How many electrons in an atom can have each of the following quantum number or sublevel designations?
(a) n = 2, l = 1
(b) 3d
(c) 4s

Answer:

[AB] is 0.65 M

Explanation:

Let the concentration of AB after 10.3s be y

Rate = ky^2 = change in concentration of AB/time

k = 0.2 L/mol.s.

Change in concentration of AB = 1.5 - y

Time = 10.3s

0.2y^2 = 1.5-y/10.3

0.2y^2 × 10.3 = 1.5 - y

2.06y^2 = 1.5 - y

2.06y^2 + y - 1.5 = 0

The value of y must be positive and is obtained using the quadratic formula

y = [-1 + sqrt(1^2 -4×2.06×-1.5)]/2(2.06) = [-1 + sqrt(13.36)]/4.12 = 2.66/4.12 = 0.65 M

Final answer:

To find the concentration of AB after 10.3 seconds, the second-order integrated rate law was used with an initial concentration of 1.50 M and a rate constant of 0.20 L/mol·s. The final concentration of AB was calculated to be approximately 0.32 M.

Explanation:

To calculate the concentration of AB after 10.3 seconds in the given decomposition reaction, we need to use the integrated rate law for a second-order reaction, which is as follows:

1/[AB] - 1/[AB]₀ = kt

Where:

[AB] is the concentration at time t,

[AB]₀ is the initial concentration,

k is the rate constant, and

t is the time elapsed.

Given that k = 0.20 L/mol·s, [AB]₀ = 1.50 M, and t = 10.3 s, we can rearrange the equation and solve for [AB] as follows:

1/[AB] = 1/1.50 M + (0.20 L/mol·s)(10.3 s)

1/[AB] ≈ 1/1.50 M + 2.06 L/mol

[AB] ≈ 1/(1/1.50 + 2.06) L/mol

[AB] ≈ 0.32 M

After 10.3 seconds, the concentration of AB is approximately 0.32 M.

Answer:

pKa = 7.2

Explanation:

When pH=pKa=2.0, there are equal amounts of the X carboxyl group and its ionized form.

Then, (75 mL * 0.10 M) 7.5 mmol of OH⁻ are added, so all 5 mmol of X-COOH converts into X-COO⁻. Then the remaining (7.5 - 5) 2.5 mmol of OH⁻ react with the second ionizable group of X

Number of X-COO⁻ = 5mmol (from the beginning) + 5mmol (from X-COOH that reacted) - 2.5 mmol (from the OH⁻ remaining) = 7.5 mmol

Because the total number of X compound moles did not change, we have (10 - 7.5) 2.5 mmol of the conjugate base of X-COO⁻.

Now we have all required data to solve this problem using Henderson-Hasselbach's equation:

pH = pKa + log [A⁻]/[HA]

6.72 = pKa + log (2.5/7.5)

pKa = 7.2

Answer:

[tex]pK_a[/tex]  = 7.20

Explanation:

Given that our Molarity of X(unknown compound)= 0.10 M

Volume of X = 100 ml = 0.1  L

Molarity  = [tex]\frac{number of moles of X}{Volume of X}[/tex]

Number of moles of X = Molarity × Volume of X

= 0.1 M × 0.1 L

= 0.01 mol

[tex]pK_a[/tex] = [tex]pH[/tex]

∴[tex][H^+][/tex] =[tex][HA][/tex]    (this literaly implies and point out that the beginning the amount of carboxyl group and the second ionizable group must be equal.)

Having said that,

Number of moles of carboxyl group in X = [tex]\frac{0.01mol}{2}[/tex]

= 0.005 mol

When 75 ml of 0.10 M NaOH is added to 100 ml of a 0.10 M solution of X;

we have the number of moles of NaOH that is being added as:

Molarity × volume of NaOH

= 0.1 M × 0.075 L

= 0.0075 mol

From the question, if NaOH molecules thoroughly dissociate the carboxyl group of X.

The excess NaOH can be calculated as:

Excess of (NaOH) = Number of moles of NaOH added - Number of moles of carboxyl group in X

Excess of (NaOH) = 0.0075 -0.005 = 0.0025 mol

∴ Applying Henderson-Hesselbalch equation; it will be easier to determine the [tex]pK_a[/tex] for second group:

[tex]pK_a[/tex]  = [tex]pH[/tex] [tex]-\frac{log[A]}{HA}[/tex]

       = 6.72 - log[tex](\frac{0.0025}{0.0075})[/tex]

       = 6.72 - log (0.333)

       = 6.72 + 0.477

[tex]pK_a[/tex]  = 7.197

≅ 7.20

Answer:

Molality = 18.5 m

Explanation:

Let's analyse data. We want to determine molality which means mol of solute / 1kg of solvent. (Hence we need, the moles of solute and the mass of solvent in kg)

12.83 M means molarity → mol of solute in 1L of solution

Density refers always to solution → Mass of solution / Volume of solution

1L = 1000 mL

We can determine the mass of solution with density

0.9102 g/mL = Mass of solution / 1000 mL

Mass of solution = 0.9102 g/mL . 1000 mL → 910.2 g

Let's convert the moles of solute (NH₃) to mass

12.83 mol . 17.03 g/ 1 mol = 218.5 g

We can apply this knowledge:

Mass of solution = Mass of solvent + Mass of solute

910.2 g = Mass of solvent + 218.5 g

910.2 g - 218.5 g = 691.7 g → Mass of solvent.

Let's convert the mass in g to kg

691.7 g . 1kg / 1000 g = 0.6917kg

We can determine molalilty now → 12.83 mol / 0.6917kg

Molality = 18.5 m

To determine the molality of a 12.83 M NH3 solution with a density of 0.9102 g/mL, calculate the mass of NH3 per liter using molarity and molar mass, find the mass of the solution using volume and density, subtract the mass of NH3 to find the mass of water, and finally divide the moles of NH3 by the mass of water in kilograms.

To find the molality (m) of an aqueous solution of ammonia (NH3) with a molarity (M) of 12.83 and a density of 0.9102 g/mL, we need to calculate the number of moles of NH3 per kilogram of water. Molality is defined as moles of solute per kilogram of solvent (water). First, we determine the mass of NH3 in 1 liter of solution, considering the molarity and the molar mass of NH3 (17.03 g/mol). Using the density of the solution, we then calculate the mass of the solution and subtract the mass of the ammonia to find the mass of water.

Here's the step-by-step calculation:

This approach allows us to determine molality, which is essential for understanding the colligative properties of the solution.

Answer:

[tex]\large \boxed{\text{55.8 u}}[/tex]

Explanation:

1. Calculate the volume of the unit cell

V = l³ = (2.866 × 10⁻⁸ cm)³ = 2.354 × 10⁻²³ cm³

2. Calculate the mass of a unit cell

[tex]\text{Mass} = 2.866 \times 10^{-23}\text{ cm}^{3} \times \dfrac{\text{7.87 g}}{\text{1 cm}^{3}} = 1.853 \times 10^{-22} \text{ g}[/tex]

3. Calculate the mass of one atom

A body-centred unit cell contains two atoms.

[tex]\text{Mass of 1 atom} = \dfrac{1.853 \times 10^{-22} \text{ g}}{\text{2 atoms}} \times \dfrac{\text{1 u}}{1.661 \times 10^{-24}\text{ g}} = \textbf{55.8 u}\\\\\text{The molar mass of Fe from the Periodic Table is $\large \boxed{\textbf{55.845 g/mol}}$}[/tex]

The number of grams of sodium chloride in the solution is 3.43 g.

The number of grams of sodium chloride in the solution can be calculated using the molar mass and molarity.

First, we need to calculate the number of moles of NaCl in the solution. Using the given molarity (0.470 M) and volume (125.0 mL) of the solution, we can use the formula:

moles = molarity × volume (in L)

Therefore, moles of NaCl = 0.470 mol/L × 0.125 L = 0.05875 mol NaCl

Next, we can use the formula mass of NaCl (58.44 g/mol) to calculate the mass:

mass = moles × formula mass

Therefore, mass of NaCl = 0.05875 mol × 58.44 g/mol = 3.43 g NaCl

Answer:

∆=2dn/n

Explanation:

Where∆ = lamda

the construction of the 10.20 cm into feet so which which Alina dispersion form 400-700 NM assuming the dielectric has a refractive index of 1.32.

construction of the interference which is described by

∆=2dn/n

the concept of interference which can be described by the calculation of the thickness of the wage at both end to do this report you to the wavelengths of absorption band and the thickness of dielectric constant.

That is

∆=2dn/n

d=∆R/2n

Where ∆ is equal wavelength d = thickness

n = interference order

R = refractive index of dielectric medium

range of linear dispersion is 400nm to 700nm

∆1 = 400nm

∆2 = 700nm

n = 1.32

d = ∆1R/2n

d = 400*10^-9m*1.32/2*1

d = 2.64*10^-7m

For ∆2

d = 700*10^-9m*1.32/2*1

d=4.62*10^-7m

Details of the construction is simply calculation of the thickness of the wedge at both ends

Final answer:

To build a 10.0-cm interference wedge with linear dispersion from 400 to 700 nm using a dielectric with refractive index 1.32 involves precision crafting of a wedge where the thickness incrementally increases, causing a linear change in path length and hence wavelength dispersion.

Explanation:

Constructing a 10.0-cm interference wedge with a linear dispersion from 400 to 700 nm involves creating a transparent object with a slight and uniform increase in thickness from one end to the other, using a dielectric material with a known refractive index, which in this case is 1.32.

The wedge must be carefully designed so that the path length difference between the top and bottom surfaces changes by exactly 300 nm (the range from 400 to 700 nm) over the 10 cm length. This incremental change in path length creates a linear dispersion of wavelengths when light is shone through the wedge.

At the thinnest point, light with a wavelength of 400 nm should experience constructive interference, whereas at the thickest point, light with a wavelength of 700 nm should interfere constructively. The process involves precision cutting and polishing to ensure consistent graduation of thickness across the wedge.

Answer:

The mole fraction of hexane is 0.6204

Explanation:

Obs: X hex = 1 - X pen

Ptot = P pen + P hex

Ptot = X pen P pen + X hex P hex ---> applying X hex = 1 - X pen

X pen = (Ptot - P hex) / (P pen - P hex)

X pen = (255 - 151) / (425 - 151) = 0.3795

X hex = 1 - X pen

X hex = 1 - 0.3795 = 0.6204

Answer:

Make a graph

Explanation: edge 2021

Answer:

Explanation:

Science is sustainable and reliable, as it is thoroughly tested and updated, impartial individuals will be observing scientific proof, and science only alters when new research justifies shift.The main concerns of many, if not most are scientific and engineering activities stability and transition. Efforts are made to ensure whatever theory is totally valid after a thorough investigation.

Answer:

c. an element.

Explanation:

An element -

It refers to the substance , which has same type of atoms , with exactly same number of protons , is referred to as an element .

In term of chemical species , elements are the smallest one , and can not be bifurcated down to any further small substance by the means of any chemical reaction .

Hence , from the given information of the question ,

The correct term is an element  .

Answer:

C. an element.

Explanation:

Answer:

This experiment requires two 3-h lab sessions: reduction of p-isobutylacetophenone to an alcohol and then convert this alcohol to the corresponding chloride

-convertion of chloride to a Grignard reagent

Explanation:

A method for the synthesis of ibuprofen in introductory organic chemistry laboratory .This experiment requires two 3-h lab sessions. All of the reactions and techniques are a standard part of any introductory organic chemistry course. In the first lab session, reduction of p-isobutylacetophenone to an alcohol and then convert this alcohol to the corresponding chloride. In the second session, convert this chloride to a Grignard reagent, which is then carboxylated and protonated to give ibuprofen. Although the final yield is modest, this procedure offers both practicability and reliability. Permanent-magnet 60 MHz 1H NMR spectra of the final product and the two intermediates are clean and are easily interpreted by the students. Because, as previously reported, the benzylic methylene and the benzylic methine of ibuprofen have virtually identical 13C NMR chemical shifts and cancel or nearly cancel each other in the DEPT spectrum, this synthesis provides a fitting opportunity for the introduction of HETCOR even with a permanent-magnet Fourier transform instrument.

Final answer:

To convert benzene into p-isobutylacetophenone, a Friedel-Crafts Acylation followed by a Friedel-Crafts Alkylation reaction is performed, with careful control of reaction conditions for optimal yield.

Explanation:

To convert benzene into p-isobutylacetophenone, a series of organic reactions must be performed. This process starts with the conversion of benzene into an acetophenone derivative. Here's a general approach that one might take, starting with benzene:

Friedel-Crafts Acylation: To introduce the acetyl group, you perform a Friedel-Crafts acylation reaction with acetyl chloride in the presence of a Lewis acid catalyst like aluminum chloride (AlCl3). This gives you acetophenone.

Friedel-Crafts Alkylation: Next, to add the isobutyl group at the para position, perform a Friedel-Crafts alkylation using isobutyl chloride with again AlCl3 as the catalyst.

Additional purification steps may be necessary to isolate the desired p-isobutylacetophenone.

Throughout the synthesis, reaction conditions such as temperature and solvent used will need to be carefully controlled for optimal yield and product purity. This compound is a synthetic intermediate potentially used in the synthesis of the anti-inflammatory agent ibuprofen, which highlights the significance of atom economy and green chemistry principles in pharmaceutical manufacturing.

Answer:

-2.80 × 10³ kJ/mol

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.

Qcal + Qcomb = 0

Qcomb = - Qcal [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ

where,

C: heat capacity of the calorimeter

ΔT: change in the temperature

From [1],

Qcomb = - Qcal = -29.2 kJ

The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:

ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol

Answer: The partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm

Explanation:

To calculate the pressure of the gas, we use the equation given by ideal gas, which follows:

[tex]PV=nRT[/tex]       ......(1)

where,

P = pressure of the gas

V = Volume of the gas

T = Temperature of the gas

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles of gas

We are given:

[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=1.80mol[/tex]

Putting values in equation 1, we get:

[tex]p_A\times 7.07L=1.80mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{A}=\frac{1.80\times 0.0821\times 303.4}{7.07}=6.34atm[/tex]

We are given:

[tex]V=7.07L\\T=30.4^oC=[30.4+273]K=303.4K\\n=4.86mol[/tex]

Putting values in equation 1, we get:

[tex]p_B\times 7.07L=4.86mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 303.4K\\\\p_{B}=\frac{4.86\times 0.0821\times 303.4}{7.07}=17.1atm[/tex]

Hence, the partial pressure of gas A is 6.34 atm and that of gas B is 17.1 atm

Answer:

The partial pressure of gas A is 6.34 atm

The partial pressure of gas B is 17.12 atm

Explanation:

Step 1 :Data given

Volume of cylinder = 7.07 L

Number of moles gas A = 1.80 moles

Number of moles gas B = 4.86 moles

Temperature =30.4 ° C = 303.55 K

Step 2: Calculate pressure of gas A

p*V = n*R*T

p =(n*R*T)/V

⇒ with p = the partial pressure of gas A

⇒ with V = The volume of the cylinder = 7.07 L

⇒ with n = the number of moles gas A = 1.80 moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 303.55 K

p = (1.80 *0.08206 *303.55)/7.07

p = 6.34 atm

Step 3: Calculate pressure of gas B

p*V = n*R*T

p =(n*R*T)/V

⇒ with p = the partial pressure of gasB

⇒ with V = The volume of the cylinder = 7.07 L

⇒ with n = the number of moles gas B = 4.86 moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 303.55 K

p = (4.86 *0.08206 *303.55)/7.07

p = 17.12 atm

The partial pressure of gas A is 6.34 atm

The partial pressure of gas B is 17.12 atm

A first-order decomposition reaction has a rate constant of 0.00440 yr−1. Hence the correct answer is 473.418years. Rounded to the correct number of significant figures, the time it takes for the reactant to reach 12.5% of its original value is approximately 470 years.

A first-order reaction follows the exponential decay equation:

[tex]\[ [A][/tex] = [tex][A]_0 \times e^{-kt}[/tex]

Where:

[tex]\([A]\)[/tex] is the concentration of reactant at time [tex]\(t\)[/tex]

[tex]\([A]_0\)[/tex] is the initial concentration of the reactant.

[tex]\(k\)[/tex] is the rate constant.

[tex]\(t\)[/tex] is time.

It is given that the rate constant[tex]\(k\)[/tex] is 0.00440 y[tex]r^(^-^1^)[/tex] and we want to find out how long it takes for the reactant concentration to reach 12.5% of its original value, which means [tex]\([A][/tex] = [tex]0.125 \times [A]_0\).[/tex]

The equation to solve for time[tex]\(t\)[/tex]:

[tex]\[ t[/tex] = [tex]-\frac{1}{k} \ln\left(\frac{[A]}{[A]_0}\right)[/tex]

Substitute the given values:

[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \ln\left(\frac{0.125 \times [A]_0}{[A]_0}\right)[/tex]

Simplifying:

[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \ln(0.125) \][/tex]

Now calculate the value:

[tex]\[ t = -\frac{1}{0.00440 \, \text{yr}^{-1}} \times (-2.07944) \][/tex]

[tex]\[ t \approx 473.418 \, \text{years} \][/tex]

Rounded to the correct number of significant figures, the time it takes for the reactant to reach 12.5% of its original value is approximately 470 years.

Learn more about the first order decomposition here.

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A first-order decomposition reaction is when the rate of the reaction is proportional to the concentration of the reactant. To find the time it takes for the reactant to reach 12.5% of its original value, we can use the equation ln([reactant] / [initial]) = -kt, where t represents time and k is the rate constant. Using the given rate constant, we find that it would take approximately 32.48 years for the reactant to reach 12.5% of its original value.

A first-order decomposition reaction is one in which the rate of the reaction is proportional to the concentration of the reactant. The rate equation for a first-order reaction is given by: rate = k[reactant], where k is the rate constant and [reactant] is the concentration of the reactant.

In this case, the rate constant is given as 0.00440 yr-1. To find the time it takes for the reactant to reach 12.5% of its original value, we can use the equation ln([reactant] / [initial]) = -kt, where [initial] is the initial concentration. Rearranging the equation, we have t = ln([reactant] / [initial]) / -k. Plugging in the percentage values, we get t = ln(0.125) / -0.00440 yr-1.

Calculating this value, we find that t ≈ 32.48 years.

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Answer:

a.A pure substance is made up of only 1 type of particle

Explanation:

When a substance is pure, it has only one type of particle. These particles maybe molecules, ions or atoms linked in a definite way throughout the substance. If a substance contains different particles, it cannot be regarded as a pure substance because its properties will be observed as a compromise of the individual properties of its different components.

The statement that accurately describes a pure substance is that it is made up of only one type of particle. Both elements and compounds can be considered as pure substances.

The correct statement to describe a pure substance is:

a. A pure substance is made up of only 1 type of particle.

This means that a pure substance is only made up of identical atoms if it is an element, or identical molecules if it is a compound. Therefore, both elements and compounds can be considered as pure substances, contradicting options c and d. It also contradicts option b because a pure substance is not made up of more than one type of particle.

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The question is incomplete, here is the complete question:

A chemist determines by measurements that 0.0850 moles of fluorine gas participate in a chemical reaction. Calculate the mass of fluorine gas that participates. Be sure your answer has the correct number of significant digits.

Answer: The mass of fluorine gas that is precipitated is 3.23 grams

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Moles of fluorine gas = 0.0850 moles

Molar mass of fluorine gas = 38.0 g/mol

Putting values in above equation, we get:

[tex]0.0850mol=\frac{\text{Mass of fluorine gas}}{38.0g/mol}\\\\\text{Mass of fluorine gas}=(0.0850mol\times 38.0g/mol)=3.23g[/tex]

Hence, the mass of fluorine gas that is precipitated is 3.23 grams

To calculate the mass of fluorine gas that participates in a chemical reaction, multiply the number of moles by the molar mass of fluorine. The mass is approximately 76 grams with 2 significant digits.

To calculate the mass of fluorine gas that participates in a chemical reaction, you need to know the number of moles of fluorine gas involved and the molar mass of fluorine (F2). The molar mass of fluorine is approximately 38 grams per mole. Multiply the number of moles by the molar mass to obtain the mass of fluorine gas participating in the reaction.

Example:

If the chemist determines that 2 moles of fluorine gas participate in the reaction, the mass of fluorine gas can be calculated as follows:

Mass = number of moles x molar mass

Mass = 2 moles x 37.996 grams/mole

Mass = 75.992 grams

Therefore, the mass of fluorine gas that participates in the reaction is approximately 76 grams . Remember to use the correct number of significant digits in your final answer, which in this case is 2 significant digits.

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Answer:

option D is correct

D. This solution is a good buffer.

Explanation:

TRIS (HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]CNH[tex]_{2}[/tex]

if TRIS is react with HCL it will form salt

(HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]CNH[tex]_{2}[/tex] + HCL ⇆   (HOCH[tex]_{2}[/tex])[tex]_{3}[/tex]NH[tex]_{3}[/tex]CL

Let the reference volume is 100

Mole of TRIS is =  100 × 0.2 = 20

Mole of HCL is =  100 × 0.1 = 10

In the reaction all of the HCL will Consumed,10 moles of the salt will form

and 10 mole of TRIS will left

hence , Final product will be salt +TRIS(9 base)

H = Pk[tex]_{a}[/tex] + log (base/ acid)

8.3 + log(10/10)

8.3

The correct statements are that the resulting solution is a good buffer and the ratio of [conjugate base]/[conjugate acid] is [0.1 M]/[0.1 M]. The majority of TRIS is not in the acid form in the solution.

If equal volumes of 0.1 M HCl and 0.2 M TRIS are mixed together, the excess TRIS will react with HCl forming its conjugate acid (TRIS-HCl) and reducing the concentration of HCl. Since we started with more TRIS, half of it will remain unreacted and exist as the base (TRIS), while the other half will form the conjugate acid (TRIS-HCl). Therefore, the concentrations of the conjugate base and the conjugate acid will both be 0.1 M.

So, the statement C. The ratio of [conjugate base]/[conjugate acid] is [0.05 M]/[0.05 M] is incorrect. The correct ratio is 0.1 M/0.1 M.

The pKa of TRIS is 8.30 meaning that it can act as a buffer at around pH 8.30. Since the solution consists of equal concentrations of a weak base and its conjugate acid, it will indeed create a buffer solution. Consequently, D. This solution is a good buffer is correct.

The statement E. The majority of TRIS will be in the acid form in the solution is not correct. The base and acid forms are in equal amounts.

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To calculate the percent dissociation of benzoic acid, use the ionization constant and the initial concentration, then quantifying the degree of dissociation by how much H+ ion concentration is produced. A simplified approach is given assuming low degree of ionization, typical of weak acids.

The question asks for the percent dissociation of benzoic acid, which is a chemistry concept. To calculate percent dissociation, you need the ionization constant of the acid (Ka), which you can lookup in the ALEKS Data resources, and the concentration of the acid, which is given as 2.4mM.If the Ka for benzoic acid is x, then the equilibrium expression for the dissociation of C6H5CO2H into ions is [C6H5CO2-][H+]/[C6H5CO2H] = x. Solve this equation to find the concentration of H+ ions. Once you calculate the concentration of H+ ions, the percent dissociation is ([H+]/initial concentration of the acid) x 100%.This is a simplified approach, assuming the degree of ionization is less than 5%, as it is with weak acids in relatively dilute solutions. If this assumption is not valid, a quadratic equation would need to be used.

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Answer:

1.82 cm

Explanation:

Utilize the equation [tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex] to calculate the change in volume and size of an air bubble.

P1 = pressure at 50m = [tex]P_{A}[/tex] + ρ*g*h  (where [tex]P_{A}[/tex] = atmospheric pressure, ρ = density of water, g = acceleration due to gravity, h =  height/depth)

P1 = 1.01 x 10⁵ Pa + (ρ x g x h)

    = 1.01 x 10⁵ Pa + (1000 kg/m³ x  9.8 m/s² x 50 m )

    =  1.01 x 10⁵ Pa + 4.9 x 10⁵ Pa

    = 5.91 x 10⁵ Pa

V1 = [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex]     [tex]r_{1}[/tex] = 10 cm =  1 x 10⁻² m

T1 = 10 °C = 10 + 273 = 283 K

P2 = [tex]P_{A}[/tex] = 1.01 x 10⁵ Pa because at the surface, pressure is equal to atmospheric pressure

V2 = [tex]\frac{4}{3}\pi r_{2} ^{3}[/tex]  [tex]r_{2}[/tex] = ??

T2 = 20 °C = 20 + 273 = 293 K

[tex]\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }[/tex]

V₂ = P₁V₁T₂

         P₂T₁

[tex]\frac{4}{3}\pi r_{2} ^{3}[/tex] = P₁ x [tex]\frac{4}{3}\pi r_{1} ^{3}[/tex]  x T₂

                P₂T₁

cancel out common terms

[tex]r_{2}[/tex]³ = 5.91 x 10⁵ Pa x (1 x 10⁻² m)³ x  293 k

              1.01 x 10⁵ Pa x 283 k

 [tex]r_{2}[/tex]³ = 757.9 x 10⁻⁹

   [tex]r_{2}[/tex] = 9.1 x 10⁻³ m

   [tex]r_{2}[/tex] = 0.91 cm  

Therefore, bubbles diameter = 2r = 1.82 cm

Answer:

1.82 cm

Explanation:

The pressure done by a column of a liquid is called the hydrostatic pressure (Ph) and it can be calculated by:

Ph = Patm + ρgh

Where Patm is the atmospheric pressure under the column (101325 Pa), ρ is the density of the liquid (1000 kg/m³ for water), g is the gravity acceleration (9.8 m/s²), and h is the depth (50 m), so:

Ph = 101325 + 1000*9.8*50

Ph = 591325 Pa

Because the bubble is in equilibrium with the surroundings, its pressure is the same as the surroundings. Supposing a perfect sferic bubble, its volume is:

V = (4/3)*π*r³

Where r is the radius, which is half of the diameter, so r = 0.5 cm.

V = (4/3)*π*(0.5)³

V = 0.52 cm³

According to the ideal gas law, the multiplication of the pressure (P) by the volume (V) divided by the temperature (T) of a gas is constant, so if 1 is the state where the bubble is 50 m depth, and 2 the state at the surface:

P1*V1/T1 = P2*V2/T2

P1 = Ph = 591325 Pa

V1 = 0.52 cm³

T1 = 10°C + 273 = 283 K

P2 = 101325 Pa (atmosferic pressure)

T2 = 20°C + 273 = 293 K

591325*0.52/283 = 101325*V2/293

101325V2 = 318,354.3357

V2 = 3.14 cm³

V2 = (4/3)*π*r³

(4/3)*π*r³ = 3.14

r³ = 0.75

r = ∛0.75

r = 0.91 cm

The diameter is then 2*r = 1.82 cm.

Answer:

[tex]n=2.59\times 10^{16}[/tex] photons

Explanation:

[tex]E=n\times \frac{h\times c}{\lambda}[/tex]

Where,  

n is the number of photons

h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

[tex]\lambda[/tex] is the wavelength of the light

Given that, wavelength = 514 nm = [tex]514\times 10^{-9}\ m[/tex]

Energy = 10.0 mJ = 0.01 J ( 1 mJ = 0.001 J )

Applying the values as:-

[tex]0.01=n\times \frac{6.626\times 10^{-34}\times 3\times 10^8}{514\times 10^{-9}}[/tex]

[tex]\frac{19.878n}{10^{17}\times \:514}=0.01[/tex]

[tex]n=2.59\times 10^{16}[/tex] photons

A couple purchases a house for $400,000.00. They pay 20% down at closing, and take out a mortgage of $320,000.00. The mortgage company offers them a 4.80% annual rate with monthly compounding. The mortgage will require monthly payments for the next 30 years.

What will be the monthly payment on this mortgage?

Answer:

$ 1678.91

Explanation:

Given that;

Cost of purchasing a house = $400,000.00

Down payment =20%

Mortgage Value (MV) = $320,000.00

Annual rate offered by the mortgage company = 4.80% yearly i.e 0.4 per month

Duration of the Mortgage Loan (n) =  30 years which is equivalent to 360 months.

if we represent the monthly repayment with MR ,To calculate the monthly repayment MR;we have;

MV = MR × [tex](\frac{1}{i})*[1-(\frac{1}{(1+i)^{n}} )}][/tex]

where i = 0.004 (4.8 % annually expressed as 0.48, divided by 12 monthly payments per year)

∴

[tex]320,000.00[/tex] = [tex]MR[/tex] [tex]*(\frac{1}{0.004})*[1-(\frac{1}{1+0.004)^{360}}})][/tex]

320,000.00 = MR × 190.60

Monthly repayment (MR) = $ 1678.90870933

Monthly repayment (MR) ≅ $ 1678.91

The ratio of acetate to acetic acid is approximately 1:1, while the ratio of ethylamine to ethylammonium ion is close to 0:1.

The ratio of acetate to acetic acid can be determined using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Since the pH is 7.4, which is close to the pKa of acetic acid (4.76), the ratio of acetate (A-) to acetic acid (HA) will be close to 1:1. The ratio of ethylamine to ethylammonium ion can be determined using the same equation. In this case, the pKa is the pKa of ethylamine (10.64). If the pH is 7.4, the ethylamine (C2H5NH2) will be mostly in its protonated form, ethylammonium ion (C2H5NH3+), giving a ratio close to 0:1.

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Answer:

a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰

b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰

Explanation:

The condensed electron configurations of given elements are below

a) Ba (Z = 56), [Xe].6s²

b) O (Z = 8), [He].2s².2p⁴

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p²

Since atoms tend to donate/receive more electrons to achieve the saturated or half-saturated orbital. So in our case it happens as below

a) Ba (Z = 56), [Xe].6s² - 2e ⇒ Ba²⁺, [Xe].6s⁰

b) O (Z = 8), [He].2s².2p⁴ + 2e ⇒ O²⁻, [He].2s².2p⁶

c) Pb (Z = 82), [Xe].4f¹⁴.5d¹⁰.6s².6p² - 2e ⇒ Pb²⁺, [Xe].4f¹⁴.5d¹⁰.6s².6p⁰

Answer:

13.301

Explanation:

To calculate the pH of the solution, we must obtain the pOH of the solution as illustrated below:

The dissociation equation is given below

Ba(OH)2 <==> Ba^2+ + 2OH^-

Since Ba(OH)2 dissociate to produce 2moles of OH^-, the concentration of OH^- = 2x0.1 = 0.2M

pOH = - Log[OH^-]

pOH = - Log 0.2

pOH = 0.699

But

pH + pOH = 14

pH = 14 — pOH

pH = 14 — 0.699

pH = 13.301

Answer:

The pH of this barium hydroxide solution is 13.30

Explanation:

Step 1: Data given

Concentration Ba(OH)2 = 0.10 M

Step 2: Calculate [OH-]

Ba(OH)2 ⇒ Ba^2+ + 2OH-

[OH-] = 2*0.10 M

[OH-] = 0.20 M

Step 3: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.20)

pOH = 0.70

Step 4: Calculate pH

pH + pOH = 14

pH = 14 -pOH

pH = 14 - 0.70

pH = 13.30

The pH of this barium hydroxide solution is 13.30

Answer:

Molar concentration of Fe(NO3)3 . 9H2O = 0.12M

Explanation:

Fe(NO3).9H2O --> Fe(NO3)3 + 9H2O

By stoichiometry,

1 mole of Fe(NO3)3 will be absorb water to form 1 mole of Fe(NO3)3 . 9H2O

Therefore, calculating the mass concentration of Fe(NO3)3;

Molar mass of Fe(NO3)3 = 56 + 3*(14 + (16*3))

= 242 g/mol

Mass concentration of Fe(NO3)3 = molar mass * molar concentration

= 242 * 0.2

= 48.4 g/L

Molar mass of Fe(NO3)3 . 9H2O = 56 + 3*(14 + (16*3)) + 9* ((1*2) + 16)

= 242 + 162 g/mol

= 404g/mol

Concentration of Fe(NO3)3 . 9H2O = mass concentration/molar mass

= 48.4 /404

= 0.12 mol/l

Molar concentration of Fe(NO3)3 . 9H2O = 0.12M

Answer: The number of molecules of carbon dioxide gas are [tex]2.815\times 10^{21}[/tex]

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]0.034mol/L.atm[/tex]

[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas

[tex]p_{CO_2}[/tex] = pressure of carbon dioxide gas = 0.250 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M[/tex]

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Molarity of carbon dioxide = [tex]8.5\times 10^{-5}M[/tex]

Volume of solution = 0.550 L

Putting values in above equation, we get:

[tex]8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol[/tex]

According to mole concept:

1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules

So, [tex]4.675\times 10^{-3}[/tex] moles of carbon dioxide will contain = [tex](6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}[/tex] number of molecules

Hence, the number of molecules of carbon dioxide gas are [tex]2.815\times 10^{21}[/tex]

Answer:

2.8 *10^21 molecules CO2 are dissolved in the 0.550 L water

Explanation:

Step 1: Data given

Volume of water = 0.550 L

Temperature = 25.0 °C

Pressure of CO2 = 0.250 atm

The Henry’s constant for CO2 and water at 25 °C = 0.034 M/atm

Step 2: Henry's law

C(CO2) = Kh * p(CO2)

⇒ with C(CO2) = the molar solubility of CO2

⇒ with Kh = Henry's constant = 0.034 M/atm = 0.034 mol/(L * atm)

⇒ with p(CO2) = the pressure of CO2 = 0.250 atm

C(CO2) = 0.034 mol/(L*atm) * 0.250 atm

C(CO2) = 0.0085 mol /L

Step 3: Calculate moles CO2

Moles CO2 = volume * molar solubility CO2

Moles CO2 = 0.550 L * 0.0085 mol/L

Moles CO2 = 0.004675 moles

Step 4: Calculate molecules of CO2

Molecules CO2 = moles * Number of Avogadro

Molecules CO2 = 0.004675 * 6.022 *10^23 / mol

Molecules CO2 = 2.8 *10^21 molecules

2.8 *10^21 molecules CO2 are dissolved in the 0.550 L water

Answer:

4.4 × 10⁴ g/mol

Explanation:

The osmotic pressure (π) is a colligative property that can be calculated using the following expression.

π = M × R × T

where,

M: molarity

R: ideal gas constant

T: absolute temperature (27°C + 273.15 = 300 K)

Let's use it to find the molarity of the protein.

M = π / R × T

M = 0.053 atm / (0.082 atm.L/mol.K) × 300 K

M = 2.2 × 10⁻³ M

The molarity of the protein is:

M = mass of the protein / molar mass of the protein × liters of solution

molar mass of the protein = mass of the protein / M  × liters of solution

molar mass of the protein = 22 g / 2.2 × 10⁻³ mol/L  × 0.226 L

molar mass of the protein = 4.4 × 10⁴ g/mol

The molar mass of the protein is 4.4 * 10⁴ g/mol

It  is a colligative property that can be calculated using the following expression.

π = M × R × T

where,

[tex]M = \frac{\pi}{R * T} \\\\M = \frac{0.053 atm}{(0.082 atm.L/mol.K) * 300 K}\\\\ M = 2.2 * 10^{-3} M[/tex]

M = mass of the protein / Molar mass of the protein * liters of solution

Molar mass of the protein = mass of the protein / M  * liters of solution

Molar mass of the protein = [tex]\frac{22 g}{2.2 * 10^{-3} mol/L * 0.226 L}[/tex]

Molar mass of the protein = [tex]4.4 * 10^4 g/mol[/tex]

Thus, the molar mass of the protein is 4.4 * 10⁴ g/mol.

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The amount of ice added was approximately 38.5 grams.

To calculate this, we can use the principle of conservation of energy. The heat lost by the water as it cools down to the final temperature (15 °C) is equal to the heat gained by the ice as it melts and then warms up to the final temperature.

First, we calculate the heat lost by the water:

[tex]\[ Q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T \][/tex]

Where:

[tex]\( m_{\text{water}} = 64.3 \, \text{g} \)[/tex] (mass of water)

[tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex] (specific heat of water)

[tex]\( \Delta T = 55°C - 15°C = 40°C \)[/tex] (change in temperature)

[tex]\[ Q_{\text{water}} = 64.3 \, \text{g} \times 4.184 \, \text{J/g°C} \times 40°C = 10707.712 \, \text{J} \][/tex]

Next, we calculate the heat gained by the ice:

[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \times L_f + m_{\text{ice}} \times c_{\text{water}} \times \Delta T \][/tex]

Where:

[tex]\( L_f = 334 \, \text{J/g} \)[/tex] (heat of fusion of water)

[tex]\( m_{\text{ice}} \)[/tex] is the mass of ice we want to find

[tex]\( c_{\text{water}} = 4.184 \, \text{J/g°C} \)[/tex]  (specific heat of water)

[tex]\( \Delta T = 15°C \)[/tex] (change in temperature, from 0 °C to 15 °C)

Let's set up the equation using [tex]\( m_{\text{ice}} \):[/tex]

[tex]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times 334 \, \text{J/g} + m_{\text{ice}} \times 4.184 \, \text{J/g°C} \times 15°C \][/tex]

Now, we solve for [tex]\( m_{\text{ice}} \):[/tex]

[tex]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times (334 \, \text{J/g} + 62.76 \, \text{J/g}) \]\[ 10707.712 \, \text{J} = m_{\text{ice}} \times 396.76 \, \text{J/g} \]\[ m_{\text{ice}} = \frac{10707.712 \, \text{J}}{396.76 \, \text{J/g}} \approx 27.0 \, \text{g} \][/tex]

However, this is the amount of ice needed to cool the water to 0 °C. To find the total amount of ice needed to cool the water to 15 °C, we add the ice that will melt at 0 °C to the ice that will further cool down to 15 °C:

[tex]\[ m_{\text{total ice}} = m_{\text{ice}} \text{ at 0 °C} + m_{\text{ice}} \text{ cooling from 0 °C to 15 °C} \]\[ m_{\text{total ice}} = 27.0 \, \text{g} + (27.0 \, \text{g} \times 15/334) \approx 38.5 \, \text{g} \][/tex]

So, approximately 38.5 grams of ice were added to the water.

Complete Question:

A quantity of ice at 0 °C was added to 64.3 g of water in a glass at 55 °C. The final temperature of the system was 15 °C. How much ice was added?

A)The melting point of water is 0 °C.

B)The heat of fusion of water is 334 J g–1 .

C)The specific heat of liquid water is 4.184 J g–1 °C –1

Answer:

(a) n = 2, l = 0 ⇒ sublevel s, ⇒ ml = 0, number of orbitals = 1

(b) n = 3, l = 2 ⇒ sublevel d, ⇒ ml = 0, ±1, ±2, number of orbitals = 5

(c) n = 5, l = 1 ⇒ sublevel p, ⇒ ml = 0, ±1, number of orbitals = 3

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n,

2. Subshell number, 0 ≤ l ≤ n − 1, from s, p, d, f, g, h...

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

In our case

(a) n = 2, l = 0 ⇒ sublevel s

-l ≤ ml ≤ l ⇒ ml = 0, number of orbitals = 1

(b) n = 3, l = 2 ⇒ sublevel d

-l ≤ ml ≤ l ⇒ ml = 0, ±1, ±2, number of orbitals = 5

(c) n = 5, l = 1 ⇒ sublevel p

-l ≤ ml ≤ l ⇒ ml = 0, ±1, number of orbitals = 3

The first pair of quantum numbers (n, l) represents the 2s sublevel with 1 orbital. The second pair represents the 3d sublevel with 5 orbitals. The third pair represents the 5p sublevel with 3 orbitals.

The information given refers to quantum numbers in the quantum mechanical model of the atom, a fundamental concept in high school physics and chemistry. This model explains the behavior of electrons in atoms.

(a) For n = 2 and l = 0, the sublevel designation is 2s. The permissible ml value is 0 and there is 1 orbital.
(b) For n = 3 and l = 2, the sublevel designation is 3d. The permissible ml values range from -2, -1, 0, 1, 2 and there are 5 orbitals.
(c) For n = 5 and l = 1, the sublevel designation is 5p. The permissible ml values are -1, 0, 1 and there are 3 orbitals.

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Answer and Explanation:

a) The direction is shown in the cube diagram attached to this solution.

b) the angle between two planes (h₁, k₁, l₁) and (h₂, k₂, l₂) is given by the formula,

Cos Φ = (h₁h₂ + k₁k₂ + l₁)/√((h₁² + k₁² + l₁²)(h₂² + k₂² + l₂²))

For (111) and (112)

Cos Φ = (1.1 + 1.1 + 1.2)/√((1² + 1² + 1²)(1² + 1² + 2²))

Cos Φ = (1 + 1 + 2)/√((1+1+1)(1+1+4))

Cos Φ = 4/√(3×6)

Cos Φ = 4/√18

Φ = cos⁻¹ (4/√18) = 19.56°

c) equation 3.3 is missing from the question, I would be back to provide the answers to that as soon as the equation is provided!

Hope this Helps!!

A simple cubic lattice has different lattice directions represented by [111] and [112]. The angle between these two directions can be determined using trigonometry or Equation 3.3. The [111] direction passes through the corner atoms along the body diagonal of the unit cell, while the [112] direction passes through the edges of the cubic unit cell.

In a simple cubic lattice, the [111] lattice direction passes through the corner atoms along the body diagonal of the unit cell. This lattice direction is represented by a line passing through the center of opposite face diagonals, as shown in Figure 10.50. On the other hand, the [112] lattice direction passes through the edges of the cubic unit cell.

To determine the angle between the [111] and [112] lattice directions using trigonometry, we can use the formula:

cos(θ) = A · B / (|A| · |B|)

where A and B are the [111] and [112] lattice directions as vectors. By substituting the values, we can calculate the angle between these two directions.

Alternatively, Equation 3.3 in the reference material can be used to calculate the angle between [111] and [112] directions:

cos(θ) = (h1 · h2 + k1 · k2 + l1 · l2) / (sqrt(h1^2 + k1^2 + l1^2) · sqrt(h2^2 + k2^2 + l2^2))

where h1, k1, and l1 are the Miller indices for the [111] direction, and h2, k2, and l2 are the Miller indices for the [112] direction.

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Answer : The amount of formaldehyde permissible are, [tex]5.4\times 10^{-6}g[/tex]

Explanation : Given,

Density of air = [tex]1.2kg/m^3=1.2g/L[/tex]     [tex](1kg/m^3=1g/L)[/tex]

First we have to calculate the mass of air.

[tex]\text{Mass of air}=\text{Density of air}\times \text{Volume of air}[/tex]

[tex]\text{Mass of air}=1.2g/L\times 6.0L[/tex]

[tex]\text{Mass of air}=7.2g[/tex]

Now we have to calculate the amount of formaldehyde.

Permissible exposure level of formaldehyde = 0.75 ppm = [tex]\frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}[/tex]

Amount of formaldehyde in 7.2 g of formaldehyde = [tex]7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}[/tex]

Amount of formaldehyde in 7.2 g of formaldehyde = [tex]5.4\times 10^{-6}g[/tex]

Thus, the amount of formaldehyde permissible are, [tex]5.4\times 10^{-6}g[/tex]

The study of the chemical and bonds is called chemistry.

The correct answer is  [tex]5.4*10^{-6[/tex]

All the data is given in the question. therefore

Hence, the correct answer to the question is [tex]5.4*10^{-6[/tex].

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