If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of TES is 7.55.)

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:

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:

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:

[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: 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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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:

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 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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Answer: 288.32g/mol

Explanation:Please see attachment for explanation

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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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:

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.

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:

- 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

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:

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 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: 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

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:

Answer: option A. acidic

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

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

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: 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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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:

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 :

(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:

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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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