Compare Newton's law of viscosity and Hooke's law of elasticity. What is the origin of these "laws"?

Explanation:

- Simple Distillation: its a separation method that can be used when the two or more liquids in the mix have at least 50 degrees of difference between their boiling points.

-Azeotropic distillation: is a technique to break an azeotrope (constant boiling point mixtures), that can't be separated by simple distillation, by adding another component to generate a new azeotrope (between one initial component and the new one added) with lower boiling point.

-Extractive distillation: is a process to separate mixtures with close boiling points by adding a miscible, high boiling or none volatile solvent to increase the relative volatility of the liquids in the mix, this increases the separation factor. It differences from the azeotropic method because it doesn't form an azeotrope.

-Liquid-liquid extraction: is a method to separate compounds based on their relative solubilities in two different immiscible liquids.

After describing all the methods we can conclude that all of them are methods to separate substances based on their physical properties, this is their similarity. The difference between this method is the property it uses to separate (solubility in the case of extraction and boiling point in the case of destinations), the cases in which they bare used (when the liquids difference in boiling points is bigger [simple] or close [attractive and azeotropic]) and the formation of azeotropes (present in azeotropic and absent in extractive).

I hope you find this information useful and interesting! Good luck!

Answer:

The glycosylation reaction or glycoside formation is an organic reaction in which the hemiacetal group of cyclists ketoses or aldoses turns into acetals, named glycosides. Reaction in the attached picture.

Explanation:

Carbohydrates can be found in an open-chain form or a cyclic form. For the second one, the carbonyl group of the aldehyde could react with the alcohol group of the molecule to form the cycle. As shown in the attached picture, the alcohol group of this cyclic form could react with an alcohol (like methanol) in acidic conditions to form an acetal. These compounds are stable at neutral and acidic conditions, but they hydrolyze at basic conditions. This reaction produces both acetals anomers (α and β) because the attack of the nucleophile (alcohol) could be from both sides. However, the most stable anomer will predominate.

Answer : The 3D representation of [tex]CH_3CHCHI[/tex] are shown below.

Explanation :

The configuration of the geometrical isomers is designated by two system which are, Cis-trans system and E-Z system.

The rules for E-Z system are :

The atoms or groups attached to each olefinic carbon are given priority as per the sequence rule.

If the higher priority groups are present on same sides across the double bond, the geometrical isomer is said to have Z-configuration or 'cis'.

If the higher priority groups are present on opposite sides across the double bond, the geometrical isomer is said to have E-configuration or 'trans'.

In the given compound, higher priority groups are [tex]CH_3[/tex] and [tex]I[/tex].

When [tex]CH_3[/tex] and [tex]I[/tex] are present on same sides across the double bond, the geometrical isomer is said to have 'cis'.

When [tex]CH_3[/tex] and [tex]I[/tex] are present on opposite sides across the double bond, the geometrical isomer is said to have 'trans'.

The 3D representation is shown by dash and wedge bonds in which dash shows that the molecules are below the plane and wedge shows that the molecules are above the plane.

21.077 × 10²¹ silver atoms are there in 3.86 g of silver.

Number of atoms = Number of moles × Avogadro's Number

Avogadro's number is the number of particles in one mole of substance. 6.022 × 10²³ is known as Avogadro's constant / Avogadro's number.

Avogadro's Number = 6.022 × 10²³

To calculate the number of moles of a substance we have to divide the given mass/weight of the substance by the molar mass of the substance.

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

                             = [tex]\frac{3.86\ g}{107.87\ g/mol}[/tex]

                             = 0.035 mol

Now put the value of number of moles in above formula, we get

Number of atoms = Number of moles × Avogadro's Number

                             = 0.035 × 6.022 × 10²³

                             = 0.21077  × 10²³

                             = 21.077 × 10²¹

Thus, from the above conclusion we can say that 21.077 × 10²¹ silver atoms are there in 3.86 g of silver.

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Answer:

0,12 μmol/L of MgF₂

Explanation:

Preparation of solutions is a common work in chemist's life.

In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in  μmol/L

You have 0,00598 μmol but not Liters.

To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:

50,0 mL (1L/1000mL) = 0,05 L of water.

Thus, concentration in  μmol/L is:

0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-

I hope it helps!

The concentration of the chemist's magnesium fluoride solution is 0.12 μmol/L when rounded to two significant digits, calculated by dividing the moles of solute by the volume of the solution in liters.

To calculate the concentration of the magnesium fluoride solution in μmol/L, we use the formula C = (n/V), where 'C' is the concentration, 'n' is the number of moles of solute, and 'V' is the volume of solution in liters.

First, we convert the volume from milliliters to liters: 50 mL = 0.050 L.

The number of moles (n) of MgF2 given is 0.00598 μmol, which is already in the correct units for our calculation.

Therefore, the concentration (C) of the solution is:

C = n/V

C = 0.00598 μmol / 0.050 L

C = 0.1196 μmol/L

When rounded to two significant digits, the concentration of the magnesium fluoride solution is 0.12 μmol/L.

Answer:

Tube 1: 10⁻¹, tube 2: 10⁻², tube 3: 10⁻³, tube 4: 10⁻⁴, tube 5: 10⁻⁵.

Explanation:

Serial dilutions are dilutions that the concentration decreases by the same quantity in each successive step. It means that the undiluted will be used for the first step, then the first will be used for the second, and successively. So, for a 10-fold, the concentration must decrease 1/10 in each step, it means that the dilution will be 1/10 in the first one (because it's 1 in tube 0).

In tube 1, the dilution is 1/10 = 0.1 = 10⁻¹;

In tube 2, the dilution will decrease more 1/10, so it will be 1/100x10 = 1/100 = 0.01 = 10⁻²;

In tube 3, it will be 1/1000x1/10 = 1/1000 = 0.001 = 10⁻³

In tube 4, it will be 10⁻⁴, and

In tube 5, it will be 10⁻⁵.

Answer:

49.95 g of HCl

Explanation:

Let's formulate the chemical equation involved in the process:

Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O

This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2

1 mole Ca(OH)2 ---- 2 moles HCl

0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl

Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:

1 mole of HCl ---- 36.46094 g

1.37 moles ------ x = 49.95 g of HCl

Answer:

The ethanol has 21 vibrational modes.

Explanation:

A molecule can show 3 types of motions: one external called translational and two internal called rotational and vibrational.

In order to calculate the vibrational modes of a molecule we need to know the degrees of freedom of this molecule, it means the number of variables that are involved in the movement of this particle.

If we know that atoms are three dimensional we will know that they have 3 coordinates expressed as 3N. But the atoms are bonded together so they can move not only in translational but also rotational and vibrational. So, the rotational move can be described in 3 axes and the other vibrational move can be described as

3N-5 for linear molecules

3N-6 For nonlinear molecules like ethanol

So using the formula for nonlinear molecules where N is the amount of atoms in the chemical formula, so ethanol has 9 atoms

3(9)-6= 21

Thus, ethanol has 21 vibrational modes.

Final answer:

Ethanol has 21 vibrational modes, which can be calculated using the formula 3N-6 for non-linear molecules, where N equals the number of atoms in the molecule. The vibrational modes consist of various bond stretching and bending movements, and these can only occur at certain quantized frequencies.

Explanation:

The number of vibrational modes in a molecule depends on the number of atoms it contains and the types of bonds present. In the case of ethanol (C2H5OH), which is a relatively complex molecule, the actual number of vibrational modes can be calculated using the formula 3N-6 for non-linear molecules, where N is the number of atoms. Considering ethanol has 9 atoms, we can calculate its vibrational modes as 3(9)-6, resulting in 21 vibrational modes.

The reason for the 3N-6 rule is that each atom contributes three degrees of freedom (x, y, and z axes), but we subtract the six degrees corresponding to the translational and rotational movements of the molecule as a whole. The vibrational modes in a molecule are the various ways that the molecule's bonds can stretch, bend, and twist. These modes consist of various stretching and bending movements of the carbon-hydrogen, carbon-carbon, and carbon-oxygen bonds within the ethanol molecule.

It's important to note that molecular vibrations are quantized, meaning a molecule can only access these modes at specific frequencies. When a molecule like ethanol is exposed to infrared radiation that matches the frequency of one of its vibrational modes, it can absorb that radiation, causing a transition to a higher energy vibrational state.

Answer:

b.

Explanation:

A conjugate acid-base pair are related to each other through the addition or loss of a proton. Acids react by giving up a proton, and the resulting product of that reaction is the conjugate base. Thus, the conjugate base will have one proton less than the acid. Similarly, bases react by receiving protons, so the conjugate acid will have one proton more than the conjugate base that reacted.

In the following example, HA is the acid while A⁻ is the conjugate base:

HA + H₂O ⇒ H₃O⁺ + A⁻

Answer option a. and c. are not correct since bases to not give up protons and acids to not receive protons. Base on the above information, a conjugate base reacts to produce the conjugate acid and vice versa. Thus, to produce the conjugate acid, the conjugate base received a proton. The correct answer is b.

A conjugate acid is formed by the addition of a proton to a base, and has the potential to donate that proton back, reverting to its original base form.

A conjugate acid is the species that is formed by the addition of a proton to a base. When we describe acid-base reactions in terms of conjugate acids and conjugate bases, we are applying the Brønsted-Lowry Acid-Base Theory. In this context, a conjugate acid/base pair is defined by the gain or loss of a single hydrogen ion. For example, when a base accepts a proton (H+), it forms its conjugate acid, and this conjugate acid has the potential to donate that proton back, essentially reverting to its original base form.

Conversely, when an acid donates a proton, it forms a conjugate base, which can accept a proton in a reverse reaction, thus potentially reforming the original acid. These relationships are fundamental to understanding acid-base chemistry and reactions where proton transfer occurs.

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Answer:

The reaction rate k is 0.0012563 (1/hour).

Explanation:

We considered the reactions occurring in the plant as first order, and represented by this equation:

[tex]y = L (1- e^{-kt})[/tex]

where y is the BOD at time t, L is the initial value of BOD and k is the reaction rate.

If we replaced with the values

y = 2 mg O2/l (1% of the initial value)

L = 200 mg 02/l

t = 8 hr

We can calculate k

[tex]y=L(1-e^{-kt})\\\\k=-(1/t)*ln(1-y/L) = -(1/8)*ln(1-2/200)=-(1/8)*(-0.01)=0.0012563 \, hour^{-1}[/tex]

The reaction rate k is 0.0012563 1/hour.

Final answer:

The rate of reaction for a municipal wastewater treatment plant can be found by calculating the total oxygen consumed to lower the BOD, using the volume of wastewater and the residence time. The initial BOD is 200 mg O2/liter and is reduced to a negligible level, indicating a complete reaction over the course of the 8-hour residence time.

Explanation:

Finding the Rate of Reaction in Wastewater Treatment Tanks

The student is tasked with finding the rate of reaction for a municipal wastewater treatment plant where wastewater is treated with aerobic bacteria that decompose organic material. The process significantly lowers the biochemical oxygen demand (BOD) from an entering feed of 200 mg O2/liter to a negligible level in the effluent. The key to solving for the rate of reaction is utilizing the given information: the wastewater volume (32,000 m3/day) and the mean residence time (8 hours)

To find the rate of reaction, we need to calculate the amount of oxygen consumed by the microorganisms to reduce the BOD to negligible levels. The initial BOD is given as 200 mg O2/liter, and since it's reduced to a negligible level, we can assume that virtually all the initial oxygen demand is consumed by the reaction. The volume converted to liters (since BOD is in mg/liter) and residence time in days must be considered to calculate the daily rate of reaction:

Once the daily rate is found, the rate of reaction will be the amount of oxygen consumed per day divided by the volume of wastewater treated in one day (in liters), providing the rate in mg O2/liter/day.

Answer: The mass of air is 1.18 g

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

[tex]PV=nRT[/tex]

Or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of the gas = 150 psia = 10.2 atm   (Conversion factor:  1 psia = 0.068 atm)

V = Volume of gas = [tex]10in^3=0.164L[/tex]    (Conversion factor:  [tex]1in^3=0.0164L[/tex] )

m = mass of air = ?

M = Average molar mass of air = 28.97 g/mol

R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]440^oF=499.817K[/tex]  (Conversion factor: [tex](T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}[/tex]  )

Putting values in above equation, we get:

[tex]10.2atm\times 0.164L=\frac{m}{28.97g/mol}\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 499.817K\\\\m=1.18g[/tex]

Hence, the mass of air is 1.18 g

Answer:

The value of y = 5.1478

Explanation:

The linear equation is an equation obtained when a linear polynomial is equated to zero. When the solution obtained on solving the equation is substituted in the equation in place of the unknown, the equation gets satisfied.

The given equation: 5.3 x 10- (y)(2y) = 0

⇒ 53 - 2y² = 0

⇒ 2y² = 53

⇒ y² = 53 ÷ 2 = 26.5

⇒ y = √26.5 = 5.1478

Answer:

The compound is [tex]Mn_{3}O_{4}[/tex]

Explanation:

The mass percentage of Mn is 72.1% and the mass percentage of O is 27.9%.

The mass percentage of a compound is given by:

[tex]percentage_{A}=\frac{n*MM_{A}}{MM_{C}} *100[/tex]

where:

n is its coefitient in the compund formula

MMa=Molar mass of the element A

MMc=Molar mass of the compound

So, we can figure out which compound is by dividing the percentage by its molar mass

Mn=72.1÷54.938045=1.31239

O=27.9÷15.9994=1.74382

Then, we divide each result by the smaller one (Mn)

Mn=1.31239÷1.31239=1

O=1.74382÷1.31239=1.3287

Each the realation of Mn:O is 1:1.3287

Then we multiply each result by 3:

Mn=1×3=3

O=1.3287×3=3.986≈4

Finally we figure out that the compound has 3 atoms of Mn and 4 atoms O. Result= [tex]Mn_{3}O_{4}[/tex]

Mn3O4 is sometimes used as a starting material in the production of soft ferrites e.g. manganese zinc ferrite, and lithium manganese oxide, used in lithium batteries.

The empirical formula of a compound with 72.1% Mn and 27.9% O is MnO.

The empirical formula of a compound can be determined based on its mass percentage composition of different elements. In this case, the compound is composed of 72.1% Mn (manganese) and 27.9% O (oxygen).

To find the empirical formula, we need to consider the relative number of atoms in the compound. Assuming we have 100g of the compound, we have 72.1g of Mn and 27.9g of O.

Using the molar masses of Mn (54.938 g/mol) and O (15.999 g/mol), we can calculate the number of moles of each element:

Now, we divide the number of moles by the smallest value to get the mole ratio:

Since we want whole-number ratios, we round the ratios to the nearest whole number:

Therefore, the empirical formula of the compound with 72.1% Mn and 27.9% O is MnO.

The mole fraction of NaCl in the solution is 3.15 × 10-2.

The mole fraction of NaCl in a solution prepared by dissolving 117 g NaCl in 1.11 kg H2O can be calculated by dividing the moles of NaCl by the total moles of solute and solvent. First, we convert the mass of NaCl to moles using its molar mass. Then, we calculate the moles of water by dividing its mass by its molar mass. Finally, we divide the moles of NaCl by the sum of moles of NaCl and moles of water to get the mole fraction of NaCl in the solution.

Given:

Step 1: Convert mass of NaCl to moles:

Moles of NaCl = (mass of NaCl / molar mass of NaCl) = (117 g / 58.44 g/mol) = 2 moles

Step 2: Convert mass of H2O to moles:

Moles of H2O = (mass of H2O / molar mass of H2O) = (1110 g / 18.02 g/mol) = 61.55 moles

Step 3: Calculate mole fraction of NaCl:

Mole fraction of NaCl = (moles of NaCl / (moles of NaCl + moles of H2O)) = (2 moles / (2 moles + 61.55 moles)) = 3.15 × 10-2

Answer:

We use coolant instead of water because when heated, water can start to accumulate the minerals that it has, and this can corrode the metal on your car. Other solution could be to use distilled water, but is will also corrode the metal of your radiator. One advantage of ethylene glycol is that prevents the corrosion in your radiator providing a protective coating to the metals.

Explanation:

The given data is as follows.

             P = 3 atm

                = [tex]3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}[/tex]  

                 = [tex]3.03975 \times 10^{5} Pa[/tex]

    [tex]V_{1}[/tex] = 9 L = [tex]9 \times 10^{-3} m^{3}[/tex]    (as 1 L = 0.001 [tex]m^{3}[/tex]),  

        [tex]V_{2}[/tex] = 15 L = [tex]15 \times 10^{-3} m^{3}[/tex]

            Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

                  W = [tex]P \times \Delta V[/tex]

or,                W = [tex]P \times (V_{2} - V_{1})[/tex]

Therefore, putting the given values into the above formula as follows.

                  W = [tex]P \times (V_{2} - V_{1})[/tex]

                      = [tex]3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})[/tex]

                      = 1823.85 Nm

or,                   = 1823.85 J

As internal energy of the gas [tex]\Delta E[/tex] is as follows.

                     [tex]\Delta E[/tex] = Q - W

                                  = 800 J - 1823.85 J

                                  = -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

Answer:

[tex]\boxed{\text{115.2 torr}}[/tex]

Explanation:

Let’s call water Component 1 and lactose Component 2.

According to Raoult’s Law,  

[tex]p_{1} = \chi_{1}p_{1}^{\circ} \text{ and}\\p_{2} = \chi_{2}p_{2}^{\circ}[/tex]

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

Data:

m₁ = 110.0 g;    p₁° = 118.0 torr

m₂ = 50.00 g; p₂° =    0    torr

1. Calculate the moles of each component

[tex]n_{1} = \text{110.0 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{6.104 mol}\\\\n_{2} = \text{50.00 g} \times \dfrac{\text{1 mol}}{\text{342.3 g}} = \text{0.1461 mol}[/tex]

2. Calculate the mole fraction of each component

[tex]\begin{array}{rcl}\chi_{2} & = & \dfrac{n_{2}}{n_{1} + n_{2}}\\\\&= & \dfrac{0.1461}{6.104 + 0.1461}\\\\& = &\dfrac{0.1461}{6.250}\\\\& = & \mathbf{0.023 37}\\\chi_{1}& = &1 - \chi_{2}\\& = &1 - 0.023 37\\& = & \mathbf{0.976 63}\\\end{array}[/tex]

3. Calculate the vapour pressure of the mixture

[tex]p_{1} = 0.97663 \times \text{118.0 torr} = \text{ 115.2 torr}\\p_{2} = 0\\p_{\text{tot}} = p_{1} + p_{2} = \text{115.2 torr + 0} = \textbf{115.2 torr}\\\text{The partial pressure of water above the solution is $\boxed{\textbf{115.2 torr}}$}[/tex]

To find the partial pressure of water above the solution, calculate the mole fraction of water in the solution using the given masses of lactose and water. Multiply the mole fraction of water by the vapor pressure of pure water at 55 °C to find the partial pressure of water above the solution.

To find the partial pressure of water above the solution, we need to calculate the mole fraction of water in the solution using the given masses of lactose and water. First, find the moles of lactose by dividing its mass by its molecular weight. Next, find the moles of water by dividing its mass by its molecular weight. Then, calculate the mole fraction of water by dividing the moles of water by the total moles in the solution. Finally, multiply the mole fraction of water by the vapor pressure of pure water at 55 °C to find the partial pressure of water above the solution.

Moles of lactose = 50.00 g / 342.3 g/mol = 0.146 mol

Moles of water = 110.0 g / 18.015 g/mol = 6.105 mol

Mole fraction of water = moles of water / (moles of lactose + moles of water) = 6.105 mol / (0.146 mol + 6.105 mol) ≈ 0.976

Partial pressure of water = mole fraction of water * vapor pressure of pure water at 55°C = 0.976 * 118.0 torr ≈ 115.2 torr

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Answer:

The answer to your question is:

mass = 3.74 g

Explanation:

Data

V = (4/3) πr³

density = 1.74 g/cm³

radius = r = 0.80 cm

Process

V = (4/3) π(0.8)³             Substitution

V = 2.1446 cm³

mass = density x volume

mass = 1.74 x 2.1446      Substitution

mass = 3.74 g

I don't understand if the second section is also a question.

Answer:

9.2584 mol/L is the solubility of helium in water at 25 °C.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{He}=K_H\times p_{H_2O}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]3.26\times 10 mol/L atm[/tex]

[tex]p_{He}[/tex] = partial pressure of carbonated drink = 0.284 atm

Putting values in above equation, we get :

[tex]C_{He}=3.26\times 10 mol/L atm\times 0.284 atm=9.2584 mol/L[/tex]

9.2584 mol/L is the solubility of helium in water at 25 °C.

Answer: The change in volume is 2.05 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=3.25L\\T_1=250.0K\\V_2=?\\T_2=406.8K[/tex]

Putting values in above equation, we get:

[tex]\frac{3.25L}{250.0K}=\frac{V_2}{406.8K}\\\\V_2=5.30L[/tex]

Change in volume = [tex]V_2-V_1=(5.30-3.25)L=2.05L[/tex]

Hence, the change in volume is 2.05 L

Answer:

We need  2.933 L of 0.15 mg /mL of protein solution.

Explanation:

Concentration of given solution[tex]C_1 = 0.15 mg/mL[/tex]

1 mg = 0.001 g , 1 mL = 0.001 L

[tex]C_1=\frac{0.15\times 0.001 mg}{1\times 0.001 L}=0.15 g/L[/tex]

Molecular weight of protein = 22,000 Da =22,000 g/mol

Initial concentration in moles/liter:

[tex]C_1=\frac{0.15 g/L}{22,000 g/mol}=6.8182\times 10^{-6} mol/L[/tex]

Initial concentration in micromoles/mL :

1 L = 1000 mL

[tex]C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000 mL}=6.8182\times 10^{-3} \mu mole/ mL[/tex]

Initial concentration in micromoles/microLiter :

1 L = 1000,000 μL

[tex]C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000000 \mu L}=6.8182\times 10^{-6}\mu mol/\mu L[/tex]

Moles of protein required = 20 μmoles

n(Moles)=C(concentration) × V(Volume of solution)

[tex]20 \mu mol=6.8182\times 10^{-6}\mu mol/\mu L\times V[/tex]

[tex]V =\frac{20 \mu mol}{6.8182\times 10^{-6}\mu mol/\mu L}[/tex]

[tex]V=2.933\times 10^6 \mu L = 2.933 L[/tex]

We need  2.933 L of 0.15 mg /mL of protein solution.

Explanation:

The volumetric flow rate of water will be as follows.

       q = [tex]600 gpm \times \frac{0.000063 m^{3} s^{-1}}{1 gpm}[/tex]

         = 0.0378 [tex]m^{3}/s [/tex]

    Diameter = [tex]8 in \times \frac{0.0254 m}{1 in}[/tex]

                   = 0.2032 m

Relation between area and diameter is as follows.

           A = [tex]\frac{\pi}{4} \times D^{2}[/tex]

               = [tex]\frac{3.14}{4} \times (0.2032 m)^{2}[/tex]

               = 0.785 x 0.2032 x 0.2032

               = 0.0324 [tex]m^{2}[/tex]

Also,     q = A × V

or,         V = [tex]\frac{q}{A}[/tex]

                = [tex]\frac{0.0378 m^{3}/s}{0.0324 m^{2}}[/tex]

                = 1.166 m/s

As, viscosity of water = 1 cP = [tex]10^{-3}[/tex] Pa-s

Density of water = 1000 [tex]kg/m^{3}[/tex]

Therefore, we will calculate Reynolds number as follows.

 Reynolds number = [tex]\frac{D \times V \times density}{viscosity}[/tex]

                                 = [tex]\frac{0.2032 m \times 1.166 m/s  \times 1000}{10^{-3}}[/tex]  

                                = 236931.2

Hence, the flow will be turbulent in nature.

Thus, we can conclude that the Reynolds number is 236931.2 and flow is turbulent.

Answer: Option (d) is the correct answer.

Explanation:

The given data is as follows.

Tube diameter d = 10 mm = 0.01 m

Velocity of glycerol, v = 0.5 m/s

Density of glycerol ([tex]\rho[/tex]) = 1240 kg/m3

Dynamic viscosity of glycerol ([tex]\mu[/tex]) = 0.0813 pa.s

Reynolds number (Re) = [tex]\rho \times velocity \times \frac{density}{\mu}[/tex]

                                     = [tex]1240 \times 0.5 \times \frac{0.01}{0.0813}[/tex]

                                     = 76.26

Therefore, according to Reynolds number we can say that flow is laminar.

                     Lt = [tex]0.05 \times Re \times d[/tex]

                         = [tex]0.05 \times 76.26 \times 0.01[/tex]

                         = [tex]0.03813 m[/tex]

As it is known that 1 m = 1000 mm. Hence, in 0.03813 m will be equal to [tex]0.03813 m \times \frac{1000 mm}{1 m}[/tex]

                         = 38.13 mm

Thus, we can conclude that the transition length of glycerol is 38.13 mm.

Answer:

Molar mass of unknown solute is 679 g/mol

Explanation:

Let us assume that the solute is a non-electrolyte.

For a solution with non-electrolyte solute remains dissolved in it -

Depression in freezing point of solution, [tex]\Delta T_{f}=K_{f}.m[/tex]

where, m is molality of solute in solution and [tex]K_{f}[/tex] is cryogenoscopic constant of solvent.

Here [tex]\Delta T_{f}=(-22.9^{0}\textrm{C})-(-28.7^{0}\textrm{C})=5.8^{0}\textrm{C}[/tex]

If molar mass of unknown solute is M g/mol then-

                 [tex]m=\frac{\frac{2.2}{M}}{0.0167}mol/kg[/tex]

So, [tex]5.8^{0}\textrm{C}=29.9^{0}\textrm{C}/(mol/kg)\times \frac{\frac{2.2g}{M}}{0.0167}mol/kg[/tex]

so, M = 679 g/mol

The  attention measures for a CaCl2  result with 1000 mg in 500 mL are0.2 w/ v, a  rate strength of 2 g/ L,0.01818 M, and molality isn't directly  reliable from the given data. For1.5 L of this  result, there are0.05454 coequals of CaCl2.

Computations for attention of CaCl2 result    In order to calculate the  colorful  attention measures of the given  result containing calcium chloride( CaCl2), we should first understand the  ensuing delineations     w/ v- Percent weight per volume expresses  attention as grams of solute per 100 mL of  result.  rate strength- The  quantum of solute in grams per 1000 mL( 1 L) of  result.  Molarity( M)- The number of  intelligence of solute per liter of  result.  Molality( m)- The number of  intelligence of solute per kilogram of detergent.    Given that the molecular weight of CaCl2 is 110 g/  spook, and we've 1000 mg( 1 g) in 500 mL, we can calculate the following     w/ v( 1 g/0.5 L) * 100 = 0.2 w/ v  rate strength 1 g in 500 mL equals 2 g in 1000 mL or 2 g/L.  Molarity( M)( 1 g/ 110 g/  spook)/(0.5 L) = 0.01818 M  Molality( m) can not be calculated without the mass of the detergent;  still, assuming the specific  graveness is 1, molality would be equal to molarity for  veritably dilute  results like this bone

Answer:

K.E. = 5.4362 × 10⁻¹⁹ J

Explanation:

The expression for Bohr velocity is:

[tex]v=\frac{Ze^2}{2 \epsilon_0\times n\times h}[/tex]

Applying values for hydrogen atom,  

Z = 1

Mass of the electron ([tex]m_e[/tex]) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

[tex]\epsilon_0[/tex] = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

[tex]v=\frac {2.185\times 10^6}{n}\ m/s[/tex]

Given, n = 2

So,

[tex]v=\frac {2.185\times 10^6}{2}\ m/s[/tex]

[tex]v=1.0925\times 10^6\ m/s[/tex]

Kinetic energy is:

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

So,

[tex]K.E.=\frac {1}{2}\times 9.1093\times 10^{-31}\times ({1.0925\times 10^6})^2[/tex]

K.E. = 5.4362 × 10⁻¹⁹ J

Explanation:

Inexhaustible resource -

It is the resource which never get reduced or deplete on constant usage , they are mainly the type of naturally occurring resources , hence they reappear in nature again .

The example of inexhaustible resources are - the wind energy , tides , solar energy and the geothermal energy .

Hence , the solar energy is also a form of inexhaustible resource .

The kinetic energy of an electron can be calculated utilizing the equation KE = hf - BE, but this specific equation cannot be solved without additional information such as the binding energy, frequency of the light radiation, or Plank's constant. Energy can also be calculated using Planck's equation,E = hf, or for a specific orbital using 13.6 eV / n², where n refers to the level of the orbital.

The kinetic energy acquired by an electron in a hydrogen atom after absorbing light radiation can be found by utilizing the equation KE = hf - BE (kinetic energy equals energy of radiation minus binding energy). In this case, if the electron absorbs a light radiation of energy 1.08x101 J, it's crucial to determine the binding energy first.

However, given the information available, further clarification is needed since binding energy, frequency of the radiation, or Plank's constant (h) are not specified in the question.

Similar energy calculations involve using Planck's equation E = hf, where E is energy, h is Planck’s constant, and f is the frequency of radiation. Furthermore, energy can also be calculated for a specific orbital of a hydrogen atom using the equation 13.6 eV / n², where n refers to the level of the orbital.

Remember, when using these equations you may need to convert your units appropriately to reach the correct answer.

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Answer:

c. 1.4 x 10²³ oxygen atoms

Explanation:

The number of oxygen atoms in one molecule of CH₃COOH is 2.

Avogadro's constant relates the number of molecules in one mole:

6.022 × 10²³ mol⁻¹

Thus, the number of oxygen atoms in one molecule of acetic acid can be converted to the number of oxygen atoms in one mole of acetic acid:

(2 oxygen atoms / molecule)(6.022 × 10²³ molecule / mol) = 1.204 x 10²⁴ atoms per mole

Finally, the number of oxygen atoms in 0.12 moles of acetic acid are calculated:

(1.204 x 10²⁴ atoms / mol)(0.12 mol) = 1.4 x 10²³ atoms/mol

Explanation:

Monosaccharides are the elementary form of the sugar and most basic units of  the carbohydrates. These sugars cannot be further hydrolyzed to form the simpler chemical compounds. The general formula is [tex]C_nH_{2n}O_n[/tex]. Example: Glucose and fructose.

Disaccharide is sugar which is formed when the two monosaccharides are joined by the glycosidic linkage. Disaccharides are soluble in the water. Examples: sucrose and lactose.

Oligosaccharide is the saccharide polymer which contains small number of the monosaccharides. They can have many functions like the cell recognition and the cell binding. Example: glycolipids which have role in immune response.

Polysaccharides are the polymeric carbohydrate molecules which are composed of the long chains of the monosaccharide units that are bound together by the glycosidic linkages which on the hydrolysis give constituent monosaccharides or th eoligosaccharides. Example: Starch.

Glycoconjugates is general classification for the carbohydrates which are covalently linked with the other chemical species such as peptides, proteins, saccharides and lipids.  Example: Blood proteins