Construct a simulated proton-decoupled 13C NMR spectrum for tert-butyl alcohol. Drag the resonance signal icon to the appropriate chemical shift positions. Then label the box above each resonance signal with the corresponding number of equivalent carbons. (Not all chemical shift bins will be used.)

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

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

Final answer:

The 1 g equivalents using the given prefixes are: mega−: 1 Mg, kilo−: 1 kg, deci−: 1 dg, centi−: 1 cg, milli−: 1 mg, micro−: 1 μg, nano−: 1 ng, pico−: 1 pg.

Explanation:

The 1 g equivalents using the given prefixes are as follows:

(a) mega−: 1 Mg (megagram)
(b) kilo−: 1 kg (kilogram)
(c) deci−: 1 dg (decigram)
(d) centi−: 1 cg (centigram)
(e) milli−: 1 mg (milligram)
(f) micro−: 1 μg (microgram)
(g) nano−: 1 ng (nanogram)
(h) pico−: 1 pg (picogram)

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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Answer: The molality of perchloric acid in the solution is 14.95 m

Explanation:

We are given:

Molarity of perchloric acid solution = 9.2 M

This means that 9.2 moles of perchloric acid are contained in 1 L or 1000 mL of solution

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

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

Moles of perchloric acid = 9.2 moles

Molar mass of perchloric acid = 100.5 g/mol

Putting values in above equation, we get:

[tex]9.2mol=\frac{\text{Mass of perchloric acid}}{100.5g/mol}\\\\\text{Mass of perchloric acid}=(9.2mol\times 100.5g/mol)=924.6g[/tex]

To calculate mass of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of solution = 1.54 g/mL

Volume of solution = 1 L = 1000 mL

Putting values in above equation, we get:

[tex]1.54 g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.54g/mL\times 1000mL)=1540g[/tex]

We are given:

Mass of solute (perchloric acid) = 924.6 grams

Mass of solution = 1540 grams

Mass of solvent = Mass of solution - mass of solute = [1540 - 924.6] g = 615.4 g

To calculate the molality of solution, we use the equation:

[tex]\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute (perchloric acid) = 924.6 g

[tex]M_{solute}[/tex] = Molar mass of solute (perchloric acid) = 100.5 g/mol

[tex]W_{solvent}[/tex] = Mass of solvent = 615.4 g

Putting values in above equation, we get:

[tex]\text{Molality of perchloric acid}=\frac{924.6\times 1000}{100.5\times 615.4}\\\\\text{Molality of perchloric acid}=14.95m[/tex]

Hence, the molality of perchloric acid in the solution is 14.95 m

Considering the definition of molality, the molality of a 9.2 M solution of perchloric acid is  14.95 [tex]\frac{moles}{kg}[/tex].

Molality is a measure of concentration that is defined as the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]

Molality is expressed in units [tex]\frac{moles}{kg}[/tex].

Being the molarity is the number of moles of solute that are dissolved in a certain volume a molarity of 9.2 M means that 9.2 moles of perchloric acid are contained in 1 L or 1000 mL of solution.

Then, the number of moles of solute is 9.2 moles of perchloric acid.

In first place, density is the ratio of the weight (mass) of a substance to the volume it occupies. So, a density of 1.54 [tex]\frac{g}{mL}[/tex]means that you have 1.54 grams per 1 mL of solution or 1540 grams per 1 L (1000 mL) of solution. So, the mass of solution is 1540 grams in 1 L of solution

The mass of solution is the sum between the mass of solute and the mass of solvent.

Being the moles of perchloric acid 9.2 moles and the molar mass of perchloric acid 100.5 [tex]\frac{g}{mole}[/tex], the mass of the solute perchloric acid is calculated as:

mass of solute= number of moles of solute× molar mass

mass of solute= 9.2 moles× 100.5 [tex]\frac{g}{mole}[/tex]

mass of solute= 924.6 grams

Considering 1 L of solution, then, the mass of solvent is calculated as:

mass of solution= mass of solute +the mass of solvent

1540 grams = 924.6 grams +the mass of solvent

1540 grams - 924.6 grams= mass of solvent

615.4 grams= 0.6154 kg= mass of solvent

Finally, the mass of solvent is 0.6154 kg.

So, being the number of moles of solute 9.2 moles and mass of solvent 0.6154 kg, the molality is calculated as:

[tex]Molality=\frac{9.2 moles}{0.6154 kg}[/tex]

Molality= 14.95 [tex]\frac{moles}{kg}[/tex]

Finally, the molality of a 9.2 M solution of perchloric acid is  14.95 [tex]\frac{moles}{kg}[/tex].

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Answer:

The molarity of the new solution is 0.72 M

Explanation:

Step 1: Data given

Volume of the original solution = 360 mL =.360 L

Molarity = 0.87 M

We add 75 mL = 0.075 L

Step 2: Calculate moles

Moles = molarity * volume

Moles = 0.87 M * 0.360 L

Moles = 0.3132 moles

Step 3: Calculate new molarity

The number of moles stays constant

Molarity = moles / volume

Molarity = 0.3132 moles / (0.36+0.075)

Molarity = 0.3132 moles / 0.435 L

Molarity = 0.72 M

The molarity of the new solution is 0.72 M

To determine the molarity of the new solution, calculate the amount of acetic acid remaining after dilution and then use the formula Molarity = Moles / Volume.

To determine the molarity of the new solution, we need to calculate the amount of acetic acid that remains in the solution after adding water. The number of moles of acetic acid before dilution can be determined using the formula:

Moles = Concentration x Volume

Next, we need to calculate the final volume of the solution by adding the volume of water (75 mL) to the initial volume of the acetic acid solution (360 mL). Using this final volume and the number of moles of acetic acid, we can determine the molarity of the new solution using the formula:

Molarity = Moles / Volume

Substituting the values into the equation, the molarity can be calculated.

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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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To find the molality of the silver nitrate solution, we can use the mass percent and density of the solution. The molality is found to be 2.99 mol/kg.

To find the molality of the silver nitrate solution, we need to first determine the moles of silver nitrate in the solution. We can use the mass percent and density to do this.

First, let's assume we have 100 g of the solution.

This means we have 36 g of silver nitrate in the solution (since it is 36% by mass).

The molar mass of AgNO3 is 169.88 g/mol, so we can convert the mass of the silver nitrate to moles: 36 g / 169.88 g/mol = 0.2124 mol.

The density of the solution is given as 1.41 g/mL, so 100 g of the solution would have a volume of 100 g / 1.41 g/mL = 70.92 mL.

Now, we can calculate the molality using the moles of AgNO3 and the mass of the water:

Molality = moles of solute / mass of solvent (in kg)

Molality = 0.2124 mol / 0.07092 kg

= 2.99 mol/kg

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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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:

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

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:

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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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.

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:

5650 g/mol

Explanation:

The osmotic pressure (π) is the pressure needed to prevent that the osmose occur in a system. Osmose is the process that a solvent goes through a membrane where the solution is more concentrated.

This property can be calculated by:

π = M*R*T

Where M is the molarity of the solution (mol/L), R is the ideal gas gas constant (62.3 torr.L/mol.K), and T is the temperature (25°C = 298 K), so:

23 = M*62.3*298

M = 1.24x10⁻³ mol/L

So, if the concentration is mass is 7.0 g/L, the molar mass (MM) of the sugar is its concentration is mass divided by the molarity:

MM = 7/1.24x10⁻³

MM = 5650 g/mol

5650 g/mole is the molar mass of the given sugar, whose osmotic pressure of solution is is the molar mass of this sugar.

It is that pressure which is applied on the solution to stop the flow of pure solvent from low concentration to high concentration through semipermeable membrane, and it is calculated as follow:

π = CRT, where

π = osmotic pressure = 23 torr (given)

C= concentration = to find?

R =  ideal gas gas constant = 62.3 torr.L/mol.K

T = temperature = 25 degree C = 298 K

Putting all these value in the above equation, we get

23 = C × 62.3 × 298

C = 1.24x10⁻³ mol/L

Here the concentration is given in the form of molarity, which defines as no. of moles of solute present in per liter of solvent. And no. of moles is calculated as :

n = W/M, where

W = given mass = 7.0 g (given)

M = molar mass = to find?

n = no. of moles = 1.24x10⁻³ mol/L (calculated above)

Putting all values in above equation we get,

M = 7/1.24x10⁻³ = 5650 g/mol.

Hence, 5650 g/mol is the molar mass of this given sugar.

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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:

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:

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

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:

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

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.

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:

To synthesize cis-2-methylcyclopentyl from methycyclopentanol, you need to replace the acetate hydroxyl group with acetate by inverting the configuration.  

Explanation:

To understand the process, you need to understand the nucleophilic mechanism taking place in the process. This is the first stage of the process. Hydroxide is a poor leaving group, to it must be converted to a good leaving group. To effect the change, it is necessary to use p-toluenesuphate.

p-toluenesuphate is favored because this can be prepared by a reaction that alters none of the bonds attached to the stereogenic center.

The reaction of p-toluensulfonate with potassium acetate in acetic acid effects the conversion to give the final product: cis-2-methylcyclopentyl.  

Final answer:

The question involves synthesizing cis-2-methylcyclopentyl acetate from trans-2-methylcyclopentanol, requiring a reaction sequence that includes the use of a tosylate intermediate and control of the stereochemistry to retain the cis configuration.

Explanation:

The question asks how to synthesize cis-2-methylcyclopentyl acetate from trans-2-methylcyclopentanol. To achieve the transformation from trans to cis stereochemistry, one might need to carry out a series of reactions, including esterification and the use of a tosylate intermediate to facilitate inversion of stereocenter or through other stereospecific mechanisms. An appropriate synthetic route could potentially involve the protection of hydroxyl group, formation of the tosylate, followed by a nucleophilic substitution that inverts the stereochemistry, and concluding with the esterification to yield the desired acetate.

It is important to note that the 'tosylate intermediate' is a common intermediate in organic synthesis allowing for an alcohol to be converted into a better leaving group for subsequent substitution reactions. Stereoselectivity is crucial as the goal is to preserve the cis stereochemistry in the final product.

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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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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Answer:

Explanation:

PFR - Plug Flow Rate

The steps and detailed integration and appropriate substitution is as shown in the attached file.

The fractional conversion of chemical A in the PFR under the given conditions is approximately 1, indicating that the conversion is almost complete.

The question pertains to the calculation of the fractional conversion of a reactant, A, in a Plug Flow Reactor (PFR). According to the given conditions, we know that the rate equation is -rA = kCa and the value of k is 45 h⁻¹. Given that the volume of the reactor is 1000 L and the molar flowrate of A is 500 mol/h, we can calculate the residence time τ as V/F = 1000 L / 500 mol/h = 2h.

For a first order irreversible reaction in a PFR, the conversion, X, is given by X = 1 - exp(-k*τ). Plugging in the given value of k and the calculated value of τ, we find that X = 1 - exp(-45 h⁻¹ * 2 h) which gives X approximately equal to 1. Hence, the fractional conversion of A is almost complete.

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Answer:

16.5 g

Explanation:

When a nonvolatile compound is dissolved in a pure solvent, the freezing point of the solvent is reduced, because the interaction solvent-solute requires more energy to be joined, and so, it freezes. This property is called cryoscopy, and the temperature change (ΔT) can be calculated by:

ΔT = Kc*W*i

Where Kc is the cryoscopy constant of the solute X, W is the molality of the solution, and i is the van't Hoff factor, which determines the percent of the solute that is dissolved. For organic molecules, such as alanine, i = 1.

The molality is the number of moles of the solute divided by the mass (in kg) of the solvent (1200 g = 1.2 kg). The molar mass of alanine is 89.09 g/mol, and the number of moles of it is the mass divided by the molar mass:

n = 45.8/89.09

n = 0.5141 mol

W = 0.5141/1.2 = 0.4284 mol/kg

So, Kc of X is:

4.10 = Kc*0.4284*1

Kc = 9.57 °C.kg/mol

So, if now sodium chloride is added to X, and the variation temperature is the same, and i = 1.82:

4.10 = 9.57*W*1.82

W = 0.2354 mol/kg

The number of moles of the solute is then:

W = n/1.2

0.2354 = n/1.2

n = 0.2825 mol

The molar mass of sodium chloride is 58.44 g/mol, thus the mass is the molar mass multiplied by the number of moles:

m = 58.44*0.2825

m = 16.5 g

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:

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.