What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochloride and an excess of acetic anhydride in an acetate buffer? Enter only the number with two significant figures.

Here is the full question

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Express your answer in two significant figures and include the appropriate units (in cm)

Answer:

ΔH ≅ 3.73 cm

Explanation:

The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:

P =   ρ × g × h

where;

ρ is the density of the fluid

g is the gravitational constant

h is the height from the surface

From the question above;

For glycerine; we have:

density of glycerine = 1260 kg/m³

gravitational constant = 9.8 m/s²

height = ???

∴

[tex]P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g[/tex]   ----- equation (1)

On the other hand for alcohol:

density of alcohol is given as = 790 kg/m³

gravitational constant = 9.8 m/s²

height = 10 cm

∴

[tex]P_{(a)= 790kg/m^3*9.8m/s^2*10[/tex]           ----------- equation (2)

if we equate equation 1 and 2 together; we have

[tex]P_{(g)= P_{(a)[/tex]

[tex]1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm[/tex]

Making [tex]h_g[/tex] the subject of the formula, we have :

[tex]h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}[/tex]

[tex]h_g[/tex] = 6.269 cm

The difference in the height denoted  by ΔH can therefore be calculated as:

ΔH [tex]= H_a-H_g[/tex]

ΔH [tex]= 10cm - 6.269cm[/tex]

ΔH = 3.731 cm

ΔH ≅ 3.73 cm           (to two significant figures)

Answer:

0.2544 moles of water were removed from the sample by the heating process.

Explanation:

Mass of empty evaporating dish = 9.687 g

Mass of evaporating dish and sample of hydrate = 18.407 g

Mass of hydrate  = 18.407 g - 9.687 g = 8.72 g

Mass of the evaporating dish and hydrate after heating = 14.007 g

14.007 g = mass of dish + mass of dehydrate

Mass of dehydrate = 14.007 g - 9.867 g = 4.14 g

Mass of water evaporated = hydrated sample - dehydrated sample

= 8.72 g - 4.14 g = 4.58 g

Moles of water evaporated :

[tex]\frac{4.58 g}{18 g/mol}=0.2544 mol[/tex]

0.2544 moles of water were removed from the sample by the heating process.

The mass of the water removed from the sample by the heating process is determined by subtracting the mass of the evaporating dish and sample after heating from before heating. This mass is then divided by the molar mass of water (18.015 g/mol) to calculate the number of moles of water removed.

To calculate the moles of water removed from the sample by the heating process, first we have to determine the mass of the water removed. This would be equal to the mass of the evaporating dish and sample of hydrate before heating minus the mass after heating, i.e., 18.407 g - 14.007 g = 4.4 g.

Then, we calculate the number of moles, using the molar mass of water as a conversion factor from mass to moles. The molar mass of water is 18.015 g/mol. Therefore, the number of moles of water removed would be 4.4 g / 18.015 g/mol = 0.244 mol.

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Answer:

Paramagnetism is dependent on the unpaired electron in the last orbital . In this regard, Cu(I) chloride is paramagnetic whereas Cu(II) chloride is not.

Explanation:

Paramagnetism is the property of materials/components which makes them attracted them weekly to the magnetic field.

It is related to electronic configuration, such that  it depends on the unpaired electron in the last orbital possess the property.

On basis of this property, Cu(I) chloride is paramagnetic while Cu(II) chloride is non paramagnetic. This is because Cu(I) chloride contains an unpaired electron in the last orbital whereas Cu(II) chloride does not have any unpaired electron.

Answer:

Explanation:

Paramagnetism is a type of magnetism whereby materials are weakly attracted to an externally applied magnetic field and then form internal, induced magnetic fields in the direction of the applied magnetic field. They are attracted to magnetic fields and have magnetic moment induced by the applied field is linear in the field strength. Paramagnetic materials include elements such as Oxygen,

Aluminium etc. and maybe some compounds like FeO etc.

Paramagnetism occurs due to the presence of unpaired electrons in an atom, so atoms with incompletely filled atomic orbitals are paramagnetic, there are exceptions such as copper exist and this is due to their spin, unpaired electrons have a magnetic dipole moment and act like tiny magnets. They have a magnetic permeability slightly greater than 1. External magnetic field causes the electrons spin to align parallel to the field hence, causing a net attraction. Paramagnetic materials include aluminium, oxygen, titanium, and iron oxide (FeO).

From the example,

Cu(I) and Cu(II)

Electronic configuration

Cu(I) - [Ar] 3d10

Cu(II) - [Ar] 3d9

[Ar] - 1s2 2s2 2p6 3s2 3p6 4s2

Therefore, Cu(I) is Paramagnetic while Cu(II) is not Paramagnetic (diamagnetic).

Answer:

The bond angle between the p orbital is assumed to be 90 degrees But according to VSEPR theory, as molecular geometry is a little bit bent thus the angle is less than 109.5 degrees.

Explanation:

IN hydrogen sulfide, H_2 S, both hydrogen atoms consist of para-magnetic 1s orbital while sulfur consists of diamagnetic 3s and 2 para magnetic 3p orbital.

The bond angle between the p orbital is assumed to be 90 degrees But according to VSEPR theory, as molecular geometry is a little bit bent thus the angle is less than 109.5 degrees.

In hydrogen sulfide (H2S), the bonding is explained using valence bond theory. The bond angle in H2S is approximately 92 degrees.

In hydrogen sulfide (H2S), the bonding can be explained using valence bond theory. According to this theory, covalent bonds are formed by the overlap of atomic orbitals. In H2S, the sulfur atom forms a covalent bond with each hydrogen atom through overlap of its sp3 hybrid orbital and the hydrogen 1s orbital.

The predicted bond angle in H2S is approximately 92 degrees. This angle is less than the ideal tetrahedral angle of 109.5 degrees due to the presence of two lone pairs of electrons on the sulfur atom, which exert greater repulsion than the bonding pairs. This results in a distorted tetrahedral shape with a smaller bond angle.

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Answer:

Explanation:

The structural repeating unit  of beta sheet is 7 anstrom/2 aminoacids. So,

  [tex]\frac{80.5 Angstrom}{3.5} = 23 aminoacids[/tex]

If a segment of a single chain  in an antiparallel beta sheet has a length of 80.5 angstrom, then there will be 23 residues in this segment.

To find the number of residues in an 80.5 Å long segment of an antiparallel beta-pleated sheet, divide the total length by the length of one residue (3.5 Å per residue), which gives approximately 23 residues.

The student asked how many residues are in a segment of a single chain in an antiparallel beta-pleated sheet with a length of 80.5 Å. To determine the number of amino acid residues in the segment, we can use the typical amino acid residue length in a ß-pleated sheet, which is approximately 3.5 Å per residue in an extended conformation. This measurement considers the distance hydrogen bonds can span between the carbonyl oxygen and amino hydrogen along the peptide chain in the secondary structure.

By dividing the total length of the chain (80.5 Å) by the length of one residue (3.5 Å), you can calculate the number of residues:

    Number of residues = Total length ÷ Length per residue

    Number of residues = 80.5 Å ÷ 3.5 Å/residue

    Number of residues ≈ 23 residues

This calculation does not account for slight variations that may occur in different proteins or specific contexts, but it provides a general estimate for the number of amino acids in a segment of a beta sheet.

Explanation:

Molecular formula of the compound =

The carbon chain in the molecule will look like :

1. [tex]H_2C=CH-CH_2-CH_2-CH_2-OH[/tex]

2. [tex]H_3C-CH=CH-CH_2-CH_2-OH[/tex]

3. [tex]H_3C-CH-CH=CH-CH_2-OH[/tex]

4. [tex]H_3C-CH_2-CH_2-CH=CH-OH[/tex]

No branching is present, so that means the valency of the carbon will be fulfilled by 10 hydrogen atoms and 1 oxygen atom

Answer:

Boron (B) is the element whose IE matches with our data.

Electronic Configuration of boron: [tex]1s^22s^22p^1[/tex]

Explanation:

Ionization Energy (IE):

It is the minimum amount of energy which is required to remove the lose electron. If the electron is closer to the nucleus then greater amount of energy is required to remove the electron.

If we look from left to right in a period, ionization energy increases due stability of valance shell.

From the data given to us:

IE₁ = 801

IE₂ = 2427

IE₃ = 3659

IE₄ = 25,022

IE₅ = 32,822

Boron (B) is the element whose IE matches with our data.

Electronic Configuration of boron: [tex]1s^22s^22p^1[/tex]

Boron has 5 electrons (3 in valance shell) that's why it has 5 Ionization Energies.

Answer:

The volume of water measured is 10mL

Explanation:

Given;

Mass of mass of beaker and the water = 23.670 g

Mass of empty beaker = 13.712 g

Then, mass of water only = Total mass of of beaker and the water minus Mass of empty beaker

mass of water only = 23.670 g - 13.712 g = 9.958 g

Density = mass/volume

Given density of water = 0.9982071 g/mL

Density of water = Mass of water/ Volume of water

Then, Volume of water =  Mass of water/Density of water

Volume of water = 9.958 g/0.9982071 g/mL

Volume of water = 9.975886 mL ≅ 10mL

Therefore, The volume of water measured is 10mL

The reaction of 19.3 g of aluminum with 63.2 g of ferric oxide is exothermic, releasing 2395.25 kJ of heat. Aluminum is the limiting reagent in the reaction.

The limiting reagent is the reactant that is used up completely in the reaction. To determine the limiting reagent, compare the ratio of the moles of each reactant to the stoichiometric coefficients in the balanced chemical equation. The reactant with the smaller ratio is the limiting reagent.

1: Calculate the number of moles of each reactant.

Moles of aluminum = 19.3 g / 27 g/mol = 0.715 mol

Moles of ferric oxide = 63.2 g / 231.5 g/mol = 0.272 mol

2: Determine the limiting reagent.

The ratio of moles of aluminum to moles of ferric oxide is 0.715 mol / 0.272 mol = 2.63. The stoichiometric ratio of aluminum to ferric oxide in the balanced chemical equation is 8:3. Since 2.63 is less than 8/3, aluminum is the limiting reagent.

3: Calculate the amount of heat released.

q = -3350 kJ/mol * 0.715 mol = -2395.25 kJ

4: Interpret the answer.

The amount of heat released is -2395.25 kJ. This means that the reaction releases 2395.25 kJ of heat.

The amount of heat released by the reaction of 19.3 g of Al with 63.2 g of Fe_3O_4 is -2395.25 kJ.

Answer:

We can conclude that the Rf of that compound has a ratio of 4. It means that the solute has a ratio value of 4 times than that of solvent. As we can see that it has traveled 4 cm , this data is useful in determination of the compound in a mixture when compared with Rf values of other compounds.

Answer: 1.07×10^-20microlitre

Explanation:

1cm3 = 1000microlitres

1.07×10^-23 cm3 of tungsten = 1.07×10^-23 x 1000 = 1.07×10^-20microlitre

The volume of a single tungsten atom in microliters is 1.07x10^-17 µL. This is found by multiplying the given volume in cubic centimeters by the conversion factor of 1,000,000 µL/cm³.

The volume of a single tungsten atom is 1.07×10-23 cm3. One cubic centimeter (cm3) is equal to 1,000,000 microliters (µL). To convert the volume from cubic centimeters to microliters, we need to multiply the original value by the conversion factor. Therefore, the volume of a tungsten atom in microliters will be 1.07×10-23 cm3 * 1,000,000 µL/cm3, which equals to 1.07×10-17 µL. Hence, the volume of a tungsten atom in microliters is 1.07×10-17 µL.

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Answer:

First Method: Vacuum Distillation and Chromatographic separation of the remains that were precipitated out from the peel.

Second Method: Extraction of components from orange peels by help of precipitation procedures that are mostly done In Situ. Those components can be recovered using the saponification process. Then these are examined under UV light spectroscopy. Now, the existence and extent of carotenoids can be determined by checking the levels of anti-oxidants.

0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.

If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?

A. less than 20∘C

B. 20∘C

C. greater than 20∘C

Answer:

Option C is correct

Explanation:

Increase in the mass of aluminium block would increase the heat capacity of block isothermaly before immersed in water at 0° so heat available for transfer is higher so equilibrium temperature of system would increase.

Answer:

Melting of snow

Evaporation of water from desk

Explanation:

Processes that increase the entropy of the universe are those processes that have an increased disorderliness. We should note that there are three principal states of matter which are the liquid, gas and solid. The gaseous state is the most disorderly while the solid is the least disorderly.

Now. We can see that the cooling of a hot cup of coffee is a process that needs or leads to a loss in temperature which obviously decreases disorderliness of the universe.

The melting of snow however is a process that leads to an increase in the disorderliness of the universe. It entails moving from the solid state to the liquid state. It tends to move to a more disordered state indicating an increase in the entropy of the universe.

The evaporation of water from the desk is quite similar to that above. Hence since we are moving from the liquid to the gaseous state via evaporation, we can state that the entropy of the universe has increased since we have moved from a state with a lesser degree of disorderliness to a state that is more disordered I.e from liquid to gaseous state.

The processes that result in the increase in entropy of the universe would be the melting of snow above 0 °C and the evaporation of water from a desk at room temperature.

Entropy refers to the degree of disorderliness of a system. The degree of disorderliness increases when the temperature of a system increases or when there is a phase change in the order solid > liquid > gas.

Thus, when snow melts, a phase change occurs from solid to liquid leading to an increase in entropy. Also. when water evaporates from a desk, it changes phase from liquid to gas and thereby leads to an increase in entropy.

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Answer:

n = 3 for similar blue light

Explanation:

The principle applied here is energy levels and energy changes. There are different energy levels depending on the value of the integer as explained by Max planck - a german physicist in 1900, Max planck claimed that electrons in an atom were presumed to be oscillating with a frequency f, then there enrrgy will be given by the plancks equation ; E =hf, where h is the plancks constant.

In general energy of each level can be written as E =nhf

In the given gases, HF should have the largest van der Waals constant a because of its stronger intermolecular forces, and Cl2 should have the largest van der Waals constant b due to its larger molecular size.

The van der Waals constants (a) and (b) are indicative of the strength of the forces between molecules, and the physical size of the molecules in a given gas, respectively.

(a) In the selection of H2, HF, F2, we would expect the molecule with the strongest intermolecular forces to have the largest van der Waals constant a. That would be HF because when comparing these gases, HF has permanent dipole-dipole interaction which is stronger than the London dispersion forces in H2 and F2.

(b) For the gases H2, HCl, Cl2, we would expect the molecule with the largest physical size to have the largest van der Waals constant b. In this case, Cl2 is the largest molecule and thus would have the largest van der Waals constant b under normal conditions.

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Answer:

73140 kJ.

Explanation:

Equation of the hydrogenation reaction:

C2H4 + H2 --> C2H6

Molar mass = (2*12) + (6*1)

= 30 g/mol

Mass of C2H4 = 15.9 kg

= 15900 g

Number of moles = mass/molar mass

= 15900/30

= 530 mol of C2H6

By stoichiometry, 1 mole of C2H4 is hydrogenated to give 1 mole of C2H6

Number of moles of C2H4 = 530 mole

Q = n * q

= 530 * 138 kJ

= 73140 kJ.

Final answer:

To calculate the total heat released when 15.9 kg of ethane (C₂H₆) is formed, we convert the mass of C₂H₆ to moles and multiply by the heat released per mole of ethene (C₂H₄). The total heat released is 73015 kJ.

Explanation:

The student is asking about the heat released during a hydrogenation reaction where ethene (C₂H₄) and hydrogen gas (H2) are converted into ethane (C₂H₆). The reaction is exothermic, meaning that heat is released when it occurs. Given that 138 kJ is released per mole of C₂H₄ reacting, to find the total heat released when 15.9 kg of C₂H₆ is formed, we need to follow these steps:

Here's the calculation:

Therefore, the total heat released when 15.9 kg of ethane is formed is 73015 kJ (to zero decimal places).

Answer:

The relation between the shielding and effective nuclear charge is given as

[tex]Z_{eff} = Z -S[/tex]

where s denote shielding

z_{eff} denote effective nuclear charge

Z - atomic number

Explanation:

shielding is referred to as the repulsion of an outermost electron to the pull of electron from valence shell.  Higher the electron in valence shell higher will be the shielding effects.  

Effective nuclear charge is the amount of net positive charge that valence electron has.

The relation between the shielding and the effective nuclear charge is given as  

[tex]Z_{eff} = Z -S[/tex]

wheres denote shielding

z_{eff} denote effective nuclear charge  

Z - atomic number

The maximum wavelength of light that can break a carbon-carbon double bond by absorbing a single photon is calculated to be 1940 nm. This is done by converting the energy required to break the bond to J/particle, and then using this to find the wavelength using the equation E=h*c/λ.

To find the maximum wavelength of light that can break a carbon-carbon double bond by absorbing a single photon, it is necessary to convert the energy required to break the bond from kJ/mol to energy per photon and then use that to calculate the wavelength. Using the relation between energy and wavelength given by the formula E=h×c/λ, where E is Energy, h is Planck's constant (6.626 x 10⁻³⁴ Js), c is the speed of light and λ is the wavelength.

First, convert the energy required to break the bond to J/particle by converting kJ to J (1kJ = 1000J) and then dividing by Avogadro's number (6.022 x 10²³). Thus, E = 614.4 kJ/mol × 1000 J/kJ / 6.022 x 10²³ particles/mol = 1.02 x 10⁻¹⁹ J/particle.

Then, rearrange the formula to solve for λ. We get λ = h × c/E. Substituting values (h = 6.626 x 10⁻³⁴ Js, c = 3.00 x 10⁸ m/s, and E = 1.02 x 10⁻¹⁹ J) gives λ = 1.94 x 10⁻⁶ meters or 1940 nm.

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The maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon is 341 nm.

To calculate the maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon, we need to use the formula E = hc/λ, where E is the energy of the photons, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength of light in meters.

First, we need to convert the bond energy from kJ/mol to J/molecule by multiplying it by Avogadro's number (6.02 x 10^23). So, the bond energy is 614 x 10^3 J/mol. Next, we can rearrange the formula to solve for λ:

λ = hc/E = (6.63 x 10^-34 J·s)(3.0 x 10^8 m/s)/(614 x 10^3 J/mol)

Calculating this expression gives us a value of λ = 3.41 x 10^-7 m, or 341 nm. Therefore, the maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon is 341 nm.

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Explanation:

Equation of the reaction:

H2(g) + F2(g) --> 2HF(aq)

1/2H2(g) + 1/2F2(g) --> HF(aq)

D(H-H) in H2 = 436 kJ/mol

D(F-F) in F2 = 158kJ/mol

ΔH bond breakage (dissociation):

1/2 mol H-H bonds = (1/2 X 436) kJ

= 218 kJ

1/2 mol F-F bonds = (1/2 X 158) kJ = = 80 kJ

Total = 218 + 80 = 298 kJ

ΔH bond formation:

1 mol H-F bonds = - 570 kJ

= DHreactant - DHproduct

ΔH°f = 298 kJ + -570 kJ = -272 kJ

Answer:

No. of Photons (red) = 7.05 No. of Photons

No. of Photons (blue) = 4.78 No. of Photons

Explanation:

v (red) = 3 x 108/700 x 10-9 = 4.28 x 10 power 15 Hz

v (blue) = 3 x 108/475 x 10-9 = 6.31 x 10 power 15 Hz

(a) No. of Photons (red) = E/hv = 2.0 x 10 power -17 / 6.626 x 10 power -34 x 4.28 x 10 power 15

No. of Photons (red) = 7.05 No. of Photons

(b) No. of Photons (blue) = E/hv = 2.0 x 10 power -17 / 6.626 x 10 power -34 x 6.31 x 10 power 15

No. of Photons (blue) = 4.78 No. of Photons

Answer:

a) f = (1.24 × 10^15) Hz and E = (8.214 × 10^-19) J

b) f = (1.36 × 10^15) Hz; E = (9.035 × 10^-19) J

Explanation:

a) The least energetic photons have the highest wavelength. That is, the wavelength of the least energetic photons is equal to the upperlimit of the wavelength inequality given.

λ = 242nm = 2.42 × 10⁻7 m

v = fλ; f = v/λ; v = 3×10^8 m/s

f = (3×10^8)/(2.42×10^-7)

f = (1.24 × 10^15) Hz

E = hf; h = planck's constant = (6.62607004 × 10^-34) Js

E = 6.626 × 10^-34 × 1.24 × 10^15

E = (8.214 × 10^-19) J

b) The photons with the least wavelength in the range provided are the most energetic ones.

λ = (2200 × 10^-10) m = (2.2 × 10^-7) m

v = fλ; f = v/λ; v = 3×10^8 m/s

f = (3×10^8)/(2.2×10^-7)

f = (1.36 × 10^15) Hz

E = hf; h = planck's constant = (6.62607004 × 10^-34) Js

E = 6.626 × 10^-34 × 1.36 × 10^15

E = (9.035 × 10^-19) J

QED!

In ozone formation, least energetic photons having wavelength 242 nm have frequency ≈ 1.24 x 10^15 Hz and energy ≈ 8.2 x 10^-19 J. The most energetic photons that ozone absorbs, with a wavelength of 2200 Å, have a frequency of ≈ 1.36 x 10^15 Hz and energy of ≈ 9.02 x 10^-19 Joules.

(a) The frequency (ν) of a photon is given by the formula: ν = c / λ, where c is the speed of light (3.00 x 10^8 m/s) and λ is wavelength. For the least energetic photons (wavelength of 242 nm), we would convert the wavelength to meters (242 nm = 242 x 10^-9 m). Applying the formula: ν = 3.00 x 10^8 m/s / 242 x 10^-9 m, we get ν ≈ 1.24 x 10^15  Hz.

The energy (E) of a photon is given by the formula: E = hν, where h is Planck’s constant (6.63 x 10^-34 Js). So, E = 6.63 x 10^-34 Js x 1.24 x 10^15 Hz, which gives us E ≈ 8.2 x 10^-19 Joules.

(b) For the most energetic of these photons, they have the shortest wavelength (2200 Å = 2200 x 10^-10 m). Using similar calculations as above: ν ≈ 1.36 x 10^15 Hz and E ≈ 9.02 x 10^-19 Joules.

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Answer:

50.0 μL

Explanation:

When a dilution is done, the mass of the solute (in this case the ampicillin) remains constant, following the Lavoiser's law that the mass is conserved. The mass is the concentration (C) multiplied by the volume (V), so if 1 is the stock solution, and 2 is the bacterial culture after the addition of the antibiotic:

m1 = m2

C1*V1 = C2*V2

C1 = 100 mg/mL = 100000 μg/mL (1 mg = 1,000μg)

C2 = 100 μg/mL

V2 = 50 mL + V1 = 50000μL + V1 (V1 in μL)

100000*V1 = 100*(50000 + V1)

1000V1 = 50000 + V1

999V1 = 50000

V1 = 50.0 μL

Final answer:

The first four lines of the Balmer series involve electron transitions from 3p to 2s, 4p to 2s, 5p to 2s, and 6p to 2s orbitals.

Explanation:

The Balmer series involves electron transitions from higher energy levels to the second principal energy level (n=2), producing visible spectral lines. For the first four lines of the visible emission spectrum in the Balmer series, the allowed transitions must follow the selection rule Δl = ±1. Therefore, these transitions can only start from orbitals with l=1, which are the p orbitals.

For nfinal = 2, the corresponding ni initial energy levels for the first four visible emission lines are:

Answer: (a) seven orbitals, (b). 3 orbitals, (c). 5 orbitals and (d). 4 orbitals.

Explanation:

In order to solve this question we need to know how to explain the behaviour of electrons in atoms,and what we need to know is what is called the quantum numbers. There are four different kinds of quantum numbers and they are;

(1). Principal quantum numbers: the principal quantum number is denoted by the letter 'n'. It is used to describe the orbitals' energy. It has the values of n=1,2,3,4,...

(2). The spin quantum numbers: the spin quantum numbers is denoted by m(s). The 's' in the parenthesis is in subscript. It has the values of +1/2 and -1/2.

(3). Azimuthal quantum numbers: this is denoted by ℓ and it is used to explain orbital angular momentum and orbital shape. It has the values of ℓ= 0,1,2,3,....n-1.

Note that => ℓ = 0; we have a s-subshell,sphere shape.

ℓ = 1; p-subshell, dumb bell shape.

ℓ=2; d- subshell, double dumb bell shape.

ℓ= 3; f - subshell, multiple lobes.

(4). Magnetic quantum number: it is denoted by m(l) where the 'l' in the parenthesis is in subscript.

===> NOTE: there are (2ℓ + 1 ) orbitals in a subshell, also, there are n^2 number of orbitals in a shell.

Having known all that above, let us jump right in to the solution.

(a). From above we can see that; there are (2ℓ + 1 ) orbitals' in a subshell, also, f= ℓ= 3.

Therefore, the number of orbitals in 5f = 2 ℓ + 1 = (2×3) + 1 = 6+1 = 7 orbitals for 5f.

(b). 4p, the numbers of orbitals in 4p is; p= ℓ= 1=> 4 ℓ + 1 = (2×1) + 1 = 2+1 = 3 orbitals for 4p.

(c). 5d, the numbers of orbitals' in 5d is; d= ℓ= 2 = (2×2) + 1 = 4 + 1 = 5 orbitals for 5d.

(d). For n= 2, the numbers of orbital is ; n^2. Where the n given is 2. Therefore, 2^2= 2×2 = 4 orbitals in n=2.

An orbital refers to a region in space where electrons can be found.

An orbital refers to a region in space where there is a high probability of finding an electron. Orbitals that posses the same amount of energy are called degenerate orbitals.

The number of orbitals in an atom that can have the following designations are shown below;

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It will cost the soccer club $68,618.50 to add the grass turf to their field for the international soccer match.

Given: Length = 0.102 km = 0.102 km  × 1000 m/km = 102 m

Width = 0.069 km = 0.069 km  × 1000 m/km = 69 m

The area of the field: Area = Length × Width

                                    Area = 102 m × 69 m = 7038 sq. meters

The area by the cost of the grass turf per square meter:

Cost = Area × Cost per square meter

Cost = 7038 sq. meters  ×$9.75/sq. meter = $68,618.50

Therefore, it will cost the soccer club $68,618.50 to add the grass turf to their field for the international soccer match.

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Answer:

Reaction I is a COMBINATION REACTION - A reaction that involves the mixing of two or more elements to form a single product.

Reaction II is COMBUSTION REACTION - A reaction that involves Oxygen to produce carbon(iv)oxide and water vapor

Reaction III is DOUBLE DISPLACEMENT REACTION - A reaction that involves the exchange of radicals.

Reaction IV is a COMBUSTION REACTION

Explanation:

Reaction I is a COMBINATION REACTION - A reaction that involves the mixing of two or more elements to form a single product.

Reaction II is COMBUSTION REACTION - A reaction that involves Oxygen to produce carbon(iv)oxide and water vapor

Reaction III is DOUBLE DISPLACEMENT REACTION - A reaction that involves the exchange of radicals.

Reaction IV is a COMBUSTION REACTION

Attached is the reactions I - 1V

The given chemical equation CH3CH2OH(l) + 3O2(g) represents a combustion reaction where CH3CH2OH reacts with oxygen to produce carbon dioxide and water.

Based on the given chemical equation, CH3CH2OH(l) + 3O2(g), the reaction is a combustion reaction. Combustion reactions involve the rapid combination of a fuel (in this case, CH3CH2OH) with oxygen (O2) to produce heat, light, and new products. In a combustion reaction, a fuel is oxidized and reacts with oxygen to form carbon dioxide and water.

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Answer : The specify the l and ml values for n = 4 are:

At l = 0,  [tex]m_l=0[/tex]

At l = 1,  [tex]m_l=+1,0,-1[/tex]

At l = 2,  [tex]m_l=+2,+1,0,-1,-2[/tex]

At l = 3,  [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]

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 m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.

Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.

As we are given, n = 4 then the value of l and ml are,

l = 0, 1, 2, 3

At l = 0,  [tex]m_l=0[/tex]

At l = 1,  [tex]m_l=+1,0,-1[/tex]

At l = 2,  [tex]m_l=+2,+1,0,-1,-2[/tex]

At l = 3,  [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]

Answer: The abundance of Li-7 isotope is higher as compared to Li-6.

Explanation:

Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]

We are given:

Two isotopes of lithium :

Li-6 and Li-7

Average atomic mass of lithium= 6.941

As, the average atomic mass of lithium is closer to the mass of isotope Li-7. This means that the relative abundance of Li-7 is higher as compared to Li-6.

Percentage abundance of Li-7> Percentage abundance of Li-6 isotope

The atomic weight of lithium is closer to the mass number of 7Li, indicating that 7Li is more abundant than 6Li in nature. Thus, the correct answer is A) The abundance of 7Li is greater than 6Li.

The atomic weight of lithium, being 6.941, is closer to 7 than 6. Consequently, this indicates that most naturally occurring lithium is of the heavier 7Li isotope. So, in terms of relative abundance, 7Li is indeed more prevalent than 6Li. This means the correct answer would be option A) The abundance of 7Li is greater than 6Li. Therefore, based on the atomic weight of lithium, we can conclude that the abundance of 7Li is greater than 6Li.

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Answer:

a) The element is Manganese (Mn)

b) The element is Zirconium (Zr)

Explanation:

The step by step analysis and explanation is as shown in the attachment