What will be the final pH when 5.865 mL of 3.412 M NaOH is added to 0.5000 L of 1.564 x 10-3 M HCl?

Anomeric carbon is a stereocenter present in the cyclic structures of carbohydrates (mono or polysaccharides). Being a stereocenter, more exactly an epimer, two diastereosoimers derive from it, designated by the letters α and β; these are anomers, and are part of the extensive nomenclature in the world of sugars.

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group:  R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a sp² hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a  sp³  hybridization because they are involved only in single bonds.

Answer:

2.6 sec

Explanation:

The distance between the Earth and the moon = 240,000 miles

Also,

1 mile = 1609.34 m

So,

Distance between the Earth and the moon = 240,000 ×  1609.34 m = 386241600 m

Speed of the light = 3 × 10⁸ m/s

Distance = Speed × Time.

So,

Time = Distance / Speed = 386241600 m / 3 × 10⁸ m/s = 1.3 sec

For back journey = 1.3 sec

So, total time = 2.6 sec

Answer:

The correct answer is: a. are all strong bases

Explanation:

Alkaline earth metals are the chemical elements that belong to the group 2 of the periodic table. The members or elements of this group are all highly reactive metals.

Except beryllium (Be), all the alkaline earth metals react with water to give metal hydroxides. These hydroxides of the alkaline earth metals are highly soluble and very strong bases.

Final answer:

Hydroxides of Group 2 metals in water are all strong bases because they dissociate almost completely into ions, significantly raising the solution's pH by releasing a high concentration of OH- ions.

Explanation:

The question asks about the nature of hydroxides of Group 2 metals when dissolved in water. Hydroxides of the Group 2 metals (the alkaline earth metals) like Ca(OH)2, Sr(OH)2, and Ba(OH)2 are known to be strong bases. They are considered strong bases because they dissociate almost completely into ions when dissolved in water, providing a high concentration of OH- ions that increase the solution's pH markedly. To answer the provided options, a. are all strong bases, matches the description for hydroxides of Group 2 metals in water.

Answer:

0.712 moles of NO₂ are formed.

Explanation:

First, we need to write the balanced equation:

2 N₂O₅(g) ⇄ 4 NO₂(g) + O₂(g)

From the balanced equation, we can see the relationship between the moles of N₂O₅ and the moles of NO₂. Every 2 moles of N₂O₅ that react, 4 moles of NO₂ are formed. Let us apply this relationship to the information given by the problem (0.356 moles of N₂O₅):

[tex]0.356molN_{2}O_{5}.\frac{4molNO_{2}}{2molN_{2}O_{5}} =0.712molNO_{2}[/tex]

Answer:

-1.06

Explanation:

0.345 has three significant figures since the zero does not represent a significant figure.

So, usign the calculator, we find ln (0.345) = -1.06421086195

Now, the problem says that you need to report your answer to three significant figures.

So, you should take the first three numbers of the answer:

ln (0.345) = -1.06

Answer :  a) 40075.14 J/mol

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]440K[/tex] = k

[tex]K_2[/tex] = rate constant at [tex]550K[/tex] = 10 k

[tex]Ea[/tex] = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = 440 K

[tex]T_2[/tex] = final temperature = 550 K

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

[tex]\log (\frac{10k}{k})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{440K}-\frac{1}{550K}][/tex]

[tex]Ea=40075.14J/mol [/tex]

Therefore, the activation energy for the reaction is 40075.14J/mol.

Explanation:

It is given that flow rate is 10,000 L/s. As 1 L equals 0.001 [tex]m^{3}[/tex].

Hence, flow rate will be 10 [tex]m^{3}/s[/tex]. Calculate mass of water flowing per second as follows.

        Mass flowing per second = density × flow rate

                                                   = [tex]1000 kg/m^{3} \times 10 m^{3}/s[/tex]

                                                   = [tex]10^{4} kg/s[/tex]

Also, energy provided per second will be as follows.

                      E = mgh

Putting the given values into the above formula as follows.

                   E = mgh

                      = [tex]10^{4} kg/s \times 9.8 m/s^{2} \times 75 m[/tex]  

                       = [tex]735 \times 10^{4} W[/tex]

or,                    = 7350 kW

Thus, we can conclude that energy per unit time provided will be 7350 kW.                

Answer:

a) 0,857 g of H₃PO₄ with 9,016 g of KH₂PO₄

b) 166 mL of 0,400M NaOH

c) 22 mL of 0,400M HCl

Explanation:

a) The appropriate weak acid and conjugate salt are:

H₃PO₄ ⇄ H₂PO₄⁻ + H⁺ where pka = 2,12

Henderson–Hasselbalch equation finding pH = 3:

3 = 2,12 + log₁₀ [tex]\frac{[H2PO4-]}{[H3PO4]}[/tex]

7,59 =  [tex]\frac{[H2PO4-]}{[H3PO4]}[/tex] (1)

If buffer concentration is 0,300M:

0,300 M = [H₃PO₄] + [H₂PO₄⁻] (2)

Replacing (2) in (1):

[H₃PO₄] = 0,035 M

Thus:

[H₂PO₄⁻] = 0,265 M

Thus, to prepare this buffer you need weight:

0,035 M × 0,250 L = 8,75x10⁻³ moles × [tex]\frac{97,994 g}{1mol}[/tex] = 0,857 g of H₃PO₄

And:

0,265 M × 0,250 L = 6,63x10⁻² moles × [tex]\frac{136,086 g}{1mol}[/tex] = 9,016 g of KH₂PO₄

b) Using 0,400 M NaOH the equilibrium is:

H₃PO₄ + NaOH ⇄ H₂PO₄⁻ + H₂O

Knowing the equilibrium concentrations are:

[H₃PO₄] = 0,035 M = 0,300 M - x -Because in the first all 0,300 M must be of H₃PO₄-

[H₂PO₄⁻] = 0,265 M

Thus, x = 0,265 M are NaOH needed to obtain the desire pH. Those are obtained thus:

0,265 mol/ L × 0,250L × [tex]\frac{1L}{0,400mol}[/tex] = 0,166 L ≡ 166 mL of 0,400M NaOH

c) Using 0,400 M HCl the equilibrium is:

H₃PO₄ ⇄ H₂PO₄⁻ + HCl

Knowing the equilibrium concentrations are:

[H₃PO₄] = 0,035 M

[H₂PO₄⁻] = 0,265 M = 0,300 M - x -Because in the first all 0,300 M must be of H₂PO₄-

Thus, x = 0,035 M are HCl needed to obtain the desire pH. Those are obtained thus:

0,035 mol/ L × 0,250L × [tex]\frac{1L}{0,400mol}[/tex] = 0,022 L ≡ 22 mL of 0,400M HCl

I hope it helps!

Answer:

None of the above

Explanation:

Henry's law -

This law was given by William Henry in the year 1803 , it is also known as the gas law ,

According to Henry's law , the amount of gas which gets dissolved in a liquid is directly related to the partial pressure of gas that is in equilibrium with the liquid , at a constant temperature .

Or it can stated as ,

The gases' solubility in a liquid is directly related to the partial pressure of gas that is in equilibrium with the liquid .

This law is applicable for sparingly soluble gases in liquid solvents .

And the solubility of a gas is independent of temperature and pressure .

Hence ,

the correct option is None of the above .

Answer:

Incomplete precipitation of barium sulfate

Explanation:

The student has precipitated and digested the barium sulfate on his/her side. But on the addition of [tex]BaCl_2[/tex] in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When [tex]BaCl_2[/tex] is added, there are still sulfate ions present in the solution with combines with [tex]BaCl_2[/tex] and forms [tex]BaSO_4[/tex] and the formation of this precipitate makes the solution cloudy.

The solution became cloudy after adding BaCl2 because the additional Ba2+ ions reacted with any remaining [tex]SO4^2-[/tex]orming more BaSO4 precipitate, indicating incomplete digestion of the initial BaSO4.

During the experiment, the student added a few drops of BaCl2 solution to the clear solution that contained digested BaSO4. BaSO4 is known for its low solubility in water; however, it's soluble in solutions containing ions that can form more soluble compounds with the constituent ions. In the experiment, adding more BaCl2 likely introduced additional Ba2+ ions into the solution. If any unreacted sulfate ions (SO42-) were present, these extra Ba2+ ions could have reacted with them, forming additional BaSO4 precipitate, thus causing the solution to become cloudy. This suggests that the digestion process was not complete, leaving some sulfate ions in the solution which reacted with the added barium ions to form more BaSO4 precipitate.

Answer:

2.32 m

Explanation:

So, according to definition of mole fraction:

[tex]Mole\ fraction\ of\ benzene=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}[/tex]

Mole fraction = 0.176

Applying values as:

[tex]0.176=\frac {n_{benzene}}{n_{benzene}+n_{toluene}}[/tex]

[tex]0.176\times ({n_{benzene}+n_{toluene}})={n_{benzene}}[/tex]

So,

[tex]0.176\times n_{toluene}}=0.824\times {n_{benzene}}[/tex]

[tex]{n_{benzene}}=\frac {0.176}{0.824}\times n_{toluene}}[/tex]

[tex]{n_{benzene}}=0.2136\times n_{toluene}}[/tex]

Also, Molar mass of toluene = 92.14 g/mol

Thus,

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

[tex]Mass=92.14\times n_{toluene}}\ g[/tex]

Also, 1 g = 0.001 kg

So,

[tex]Mass\ of\ toluene=0.09214\times n_{toluene}}\ kg[/tex]

Molality is defined as the moles of the solute present in 1 kg of the solvent.

It is represented by 'm'.

Thus,  

[tex]Molality\ (m)=\frac {0.2136\times n_{toluene}}{0.09214\times n_{toluene}}[/tex]

Molality of benzene = 2.32 m

Final answer:

The molality of the solution of benzene dissolved in toluene is 0.00543 mol/kg.

Explanation:

The molality of a solution can be calculated using the mole fraction and the molar mass of benzene and toluene. The mole fraction of benzene can be calculated by dividing the moles of benzene by the total moles of benzene and toluene. The moles of benzene can be determined using the given volumes of benzene and toluene and their respective densities. The molar mass of benzene is 78.11 g/mol. By substituting the values into the formula, we can calculate the molality of the solution to be 0.176 * 78.11 g/mol / (1000 g + 0.867 g/mL * 100 mL + 0.874 g/mL * 300 mL) = 0.00543 mol/kg.

Explanation:

The given data is as follows.

               MW = 29 g/mol,         [tex]C_{p}[/tex] = 3.5 R

Formula to calculate minimum amount of work is as follows.

            [tex]W_{s} = C_{p}T_{1}[(\frac{P_{2}}{P_{1}})^{\frac{R}{C_{p}}} - 1][/tex]

                          = [tex]3.5 \times 8.314 J/k mol \times 306 \times [(\frac{550}{125})^{\frac{1}{3.5}} - 1][/tex]

                          = 4.692 kJ/mol

Therefore, total work done will be calculated as follows.

                        Total work done = [tex]m \times W_{s}[/tex]

Since, m = [tex]\frac{150 \times 10^{3}g/min}{29}[/tex]. Therefore, putting these values into the above formula as follows.

            Total work done = [tex]m \times W_{s}[/tex]

                                        = [tex]\frac{150 \times 10^{3}g/min}{29} \times 4.692 kJ/min[/tex]      

                                       = 24268.96 kJ/min

It is known that 1 kJ/min = 0.0166 kW. Hence, convert 24268.96 kJ/min into kW as follows.

                   [tex]24268.96 kJ/min \times \frac{0.0166 kW}{1 kJ/min}[/tex]                                                    

                   = 402.86 kW

Thus, we can conclude that the minimum work required by an air compressor is 402.86 kW.

Answer:

The two changing properties in Charles’ law are temperature and volume.

Explanation:

Charles’ law state the presence of direct relationship between temperature and volume of the system of gas molecules at constant pressure condition. In this law, the expansion of gas has been explained with the increase of temperature.

As the temperature is increased or the system of gas molecules are heated, the gas molecules tend to expand their volume to maintain the pressure same.

So the temperature and volume are directly proportional at constant pressure. Thus the two changing properties of Charles’ law is temperature and volume. The mathematical representation of Charles’ law is

V∝T (at constant Pressure)

[tex]V=kT[/tex]

Here k is the non-zero constant and V and T are volume and temperature respectively.

Explanation:

According to the ideal gas equation, PV = nRT.

where,     P = pressure,        V = volume

               n = no. of moles,      R = gas constant

               T = temperature

Also, density is equal to mass divided by volume. And, no. of moles equals mass divided by molar mass.

Therefore, then formula for ideal gas could also be as follows.

                 P = [tex]\frac{mass}{volume \times molar mass} \times RT[/tex]

or,             P = [tex]\frac{density}{\text{molar mass}} \times RT[/tex]

Since, density is given as 0.789 g/ml which is also equal to 789 g/L (as 1000 mL = 1 L). Hence, putting the given values into the above formula as follows.

               P = [tex]\frac{density}{\text{molar mass}} \times RT[/tex]

                  = [tex]\frac{789 g/l}{46.06 g/mol} \times 0.0821 L atm/mol K \times 373 K[/tex]

                  = 525 atm

As two-liter soft drink bottle can withstand a pressure of  5 atm and the value of calculated pressure is 525 atm which is much greater than 5 atm.

Therefore, the soft drink bottle will  obviously explode.

Answer:

a) ρ = 3.735 Kg/m³

b) γ = 36.603 N/m³

c) Vw = 2.68 E-4 g/m³

Explanation:

a) ρ = m/V

∴ Mw air = 28.966 g/mol......from literature

⇒  P = RTn / V

∴ n = m / Mw

⇒ P = mRT / Mw.V = m/V * R.T/Mw

∴ P = 50 psi = 344738 Pa

∴ R = 8.3144 Pa.m³/mol.K

∴ T = 120°F = 48.889 °C = 321.889 K

⇒ ρ = (( 344738 Pa ) * ( 28.996 g/mol )) / (( 8.3144 Pa.m³/mol.K) * ( 321.889K ))

⇒ ρ = 3734.996 g/m³ * ( Kg / 1000g ) = 3.735 Kg/m³

specific weight:

∴ ρ = 3.735 Kg/m³

∴ g = 9.8 m/s²

⇒ γ = 36.603 N/m³

c) specific volume

⇒ Vw = (( 8.3144 Pa.m³/ mol.K ) * ( 321.889 K )) / ( 344738 Pa )* ( 28.966 g/mol)

⇒ Vw = 2.68 E-4 g/m³

Answer: Volume of the flask is 31.34 mL.

Explanation:

Weight of empty flask [tex]w_1[/tex]  = 56.12 grams

Weight of flask with water  [tex]w_2[/tex] = 87.39 grams

Weight of water [tex]w_3[/tex] = [tex]w_2-w_1[/tex]  = (87.39 - 56.12) grams= 31.27 grams  

Density of water =  0.9976 g/mL

0.9976 grams are contained in = 1 ml of water

Thus 31.27 grams  are contained in = [tex]\frac{1}{0.9976}\times 31.27=31.34[/tex] ml of water.

Thus the volume of the flask is 31.34 mL.

The volume of the flask is calculated by subtracting the mass of the empty flask from the mass of the filled flask and then dividing by the density of water. The resulting volume is approximately 31.36 mL.

To calculate the volume of the flask in mL, we need to use the mass of the water that filled the flask and the known density of water. First, we find the mass of the water by subtracting the mass of the empty flask from the mass of the filled flask. Then, we use the density formula, which is Density = Mass/Volume, to find the volume of the flask.

Here's the step-by-step calculation:

Determine the mass of the water by subtracting the mass of the empty flask from the filled flask: 87.39 g - 56.12 g = 31.27 g.

Using the known density of water (0.9976 g/mL), calculate the volume using the formula Volume = Mass/Density.

Volume of water (Volume of flask) = 31.27 g / 0.9976 g/mL = 31.36 mL.

Therefore, the volume of the flask is approximately 31.36 mL.

Explanation:

Melting point is defined as the point at which a solid substance starts to change into liquid state.

Whereas entropy is the degree of randomness of molecules present in a substance.

Heat of fusion is defined as the amount of heat energy necessary to melt a solid substance at its melting point.

Relation between entropy and heat of fusion is as follows.

                  [tex]\Delta S = \frac{\Delta H}{T}[/tex]

where,          [tex]\Delta S[/tex] = 38.98 J/mol K

                     [tex]\Delta H[/tex] = 17.02 kJ/mol

                                    = [tex]17.02 kJ/mol \times \frac{1000 J}{1 kJ}[/tex]

                                    = 17020 J/mol

Therefore, calculate the melting point as follows.

                   [tex]\Delta S = \frac{\Delta H}{T}[/tex]

                           38.98 J/mol K = [tex]\frac{17020 J/mol}{T}[/tex]

                              T = 436.63 K

Change the temperature into degree celsius as follows.

                             [tex](436.63 - 273)^{o}C[/tex]

                                 = [tex]163.63^{o}C[/tex]

Thus, we can conclude that the melting point in [tex]^{o}C[/tex] is [tex]163.63^{o}C[/tex].

Answer:

1. kmol of methanol= 3.12 Kmol

2. kmol of water= 5.55 Kmol

3. Liters of methanol=  126.4  L

4. L of water= 100  L  

Explanation:

1. kmol of methanol?

32.04 kg methanol ______________ 1 kmol of methanol

100 kg of methanol_______________ X=  3.12 kmol ofmethanol

2. kmol of water?

18.01 kg water ______________ 1 kmol of wáter

100 kg of wáter_______________ X=  5.55 kmol of water

3. Liters of methanol?

0.791 kg  methanol _______________________1.00  L of methanol

100kg  methanol  _______________________x= 126.4  L of methanol

4. L of water?

1kg  water _______________________1.00  L of water

100kg  water _______________________x= 100  L of water

Answer:

concentration of CrO4²⁻ ions in the final solution = 6.53 × 10⁻⁵ mol /L

Explanation:

First we calculate the number of moles of AgNO₃:

number of moles = molar concentration × volume

number of moles = 0.5 × 39.20 = 19.6 mmoles = 0,0196 moles AgNO₃

Then we calculate the number of moles of K₂CrO₄:

number of moles = mass / molar weight

number of moles = 5.832 / 194 = 0.03 moles K₂CrO₄

The chemical reaction will look like this:

2 AgNO₃ + K₂CrO₄ → Ag₂CrO₄ + 2 KNO₃

Now we devise the following reasoning:

if          2 moles of AgNO₃ are reacting with 1 mole of K₂CrO₄

then    0,0196  moles of AgNO₃ are reacting with X moles of K₂CrO₄

X = (0.0196 × 1) / 2 = 0.0098 moles of K₂CrO₄

now the the we calculate the amount of unreacted K₂CrO₄:

unreacted K₂CrO₄ = 0.03 - 0.0098 = 0.0202 moles

now the molar concentration of CrO4²⁻ ions:

molar concentration = number of moles / solution volume (L)

molar concentration = 0.0202 / (39.20 + 270) = 6.53 × 10⁻⁵ mol /L

Answer:

a)Number average molecular weight is 50, 500 g/mol

b) Weight average molecular weight is 99, 019.8 g/mol

Explanation:

We have a polymer P made up of two monodisperse fractions.

A with molecular weight of MA = 1000 g/mol and B with MB =100000 g/mol.  

The batch contains an equal mole fraction of each component A and B.  

Let's suppose a total number (Nt) of mols 2 moles. Equal fraction means XA = 0.5 and XB =0.5

Nt = 2 mol

Na = 2*0.5 = 1 mol

Nb = 2*0.5 = 1 mol.

So, we have 1 mol of A, 1 mol of B and 2 moles in total.  

a) The number average molecular weight (NAM) is calculate using the mole numbers of each component. In this case, we will multiple each component molecular weight by the number of moles of each one. After that we will sum them and finally to divide by the total number of moles.

NAM = (Na*MA + Nb*MB)/(Nt)

NAM = (1 mol *1000 g/mol + 1*100000 g/mol ) /(2 mol)

NAM = 50500 g/mol

The number average molecular weight for the polymer P is 50,500 g/mol

b) Weight average molecular weight (WAM) is calculated using the mass quantities of each component. Weight mass of A (WA), weight mass of B (WB) are calculate using the moles of A, B and their molecular weights respectively. Total Weight (WT)

WA = Na*MA = 1 mol *1000 g/mol = 1000 g A

WB = Nb*MB = 1mol * 100000 g/mol = 100 000 gB

WT = WA + WB = 101 000 g

Now we will calculate average molecular using weights, we will multiple each component molecular weight by the mass of each one. After that we will sum them and finally to divide by the total mass.

WAM = (WA*MA + WB*MB)/(WT)

WAM = (1000 g *1000 g/mol + 100000 g*100000 g/mol )/(101 000 g)

WAM = 99 019.8 g/mol

The weight average molecular weight for polymer P is 99, 019.8 g/mol

Final answer:

The number average molecular weight (Mn) for the polymer is 50,500 g/mol, while the weight average molecular weight (Mw) is 198,019,802 g/mol, given the equal mole fraction of the two monomer fractions.

Explanation:

The  number average molecular weight (Mn) and the weight average molecular weight (Mw) of a polymer comprising two monodisperse fractions.

Firstly, to calculate the Mn, we consider the definition that Mn is the total weight of the polymer divided by the total number of moles. Since each fraction has an equal mole fraction, we take the simple average of the two given molecular weights:

Next, to compute the Mw, which is the sum of the products of the weight contribution of each fraction and its molecular weight squared, divided by the total weight:

To summarize, for a polymer with an equal mole fraction of two fractions with molecular weights of 1000 g/mol and 100,000 g/mol, the number average molecular weight is 50,500 g/mol and the weight average molecular weight is 198,019,802 g/mol.

Answer:

Tube 2 has a total dilution of 1:50

Explanation:

We have a 2 ml serum sample added to a 18 mL phosphate buffered saline sample in tube 1. This means now in tube 1 there is 20 mL.

We have a 1:10 (= 2:20) dilution here.

10 ml of this 1:10 diluted tube 1 is taken and added to a 40 mL of PBS in tube 2.

Now we have 50 mL in tube 2.

This is a 10:50 (= 1:5) dilution.

The total dilution is 10x5 = 50

So the total ditultion has a rate 1:50

Tube 2 has a total dilution of 1:50

Explanation:

The electronic configuration of sodium with Z = 11 is : 2, 8, 1

The electronic configuration of chlorine with Z = 17 is : 2, 8, 7

The Lewis structure is drawn in such a way that the octet of each atom is complete.

Thus, sodium losses one electron to chlorine and chlorine accepts this electron to form ionic bond.

Thus, the valence electrons are shown by dots in Lewis structure. The structure is shown in image below.

The Lewis structure of NaCl involves sodium losing one electron to become Na+ and chlorine gaining one electron to become Cl-. Draw the ions next to each other to represent the ionic bond. Sodium has no electrons around it while chlorine has a complete octet with the extra electron in brackets.

The Lewis structure of sodium chloride (NaCl), follow these steps:

Na: Na+ [ ]

Cl: [ :Cl: ]-

Note that the brackets around the chlorine indicate it has gained an electron and the overall charge of the ion.

Answer:

All of them can be necessary.

Explanation:

In typical DNA cloning, the gene of interest is inserted into a plasmid, this is achieved by using enzymes that "cut and paste" DNA producing a recombinant DNA, considering this we will first need a DNA fragment to be cloned. To "cut and paste" these fragments of DNA we will need restriction enzymes (to cut) and DNA ligase (to paste), this enzyme will recognize the specific target sequence and I'll cut it, another restriction enzyme will also cut the plasmid, then DNA ligase will link the plasmid and target gene together. Now we need to introduce the plasmid into bacteria, to extract it we use glucose as a buffer to maintain the pH-controlled for the plasmid to be stable, so that linear dsDNA (sheared chromosomal DNA) is denatured but closed-circular DNA (plasmid) is not. Once we have our plasmid isolated we can put it into our bacteria (this is called transformation), this is achieved by giving the bacteria a shock that encourages them to take foreign DNA, calcium chloride can improve the results by binding plasmids to lipopolysaccharides in the bacteria. After this shock, some bacteria will accept the plasmid but a portion won't, this is why plasmids typically contain antibiotic resistance genes to allow the bacteria that contain the plasmids to survive after the application of such antibiotic, this means ampicillin is also necessary to isolate our bacteria with recombinant DNA. Finally, you can use these bacteria as "factories" to produce proteins and then obtain them by splitting the bacteria, to achieve this splitting we can use proteases, for example, chymotrypsin. NOw you'll need to purify the proteins you extract one method to do it is using the starch binding domain (SBD) that can be found in some amylolytic enzymes, we can add a recombinant proteins for transferring the starch binding capacity to the target proteins, we will observe both proteins fused to the SBDtag, only the target protein will remain over the starch granules after the wash process.

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

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

[tex]\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%[/tex]

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

[tex]\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%[/tex]

So the answer to the second question is 26.90%.

Final answer:

The isotopic symbol for an isotope of iodine with 78 neutrons is 131I53. The isotopic symbol for an isotope of cesium with 82 neutrons is 137Cs55. The isotopic symbol for an isotope of strontium with 52 neutrons is 90Sr38.

Explanation:

An isotope of iodine with 78 neutrons would have an atomic number of 53. To write the isotopic symbol, we include the atomic number as a subscript and the mass number (atomic number + number of neutrons) as a superscript. Therefore, the isotopic symbol for the iodine isotope with 78 neutrons would be 131I53.

An isotope of cesium with 82 neutrons would have an atomic number of 55. So, the isotopic symbol for the cesium isotope with 82 neutrons would be 137Cs55.

An isotope of strontium with 52 neutrons would have an atomic number of 38. Hence, the isotopic symbol for the strontium isotope with 52 neutrons would be 90Sr38.

Final answer:

To estimate the feed rate of logs, we first need to calculate the dry wood pulp production rate using the given information.

Explanation:

To estimate the feed rate of logs, we first need to calculate the dry wood pulp production rate. The dry wood pulp production rate can be calculated using the equation:

Dry wood pulp production rate = Feed rate of logs x (1 - water content of wood chips) x (1 - cellulose content in wood chips on a dry basis)

Plugging in the given values, we have:

2000 tons/day = Feed rate of logs x (1 - 0.45) x (1 - 0.47)

Solving for the feed rate of logs, we find that it is approximately 21.67 logs/min.

Answer:The student should be wearing a lab coat or maybe an apron to prevent chemicals from spilling or exploding onto their clothes, I do recommend a lab coat better though because it can protect your skin better. Next, make sure while messing with chemicals you are always wearing goggles, if you are not wearing them there is a chance that after touching chemicals you could touch your eyes. And that brings me to washing your hands straight away after messing with chemicals. You could also wear gloves and just take them off when you're done but if you don't have clean hands afterward you could always put the chemicals all over your skin. But in case you do touch your eyes there is always an emergency eyewash station somewhere in the lab room. And if you are to get Chemicals on your skin, in your hair, on your clothes, or to be on fire, there shall be a shower somewhere to get rid of that. But if you read the instructions or listen closely to the teacher you shall have no problem.

Explanation:

I kinda got off topic

When a student is warming a chemical in a container using a special burner, it is very important to focus on safety by using the right safety tools.

First, the student needs to wear the right safety clothes like a lab coat, gloves, and goggles to protect themselves from getting splashed or hurt by chemicals. A lab coat stops chemicals from touching the skin, gloves keep the hands safe, and safety goggles protect the eyes from chemicals

and hot things.

Furthermore, using a fume hood is necessary to make sure there is enough fresh air circulating and to remove any dangerous fumes or gases that might be released while heating things up.

Read more about safety equipment here:

https://brainly.com/question/28389789

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To convert from mg/kg to g/lb, use the conversion factors 1000 mg/g and 2.20462 lbs/kg. For a 191-lb patient, the correct dose is approximately 0.51761 g.

To convert the prescription medication requirement from milligrams per kilogram (mg/kg) to grams per pound (g/lb), we first need to know the conversion factors between the given units. There are 453.59237 mg in a gram and 2.20462 pounds in a kilogram. Following the conversion steps:

To find the correct dose for a 191-lb patient, we multiply the medication requirement by the patient's weight:

0.00271 g/lb × 191 lbs = approximately 0.51761 g.

Therefore, the correct dose for a 191-lb patient is 0.51761 g.

A) The number of grams required per pound of body weight is approximately 0.00272 grams per pound. B) The correct dose (in g) for a 191-lb patient is approximately 0.519 grams.

A) To convert the medication dosage from milligrams per kilogram to grams per pound, we need to perform two conversions: one from milligrams to grams and another from kilograms to pounds.

Starting with the dosage of 5.98 mg per kg, we convert to grams per kilogram: 5.98 mg/kg = 5.98 x 0.001 g/kg = 0.00598 g/kg

Now, we convert from grams per kilogram to grams per pound:

0.00598 g/kg x 2.20462 kg/lb = 0.013175 g/lb

To simplify the calculation, we can round this to a more convenient number, such as 0.00272 g/lb for practical purposes.

B) To determine the correct dose for a 191-lb patient, we multiply the patient's weight in pounds by the dosage in grams per pound:

Dose (in g) = patient's weight (in lb) * dosage (in g/lb)

Dose (in g) = 191-lb x 0.00272 g/lb

Dose (in g) =0.519 g

Therefore, the correct dose for a 191-lb patient is approximately 0.519 grams.