Gold has a density of 0.01932 kg/cm3. What is the mass (in kg) of a 92.5 cm3 sample of gold?A) 1.79B) 0.560C) 92.5D) 0.000209E) 4790

Answer:

the flow rate for steam from the boiler plant = [tex]0.099kg/s[/tex]

the flow rate from the city water = 0.901 kg/s

the rate of entropy production in the mixing tank = 0.2044 kJ/k

Explanation:

In a well-insulated mixing tank where:

[tex]Q_{cv} = 0[/tex]  & [tex]W_{cv}=0[/tex]

The mass flow rates can be calculated using the formula:

[tex]Q_cv+m_1h_1+m_2h_2=m_3h_3+W_{cv[/tex]      ------ equation (1)

so;

[tex]0+m_1h_1+m_2h_2=m_3h_3+0[/tex]               ------- equation (2)

Given that:

From the steam in the boiler plant;

The temperature (T₁) = 200°C

Pressure (P₁) = 10 bar

The following data from compressed water and super-heated steam tables were also obtained at: T₁ = 200°C

h₁ = 2828.27 kJ/kg

s₁ = 6.95 kJ/kg K

m₁ (flow rate for steam in the boiler plant) = ????

Also, for city water

The temperature (T₂) = 20°C

Pressure (P₂) = 1 bar

Data obtained from compressed water and super-heated steam tables are as follows:

h₂ = 84.01 kJ/kg

s₂ = 0.2965 kJ/kg K

m₂ (flow rate for city water) = ???

For  stream of hot wat at 85 deg C

Temperature (T₃) = 85°C

h₃([tex]h_f[/tex]) = 355.95 kJ/kg

s₃([tex]s_f[/tex]) = 1.1344 kJ/kg K

m₃ = 1 kg/s

so since:

m₁ + m₂ = m₃       (since m₃  = 1)

m₂ = 1 -  m₁

From equation (2);

[tex]0+m_1h_1+m_2h_2=m_3h_3+0[/tex]    

= [tex]m_1(2828.27)+(1-m_1)(84.01)=1(355.95)[/tex]

= [tex]2828.27m_1+(84.01-84.01m_1)=(355.95)[/tex]

= [tex]2828.27m_1-84.01m_1=355.95-84.01[/tex]

= [tex]m_1(2828.27-84.01)=355.95-84.01[/tex]

[tex]m_1 = \frac{355.95-84.01}{2828.27-84.01}[/tex]

[tex]m_1 = \frac{271.94}{2744.26}[/tex][tex]m_1 = 0.099 kg/s[/tex]

∴ the flow rate for steam from the boiler plant = [tex]0.099kg/s[/tex]

since; m₂ = 1 -  m₁

m₂ = 1 -  0.099 kg/s

m₂ = 0.901 kg/s

∴ the flow rate from the city water = 0.901 kg/s

b)

rate of entropy production in the mixing tank can be determined using the formula:

Δ[tex]S_{production} = m_3}s_3-(m_1s_1+m_2s_2)[/tex]

Δ[tex]S_{production}[/tex] [tex]= (1)(1.1344)-(0.099)(6.6955)-0.901(0.2965)[/tex]

Δ[tex]S_{production}[/tex] [tex]= 1.1344-0.6628545-0.2671465[/tex]

Δ[tex]S_{production}[/tex] [tex]= 1.1344 - 0.930001[/tex]

Δ[tex]S_{production}[/tex] [tex]= 0.204399[/tex]

Δ[tex]S_{production}[/tex] ≅ 0.2044 kJ/k

∴ the rate of entropy production in the mixing tank = 0.2044 kJ/k

Answer:

entropy change of the system = +71 J/K

Explanation:

for a reversible process ;

ΔSsurrounding + ΔSsystem = 0 ............(1)

but ΔSsurrounding = - 71 J/K

substitute in equation (1)

ΔSsystem = -ΔSsurrounding

ΔSsystem = -(- 71 J/K)

= +71 J/K

For a reversible process where the surroundings undergo an entropy change of -71 J/K, the entropy change of the system is 71 J/K.

Since the process is reversible, the total entropy change for the universe is zero. This implies that the sum of the entropy changes for the system and the surroundings must be zero.

ΔSuniverse = ΔSsys + ΔSsurr = 0

Given that the entropy change of the surroundings, ΔSsurr, is -71 J/K, we can solve for the entropy change of the system, ΔSsys:

ΔSsys = - ΔSsurr

ΔSsys = - (-71 J/K)

ΔSsys = 71 J/K

Therefore, the entropy change of the system for this process is 71 J/K.

Answer:

P(total) = 1329 torr

Explanation:

Given data:

Pressure of nitrogen = 984 torr

Pressure of carbon dioxide = 345 torr

Total pressure of system = ?

Solution:

The given problem will be solve through the Dalton law of partial pressure of gases.

According to the this law,

The total pressure exerted by the mixture of gases is equal to the sum of partial pressure of individual gas.

Mathematical expression,

P(total) = P₁ + P₂ +.......+ Pₙ

Here we will put the values in formula,

P₁ = partial pressure of nitrogen

P₂ = partial pressure of carbon dioxide

P(total) = 984 torr + 345 torr

P(total) = 1329 torr

Final answer:

The total pressure of the system is calculated by adding the partial pressures of the individual gases, resulting in a total pressure of 843.42 torr.

Explanation:

To calculate the total pressure of a mixture of gases, you use Dalton's Law of Partial Pressures which implies that the overall pressure exerted by a mixture of non-reacting gases is equal to the sum of their partial pressures.

In this case,

The partial pressure of nitrogen (N₂) is given as 78% of 984 torr, which is 767.52 torr.

The partial pressure of carbon dioxide (CO₂) is given as 22% of 345 torr, which is 75.9 torr.

The total pressure is the sum of these partial pressures:

767.52 torr (N₂) + 75.9 torr (CO₂) = 843.42 torr

The total pressure of the system is 843.42 torr.

Answer:

It takes 151.9 s for the reactant concentration to decrease to 0.0063 M

Explanation:

k = ln2 / t1/2

k = ln2 / 35.5 s = 0.0195 s^-1

ln [A]t = -kt + ln [A]o

ln (0.0063) = -0.0195t + ln (0.0700)

t = 123.32 s

Since t is counted from 28.6 s as the initial time, the time after start of reaction to reach 0.0063 M is

28.6  + 123.32 s = 151.9 s

Answer:

1.98x10⁻¹² kg

Explanation:

The energy of a photon is given by:

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

Now we multiply that value by Avogadro's number, to calculate the energy of 1 mol of such protons:

Finally we calculate the mass equivalence using the equation:

The mass equivalence of 1 mole of photons of this wavelength is equal to [tex]1.78 \times 10^{-11} \;kg[/tex]

Given the following data:

We know that the speed of light is [tex]3 \times 10^8[/tex] m/s.

Planck constant = [tex]6.626 \times 10^{-34}\;J.s[/tex]

Avogadro constant = [tex]6.02 \times 10^{23}[/tex]

To determine the mass equivalence of 1 mole of photons of this wavelength:

First of all, we would calculate the energy associated with the photons of this wavelength by using the Planck-Einstein relation:

Mathematically, Planck-Einstein relation for photon energy is given by the formula:

[tex]E = hf = h\frac{v}{\lambda}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]E = \frac{6.626 \times 10^{-34} \; \times \;3 \times 10^8 }{6.71 \times 10^{-7}} \\\\E = \frac{1.99 \times 10^{-25}}{6.71 \times 10^{-7}} \\\\E=2.96 \times 10^{-19} \;Joules[/tex]

For the energy of 1 mole of photons:

[tex]E = 2.96 \times 10^{-19} \times 6.02 \times 10^{23}\\\\E = 1.78 \times 10^5 \; Joules[/tex]

Now, we can find the mass equivalence of 1 mole of photons by using the formula:

[tex]E = mc^2\\\\m = \frac{E}{c^2} \\\\m = \frac{1.78 \times 10^5}{(3.0 \times 10^8)^2} \\\\m = \frac{1.78 \times 10^5}{(9.0 \times 10^{16})}\\\\m = 1.78 \times 10^{-11} \;kg[/tex]

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The activation energy is 124.776[tex]\frac{kJ}{mole}[/tex]

b) The frequency factor is 1.77 ×[tex]10^{18}[/tex]

c) The rate constant is 0.00033 [tex](\frac{dm^{3} }{mole} )^{2}[/tex][tex]\frac{1}{s}[/tex]

Explanation:

From the question the elementary reaction for A and B is given as

      2A + B → 4C

The rate equation the elementary reaction is

      -[tex]r_{A}[/tex] = [tex]k[/tex][tex][A]^{2}[/tex][tex][B][/tex]

            =  [tex]k[2]^{2}[1.5][/tex]

            = 6k

     [tex]k = \frac{-r_{A} }{6}[/tex]

      When temperature changes, the rate constant change an this causes the rate of reaction to change as shown on the second uploaded image.

The relationship between temperature and rate constant can be deduced from these equation

                    [tex]k = Aexp(-\frac{E_{a} }{RT} )[/tex]

             taking ln of both sides we have

                   [tex]lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}[/tex]

        Considering the graph for the rate constant [tex]ln k[/tex] and [tex](\frac{1}{T} )[/tex] the slope from the equation is [tex]-(\frac{E_{a} }{R})[/tex] and the intercept is [tex]ln A[/tex]

From the given table we can generate another table using the equation above as shown on the third uploaded image

The graph of [tex]ln k[/tex]  vs [tex](\frac{1}{T} )[/tex]  is shown on the fourth uploaded image

  From the graph we can see that the slope is [tex]-(\frac{E_{a} }{R} ) = - 15008[/tex]

Now we can obtain the activation energy [tex]E_{a}[/tex] by making it the subject in the equation also generally R which is the gas constant is [tex]8.145 \frac{J}{kmole}[/tex]

                [tex]E_{a} = 15008 × 8,3145\frac{J}{molK}[/tex]  

                     [tex]= 124\frac{KJ}{mole}[/tex]

    Hence the activation energy is [tex]= 124\frac{KJ}{mole}[/tex]

b) From the graph its intercept is [tex]ln A = 42.019[/tex]

                                                          [tex]A = exp(42.019)[/tex]

                                                             [tex]=1.77 × 10^{18}[/tex]

Hence the frquency factor A is  [tex]=1.77 × 10^{18}[/tex]

c) From the equation of rate constant

                                          [tex]lnk =ln A - (\frac{E_{a} }{R}) \frac{1}{T}[/tex]

We have

                [tex]ln k = 42.019 - 15008 * (\frac{1}{300} )[/tex]

                      [tex]k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}[/tex]

Hence the rate constant is [tex]k = 0.00033(\frac{dm^{3} }{mole} )^{2} \frac{1}{s}[/tex]    

The quantum numbers serve for experimenting with electron positions within an atom. For each given atom or ion - Li, Br-, Cs, and B - the quantum numbers for the outermost electron exhibit the electron's location in different shells, subshells and spins.

The quantum numbers provide a unique address for each electron in an atom, listed as (n, l, ml, ms), where n is the principal quantum number, l is the azimuthal quantum number, ml is the magnetic quantum number, and ms is the spin quantum number.

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For these cases, the quantum numbers have been determined based on electron configurations and orbital positions. The solutions cover Li, Br⁻, Cs, and B electrons specifically.

Quantum numbers are used to describe the position and energy of an electron in an atom. Let's break down the quantum numbers for each of the given scenarios:

Answer:

Part A:

For electron:

[tex]\lambda_e=2.1392*10^{-10} m[/tex]

For Proton:

[tex]\lambda_p=1.16696*10^{-13} m[/tex]

Part B:

For electron:

[tex]\lambda_e=9.44703*10^{-12} m[/tex]

For Proton:

[tex]\lambda_p=2.20646*10^{-13} m[/tex]

Explanation:

Formula for wave length λ is:

[tex]\lambda=\frac{h}{mv}[/tex]

where:

h is Planck's constant=[tex]6.626*10^{-34}[/tex]

m is the mass

v is the velocity

Part A:

For electron:

[tex]\lambda_e=\frac{6,626*10^{-34}}{(9.11*10^{-31})*(3.4*10^6)} \\\lambda_e=2.1392*10^{-10} m[/tex]

For Proton:

[tex]\lambda_p=\frac{6,626*10^{-34}}{(1.67*10^{-27})*(3.4*10^6)} \\\lambda_p=1.16696*10^{-13} m[/tex]

Wavelength of proton is smaller than that of electron

Part B:

Formula for K.E:

[tex]K.E=\frac{1}{2}mv^2\\v=\sqrt{2 K.E/m}[/tex]

For Electron:

[tex]v_e=\sqrt{\frac{2*2.7*10^{-15}}{9.11*10^{-31}}} \\v_e=76990597.74\ m/s[/tex]

Wavelength for electron:

[tex]\lambda_e=\frac{6,626*10^{-34}}{(9.11*10^{-31})*(76990597.74)} \\\lambda_e=9.44703*10^{-12} m[/tex]

For Proton:

[tex]v_p=\sqrt{\frac{2*2.7*10^{-15}}{1.67*10^{-27}}} \\v_p=1798202.696\ m/s[/tex]

Wavelength for proton:

[tex]\lambda_p=\frac{6,626*10^{-34}}{(1.67*10^{-27})*(1798202.696)} \\\lambda_p=2.20646*10^{-13} m[/tex]

Wavelength of electron is greater than that of proton.

Explanation:

(a)    The given data is as follows.

     Temperature (T) = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K

      [tex]\Delta H_{vap}[/tex] = 43000 J/mol

Since, both the liquid and vapors are at equilibrium. Therefore, change in free energy will be calculated as follows.

      [tex]\Delta G_{vap} = \Delta H_{vap} - T \Delta S_{vap}[/tex] = 0        

            43000 - [tex](298 \times \Delta S_{vap})[/tex] = 0

                      [tex]\Delta S_{vap}[/tex] = -144 J/mol K

Negative sign indicates an increase in entropy of the system.

Now, for 1 mole of [tex]CCl_{4}[/tex] is as follows.

     = 144 J/K

So,     [tex]S_{vapor}[/tex] - 214 = 144 J/k

                         = 358 J/K

Therefore, we can conclude that entropy of [tex]CCl_{4}[/tex] vapor is 358 J/K.

(b)  As we know that intensive variable are the variables which do not depend on the amount of a substance.

So, in the given situation only temperature will act as an intensive variable that will be required to completely specify the vapor-liquid mixture of [tex]CCl_{4}[/tex].

(a) The entropy of 1 mole of the vapor at 25°C is 358.3 J/K. (b) One intensive variables are required to completely specify the vapor-liquid mixture of CCl4.

Let's address each part of your question step by step.

(a) Entropy of 1 mole of CCl₄ vapor at 25°C (298 K)

Given data:

- The heat of vaporization [tex](\(\Delta H_{\text{vap}}\)) of CCl_4 \ is\ 43000 J/mol\ at\ 25\°C.[/tex]

- The entropy of 1 mol of liquid CCl₄ [tex](\(S_{\text{liquid}}\))[/tex] is 214 J/K at 25°C.

We need to find the entropy of 1 mole of the vapor [tex](\(S_{\text{vapor}}\))[/tex] at 25°C (298 K).

The change in entropy during vaporization [tex](\(\Delta S_{\text{vap}}\))[/tex] can be calculated using the formula:

[tex]\[ \Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T} \][/tex]

Where:

- [tex]\(\Delta H_{\text{vap}}\)[/tex] = 43000 J/mol

- T = 298 K

So,

[tex]\[ \Delta S_{\text{vap}} = \frac{43000 \, \text{J/mol}}{298 \, \text{K}} \]\[ \Delta S_{\text{vap}} = 144.3 \, \text{J/(mol} \cdot \text{K)} \][/tex]

Now, using the relation:

[tex]\[ S_{\text{vapor}} = S_{\text{liquid}} + \Delta S_{\text{vap}} \][/tex]

Substitute the values:

[tex]\[ S_{\text{vapor}} = 214 \, \text{J/(mol} \cdot \text{K)} + 144.3 \, \text{J/(mol} \cdot \text{K)} \]\[ S_{\text{vapor}} = 358.3 \, \text{J/(mol} \cdot \text{K)} \][/tex]

So, the entropy of 1 mole of CCl₄ vapor at 25°C is:

[tex]\[ S_{\text{vapor}} = 358.3 \, \text{J/(mol} \cdot \text{K)} \][/tex]

(b) Number of intensive variables to specify the vapor-liquid mixture of CCl₄

For a system in equilibrium, the number of intensive variables required to completely specify the state of the system is given by the Gibbs phase rule:

F = C - P + 2

Where:

- F  is the number of degrees of freedom (intensive variables needed).

- C is the number of components.

- P is the number of phases.

In this case:

- C  (number of components) = 1 (CCl₄).

- P (number of phases) = 2 (liquid and vapor).

Using the Gibbs phase rule:

F = 1 - 2 + 2

F = 1

Thus, one intensive variable (e.g., temperature or pressure) is required to completely specify the vapor-liquid mixture of CCl₄.

Answer: 6.26atm

Explanation:Please see attachment for explanation

Answer:

The osmotic pressure is 6.26 atm

Explanation:

Step 1: Data given

Mass of testosterone = 12.4 grams

Volume of benzene = 168 mL

Temperature = 298 Kelvin

Step 2: Calculate moles of testosterone

Moles testosterone = mass / molar mass

Moles testosterone = 12.4 grams / 288.42 g/mol

Moles testosterone = 0.0430 moles

Step 3: Calculate molarity

Molarity = moles / volume

Molarity = 0.0430 moles / 0.168 L

Molarity = 0.256 M

Step 4 : Calculate the osmotic pressure

π = iMRT

⇒ with i = The Van't hoff factor = 1 (since testosterone is nonelectrolyte)

⇒ with M = the molair concentration = 0.256 M

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

⇒ with T = the temperature = 298 K

π = 1*0.256*0.08206*298

π = 6.26 atm

The osmotic pressure is 6.26 atm

Answer:

Explanation:

a ) All the electronic levels below n = 3 have same energy so there will not be any evolution of energy in electronic transition from n=3 to n=2 or to  n= 1 .

b ) Ionisation energy of ground state = 15 x 10⁻¹⁹ J per electron

= 15 x 10⁻¹⁹ x 6.02 x 10²³ J / mol

= 90 x 10⁴ J/mol

= 900 kJ / mol

c )  the shortest wavelength (in nm) of radiation that could be absorbed without causing ionization will correspond to n = 4 to n = 6

= (11 - 2 ) x 10⁻¹⁹ J

= 9 X  10⁻¹⁹ J

= 9 X  10⁻¹⁹ J / 1.6 X 10⁻¹⁹

= 5.625 eV

= 1244 / 5.625 nm

= 221.155 nm

Answer: The volume of sodium bicarbonate needed is 196.6 mL

Explanation:

To calculate the volume of base, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]NaHCO_3[/tex]

We are given:

[tex]n_1=2\\M_1=1.9M\\V_1=98.3mL\\n_2=1\\M_2=1.9M\\V_2=?mL[/tex]

Putting values in above equation, we get:

[tex]2\times 1.9\times 98.3=1\times 0.19\times V_2\\\\V_2=\frac{2\times 1.9\times 98.3}{1\times 1.9}=196.6mL[/tex]

Hence, the volume of sodium bicarbonate needed is 196.6 mL

Answer: 288.32g/mol

Explanation:Please see attachment for explanation

Answer:

- 2977 kJ

Explanation:

Let's consider the following reaction.

3 CH₄(g) + 4 O₃(g) → 3 CO₂(g) + 6 H₂O(g)

We can find the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = [3 mol × ΔH°f(CO₂(g)) + 6 mol × ΔH°f(H₂O(g))] - [3 mol × ΔH°f(CH₄(g)) + 4 mol × ΔH°f(O₃(g))]

ΔH°r = [3 mol × (-393.5 kJ/mol) + 6 mol × (-241.8 kJ/mol)] - [3 mol × (-74.87 kJ/mol) + 4 mol × (142.7 kJ/mol)]

ΔH°r = - 2977 kJ

Answer:

Solubility of CO2 = 0.045 g/l

Explanation:

Using Henry’s law which states that the quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.

C = k * Pg

Where C = concentration of the gas

Pg = pressure of the gas

k = proportionality constant

Calculating k,

k = C/Pg

Converting g/100ml to g/l,

0.00412 g/100ml; 0.00412g/100ml * 1000ml/1l

= 0.0412 g/l

Pressure in bar,

Pg = 1.01 bar

k = 0.0412/1.01 bar

= 0.0412 g/l.bar

Molar mass of CO2 = 12 + (16*2)

= 44 g/mol

Molar concentration of CO2 = k/molar mass

= 0.000936 mol/l.bar

Pressure of oxygen = 1.533 bar

Molar mass of O2 = 32 g/mol

Solubility of O2 = molar mass * molar concentration * pressure (bar)

= 32 * 0.000936 * 1.533

= 0.0459 g/l

Pressure in mmHg,

Pg = 760 mmHg

k = 0.0412/760

= 0.000054 g/l.mmHg

Molar mass of CO2 = 12 + (16*2)

= 44 g/mol

Molar concentration of CO2 = k/molar mass

= 0.00000123 mol/l.mmHg

Pressure of oxygen = 1150 mmHg

Molar mass of O2 = 32 g/mol

Solubility of O2 = molar mass * molar concentration * pressure (mmHg)

= 32 * 0.00000123 * 1150

= 0.0453 g/l

Final answer:

To calculate the solubility of oxygen in water at a new pressure, apply Henry's Law using the given solubility at a standard pressure, adjusting for the new pressure using a simple formula.

Explanation:

The student is asking about calculating the solubility of oxygen gas in water at a different pressure using Henry's law. Henry's Law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Therefore, to calculate the solubility of oxygen at a pressure of 1150 mmHg, we use the known solubility at a standard pressure (760 mmHg or 1 atm) and adjust for the new pressure.

Given that the solubility of oxygen is 0.00412 g/100 mL at 20°C and 760 mmHg, and we are asked to find its solubility at 1150 mmHg. We can use the formula from Henry's Law:

S1 / P1 = S2 / P2

Where S1 is the solubility at the initial pressure (0.00412 g/100 mL), P1 is the initial pressure (760 mmHg), S2 is the solubility at the new pressure, and P2 is the new pressure (1150 mmHg).

Rearranging the equation to solve for S2 gives us:

S2 = S1 * (P2 / P1) = 0.00412 * (1150 / 760)

Solving this gives us the solubility of oxygen in water at 20°C and a pressure of 1150 mmHg. Remember, the actual calculations were omitted here, but applying this formula with the given values will provide the answer.

Answer: The molar mass of the unknown protein is 6387.9 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

[tex]\pi=iMRT[/tex]

or,

[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution = 0.0766 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of protein = 100. mg = 0.100 g   (Conversion factor:  1 g = 1000 mg)

Molar mass of protein = ?

Volume of solution = 5.00 mL

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

T = temperature of the solution = [tex]25^oC=[25+273]K=298K[/tex]

Putting values in above equation, we get:

[tex]0.0766atm=1\times \frac{0.100\times 1000}{\text{Molar mass of protein}\times 5}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=\frac{1\times 0.100\times 1000\times 0.0821\times 298}{0.0766\times 5}=6387.9g/mol[/tex]

Hence, the molar mass of the unknown protein is 6387.9 g/mol

The element with the electron configuration 1s²2s²2p⁴ is **oxygen (O)**.

Here's its orbital diagram:

     ↑      ↑

1s:    |      |

     ↓      ↓

     ↑      ↑

2s:    |      |

     ↓      ↓

     ↑      ↓      ↑      ↓

2p:  -----|-----|-----|-----

     ↓      ↑      ↓      ↑

**Quantum numbers of the sixth electron:**

* Principal quantum number (n): 2 (second energy level)

* Azimuthal quantum number (l): 1 (p orbital)

* Magnetic quantum number (m_l): 0 (p_z orbital)

* Spin quantum number (m_s): +½ or -½ (two possible spin states)

**Explanation:**

- The electron configuration follows the Aufbau principle, filling orbitals from lowest energy to highest energy.

- The first two electrons fill the 1s orbital, the next two fill the 2s orbital, and the sixth electron goes into one of the three 2p orbitals (2p_x, 2p_y, or 2p_z).

- The magnetic quantum number (m_l) specifies the specific 2p orbital, and in this case, it's 0 for the 2p_z orbital.

- The spin quantum number (m_s) represents the electron's spin, which can be either +½ or -½.

Explanation:

It is known that charge on xenon nucleus is [tex]q_{1}[/tex] equal to +54e. And, charge on the proton is [tex]q_{2}[/tex] equal to +e. So, radius of the nucleus is as follows.

            r = [tex]\frac{6.0}{2}[/tex]

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = [tex]5.5 \times 10^{-15} m[/tex]     (as 1 fm = [tex]10^{-15}[/tex])

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = [tex]\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}[/tex]

Putting the given values into the above formula as follows.

           F = [tex]\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}[/tex]

              = [tex](9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}[/tex]

              = [tex](9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}[/tex]

              = 411.2 N

or,           = [tex]4.1 \times 10^{2}[/tex] N

Thus, we ca conclude that [tex]4.1 \times 10^{2}[/tex] N is the electric force on a proton 2.5 fm from the surface of the nucleus.

Answer: The concentration of chloride ions in the solution obtained is 1.27 M

Explanation:

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

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]     .....(1)

Molarity of calcium chloride solution = 0.439 M

Volume of solution = 194.4 mL

Putting values in equation 1, we get:

[tex]0.439=\frac{\text{Moles of calcium chloride}\times 1000}{194.4}\\\\\text{Moles of calcium chloride}=\frac{(0.439\times 194.4)}{1000}=0.085mol[/tex]

1 mole of calcium chloride produces 2 moles of chloride ions and 1 mole of calcium ion

Moles of chloride ions in calcium chloride = [tex](2\times 0.085)=0.170mol[/tex]

Molarity of aluminum chloride solution = 0.497 M

Volume of solution = 363 mL

Putting values in equation 1, we get:

[tex]0.497=\frac{\text{Moles of }AlCl_3\times 1000}{363}\\\\\text{Moles of }AlCl_3=\frac{(0.497\times 363)}{1000}=0.180mol[/tex]

1 mole of aluminum chloride produces 3 moles of chloride ions and 1 mole of aluminum ion

Moles of chloride ions in aluminum chloride = [tex](3\times 0.180)=0.540mol[/tex]

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.170 + 0.540) moles = 0.710 moles

Total volume of the solution = (194.4 + 363) mL = 557.4 mL

Putting values in equation 1, we get:

[tex]\text{Concentration of chloride ions}=\frac{0.710mol\times 1000}{557.4}\\\\\text{Concentration of chloride ions}=1.27M[/tex]

Hence, the concentration of chloride ions in the solution obtained is 1.27 M

To prepare a 0.00200 M NaCl solution from a 0.500 M stock, dilute 0.4 mL of the stock to 100 mL with water using volumetric pipets and a flask.

To prepare 100 mL of a 0.00200 M NaCl solution from a 0.500 M stock solution, you need to perform a dilution. First, use the dilution equation C1V1 = C2V2, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the concentration of the diluted solution, and V2 is the volume of the diluted solution. Plugging in the values, you get (0.500 M) V1 = (0.00200 M)(100 mL), which simplifies to V1 = (0.00200 M)(100 mL) / (0.500 M) = 0.4 mL. This is the volume of the stock solution you need to pipet into the 100-mL volumetric flask. As we only have 1-mL, 5-mL, and 10-mL volumetric pipets available, you can use the 1-mL pipet to add 0.4 mL of the stock solution to the volumetric flask, then fill the flask with water up to the 100-mL mark to achieve the desired concentration.

Final answer:

The volume of the diver's lungs would increase by a factor of three when ascending from a depth of 68 feet to the surface, due to the pressure change from 3 ATM to 1 ATM.

Explanation:

The question asks us to calculate the change in volume of a diver's lungs when ascending from a depth of 68 feet in fresh water to the surface, where the pressure is 1 ATM, assuming they do not exhale. According to the information given, in fresh water, pressure increases by 1 ATM every 34 feet. Therefore, at 68 feet, the pressure would be 3 ATM (including 1 ATM of atmospheric pressure at the surface). Using Boyle's Law, which states that pressure is inversely proportional to volume when temperature is constant (P1V1 = P2V2), the volume of the diver's lungs at the surface would be three times the volume underwater because the pressure decreases from 3 ATM to 1 ATM during ascent.

Answer:

copper(I) bromide: CuBr

copper(I) oxide: Cu₂O

copper(II) bromide: CuBr₂

copper(II) oxide: CuO

iron(III) bromide: FeBr₃

iron(III) oxide: Fe₂O₃

lead(IV) bromide: PbBr₄

lead(IV) oxide: PbO₂

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The formulas of the ionic compounds are as follows;

copper(I) bromide ----- CuBr

copper(I) oxide ------ Cu2O

copper(II) bromide ----- CuBr2

copper(II) oxide   ------- CuO

iron(III) bromide ------ FeBr3

iron(III) oxide ------ Fe2O3

lead(IV) bromide ---- PbBr4

lead(IV) oxide  ------ PbO2

The formula of an ionic compound is written with knowledge of the oxidation state of the metal in the ionic compound.

The oxidation state of each atom determines the subscript it carries in the formula as shown in the chemical formulas written above.

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Answer:

The pH of the resultant solution is 2.4

Explanation:

From the data

          V_total=V_w+V_a

          V_total=400+200 ml

Now the concentration of solution is given as

[tex]M_1V_1=M_2V_2\\M_2=\frac{M_1V_1}{V_2}\\M_2=\frac{0.30 \times 0.2}{0.6}\\M_2=0.1 M\\[/tex]

The reaction equation is given as

Equation                            [tex]HCOOH + H_2O \rightarrow HCOO^- +H_3O^+\\[/tex]

Initial Concentration is        0.1                             0               0

Change is                            -x                                x               x

_______________________________________________

At Equilibrium                  0.1-x                              x                x

Now the equation for K_a is given as

[tex]K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}[/tex]

Here Ka, for pKa-=3.75,  is given as

                   [tex]K_a=10^{-pKa}\\K_a=10^{-3.75}\\K_a=1.78 \times 10^{-4}[/tex]

Substituting this and concentrations at equilibrium in equation of Ka gives

[tex]K_a=\frac{[HCOO^-][H_3O^+]}{[HCOOH]}\\1.78 \times 10^{-4}=\frac{xx}{0.1-x}\\[/tex]

Solving for x

[tex]1.78 \times 10^{-4} (0.1-x)=x^2\\[/tex]

Here 0.1-x≈0.1 so

[tex]1.78 \times 10^{-5} =x^2\\x=\sqrt{1.78 \times 10^{-5}}\\x=4.219 \times 10^{-3}[/tex]

As [tex][H_3O^+]=[H^+][/tex]=x so its value is 4.219 x10^(-3) so pH is given as

[tex]pH=-log[H^+]\\pH=-log (4.219 \times 10^{-3})\\pH=2.37 \approx 2.4[/tex]

So the pH of the resultant solution is 2.4

Answer:

2.4

Explanation:

Given that:

Initial Concentration (M₁) of our formic acid from the question= 0.30 M

Initial volume (V₁) of the formic acid = 200 mL

Volume of water added = 400 mL

pKa= 3.75

∴ we can determine the total volume of the final solution by the addition of (Initial volume (V₁) of the formic acid) + (Volume of water added)

= 200 mL + 400 mL

= 600 mL

We can find the final concentration using the formula;

M₁V₁ = M₂V₂

M₂= [tex]\frac{0.30*200}{600}[/tex]

M₂= 0.1M

∴ The concentration of the diluted formic acid(since water is added to the initial volume) = 0.1M

From our pKa which is = 3.75

Ka of formic acid (HCOOH) = [tex]10^{-3.75}[/tex]

                                             =  1.78 × 10⁻⁴

If water is added to the formic acid; the equation for the reaction can be represented as;

                                  HCOOH + H₂O    ⇒    HCOO⁻    + H₃O

Initial                            0.1                                   0              0

Change                        -x                                    +x             +x

Equilibrium                  0.1-x                                 x              x

Ka = [tex]\frac{x^2}{0.1-x}[/tex] = 1.78 × 10⁻⁴

x² = (0.1 × 1.78 × 10⁻⁴)      since x = 0

x² =  1.78 × 10⁻⁵

x = [tex]\sqrt{1.78*10^{-5}}[/tex]

x = 0.004217

x = [tex][H_3O^+]=[H^+][/tex] =  0.004217 M

pH = [tex]-log[H^+][/tex]

= -log (0.004217)

= 2.375

≅ 2.4 (to  one decimal place)

We can thereby conclude that the final pH of the formic acid solution = 2.4

High ionization energies are common in elements found on the right side of the periodic table, notably the Noble Gases due to their full electron shells. In contrast, low ionization energies are associated with elements on the left side of the periodic table, particularly metals like those of the Alkali Metal group with just one or two electrons in their outermost shell.

In the periodic table, elements with relatively high IEs (Ionization Energies) are generally located on the right side, specifically in the group of Noble Gases. Noble gases like Helium or Neon have high IEs because they have full electron shells, and thus it requires a large amount of energy to remove an electron. On the other hand, the elements with relatively low IEs are located on the left side of the periodic table. This is because metals, such as those in the Alkali Metal group, have only one or two electrons in their outermost shell. These electrons are relatively easy to remove, resulting in a low ionization energy.

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Answer:

1.71 × 10²⁵ atoms

Explanation:

A pure titanium cube has an edge length of 2.64 in. In centimeters,

2.64 in × (2.54 cm/ 1 in) = 6.71 cm

The volume of the cube is:

V = (length)³ = (6.71 cm)³ = 302 cm³

Titanium has a density of 4.50 g/cm³. The mass corresponding to 302 cm³ is:

302 cm³ × 4.50 g/cm³ = 1.36 × 10³ g

The molar mass of titanium is 47.87 g/mol. The moles corresponding to 1.36 × 10³ g are:

1.36 × 10³ g × (1 mol/47.87 g) = 28.4 mol

1 mol of Ti contains 6.02 × 10²³ atoms of Ti (Avogadro's number). The atoms in 28.4 moles are:

28.4 mol × (6.02 × 10²³ atoms/1 mol) = 1.71 × 10²⁵ atoms

Answer:

Explanation:

 The Ce metal has electronic configuration as follows

[Xe] 4f¹5d¹6s²

After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.

Eu  has electronic configuration as follows

[ Xe ] 4 f ⁷6s²

[ Xe ] 4 f ⁷

Its outermost orbit contains 2 electrons so  Eu²⁺ is stable. Its +3 oxidation state is also stable.

Ce⁺²

Answer:

The drawing of the mechanism and the organic product is shown below.

Explanation:

Alkynes react with bromine to form a dihaloalkene (with an equivalent of the halogen) or a tetrahaloalkane derivative (with two equivalents of the halogen), which is this case. The triple bond adds two halogen molecules as shown in the drawing, an anti addition occurring.

Answer:

Molarity = 0.5 M

Osmolarity = 0.5 x 2 = 1 Osmpl.

Molecules of Cl2 = 6.02 x [tex]10^{23}[/tex] / 4= 1.505 x [tex]10^{23}[/tex] no. of molecules

Explanation:

If we add half mole in 1L volume than molarity will obviously be 0.5 M.

The osmolarity is molarity multiplies by number of dissociates of solute that for CaCl2 are 2. So, 2 x 0.5 = 1

Half will be molecules of Ca and half will be of Cl2 for 0.5M.

Final answer:

The molarity of a 1-liter solution with half a mole of calcium chloride is 0.5 M. The osmolarity, considering the dissociation of calcium chloride into three ions, is 1.5 osmol/L. The solution would contain 6.022 x 10²³chloride ions.

Explanation:

The molarity of a solution is defined as the number of moles of a solute per liter of solution. In this case, to calculate the molarity of calcium chloride, CaCl₂, in a 1-liter solution containing half a mole, we simply divide the moles of solute by the volume of the solution in liters:

Molarity (M) = moles of solute / liters of solution

Molarity (M) = 0.5 mol / 1 L = 0.5 M

The osmolarity of a solution is the total concentration of solute particles dissolved in a solution. Since calcium chloride dissociates into three ions (one Ca²⁺ ion and two Cl
- ions), the osmolarity is as follows:

Osmolarity = molarity  x number of particles per formula unit

Osmolarity = 0.5 M  x 3 = 1.5 osmol/L

For the number of chloride ions in the solution, we must look at the stoichiometry of calcium chloride: each formula unit of CaCl₂ yields two Cl⁻ions. So, we have:

Number of Cl⁻ ions = 0.5 mol CaCl₂ x (2 mol Cl⁻ / 1 mol CaCl₂) x (6.022 x 10²³ ions/mol)

Number of Cl⁻ ions = 1 mol Cl⁻ x 6.022 x 10²³ ions/mol = 6.022 x 10²³ ions

Answer:

a)

From VSPER theory, the compound TeF₄ and TeCl₄ AX₄E type of models. Hence, the molecular geometry of this molecule is trigonal bipyramidal. The plausible structure of this molecule are shown on the first uploaded image

Looking at the image we see that  structure 1 has three lone pairs repulsion with 90° angle. From  VSPER theory, the repulsive force between the electron  pair is given as

lone pair - lone pair > lone pair - bond pair > bond pair - bond pair

The above shows that bond pair -bond pair has less repulsion than lone pair- bond pair, and next is lone pair - lone pair

This means that structure 2 is more stable structure for TeF₄ or TeCl₄ .,because the lone pair is present at equatorial position.

This stable structure is shown on the second uploaded image

A lone pair of electron occupy more space around the central atom than a bond pair electron that are present at axial positions, the two atoms that are present at axial position are tilt away from the lone pair

       We can see the structure on the third uploaded image

Due to the repulsive interaction , the bond distance between Te - X(axial) increases, which means that the Te - X(axial) distance is longer than  Te - X(equitorial)  distance.

b)

It is a general concept that TeX₄ ha a bond angle of 90° for axial position and  120° in between equitorial position. This is shown on the fourth uploaded image

Considering TeF₄ and TeCl₄ , F is more electro-negative element  Than Cl.This mean that the fluorine would pull strongly the  shared electron away from the Te which reduces the electron density near the central Te atom.This would result in decrease bond angle be TeCl₄

So , F(axial) - Te - F(axial) has smaller bond angle than Cl(axial) - Te - Cl(axial)

Considering the equitorial position we see that because of the highly electro - negative fluorine atom, the bond angle decrease much in  TeF₄ than in  TeCl₄

      This means that  F(equitorial) - Te - F(equitorial)  has smaller bond angle than Cl(equitorial) - Te - Cl(equitorial)

The molecules TeCl4 and TeF4 both have the axial bonds longer than the equatorial bonds due to repulsion. Secondly TeF4 and TeCl4 have axial bond angles less than equatorial bond angles due to their greater electronegativity.

The molecule TeCl4  has a see - saw shape. There are four bonds pairs and one lone pair in the molecule. We have to note that the axial bonds are longer than the equatorial bonds owing to the greater repulsion from the equatorial bonds leading to elongation of bonds.

Looking at the molecules TeF4 and TeCl4, we know that F and Cl are more electronegative than Te hence they pull more on the shared electron pair hence the axial bond angles are smaller than the equatorial bond angles.

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Answer :

(a) Number of electrons in an atoms is, 6

(b) Number of electrons in an atoms is, 2

(c) Number of electrons in an atoms is, 14

Explanation :

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as . The value of this quantum number ranges from . When l = 2, the value of

Spin Quantum number : It describes the direction of electron spin. This is represented as . The value of this is  for upward spin and  for downward spin.

Number of electrons in a sublevel = 2(2l+1)

(a) 4p

n = 4

Value of 'l' for 'p' orbital : l = 1

At l = 1,  [tex]m_l=+1,0,-1[/tex]

Number of electrons in an atoms = 2(2l+1) = 2(2×1+1) = 6

(b) n = 3, l = 1, ml = +1

As we know that, a sublevel of 'p' orbital can accommodate 6 electrons but 1 orbital can accommodate only 2 electrons. So,

Number of electrons in an atoms for (ml = +1) = 2

(b) n = 5, l = 3

Number of electrons in an atoms = 2(2l+1) = 2(2×3+1) = 14

There are a given specific number of electrons that can be found in a particular orbital or energy level.

In an atom, electrons are arranged in orbitals. These orbitals lie within specific energy levels. There is a maximum number of electrons that can be found in a particular energy level as well as in a particular orbital.

The maximum number of electrons that can be found in the following orbitals and energy levels are shown below;

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