Draw the structures of the 4 isomers of C8H18 that contain 2 methyl branches on separate carbons of the main chain.

Answer: 67.8 %.

Explanation:

Okay, let us delve right into the solution to the question;

The balanced chemical reaction is given by the equation (1) below;

2 HCOONa + H2SO4 ---------> 2 CO + 2 H2O + Na2SO4. ----------------------------------------------------------------------------(1).

From the balanced chemical reaction in equation (1) above we can see that; 2 moles of HCOONa reacts with one moles of tetraoxosulphate acid, H2SO4 to produce 2 moles of carbonmonoxide,CO; 2 moles of water, H2O and 1 mole of sodium tetraoxosulphate, Na2SO4.

The parameters given from the question are; total atmospheric pressure, P(t) = 752 torr, volume of CO= 242 mL = 0.242 Litres.

STEP ONE : find the carbon monoxide,CO pressure; P(CO).

Using the formula below;

P(t) = P(CO) + P(H2O). Hence;

P(CO) = P(t) - P(H2O). Note that P(H2O)= 19.8 torr.

==>P(CO)= 752 torr - 19.8 torr = 732.2 torr.

STEP TWO: calculate the number of moles of Carbonmonoxide,CO.

Using the formula below;

Number of moles= pressure(P) × volume(v) / gas constant(R) × temperature (T).

That is, n= PV/RT.

n= 732 torr × 0.242 Litres/ 62.4 × 295.15.

= 9.62 × 10^-3 mol of CO.

STEP THREE:

2 moles of HCOONa = 2 moles of CO.

=> 2 moles of HCOONa = 2 moles of CO/ 2 moles of CO = 1 mol( HCOONa/ CO).

Then, 9.62 × 10^-3 mol of CO × 1 mol( HCOONa/ CO).

==> 9.62 × 10^-3 mol HCOONa × molar masss of HCOONa(68 grams/mol)

= 0.654 grams.

Therefore, the percentage of sodium formate in the original mixture = 0.654 grams/ 0.964 gram × 100 = 67.8 %.

The equation for the reaction of sodium with sulfuric acid is 2Na + H₂SO₄ → Na₂SO₄ + H₂.

The equation for the reaction of sodium with sulfuric acid is:

2Na + H₂SO₄ → Na₂SO₄ + H₂

In this reaction, sodium reacts with sulfuric acid to form sodium sulfate and hydrogen gas. This reaction is a displacement reaction where sodium replaces hydrogen in the sulfuric acid to form sodium sulfate and liberate hydrogen gas.

The balanced equation for the reaction is:

2Na(s) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂(g)

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The change in energy of the system in the given reaction is -1045 kJ. This value is calculated using the first law of thermodynamics taking in account that the energy absorbed is lost by the system and the work done on the system is obtained.

In this chemical reaction, we are dealing with a process that involves heat absorption and work done on the system. Both these elements contribute to the change in energy of the system, denoted by ΔE.

The first law of thermodynamics states that ΔE = q + w, where 'q' represents heat and 'w' represents work. However, notice that the heat is absorbed by the surroundings, which means the system is losing that amount of heat, so q = -1.1 MJ = -1100 kJ (as 1MJ = 1000 kJ).

Also, the work is done on the system, so it's positive, and it's given in calories, we need to convert it into kilojoules (kJ), for that, use the conversion factor 1 cal= 0.004184 kJ, so w = 13.2 kcal * 4.184 = 55.23 kJ.

Plugging these values into thermodynamics equation, we get ΔE = -1100 kJ + 55.23 kJ = -1044.77 kJ. Thus, the change in energy of the system, ΔE, is -1045 kJ (rounded to 0 decimal places).

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Final answer:

To calculate the change in internal energy (ΔE) for the reaction, convert all values to kJ, then apply the formula ΔE = q + w. The answer is +1155 kJ, indicating energy absorption.

Explanation:

The question asks to calculate the change in internal energy (ΔE) for a chemical reaction where gaseous reactants form a liquid product. The energy absorbed by the surroundings is given as 1.1 MJ, and the work done on the system is 13.2 kcal. To find ΔE, we use the formula ΔE = q + w, where q is the heat absorbed by the system and w is the work done on the system.

First, convert all values to the same unit (kJ): Heat absorbed, q = 1.1 MJ = 1100 kJ; Work done, w = 13.2 kcal × 4.184 kJ/kcal = 55.23 kJ. Thus, ΔE = 1100 kJ + 55.23 kJ = 1155.23 kJ. Therefore, the change in internal energy for the reaction is +1155 kJ, indicating that the system absorbed energy.

The density of the unknown gas in a 2.85 L flask at 33 °C and 1.00 atm is approximately 0.93 g/L.

The question asks for the density of an unknown gas contained in a flask at given temperature and pressure conditions. To find the density (d) of the gas, we use the formula d = mass/volume. The mass of the gas is given as 2.65 g, and the volume of the flask is given as 2.85 L. Therefore, the density of the gas can be calculated as:

d = mass / volume

d = 2.65 g / 2.85 L

d = 0.9298 g/L

After performing the calculation, we find that the density of the gas is approximately 0.93 g/L at the given conditions of 33 °C and 1.00 atm.

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 and Explanation :

Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.

(a) V³⁺

The electronic configuration is -  

[tex][Ar]3d^1[/tex]

The electrons in 3d orbital = 1 (Unpaired)

Thus, the ion is paramagnetic as the electrons are unpaired.

(b) Cd²⁺

The electronic configuration is -  

[tex][Kr]4d^{10}[/tex]

The electrons in 4d orbital = 10 (paired)

Thus, the ion is diamagnetic as the electrons are paired.

(c) Co³⁺

The electronic configuration is -  

[tex][Ar]3d^6[/tex]

The electrons in 3d orbital = 6 (Unpaired)

Thus, the ion is paramagnetic as the electrons are unpaired.

(d) Ag⁺

The electronic configuration is -  

[tex][Kr]4d^{10}[/tex]

The electrons in 4d orbital = 10 (paired)

Thus, the ion is diamagnetic as the electrons are paired.

Answer:

a) Approximate boiling point of HBr = -60.15 °C

b) Approximate boiling point of AsH₃ = -52.25 °C

Explanation:

Döbereiner stated that some elements could be arranged in groups of 3 similar elements ( known as "triads) , and the element of the middle ( elements are ordered with respect to their atomic mass) would have properties between the other 2 ( the average value)

a) In the first case the triad would be the halogen triad ( Cl , Br and I ) . And according to Döbereiner , the boiling point of HBr should be the average of HCl and HI . Therefore

Approximate boiling point of HBr = [(- 84.9°C) + (-35.4°C)]/2 = -60.15 °C

b) Simmilarly for  AsH₃ , PH₃ and SbH₃ , the boiling point of AsH₃ would be

Approximate boiling point of AsH₃ = [(- 87.4°C) + (-17.1°C)]/2 = -52.25 °C

Answer: The thickness of metal sheet is 1.93 mm

Explanation:

The metals sheet is in the form of cuboid.

To calculate the width of the metal sheet for the given volume, we use the equation to calculate the volume of cuboid, which is:

[tex]V=lbh[/tex]

where,

V = volume of metal sheet = [tex]45.5cm^3[/tex]  

l = length of metal sheet = 15.9 cm

b = width of metal sheet = 14.8 cm

h = height of metal sheet = ? cm

Putting values in above equation, we get:

[tex]45.5cm^3=15.9\times 14.8\times h\\\\h=\frac{45.5}{15.9\times 14.8}=0.193cm[/tex]

Converting this thickness into millimeters, we use the conversion factor:

1 cm = 10 mm

So, [tex]0.193cm\times \frac{10mm}{1cm}=1.93mm[/tex]

Hence, the thickness of metal sheet is 1.93 mm

To find the thickness (height) of the metal sheet, use the formula for volume and rearrange it to solve for the height.

To find the thickness (height) of the metal sheet, we can use the formula for volume. The formula for volume of a rectangular solid is:

Volume = Length * Width * Height

Given that the volume is 45.5 cm3, the length is 15.9 cm, and the width is 14.8 cm, we can rearrange the formula to solve for the height:

Height = Volume / (Length * Width)

Substituting the values, we have:

Height = 45.5 cm3 / (15.9 cm * 14.8 cm)

We can now calculate the height of the metal sheet in millimeters.

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Answer:

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

Explanation:

Side of the cubic box = s = 20.0 cm

Volume of the box ,V= [tex]s^3[/tex]

[tex]V=(20.0 cm)^3=8000 cm^3=8\times 10^{-3} m^3[/tex]

Root mean square speed of the of helium molecule : 200m/s

The formula used for root mean square speed is:

[tex]\mu=\sqrt{\frac{3kN_AT}{M}}[/tex]

where,

= root mean square speed

k = Boltzmann’s constant = [tex]1.38\times 10^{-23}J/K[/tex]

T = temperature = 370 K

M = mass helium = [tex]3.40\times 10^{-27}kg/mole[/tex]

[tex]N_A[/tex] = Avogadro’s number = [tex]6.022\times 10^{23}mol^{-1}[/tex]

[tex]T=\frac{\mu _{rms}^2\times M}{3kN_A}[/tex]

Moles of helium gas = n

Number of helium molecules = N =[tex]2.00\times 10^{23}[/tex]

N = [tex]N_A\times n[/tex]

Ideal gas equation:

PV = nRT

Substitution of values of T and n from above :

[tex]PV=\frac{N}{N_A}\times R\times \frac{\mu _{rms}^2\times M}{3kN_A}[/tex]

[tex]PV=\frac{N\times R\times \mu ^2\times M}{3k\times (N_A)^2}[/tex]

[tex]R=k\times N_A[/tex]

[tex]PV=\frac{N\times \mu ^2\times M}{3}[/tex]

[tex]P=\frac{2.00\times 10^{23}\times (200 m/s)^2\times 3.40\times 10^{-27} kg/mol}{3\times 8\times 10^{-3} m^3}[/tex]

[tex]P=1133.33 Pa =1.133 kPa[/tex]

(1 Pa = 0.001 kPa)

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

The question asks for the average pressure exerted by helium gas molecules on the walls of a cubic container. Using the equation PV = Nmv^2, we can calculate pressure by substituting the given values for volume, number of molecules, mass of one molecule, and root-mean-square speed.

The question is asking to calculate the average pressure exerted by helium gas molecules on the walls of a cubic container. The important formula relating pressure (P), volume (V), number of molecules (N), mass of a molecule (m), and the square of the rms speed (v2) of the molecules in a gas is:

PV = Nmv2,

First, we need to determine the volume of the container, which is the cube of one side, so V = (20 cm)3 = (0.2 m)3. Inserting the given values into the equation and solving for P gives us the desired answer. Recall that the rms speed is given, so no temperature calculations are needed.

Therefore, using all given data points:

Volume (V) = (0.2 m)3

Number of molecules (N) = 2.00 × 1023

Mass of one helium molecule (m) = 3.40 × 10-27 kg

Root-mean-square speed (vrms) = 200 m/s

By substituting these values, we can find the pressure exerted by the gas. This represents an application of kinetic theory of gases which assumes the behavior of an ideal gas.

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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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:

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:

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 mass of urea needed is 12.89 grams

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{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution = 24.3 atm

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

[tex]m_{solute}[/tex] = mass of urea = ? g

[tex]M_{solute}[/tex] = molar mass of urea = 60.1 g/mol

[tex]V_{solution}[/tex] = Volume of solution = 216 mL

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

T = temperature of the solution = 298 K

Putting values in above equation, we get:

[tex]24.3atm=1\times \frac{m_{solute}\times 1000}{60.1\times 216}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\m_{solute}=\frac{24.3\times 60.1\times 216}{1\times 1000\times 0.0821\times 298}=12.89g[/tex]

Hence, the mass of urea needed is 12.89 grams

By using the formula for osmotic pressure and solving for the number of moles, we establish that approximately 12.74 grams of urea are required to generate an osmotic pressure of 24.3 atm in 216 mL of water at 298 K.

The problem deals with finding the mass of urea needed to generate a certain osmotic pressure. One can utilize the formula for osmotic pressure (Π = n/V * R * T) which is similar to the ideal gas law. In this formula, Π is your osmotic pressure (24.3 atm), n is the number of moles, V is the volume in liters (216 mL = 0.216 liters), R is the ideal gas constant (0.0821 L*atm/(K*mol)) and T is the temperature in Kelvin (298 K).

By solving for n (the number of moles), we obtain n = ΠV / RT = (24.3 atm * 0.216 L) / (0.0821 L atm/(K mol) * 298 K) which results in n = 0.212 moles.

Urea has a molar mass of 60.1 g/mol, so the mass of urea required is n * molar mass = 0.212 moles * 60.1 g/mol = 12.74 g. Thus, approximately 12.74 grams of urea are needed to generate an osmotic pressure of 24.3 atm in the given conditions.

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Answer: option A. acidic

Explanation: acidic solution is characterized by the presence of Hydrogen ion.

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion ([tex]Li^+[/tex])

Explanation:

First of all

v=f*λ

In our case v=c

c=f*λ

λ=c/f

where:

c is the speed of light

f is the frequency

[tex]\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m[/tex]

Using Rydberg's Formula:

[tex]\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]

Where:

R is Rydberg constant=[tex]1.097*10^7[/tex]

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

[tex]\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99[/tex]

z≅3

Atomic number is 3, So ion is Lithium ion ([tex]Li^+[/tex])

Answer:

The answer to the question is

The mass percentage of NaCl(s) in the mixture is 49.7%

Explanation:

The given variables are

mass of sample of mixture = 0.3146 g

Volume of AgNO₃ required to react comletely with the chloride ions = 45.70 mL

Concentration of the AgNO₃ added = 0.08765 M

The equations for the reactions oare

NaCl(aq) + AgNO₃ (aq) = AgCl(s) + NaNO₃(aq)

AgNO₃ (aq) + KBr (aq) → AgBr (s) + KNO₃

The equation for the reaction shows one mole of NaCl reacts with one mole of AgNO₃ to form one mole of AgCl

Thus 45.70 mL of 0.08765 M solution of AgNO₃ contains[tex]\frac{45.7}{1000} (0.08765) = 0.004 moles[/tex]

Therefore the sum of the number of moles of Br⁻ and Cl⁻

precipitated out of the solution =  0.004 moles

Thus if the mass of NaCl in the sample = z then the mass of KBr = y

However the mass of the sample is given as 0.3146 g which  means the molarity of the solution is 0.004 moles

given by

[tex]\frac{z}{58.44} + \frac{y}{119} = 0.004 moles[/tex]  and z + y = 0.3146

Therefore z = 0.3146 - y which gives

[tex]\frac{(0.3146-y)}{58.44} + \frac{y}{119} = 0.004 moles[/tex]

-8.7×10⁻³y +0.54×10⁻³ = 0.004

or 8.7×10⁻³y = 1.37769× 10⁻³

y = 0.158 g and z = 0.156 Thus the mass of NaCl = 0.156 g and the mass percentage = 0.156/0.3146×100 = 49.7% NaCl

The mass percentage of NaCl(s) in the mixture is 49.7%

Answer:

The age of the fossil be [tex]2.1987\times 10^{4} years[/tex].

Explanation:

Formula used :

[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]

where,

[tex]N_o[/tex] = initial mass of isotope  C-14 = x

N = mass of the parent isotope left after the time, (t)  = 70.0% of x=0.07x

[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope  C-14 = 5730 years

[tex]\lambda[/tex] = rate constant

Let the age of the fossil be t.

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

[tex]0.07x=x\times e^{-(\frac{0.693}{5730 years})\times t}[/tex]

[tex]t=2.1987\times 10^{4} years[/tex]

The age of the fossil be [tex]2.1987\times 10^{4} years[/tex].

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

Quantum numbers n, l, and ml are used to describe an electron's location in an atom. For n=2, l can be 0 or 1, and four orbitals exist. For l=3, ml can range from -3 to 3, providing seven orbitals.

In the realm of quantum mechanics, the values of the quantum numbers n, l, and ml tell us a lot about the electron's location in an atom. If the principal quantum number, n = 2, the angular momentum quantum number, l, can have values ranging from 0 to n-1. In this case, that means l can be 0 or 1. With this, the total number of orbitals (regions where you can most likely find an electron) possible at the n = 2 energy level is 4 (calculated as n2).

If the magnetic quantum number, l = 3, then, the magnetic quantum number ml can have values ranging from -l to +l, which means ml can range from -3 to 3. This produces a total of 7 orbitals possible at the l = 3 sublevel, each of which can hold two electrons.

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For n = 2, l can have values 0 and 1, with a total of 4 orbitals. For l = 3, ml ranges from -3 to 3, with 7 orbitals at the l = 3 sublevel.

The problem involves determining the possible values of different quantum numbers. When the principal quantum number n is 2, the angular momentum quantum number l can range from 0 to n-1, meaning l can have the values 0 and 1. The total number of orbitals possible at the n = 2 energy level is calculated by the formula n², thus the number is 4 orbitals.

If the angular momentum quantum number l is 3, the magnetic quantum number ml can range from -l to +l, meaning it can have values from -3 to 3 (-3, -2, -1, 0, 1, 2, 3). The total number of orbitals possible at the l = 3 sublevel is 2l+1, which is 7.

Answer:

 a) 1.866 × 10 ⁻¹⁹ J      b)   3.685 × 10⁻¹⁹ J

Explanation:

the constants involved are

h ( Planck constant) = 6.626 × 10⁻³⁴ m² kg/s

Me of electron = 9.109 × 10 ⁻³¹ kg

speed of light = 3.0 × 10 ⁸ m/s

a) the Ek ( kinetic energy of the dislodged electron) = 0.5 mu²

Ek = 0.5 × 9.109 × 10⁻³¹ × ( 6.40 × 10⁵ )² = 1.866 × 10 ⁻¹⁹ J

b) Φ ( minimum energy needed to dislodge the electron ) can be calculated by this formula

hv =   Φ + Ek

where Ek = 1.866 × 10 ⁻¹⁹ J

v ( threshold frequency ) = c / λ where c is the speed of light and λ is the wavelength of light = 358.1 nm = 3.581 × 10⁻⁷ m

v = ( 3.0 × 10 ⁸ m/s ) / (3.581 × 10⁻⁷ m ) = 8.378 × 10¹⁴ s⁻¹

hv = 6.626 × 10⁻³⁴ m² kg/s ×  8.378 × 10¹⁴ s⁻¹ = 5.551 × 10⁻¹⁹ J

5.551 × 10⁻¹⁹ J = 1.866 × 10 ⁻¹⁹ J + Φ

Φ = 5.551 × 10⁻¹⁹ J - 1.866 × 10 ⁻¹⁹ J = 3.685 × 10⁻¹⁹ J

Final answer:

Paramagnetic ions are those with unpaired electrons, which can be found by looking at the electron configuration of Fe+ to Fe14+. The more unpaired electrons, the stronger the attraction to a magnetic field.

Explanation:

The student has asked which ions from Fe+ to Fe14+ are paramagnetic and which would be most strongly attracted to a magnetic field. An ion is considered paramagnetic if it has one or more unpaired electrons. To determine this, we can look at the electron configuration of each iron ion. Iron (Fe) has an electron configuration of [Ar] 4s2 3d6. When it loses electrons to become ionized (Fe+ to Fe14+), it loses them from its outermost shell first, which is the 4s shell, and then from the 3d shell. Paramagnetism increases with the number of unpaired electrons, so the ions with the highest number of unpaired electrons in the d shell will be most strongly attracted to a magnetic field.

Answer:

The difference in the magnetic orientation influences the thermal stability of the allotropes of iron.

Explanation:

It is known that the allotropes of iron exist in three phases: α - phase, β- phase, and γ-phase. However, two prominent structures are the  α - phase and γ-phase. Now, let us look at the two phrases:

α - phase

This structure is a body-centered cube. It means that the unit cell structure resembles a cube. The lattice points are in the face of the cube. This subsequently affects the magnetic structure of the iron allotrope.

γ-phase

This allotrope has a lattice structure. It simply means that the structure has lattice points on the face of the cube. The structure generally affects the magnetic properties of the transitional metal; hence the stability of the γ-phase compared to α-phase.

Allotropes stable at high temperatures have higher enthalpies than those at low temperatures because they require more energy to maintain their structural bonds at these elevated temperatures. Hess's law and observations like the enthalpy differences in the thermite reaction further support this understanding in the context of energy changes and phases of matter.

The enthalpies of allotropes that are stable at high temperatures are higher than those stable at low temperatures because more energy is required to maintain the structure and bonding in the high-temperature allotrope. For example, in the case of iron, γ-Fe (gamma iron) has a higher enthalpy than α-Fe (alpha iron). This is due to the difference in bonding and structure at different temperatures. As temperature increases, thermal energy overcomes stronger bonds, resulting in allotropes with higher enthalpies at these temperatures.

Hess's law can illustrate this concept further. Considering the thermite reaction, the heat produced during the reaction of aluminum with iron(III) oxide indicates an exothermic reaction that causes the iron to melt. In general, transformations like changing phases from solid to liquid require energy, and allotropes that must retain more complex, less stable structures at higher temperatures inherently have higher enthalpies.

Moreover, substances with high melting and boiling points usually have strong bonds and interactions to maintain those phases, which means their reactions typically involve greater changes in enthalpy. This is why the enthalpy of vaporization is much greater than the enthalpy of fusion, and the same principle can be applied to allotropes stable at different temperatures. Allotropes stable at higher temperatures have structures that require more energy to maintain, hence their higher enthalpies.

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Answer:

[tex]\lambda=459.1\times 10^{-7}\ m[/tex] = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

Explanation:

Given that:- Energy = 2.7 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.602 × 10⁻¹⁹ J

So, Energy = [tex]2.7\times 1.602\times 10^{-19}\ J=4.33\times 10^{-19}\ J[/tex]

Considering:-

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

Where,  

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

So,  

[tex]4.33\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]

[tex]4.33\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]

[tex]\lambda=459.1\times 10^{-7}\ m[/tex] = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

Gold's warm yellow color is due to the absorption of photons with an energy of about 2.7 eV, corresponding to the energy difference between the 5d and 6s electron sublevels. The absorbed light being in the range of yellow frequencies causes the reflected light to give gold its characteristic color.

The color of gold is attributed to the energy difference between its 5d and 6s electron sublevels. Photons of light that have an energy of 2.7 eV are absorbed to promote an electron from the 5d to the 6s sublevel. The photons corresponding to this energy level fall within the visible spectrum of light and are in the range that produces a warm yellow color. The absorbed photons are those that do not get reflected and therefore the color we see is the complementary color of the absorbed photons, which gives gold its distinctive yellow shine.

To understand this further, we can use the equation for energy of a photon (E = hf), where 'h' is Planck's constant and 'f' is the frequency of the light. Since the energy difference corresponds to the colors that are absorbed, the light that is not absorbed determines the color we perceive. Yellow light has the right frequency so that when it is mixed with other unabsorbed colors, it gives gold its unique luster.

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 :  The condensed ground-state electron configuration for each is:

(a) [tex][Xe]6s^2[/tex]

(b) [tex][Ar]4s^23d^7[/tex]

(c) [tex][Kr]5s^14d^{10}[/tex]

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

Noble-Gas notation : It is defined as the representation of electron configuration of an element by using the noble gas directly before the element on the periodic table.

(a) The given element is, Ba (Barium)

As we know that the barium element belongs to group 2 and the atomic number is, 56

The ground-state electron configuration of Ba is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^2[/tex]

So, the condensed ground-state electron configuration of Ba in noble gas notation will be:

[tex][Xe]6s^2[/tex]

(b) The given element is, Co (Cobalt)

As we know that the cobalt element belongs to group 9 and the atomic number is, 27

The ground-state electron configuration of Co is:

[tex]1s^22s^22p^63s^23p^64s^23d^7[/tex]

So, the condensed ground-state electron configuration of Co in noble gas notation will be:

[tex][Ar]4s^23d^7[/tex]

(c) The given element is, Ag (Silver)

As we know that the silver element belongs to group 11 and the atomic number is, 47

The ground-state electron configuration of Ag is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^14d^{10}[/tex]

So, the condensed ground-state electron configuration of Ag in noble gas notation will be:

[tex][Kr]5s^14d^{10}[/tex]

Here are the orbital diagrams and condensed ground-state electron configurations for Ba, Co, and Ag.

Orbital diagram and condensed ground-state electron configurations for (a) Ba, (b) Co, and (c) Ag:

(a) Barium (Ba) has an atomic number of 56. Its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2, with 2 valence electrons in the 5s orbital. The orbital diagram can be represented as follows:
5s:[2]

(b) Cobalt (Co) has an atomic number of 27. Its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7, with 7 valence electrons in the 3d orbital. The orbital diagram can be represented as follows:
3d:[7]

(c) Silver (Ag) has an atomic number of 47. Its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^1 4d^10, with 1 valence electron in the 5s orbital. The orbital diagram can be represented as follows:
5s:[1]

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The question is incomplete, here is the complete question:

[tex]KBr(aq.)+AgNO_3(aq.)\rightarrow KNO_3(aq.)+AgBr(s)[/tex]

Classify the type of the reaction (check all that apply)

1. Combination

2. Precipitation

3. Single replacement

4. Combustion

5. Double replacement

6. Acid-base

7. Decomposition

Answer: The given reaction is a type of precipitation and double displacement reaction.

Explanation:

Combination reaction is defined as the reaction in which smaller substances combine to form a larger substance.

[tex]A+B\rightarrow AB[/tex]

Precipitation reaction is defined as the reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

[tex]A+BC\rightarrow AC+B[/tex]

Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

[tex]\text{Hydrocarbon}+O_2\rightarrow CO_2+H_2O[/tex]

Double displacement reaction is defined as the reaction in which exchange of ions takes place.

[tex]AB+CD\rightarrow CB+AD[/tex]

An acid-base reaction is known as neutralization reaction.  This reaction is defined as the reaction in which an acid reacts with a base to produce a salt and water molecule.

[tex]HX+BOH\rightarrow BX+H_2O[/tex]

Decomposition reaction is defined as the reaction in which a large substance breaks down into smaller substances.

[tex]AB\rightarrow A+B[/tex]

For the given chemical equation:

[tex]KBr(aq.)+AgNO_3(aq.)\rightarrow KNO_3(aq.)+AgBr(s)[/tex]

As, ions are getting exchanges and also a solid salt is getting formed. The above reaction is a type of double displacement and precipitation reaction.

Hence, the given reaction is a type of precipitation and double displacement reaction.

The reaction between KBr(aq) and AgNO3(aq) is a double replacement reaction, where the bromide and nitrate anions switch cations. This results in the formation of AgBr(s), a precipitate, and KNO3(aq).

The equation represents a double replacement reaction. When potassium bromide (KBr) and silver nitrate (AgNO3) are combined in an aqueous solution, they undergo a double replacement reaction.

In a double replacement reaction, the cations and anions of the two reactants switch places, forming two new compounds. In this case, KBr(aq) + AgNO3(aq) will yield AgBr(s) and KNO3(aq), with AgBr being a precipitate.

It's called a double replacement because both K and Ag exchange their anions (bromide and nitrate respectively). So this reaction doesn't fall under the categories of combination, precipitation, single replacement, combustion, acid-base or decomposition reaction, but it's specifically a double replacement reaction.

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Using the Ideal Gas Law, missing variables (n, T, P, or V) are calculable with known values. Each set requires rearranging PV = nRT appropriately and substituting provided values, while ensuring units are consistent, to find the missing quantity.

The Ideal Gas Law, PV = nRT, allows us to calculate the missing variable (P, V, T, or n) when the other three are known. The gas constant (R) has values depending on the units used, commonly 0.08206 L.atm/(K•mol) for calculations involving liters, atmospheres, and moles. Let's solve each set:

Explanation:

First, we will calculate the feed rate of alum as follows.

   [tex]\frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}[/tex]

                  = 228000 g/day

Converting this amount into g/min as follows.

     [tex]\frac{228000 g}{1 day} \times \frac{1 day}{1440 min}[/tex]

          = 158 g/min

Now, the chemical equation will be as follows.

    [tex]Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O[/tex]

 [tex]\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}[/tex]

      = 0.151 mmol [tex]mmol SO^{2-}_{4}/L[/tex]

[tex]\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}[/tex]

           = 15.15 mg [tex]CaCO_{3}[/tex]/L

For precipitate:

[tex]Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O[/tex]

  [tex]\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}[/tex]

     = 7.88 [tex]Al(OH)_{3}/L[/tex]

  [tex]\frac{7.88 mg Al(OH)_{3}}{1 L} \times \frac{3800 m^{3}}{1 day} \times \frac{1000 L}{1 m^{3}} \times \frac{1 kg}{10^{6} mg}[/tex]

          = 29.9 [tex]Al(OH)_{3}/day[/tex]

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: The nitronium ion is a strong electrophile

Explanation:

The detailed mechanism of the reaction is shown in the images attached. The flow of electrons has been shown with arrows. HNO3 is first protonated by H2SO4 this protonated specie H2NO3+ now forms H2O and NO2+.

The nitronium ion is formed when concentrated sulfuric acid protonates nitric acid, which then loses water to produce the reactive electrophile NO2+ needed for electrophilic aromatic substitution reactions.

The formation of the nitronium ion (NO2+) from concentrated nitric acid in concentrated sulfuric acid is a crucial step in the electrophilic aromatic substitution mechanism. The process begins with the protonation of nitric acid by sulfuric acid, creating the nitronium ion and water. Specifically, sulfuric acid acts as a strong acid and protonates the nitric acid, which loses a water molecule and forms the nitronium ion. This positively charged electrophile is highly reactive and capable of attacking the electron-rich aromatic ring, initiating a substitution reaction.

In the context of aromatic substitutions, the nitronium ion is very strong and reactive enough to overcome the stability of the aromatic ring's delocalized electrons. Its formation is necessary because aromatic compounds don't readily react with partial positive electrophiles, making the full cation electrophile essential for the reaction to proceed.