKa = 14 - pKb

pKb = -log(Kb) = -log(4.004 × 10⁻⁵) = 4.397

pKa = 14 - 4.397 = 9.603

The Henderson-Hasselbalch equation can now be used to find the ratio of protonated to unprotonated forms:

pH = pKa + log([A⁻]/[HA])

9.370 = 9.603 + log([A⁻]/[HA])

-0.233 = log([A⁻]/[HA])

[A⁻]/[HA] = 10⁻⁰·²³³ ≈ 0.588

The fraction of the amine that is protonated ([HA]) can be found by considering:

fraction protonated ([HA]) = [HA] / ([HA] + [A⁻])

Let [HA] = x and [A⁻] = 0.588x, so:

fraction protonated ([HA]) = x / (x + 0.588x) = 1 / (1 + 0.588) ≈ 0.63

Therefore, the percentage protonated is approximately:

63%

Answer:

The pressure of hydrogen is 718.7 mmHg. (option a)

Explanation:

Total pressure on the system is 0.977 atm.

Vapor pressure of water is 0.0313 atm, so

H₂ pressure =  Total pressure - Vapor pressure

H₂ pressure = 0.977 atm - 0.0313 atm = 0.9457 atm

Let's convert atm to mmHg

1 atm ____ 760 mmHg

0.9457 atm ____ (0.9457 atm . 760 mmHg) / 1

= 718.7 mmHg

Final answer:

The pressure of hydrogen gas when Mg is dropped into HCl over water at 298.15 K and 0.977 atm, with a water vapor pressure of 0.0313 atm, is found to be closest to 718.7 mm Hg after using Dalton's Law of Partial Pressures and converting atm to mmHg.

Explanation:

To determine the pressure of hydrogen gas evolved when Mg is dropped into HCl over water, we need to use Dalton's Law of Partial Pressures which states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases.

Using the given data: the total pressure is 0.977 atm, and the water vapor pressure at 298.15 K is 0.0313 atm. To find the hydrogen gas pressure, we subtract the water vapor pressure from the total pressure:

Pressure of hydrogen = Total pressure - water vapor pressure
Pressure of hydrogen = 0.977 atm - 0.0313 atm
Pressure of hydrogen = 0.9457 atm

Now convert the pressure from atm to mmHg, knowing that 1 atm = 760 mmHg:

Pressure of hydrogen = 0.9457 atm × 760 mmHg/atm
Pressure of hydrogen = 718.13 mm Hg

Therefore, the closest answer to the calculated pressure of hydrogen is 718.7 mm Hg, which corresponds to option A.

Answer:

1. 4.32 h

2. 4.12 g/L

Explanation:

1. For a batch culture, the time (tb) can be calculated by:

tb =  ln (X/X0)/μmax

Where X0 is the initial mass concentration of the cells (12 g/100L = 0.12 g/L), X is the mass concentration of the cells at tb, and μmax is the maximum specific growth rate of the cells.

The biomass yield (Y) is:

Y = (X - X0)/ (S0 - S)

Where S is the mass concentration of the substrate at tb and S0 the initial mass concentration of the substrate (glucose in this case).

Reorganizing:

X = Y*(S0 -S) + X0

Let's assume that at the stationary state all substrate was consumed, so S = 0.

tb = ln[(YS0 + X0)/X0)]/μmax

tb = ln[(0.575*10 + 0.12)/0.12]/0.9

tb = 4.32 h

2.  If 70% of the substrate is consumed, S = 10 - 0.7*10 = 3 g/L

tb = ln[(0.575*(10-3) + 0.12)/0.12]/0.9

tb = 3.93 h

The initial concentration is X0 = 0.12 g/L, the X:

tb =  ln (X/X0)/μmax

3.93 = ln(X/0.12)/0.9

ln(X/0.12) = 3.537

X/0.12 = [tex]e^{3.537}[/tex]

X/0.12 = 34.36

X = 4.12 g/L

Answer:

a) to = 4.3 h

b) xf = 4.1 g/L

Explanation:

We have the following data:

So = 10 g*L^-1 = initial concentration

Ys = biomass yield = 0.575 g*g^-1

umax = maximum specific growth rate = 0.9 h^-1

The initial cell concentration is equal to:

xo = 12 g/100 L = 0.12 g/L

a) The batch time it takes to reach the stationary phase equals:

to = (1/umax)*ln(1 + (Ys*(so-sf)/xo))) = (1/0.9)*ln(1+(0.575*(10-0))/0.12)))) = 4.3 h

b) Since the fermentation stops after consuming only 70% of the glucose, we have the following:

sf = (1-07)*so = 0.3*10 = 3 g/L

tb = (1/0.9)*ln(1+(0.575*(10-3))/(0.12)))) = 3.94 h

Finally, the final cell concentration can be found by the following equation:

xf = xo*e^(umax * tb) = 0.12 * e^(0.9 * 3.94) = 4.1 g/L

Answer:

9.86 × 10²⁴ photons

Explanation:

Let's consider the photosynthesis global reaction.

6 CO₂(g) + 6 H₂O(l) →  C₆H₁₂O₆(s) +  6 O₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

ni are the moles of reactants and products

ΔG°f(p) are the standard free Gibbs energies of formation of reactants and products

For the balanced equation for 1.00 mol of glucose, ΔG° is:

ΔG° = [1 mol × ΔG°f(C₆H₁₂O₆(s)) + 6 mol × ΔG°f(O₂(g))] - [6 mol × ΔG°f(CO₂(g)) + 6 mol × ΔG°f(H₂O(l))]

ΔG° = [1 mol × (-910.6 kJ/mol) + 6 mol × 0 kJ/mol] - [6 mol × (-394.4 kJ/mol) + 6 mol × (-237.1 kJ/mol)]

ΔG° = 2878 kJ

The energy of each photon (E) can be calculated using the Planck-Einstein's equation.

E = h . c. λ⁻¹

where,

h is the Planck's constant

c is the speed of light

λ is the wavelength

E = (6.626 × 10⁻³⁴ J.s) × (3.00 × 10⁸ m/s) × (680 × 10⁻⁹ m)⁻¹ = 2.92 × 10⁻¹⁹ J

The minimum number of photons required is:

[tex]2878\times 10^{3} J.\frac{1photon}{2.92\times 10^{-19}J} =9.86 \times 10^{24}photon[/tex]

Final answer:

The net change in photosynthesis involves forming glucose and O2 from CO2 and H2O with the aid of solar energy. The minimum number of photons needed at 680 nm to form 1.00 mol of glucose can be calculated using the energy requirements for the synthesis of glucose in photosynthesis.

Explanation:

The question pertains to the light-dependent reactions of photosynthesis, where solar energy is absorbed by chlorophyll to produce glucose from CO2 and H2O. Calculating the minimum number of photons with a wavelength of 680 nm needed to prepare 1.00 mol of glucose requires understanding the energy associated with photons and the process of photosynthesis. Since solar energy is converted into chemical energy in the form of ATP and NADPH during photosynthesis, and a total of 54 ATP equivalents are required to synthesize one molecule of glucose, the number of photons must provide at least this much energy. Considering that each photon's energy is inversely proportional to its wavelength, we can use the relationship E = hc/λ (where E is energy, h is Planck's constant, c is the speed of light, and λ is wavelength) to calculate the energy of a photon of 680 nm, and subsequently, determine the total number of photons required to provide the 54 ATP equivalents needed for glucose synthesis.

Answer:

a. The phenolphthalein acts as a color changing indicator to signal the endpoint of the reaction.

Explanation:

Phenolphthalein is an organic substance with chemical formula [tex]C_{20}H_{14}O_{4}[/tex].

It is a substance commonly used in acid-base titrations to indicate the end point in the titration because phenolphthalein is colorless in acidic solutions but turns a purplish-pink color in basic solutions.

In this way it helps visually to notice when the final point of the titration has been reached.

Phenolphthalein is added to an Erlenmeyer flask in a titration to act as a color-changing indicator that signals the endpoint of the reaction by turning from colorless to pink as the solution transitions from acidic to basic conditions, close to the equivalence point.

The purpose of adding phenolphthalein to an Erlenmeyer flask prior to starting a titration is essentially to serve as a color-changing indicator that signals the endpoint of the reaction. When used in an acid-base titration, phenolphthalein shifts in color from colorless in acidic conditions to pink in basic conditions. This color transition happens around a pH of 8.2 to 10, indicating that the solution has reached an alkaline environment which, for many titrations, corresponds closely to the equivalence point.

In titrations, the correct amount of titrant (a solution of known concentration) is added slowly to a solution until the reaction reaches its endpoint, which is often near the equivalence point. An indicator like phenolphthalein undergoes a sharp color change at a specific pH range, making it easier to determine when the reaction has reached this point. The precise moment of color change is critical in determining the exact volume of titrant added, essential for calculating the concentration of the unknown solution.

Answer: The boiling point of water is an absolute constant

Explanation:

The boiling point of a compound is NOT an absolute constant. This is because at certain conditions such as a change in altitude, the boiling point of a compound changes.

As temperature increases, evaporation increases and vapour pressure increases. Compounds boils when vapour pressure equal to the atmospheric pressure.

At higher altitudes, atmospheric pressure is decreses.

When atmospheric pressure is decresed, the vapour pressure of the compound is lowered to reach boiling point. Therefore, the temperature needed for vapour pressure to equal atmospheric pressure is lower. The boiling point is lower at higher altitude.

Hence boiling point depends on atmospheric pressure

The boiling point of a compound is not an absolute constant.

The statement "The boiling point of a compound is an absolute constant" is FALSE.

The boiling point of a compound is affected by several factors, including intermolecular forces and molecular structure. Compounds with stronger intermolecular forces, such as hydrogen bonds, will generally have higher boiling points. Viscosity, which is the resistance of a liquid to flow, is not directly related to boiling point. Nonvolatile compounds, which do not easily evaporate, may have lower boiling points depending on their molecular structure.

Therefore, the correct statement is that the boiling point is not an absolute constant.

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To determine the stronger oxidizing agent in each pair of substances, compare the oxidizing strengths using data from Appendix E. The stronger oxidizing agents are O3(g), Cl2(g), Co2+(aq), and BrO3-(aq).

To determine the stronger oxidizing agent in each pair, we need to compare the oxidizing strengths of the substances. From Appendix E in the textbook, we can find the relevant data. Here are the answers:

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To determine the stronger oxidizing agent, compare the standard reduction potentials of the species in each pair. The substance with the higher standard reduction potential is the stronger oxidizing agent.

To determine the stronger oxidizing agent in each pair, we need to compare their standard reduction potentials. The substance with the higher standard reduction potential is the stronger oxidizing agent. From Appendix E, we can find the standard reduction potentials of the species in each pair and compare them to determine the stronger oxidizing agent.

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Answer:

We need 247 mL of NaOH

Explanation:

Step 1: Data given

Molarity of NaOH = 0.587 M

Volume of 0.445 M NiBr2 solution = 163 mL = 0.163 L

Step 2: The balanced equation

NiBr2(aq) + 2NaOH(aq) Ni(OH)2(s) + 2NaBr(aq)

Step 3: Calculate moles of NiBr2

Moles NiBR2 = Molarity NiBR2 * volume

Moles NiBR2 = 0.445 M * 0.163 L

Moles NiBR2 = 0.0725 moles

Step 3: Calculate moles of NaOH

For 1 mol NiBr2 consumed, we need 2 moles NaOH

For 0.0725 moles NiBR2, we need 2* 0.0725 = 0.145 moles NaOH

Step 4: Calculate volume of NaOH

Volume = moles NaOH / Molarity NaOH

Volume = 0.145 moles / 0.587 M

volume = 0.247L = 247 mL

We need 247 mL of NaOH

Answer: b. Catalysis occurs at the active site, which usually consists of a crevice on the surface of the enzyme.

c. Generally, an enzyme is specific for a particular substrate. For example, thrombin catalyzes the hydrolysis of the peptide bond between Arg and Gly.

Explanation:

The role of the enzyme is to produce a specific product, whereas the non-biological catalyst produces more than one product. Thus they have a different degree of reaction specificity.

The active site of the enzyme is the catalytic site of the enzyme. It is the region where the substrate molecule bind and undergoes a chemical reaction. The enzyme exhibit a special crevice or opening at the active site which facilitates the binding with the substrate.

Enzymes usually bind to a specific substrate as they have an active site that allows a particular substrate to bind to the active site of the enzyme. This is because of the shape of the active site of the enzyme which has a binding affinity with a particular substrate any other substrate cannot bind to the active site.

Answer:

Explanation:

To solve this problem we use PV=nRT for both gases in their containers, in order to calculate the moles of each one:

645 Torr ⇒ 645 /760 = 0.85 atm

25°C ⇒ 25 + 273.16 = 298.16 K

0.85 atm * 1.40 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0487 mol O₂

1.13 atm * 0.751 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0347 mol N₂

Now we can calculate the partial pressure for each gas in the new container, because the number of moles did not change:

P(O₂) * 2.00 L = 0.0487 mol O₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(O₂) = 0.595 atm

P(N₂) * 2.00 L = 0.0347 mol N₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(N₂) = 0.424 atm

Finally we add the partial pressures of all gases to calculate the total pressure:

Final answer:

To find the partial pressures and total pressure in the container, convert pressures to atm, apply the ideal gas law for each gas, and sum the resulting partial pressures. The partial pressures are 0.5959 atm for O₂ and 0.4249 atm for N₂, with a total pressure of 1.0208 atm.

Explanation:

The question asks to calculate the partial pressures of O₂ and N₂ as well as the total pressure in a 2.00 L container at 25 °C, after transferring 1.40 L of O₂ at 645 Torr and 0.751 L of N₂ at 1.13 atm into it. To solve this, the pressure values need to be converted to a common unit (atmospheres), and the ideal gas law used for calculations.

First, convert the O₂ pressure from Torr to atm: 645 Torr / 760 = 0.8487 atm. Then use the ideal gas law (P₁V₁ = P₂V₂) to find the new pressures in the 2.00 L container. For O₂, (0.8487 atm)(1.40 L) / 2.00 L = 0.5959 atm. The pressure of N₂ is already in atm, so, (1.13 atm)(0.751 L) / 2.00 L = 0.4249 atm.

The total pressure is the sum of the partial pressures: 0.5959 atm (O₂) + 0.4249 atm (N₂) = 1.0208 atm. Therefore, the partial pressures are 0.5959 atm for O₂ and 0.4249 atm for N₂, with a total pressure of 1.0208 atm in the container.

Answer:

d. The quantity of heat required to raise the temperature of 1 mole of a substance by 1K

Explanation:

Heat capacity is a physical property of matter that is defined as the quantity of heat supplied for a material to change in one unit its temperature. This heat depends of the amount of material.

For that reason, molar heat capacity is defined as:

d. The quantity of heat required to raise the temperature of 1 mole of a substance by 1K

Alluding molar to moles of a material or substance.

I hope it helps!

Answer:

(a) Increase the solid  PbI₂ (s)

(b) Decrease  

(c) Increase

Explanation:

In this question  we are considering the effect of adding a common ion to an equilibrium.

The problem states that the saturated solution  of PbI₂ is at equilibrium with undissolved PbI₂ (s) :

PbI₂ (s)  ⇄     Pb²⁺+ 2 I⁻  

If we increase the concentration of I⁻ by adding the highly soluble KI , the equilibrium will react according to LeChatelier´s principle by decreasing the concentration of  Pb²⁺ precipitating some  Pb²⁺ until it reaches equilibrium again. With this in mind, lets answer the question:

(a) Increase the solid  PbI₂ (s)

(b) Decrease  

(c) Increase

The addition of KI in the solution increases the settling of lead iodide at the bottom, decreases the lead ions and increases the iodide ions in the solution.

The saturated solution has been given as the one in which the salt has been divided into maximum concentration.

The equilibrium reaction of lead iodide has been:

[tex]\rm PbI_2\;\rightleftharpoons Pb^2^+\;+2\;I^-[/tex]

The addition of KI increases the product concentration, and the lead iodide has been found to be settles in increased concentration in the bottom.

The addition of KI decreases the rate of formation of product, thus the ion of lead formation will be decreased in the reaction.

The concentration of iodide ions in the solution increases with the addition of KI with common ion effect.

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Answer:

The correct answer is D sp2 and sp3

Explanation:

Acetic acid is a weak acid containg 2 carbon atoms.the first carbon is joined to 3 hydrogen atoms by 3 covalent bonds and to the second carbo atom by one covalent bond.Therefore the first carbon of CH3COOH is sp3 hybridized.

  Whereas the second carbon is joined to the first carbon by a covalent bond,the second carbon form a double bond with oxygen and thesecond carbon forms a covalent bond with the oxygen atom of -OH group of COOH moiety.Therefore the second carbon is sp2 hybridized.

The type of hybridization that the carbon atoms of acetic acid (CH3COOH) exhibit include sp² and sp³.

Acetic acid is also referred to as ethanoic acid. It is a colorless liquid and an important chemical reagent used in the production of fabrics, synthetic fibers, and photographic film.

It has 2 carbon atoms. It should be noted that the first carbon is joined to 3 hydrogen atoms by 3 covalent bonds and to the second carbon atom by one covalent bond.

The second carbon is joined to the first carbon by a covalent bond and the second carbon forms a double bond with oxygen. Therefore, the types of hybridization exhibited for the carbon atoms are sp² and sp³.

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The volume that would be occupied by oxygen gas is equal to 10 Liters

Given the following data:

Mass = 4.8 g

Pressure = 0.50 atm

Temperature = 133°C to K = [tex]273+133 =406\;K[/tex]

Scientific data:

Molar mass of oxygen gas = 32 g/mol.

Ideal gas constant, R = 0.0821L⋅atm/mol⋅K

To determine the volume that would be occupied by oxygen gas:

First of all, we would determine the number of moles of oxygen gas present.

[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{4.8}{32}[/tex]

Number of moles = 0.15 moles

Now, we can find the volume of oxygen gas by the ideal gas law equation;

[tex]V = \frac{nRT}{P} \\\\V=\frac{0.15 \times 0.0821 \times 406}{0.5} \\\\V=\frac{4.99989}{0.5}[/tex]

Volume = 10 Liters

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Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

[tex] K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x} [/tex]

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[tex] [HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M [/tex]

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

[tex] K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x} [/tex]  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[tex] [SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M [/tex]

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

When .50 moles of sulfurous acid are dissolved in a 1 L solution, the concentration that will be least at equilibrium is [H2SO3].

In the given equilibrium reactions, when .50 moles of sulfurous acid are dissolved in a 1 L solution, the concentration that will be LEAST at equilibrium is [H2SO3], which represents the original acid concentration.

To determine the amount of CO2 and H2O produced in the combustion of 1.24 gallons of octane (C8H18), convert the volume to grams using the given density. Then, balance the combustion equation and calculate the mole ratio. Finally, use the mole ratio to calculate the grams of CO2 and H2O produced.

To determine the amount of CO2 and H2O produced in the combustion of 1.24 gallons of octane (C8H18), we need to use stoichiometry and conversion factors. First, convert the volume of octane from gallons to milliliters, then to grams using the given density. Next, balance the combustion equation and determine the mole ratio between octane, CO2, and H2O. Finally, use the mole ratio to calculate the grams of CO2 and H2O produced.

Step 1: Convert gallons to milliliters:

Step 2: Convert milliliters to grams:

Step 3: Balance the combustion equation:

Step 4: Calculate mole ratio:

Step 5: Calculate grams of CO2 and H2O produced:

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To find the amount of CO2 and H2O produced in the combustion of 1.24 gallons of octane, convert the volume to grams using the density of octane. Then, use the balanced equation to calculate the moles of octane, CO2, and H2O. Finally, convert the moles to grams using their respective molar masses.

To calculate the amount of CO2 and H2O produced in the combustion of 1.24 gallons of octane, we need to convert the volume of octane to grams using its density. First, convert 1.24 gallons to quarts by multiplying it by 4 (since there are 4 quarts in a gallon). This gives us 4.96 quarts. Next, convert quarts to milliliters by multiplying by 946.35 (since there are approximately 946.35 milliliters in a quart). This gives us 4685.24 mL. Finally, multiply the volume in milliliters by the density of octane (0.703 g/mL) to get the mass of octane in grams. This comes out to be 3289.02 grams.

Now, let's consider the balanced equation for the combustion of octane:

C8H18 + 12.5O2 -> 8CO2 + 9H2O

From the balanced equation, we can see that for every 1 mole of octane, 8 moles of CO2 and 9 moles of H2O are produced. To find the number of moles of octane in 3289.02 grams, divide the mass by the molar mass of octane:

3289.02g / 114.22g/mol = 28.8 mol of octane

Now, multiply the number of moles of octane by the stoichiometric coefficients from the balanced equation to find the number of moles of CO2 and H2O:

CO2: 28.8 mol octane x 8 mol CO2 / 1 mol octane = 230.4 mol CO2

H2O: 28.8 mol octane x 9 mol H2O / 1 mol octane = 259.2 mol H2O

Finally, convert the number of moles of CO2 and H2O to grams by multiplying by their respective molar masses:

CO2: 230.4 mol CO2 x 44.01 g/mol = 10,134.74 g CO2

H2O: 259.2 mol H2O x 18.02 g/mol = 4,674.24 g H2O

Therefore, in the combustion of 1.24 gallons of octane, approximately 10,134.74 grams of CO2 and 4,674.24 grams of H2O are produced.

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The final temperature of the granite and the water will be 26.6 °C.

In a mixture between two substances of different temperatures, there is heat transfer, in which the one with a lower temperature receives this heat and the one with a higher temperature loses the heat.

In a system, the value of heat lost and received is equal.

This heat is represented by the following expression:

                                         [tex]Q = m \times c \times T[/tex]

The value of heat is given in calories, the value of temperature is given in celsius and the value of mass is given in grams

Thus, to find the value of the final temperature of the mixture, it is enough to equate the heat of the two substances.

                     [tex]1100 \times (\frac{0.803}{4,18}) \times (t_{f} - 83.7) = 2000 \times 1 \times (t_{f} - 20.3)[/tex]

                               [tex]220 (t_{f} - 83.7) = 2000 (t_{f} - 20.3)[/tex]

As the value of the granite final temperature is lower than the initial temperature, the result of the temperature variation would be negative, so the signs of the expression can be changed.

                                 [tex]18414 - 220 t_{f} = 2000 t_{f} - 40600[/tex]

                                         [tex]2220 t_{f} = 59,014[/tex]

                                              [tex]t_{f} = 26.6[/tex]

So, the final temperature of the granite and the water will be 26.6 °C.

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Using conservation of energy, the final equilibrium temperature of the granite and water system can be calculated by setting the heat lost by granite equal to the heat gained by water and solving for the final temperature.

The question is asking for the final equilibrium temperature of a two-part system initially at different temperatures when they are put in contact. This is a problem that can be solved using the concept of heat transfer and the principle of conservation of energy. The heat lost by the hot granite will equal the heat gained by the cooler water until thermal equilibrium is reached.

Using the specific heats of granite and water, and their masses and initial temperatures, we set up the equation based on the principle that heat lost = heat gained:

m1c1(Tfinal - T1(initial)) = m2c2(T2(initial) - Tfinal)

Where m1 and m2 are the masses of granite and water, c1 and c2 are their specific heats, Tfinal is the final equilibrium temperature and T1(initial) and T2(initial) are the initial temperatures of granite and water, respectively. You'll need to convert the mass of the granite to grams, as the specific heat is given in grams, and use the specific heat of water (4.184 J/g°C).

Then you can solve for Tfinal to find the temperature at which the granite and water reach equilibrium.

Answer:

50. 5mol

Explanation:

The mass of Zinc required = 3.30 kg = 3300 grams

The relation between moles and mass is:

[tex]moles=\frac{mass}{atomicmass}[/tex]

Atomic mass of Zinc = 65.38g/mol

[tex]molesofZinc=\frac{3300}{65.38}[/tex]

[tex]molesofzinc=50.47mol[/tex]

The answer to three significant figures will be 50.5 mol

To find the number of moles of zinc in 3.30 kg of zinc, divide the mass in grams by the molar mass of zinc. The number of moles of zinc is 50.5 mol.

To find the number of moles of zinc in 3.30 kg of zinc, we can use the molar mass of zinc to convert the mass to moles. The molar mass of zinc (Zn) is 65.38 g/mol. First, we convert the mass from kg to g by multiplying by 1000:

3.30 kg × 1000 g/kg = 3300 g

Then, we divide the mass in grams by the molar mass to get the number of moles:

3300 g / 65.38 g/mol = 50.5 mol

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The reaction of 50 mL of Cl2 gas with 50 mL of C2H4 gas will produce 50 mL of C2H4Cl2 gas.

In this reaction, the balanced equation tells us that 1 mole of Cl2 reacts with 1 mole of C2H4 to produce 1 mole of C2H4Cl2. Since the volumes of the gases are given, we can assume that the gases are at the same pressure and temperature. This means that the ratio of the volumes of the gases is the same as the ratio of the moles of the gases.

Therefore, if we have 50 mL of Cl2 gas and 50 mL of C2H4 gas, we would have 1 mole of Cl2 and 1 mole of C2H4. The reaction would produce 1 mole of C2H4Cl2, which is equivalent to 50 mL of C2H4Cl2 gas.

Hence, the total volume of products produced in this reaction would be 50 mL.

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The reaction of 50 mL of Cl₂ gas with 50 mL of C₂H₄ gas under constant pressure and temperature produces 50 mL of C₂H₄Cl₂ gas as the product, following Avogadro's law.

The question revolves around a chemical reaction where Cl₂ gas reacts with C₂H₄ gas to produce C₂H₄Cl₂ gas. According to Avogadro's law, at constant temperature and pressure, equal volumes of gases contain the same number of molecules. Hence, the volumes of gaseous reactants and products should be in the same ratio as their coefficients in the balanced chemical equation.

In this case, the chemical equation is:

Cl₂ (g) + C₂H₄ (g) → C₂H₄Cl₂ (g)

Each reactant has a coefficient of 1, indicating that they react on a one-to-one volume basis. This means that 50 mL of Cl₂ reacts with 50 mL of C₂H₄ to produce 50 mL of C₂H₄Cl₂. Since temperature and pressure are held constant, by Avogadro's principle, the volume of the products is the same as the volume of the reactants.

Therefore, the reaction of 50 mL of Cl₂ gas with 50 mL of C₂H₄ gas will produce a total of 50 mL of C₂H₄Cl₂ gas as the product.

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Answer:

The specific heat capacity of brass is 0.402 J/g°C

Explanation:

Step 1: Data given

Mass of the sample of brass = 52.4 grams

Mass of water = 150.0 grams

Initial temperature of brass sample = 95.1 °C

Initial temperature of water = 15.0 °C

Final temperature = 17.6°C

Pressure = 1atm

Specific heat capacity = 4.184 J/g°C

Step 2: Calculate specific heat capacity of brass sample

Heat lost by the brass sample = heat won by water

Q = m*c*ΔT

Qbrass = -Qwater

m(brass)*c(brass)*ΔT(brass) = -m(water) *c(water) * ΔT(water)

⇒mass of brass sample = 52.4 grams

⇒ c(brass) = TO BE DETERMINED

⇒ ΔT = T2 - T1 = 17.6 °C - 95.1 °C = -77.5 °C

⇒ mass of water = 150.0 grams

⇒ c(water ) = 4.184 J/g°C

⇒ ΔT(water) = 17.6 °C - 15 °C = 2.6°C

52.4g * c(brass) * (-77.5°C) = -150.0g * 4.184 J/g°C * 2.6 °C

-4061 * c(brass) = -1631,76

c(brass) = 0.402 J/g°C

The specific heat capacity of brass is 0.402 J/g°C

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Answer:

(c) Total Number of moles = 3

Explanation:

To solve this question we have to follow the stoichiometry of the reaction and the fact that at equilibrium we have 1 mol of CH₃OH using the mnemonic helper RICE ( Reaction, Initial, Change, Equilibrium) as follows:

R             CO(g) + 2 H₂ (g)   ⇄  CH₃OH (g)

I (mol)        2             3                     0

C (mol)       -x         x -2x               + x

E (mol)      2-x        3-2x                   x

But we are told that at equilibrium x = mol CH₃OH = 1 , then at equilibrium we will have

E (mol)       2-(1)      3-2(1)             1

E (mol)          1            1                 1

Therefore, the total number of moles of gas present in the container at equilibrium was 3

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

HNO2 is a weak acid whose dissociation occurs as follows;  HNO2 (aq)    ⇋       H+ (aq)  +      NO2- (aq)

The equation for the dissociation of HNO2 is;

           HNO2 (aq)    ⇋       H+ (aq)  +      NO2- (aq)

We can use the value of Ka to find the  ΔG∘ for the dissociation of nitrous acid in aqueous solution as follows;

ΔG∘ = -RTlnKa

Where;

R = Gas constant = 8.314 J/K. mol

Ka = Acid dissociation constant = 4.5×10^−4

T = temperature = 25 ∘C or 298 K

Substituting values;

ΔG∘ = -(8.314 J/K. mol × 298 K × ln 4.5×10^−4)

ΔG∘ = 19.1 KJ/mol

Given that;

Q = [ H+] [NO2- ]/[HNO2]

[ H+] =  5.9×10−2 M

[NO2- ] = 6.7×10−4 M

[HNO2] = 0.21 M

Q = [5.9×10−2 M] [6.7×10−4 M]/[0.21 M]

Q = 1.88 × 10^−4

From the formula;

ΔG = ΔG∘ + RTlnQ

ΔG =  19.1  KJ/mol + (8.314 J/K. mol  ×  298 K ×  ln 1.88 × 10^−4)

ΔG = -2.15 KJ/mol

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Answer : The work done on the gas mixture is -164 kJ

Explanation :

Formula used :

[tex]w=-p\Delta V\\\\w=-p(V_2-V_1)[/tex]

where,

w = work done  = ?

p = pressure of the gas = 58.0 atm

[tex]V_1[/tex] = initial volume = 33.0 L

[tex]V_2[/tex] = final volume = 61.0 L

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

[tex]w=-p(V_2-V_1)[/tex]

[tex]w=-(58.0atm)\times (61.0-33.0)L[/tex]

[tex]w=-1624L.atm=-1624\times 101.3J=-164511.2J=-164.5kJ\aprrox -164kJ[/tex]

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas mixture is -164 kJ

Answer:

3M

Explanation:

Because HCl and NaCl do not react with each other, the final number of the Cl⁻ ions is equal to the sum of the ions of each starting reagent:

Cl⁻ moles from HCl ⇒0.100 L * 1 M = 0.1 moles Cl⁻

Cl⁻ moles from HCl ⇒0.100 L * 5 M = 0.5 moles Cl⁻

We add up the volumes from each solution:

And finally calculate the final molarity of Cl⁻ ions:

Answer:

The false statement is: (C) The average kinetic energy of the molecules is not related to the absolute temperature

Explanation:

According to the Kinetic Theory of Gases, gases are composed of small fast moving particles that are in constant random motion. These gaseous particles are known as molecules.

It is assumed that the mass of all the gaseous molecules is the same and the intermolecular interactions (attractive or repulsive) are negligible.

Also, the total volume of all the gaseous molecules is considered to be negligible as compared to the volume of the container.

According to this theory, average kinetic energy (K) of the gaseous particles is directly proportional to the absolute temperature (T) of the gas, by the equation:

[tex]K = \frac{3}{2}k_{b}T[/tex]

Here, [tex]k_{b}[/tex] is the Boltzmann constant

Therefore, the false statement about the kinetic molecular theory is (C)

Statement C, which suggests that the average kinetic energy of the molecules is not related to the absolute temperature, is incorrect.

The question is focused on the Kinetic Molecular Theory, which primarily has four postulates that describe the behavior of gases. The incorrect statement in the list is the following:

In reality, according to the Kinetic Molecular Theory, the average kinetic energy of gas particles is directly proportional to the absolute temperature. This means that as the temperature increases, the average kinetic energy of the gas particles also increases, and vice versa. Therefore, it's clear that the temperature and average kinetic energy are related, making statement C incorrect.

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Answer:

A) 6

B) 6

C) 4

E) 1

Explanation:

Coordination number, also called Ligancy, the number of atoms, ions, or molecules bound to the metal atom in a complex ion

A) [Fe(gly)2(H2O)2]

Gly = the aminoacid glycin, and works as a  bidentate ligand,Bidentate ligands bind through two donor sites. Bidentate means "two-toothed".  It can bind to a metal via two donor atoms at once.

This makes the coordination number = 6

B) [Pb(EDTA)]2

EDTA is a hexadentate ligand and forms very stable complexes.

EDTA forms 6 bonds to the Pb atom (2 Pb-N bonds and 4 Pb-O bonds). The coordination number = 6.

C) [Pt(NH3)4]2+

It's a squareplanar complex with coordination number = 4 because 4 bindings with NH3

D) Na[Au(CN)2]

Cyanide binds only to 1 Au- atom. The coordination number = 1

A) [tex][Fe(gly)2(H2O)_2]^+[/tex]: Coordination number is 6.

B)[tex][Pb(EDTA)]_2[/tex]−: Coordination number is 6.

C) [tex][Pt(NH_3)_4]^2+[/tex]: Coordination number is 4.

D) [tex]Na[Au(CN)_2]:[/tex] Coordination number is 2.

The coordination numbers for each of the given complexes are as follows:

A) [tex][Fe(gly)2(H2O)2]+[/tex]: The coordination number of the iron (Fe) ion in this complex is 6. This is because there are two glycine (gly) ligands and two water (H2O) molecules coordinated to the iron ion. Each glycine is bidentate, meaning it binds to the iron ion with two donor atoms (usually nitrogen and oxygen), contributing a total of four donor atoms. The two water molecules each provide one donor atom (oxygen). Thus, the total number of donor atoms around the iron ion is 6, giving it a coordination number of 6.

 B) [tex][Pb(EDTA)]2[/tex]: The coordination number of the lead (Pb) ion in this complex is 6. Ethylenediaminetetraacetic acid (EDTA) is a hexadentate ligand, meaning it can bind to a metal ion with all six of its donor atoms (four oxygens and two nitrogens). Since EDTA is fully deprotonated in this complex, it forms a negatively charged species that can wrap around the lead ion, resulting in a coordination number of 6.

 C) [tex][Pt(NH_3)_4]^2+:[/tex] The coordination number of the platinum (Pt) ion in this complex is 4. There are four ammonia (NH3) molecules coordinated to the platinum ion, each acting as a monodentate ligand and providing one donor atom (nitrogen). Therefore, the platinum ion is surrounded by four donor atoms, giving it a coordination number of 4.

 D) [tex]Na[Au(CN)_2][/tex]: The coordination number of the gold (Au) ion in this complex is 2. There are two cyanide (CN) ligands coordinated to the gold ion, each acting as a monodentate ligand and providing one donor atom (carbon). Thus, the gold ion is surrounded by two donor atoms, giving it a coordination number of 2.

 To summarize:

A) [tex][Fe(gly)2(H2O)_2]^+[/tex]: Coordination number is 6.

B)[tex][Pb(EDTA)]_2[/tex]−: Coordination number is 6.

C) [tex][Pt(NH_3)_4]^2+[/tex]: Coordination number is 4.

D) [tex]Na[Au(CN)_2]:[/tex] Coordination number is 2.