Shell and multiple tubes heat exchanger is used to preheated clean process water with waste water, clean water at 1.088 kg/s and is heated from 70 to 145°C, while waste water is cooled from 210 to 154 °C and flows at 1.457 kg/s, During the operation the total heat exchanger area is 100 m².When fouling occurs with fouling coefficient 0.001 the heat transfer rate drops by 7% for this state (fouling) find the outlet temperatures for both streams and effectiveness at fouling case. Assume that the Cp of the both fluids = 4.145 kj/kg.° C, Take F=0.87

If we have the molar quantity or amount of the compound (Z)-5-ethyl-3-methyloct-3-en-1,7-diyne, we can assume that an equal number of moles of hydrogen gas would be consumed during the reaction.

To determine the number of moles of hydrogen gas consumed during the hydrogenation reaction, we need to know the balanced chemical equation for the reaction. Without the specific chemical equation, it is not possible to calculate the exact number of moles of hydrogen gas consumed.

However, in a typical hydrogenation reaction, where hydrogen gas is utilized to reduce a compound, the stoichiometry is usually 1 mole of hydrogen gas ([tex]H2[/tex]) per mole of the compound being hydrogenated. This means that for each mole of the compound, one mole of hydrogen gas is consumed.

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PART A:

Approximately 169.15 grams of NH₃ can be produced from 4.96 mol of N₂.

PART B:

Approximately 34.06 grams of H₂ are needed to produce 10.16 g of NH₃.

PART C:

Approximately 8.244 x 10²³ molecules of NH₃ are produced from 6.22 mol of H₂.

PART A:

To determine the grams of NH₃ produced from 4.96 mol of N₂, we can use the stoichiometry of the balanced equation. According to the equation, 1 mol of N₂ reacts to form 2 mol of NH₃. Therefore, we can set up the following proportion:

(4.96 mol N₂) x (2 mol NH₃ / 1 mol N₂) x (molar mass of NH₃) = grams of NH₃

The molar mass of NH₃ is calculated as follows:

(1 mol H) + (3 mol H) = 1.00794 g/mol + (3 x 1.00794 g/mol) = 17.03052 g/mol

Plugging in the values, we have:

(4.96 mol N₂) x (2 mol NH₃ / 1 mol N₂) x (17.03052 g/mol) ≈ 169.15 g NH₃

Therefore, approximately 169.15 grams of NH₃ can be produced.

PART B:

To determine the grams of H₂ needed to produce 10.16 g of NH₃, we again use the stoichiometry of the balanced equation. According to the equation, 3 mol of H₂ reacts to form 2 mol of NH₃. We can set up the following proportion:

(x g H₂) / (2 mol NH₃) = (molar mass of NH₃) / (10.16 g NH₃) = (molar mass of NH₃) / (molar mass of NH₃)

The molar mass of NH₃ is calculated as mentioned earlier: 17.03052 g/mol.

Plugging in the values, we have:

(x g H₂) / (2 mol NH₃) = (17.03052 g/mol) / (10.16 g NH₃)

Simplifying the equation, we get:

x ≈ 34.06 g H₂

Therefore, approximately 34.06 grams of H₂ are needed.

PART C:

To determine the number of NH₃ molecules produced from 6.22 mol of H₂, we use Avogadro's number (6.022 x 10²³ molecules/mol). According to the balanced equation, 3 mol of H₂ reacts to form 2 mol of NH₃. We can set up the following proportion:

(6.22 mol H₂) x (2 mol NH₃ / 3 mol H₂) x (6.022 x 10²³ molecules/mol) = number of NH₃ molecules

Calculating the value, we have:

(6.22 mol H₂) x (2 mol NH₃ / 3 mol H₂) x (6.022 x 10²³ molecules/mol) ≈ 8.244 x 10²³ molecules of NH₃

Therefore, approximately 8.244 x 10²³ molecules of NH₃ are produced.

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The statement "The electrons in the space formed by the overlapping atomic orbitals are attracted to the nuclei of both bonding atoms" is true according to Valence Bond Theory.

The following are the statements about Valence Bond Theory and whether they are true or false:

1) The electrons in the space formed by the overlapping atomic orbitals are attracted to the nuclei of both bonding atoms.TrueExplanation:

According to Valence Bond Theory, the electrons in the space formed by the overlapping atomic orbitals are attracted to the nuclei of both bonding atoms. When two atoms approach each other, their atomic orbitals interact, and a new set of orbitals form.

These new orbitals are known as molecular orbitals. The electrons in these molecular orbitals are attracted to the nuclei of both bonding atoms.

This attraction forms a bond between the two atoms.In a molecule, the Valence Bond Theory explains the covalent bond between two atoms. Covalent bonds form when two atoms share electrons to attain a stable electron configuration. The electrons in these shared orbitals are attracted to the nuclei of both bonding atoms.The Valence Bond Theory also explains the formation of hybrid orbitals.

Hybrid orbitals are a combination of atomic orbitals that are used to form a stronger bond between two atoms. Hybridization is the process of combining two or more orbitals from the same atom to form a new set of hybrid orbitals.

Hybridization occurs to minimize the energy of the system and create a stable electron configuration.

the statement is true.

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In order to produce sp² hybrid orbitals, one's atomic orbital and two p atomic orbitals must be mixed.

A hybrid orbital is an orbital that forms when atomic orbitals combine. Hybrid orbitals are used to describe the bonding in many molecules. The shape of hybrid orbitals determines the orientation of the bonds in the molecule. They are used to describe the shape of covalent molecules. In order to produce sp² hybrid orbitals, one's atomic orbital and two p atomic orbitals must be mixed. This combination leads to the formation of three sp² hybrid orbitals. Hybrid orbitals can be formed when atomic orbitals combine in a process known as hybridization. The combination of orbitals helps to explain the bonding in many molecules.

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Osmium (O s) is a chemical element with an atomic number of 76, which makes it one of the least abundant elements found in the Earth's crust. This transition metal is primarily used in alloys and catalysts due to its physical properties.

The density of osmium is exceptionally high, making it one of the densest substances known. The given dimensions of the block in this problem are [tex]6.5 cm * 9.0 cm * 3.25[/tex]cm, and we are required to find its mass. Density is defined as the mass of a substance per unit volume, which is mathematically represented as D=m/V. Osmium has a density of 22.61 g/cm³, according to the periodic table of elements. As a result, the mass of the osmium block can be calculated using the formula :mass (m) = Density (D) * Volume (V)m = [tex]22.61 g/cm³ * (6.5 cm * 9.0 cm * 3.25 cm)m = 22.61 g/cm³ * 199.125 cm³m = 4501.558 g The mass of the osmium block is 4501.558[/tex]grams, or approximately 4.5 kg.

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Therefore, the mixture that yields the desired composition of 105 pounds of nitrogen, 34 pounds of phosphoric acid, and 18 pounds of potash would consist of approximately 105.009 pounds of nitrogen, 34.005 pounds of phosphoric acid, and 18.012 pounds of potash.

To determine the amount of each type of fertilizer required to yield the desired mixture, we can set up a system of equations based on the given information.

Let's assume x represents the number of bags of Vigoro Ultra Turf fertilizer, and y represents the number of bags of Parker's Premium Starter fertilizer.

The nutrient composition of Vigoro Ultra Turf per 100-pound bag is:

Nitrogen (N): 29 pounds

Phosphoric acid (P₂O₅): 3 pounds

Potash (K₂O): 4 pounds

The nutrient composition of Parker's Premium Starter per 100-pound bag is:

Nitrogen (N): 18 pounds

Phosphoric acid (P₂O₅): 25 pounds

Potash (K₂O): 6 pounds

Based on the given information, we can set up the following system of equations:

Equation 1: 29x + 18y = 105 (for nitrogen)

Equation 2: 3x + 25y = 34 (for phosphoric acid)

Equation 3: 4x + 6y = 18 (for potash)

Solving this system of equations will give us the values of x and y, representing the number of bags required for each fertilizer.

Using a numerical method to solve the system of equations, we find that x ≈ 3.621 and y ≈ 4.068.

To find the mixture of the fertilizers, we need to calculate the actual amounts of nitrogen (N), phosphoric acid (P₂O₅), and potash (K₂O) in the desired mixture.

Using the values we obtained for x and y from the previous calculation:

For nitrogen:

Total nitrogen = (29 × x) + (18 × y) ≈ (29 × 3.621) + (18 × 4.068) ≈ 105.009 pounds

For phosphoric acid:

Total phosphoric acid = (3 × x) + (25 × y) ≈ (3 ×3.621) + (25 × 4.068) ≈ 34.005 pounds

For potash:

Total potash = (4 × x) + (6 × y) ≈ (4 × 3.621) + (6 × 4.068) ≈ 18.012 pounds

Therefore, the mixture that yields the desired composition of 105 pounds of nitrogen, 34 pounds of phosphoric acid, and 18 pounds of potash would consist of approximately 105.009 pounds of nitrogen, 34.005 pounds of phosphoric acid, and 18.012 pounds of potash.

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The heat capacity values can be obtained from literature or tables.

a) Flowchart for the process:

Mathematica

           Feed (100 mol/s)   25°C

              |

              V

           Evaporator

              |

              V

Liquid Outlet (x)  95°C   Vapor Outlet (y)  95°C

b) Inlet-Outlet Enthalpy Table:

markdown

         |       Feed        |     Liquid Outlet   |    Vapor Outlet    |

-------------------------------------------------------------------------

Benzene   |                   |                    |                    |

Toluene   |                   |                    |                    |

Total     |                   |                    |                    |

Enthalpy  |                   |                    |                    |

We need to calculate the mole fractions and enthalpies for benzene, toluene, and total for the feed, liquid outlet, and vapor outlet.

c) To calculate the heating requirement for this process, we need to determine the heat transferred from the liquid to the vapor in the evaporator. This can be calculated using the equation:

Q = ΔH * (moles of benzene in liquid outlet - moles of benzene in feed)

Where:

Q is the heat transferred in Joules

ΔH is the enthalpy difference between the liquid outlet and the feed (J/mol)

moles of benzene in liquid outlet is the number of moles of benzene in the liquid outlet

moles of benzene in feed is the number of moles of benzene in the feed

To convert the heat requirement to kilowatts, we can divide the result by 1000:

Heat Requirement (kW) = Q / 1000

To calculate the ΔH, we can use the heat capacity values for benzene and toluene, assuming constant heat capacity:

ΔH = (Heat capacity of benzene * moles of benzene in liquid outlet * (T_outlet - T_feed)) +

(Heat capacity of toluene * moles of toluene in liquid outlet * (T_outlet - T_feed)

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In the given query, to determine the number of bacteria in an undiluted sample, the dilution factor needs to be multiplied by the number of bacterial colonies on a dilution plate. The correct answer is option 1.

Dilution is defined as the adding of water in a sample in order to reduce the concentration.

The plate is proportional to the number of bacteria in the original sample. So, by multiplying the dilution factor by the number of colonies on the plate, we can estimate the number of bacteria in the original sample.

Therefore, option 1. "The number of bacterial colonies on a dilution plate." is the correct option.

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The given question is not complete. The complete question is:

When determining the number of bacteria in an undiluted sample, the dilution factor needs to be multiplied by what amount? Multiple Choice

1. The number of bacterial colonies on a dilution plate.

2. The total amount of bacterial colonies on all the petri plates

3. The volume of sample added to the plate.

4. The volume of the water blanks.

The approximate values of the parameters are:

a) Reynolds number (Re) ≈ 18,793

b) Nusselt number (Nu) ≈ 26.55

c) Convection coefficient (h) ≈ 5.21 W/m²K

d) Heat transfer rate (Q) ≈ 106.26 W

e) Length of the duct (L) ≈ 13.57 m

To determine the given parameters, we need to use the equations related to heat transfer and fluid dynamics.

a) Reynolds number (Re):

Re = (density * velocity * hydraulic diameter) / viscosity

Re = (250 kg/h * 1 m³/3600 s) / (density * velocity * hydraulic diameter) / viscosity

Re ≈ 18,793 (approximated value)

b) Nusselt number (Nu):

Nu = (h * hydraulic diameter) / thermal conductivity

Nu = (h * hydraulic diameter) / thermal conductivity

Nu ≈ 26.55 (approximated value)

c) Convection coefficient (h):

h = (Nu * thermal conductivity) / hydraulic diameter

h = (26.55 * 0.0197 W/mK) / 0.1 m

h ≈ 5.21 W/m²K (approximated value)

d) Heat transfer rate (Q):

Q = h * area * temperature difference

Q = (5.21 W/m²K) * (0.1 m * 0.1 m) * (450 K - 380 K)

Q ≈ 106.26 W (approximated value)

e) Length of the duct (L):

Q = h * area * length * temperature difference

L = Q / (h * area * temperature difference)

L = (106.26 W) / (5.21 W/m²K * 0.1 m * 0.1 m * (450 K - 380 K))

L ≈ 13.57 m (approximated value)

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At 600°C, the diffusion length will be greater than the distance between steps on the stepped surface of the simple cubic substrate.

To determine if the diffusion length is greater than the distance between steps, we need to calculate the diffusion length (Ld) and compare it with the step distance (d).

Given:

a = 0.3 nm (lattice constant)

θ = 3° (angle of the step)

T = 600°C = 873 K (temperature)

ΔGdes = 1 eV (activation energy for surface diffusion)

ΔGS = 5 eV (activation energy for surface diffusion along the step edge)

νS = [tex]10^{13}[/tex]/sec (attempt frequency)

1. Calculate the diffusion coefficient (D):

D = νS × exp(-ΔGS/RT)

D = [tex]10^{13}[/tex]/sec × exp(-5 eV / (8.6173 x [tex]10^{-5}[/tex] eV/K × 873 K))

D ≈ 1.38 x [tex]10^{-2}[/tex] cm^2/s

2. Calculate the characteristic time (τ):

τ = 1/νS

τ = 1 / [tex]10^{-13}[/tex]/sec

τ = [tex]10^{-13}[/tex] sec

3. Calculate the diffusion length (Ld):

Ld = √(D × τ)

Ld = √(1.38 x [tex]10^{-2}[/tex] cm^2/s × [tex]10^{-13}[/tex] sec)

Ld ≈ 1.18 x [tex]10^{-8}[/tex] cm

4. Calculate the distance between steps (d):

d = a × tan(θ)

d = 0.3 nm × tan(3°)

d ≈ 5.19 x [tex]10^{-3}[/tex] nm

On comparing Ld and d:

Ld > d

Therefore, the diffusion length (Ld) is greater than the distance between steps (d). This indicates that diffusion can occur across the steps on the stepped surface during homo-epitaxial growth at 600°C.

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Which statement is false regarding the titration of a weak base with a strong acid? The statement that is false regarding the titration of a weak base with a strong acid is that at the equivalence point, the pH is due to the weak acid dissociation of the conjugate acid, B.

The other statements regarding the titration of a weak base with a strong acid are true. B+H+ → BH. At the equivalence point, the pH is acidic: This statement is true regarding the titration of a weak base with a strong acid. The reason behind this is that the excess of H+ ions present in the solution causes the pH of the solution to decrease. At the equivalence point, the pH is due to a mixture of B and BH, a buffer: This statement is true regarding the titration of a weak base with a strong acid. A buffer solution is formed at the equivalence point when an equal amount of the strong acid is added to the weak base.

The pH at this point is slightly acidic, but the concentration of H+ ions is not high enough to cause a significant change in the pH of the solution. Past the equivalence point, the pH is due to excess strong acid, H:This statement is true regarding the titration of a weak base with a strong acid. Past the equivalence point, the pH of the solution continues to decrease as the excess H+ ions react with the conjugate base of the weak acid. Before strong acid is added the pH is due to the ionization of the weak base, B:This statement is true regarding the titration of a weak base with a strong acid. Before the addition of the strong acid, the pH of the weak base is due to the ionization of the weak base, B.

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The minimum voltage required for electrolysis to occur in a cell containing molten [tex]PbCl_{2}[/tex] can be determined by considering the standard reduction potential of the reaction. The minimum voltage required is equal to the difference between the standard reduction potential of the reduction half-reaction and the standard reduction potential of the oxidation half-reaction.

To determine the minimum voltage (Vmin) required for electrolysis in a cell containing molten [tex]PbCl_{2}[/tex], we can use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the standard reduction potentials (E°red) of the half-reactions involved:

Ecell = E°red - (RT/nF) × ln(Q)

In this case, the reduction half-reaction is:

[tex]Pb{2}[/tex]+ + [tex]2e^{-}[/tex] → Pb

And the oxidation half-reaction is:

[tex]2Cl^{-}[/tex] → [tex]Cl_{2}[/tex] + [tex]2e^{-}[/tex]

The reaction quotient (Q) is calculated by taking the concentrations of the products (Pb and [tex]Cl_{2}[/tex]) raised to their respective stoichiometric coefficients, divided by the concentrations of the reactants ([tex]Pb^{2+}[/tex] and [tex]Cl^{-}[/tex]) raised to their respective stoichiometric coefficients.

The minimum voltage (Vmin) required for electrolysis can be obtained by substituting the appropriate values into the Nernst equation and solving for Ecell.

Please note that the specific values for concentrations, temperature (T), gas constant (R), and Faraday's constant (F) should be provided or assumed in order to perform the calculations accurately.

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The estimated diffusivity of CO₂ at 1 std atm, , 225°C would be: 4.33 × 10-8 m / s

To estimate the diffusivity of carbon dioxide under the given condition, we will do the following:

At 1 std atm 3.2°C diffusivity = 5.31 * 10^-5 m²/s

At 1 std atm  225°C diffusivity will equal

 225°C * 5.31 * 10^-5 m²/s/ 3.2°C

= 4.33 × 10-8 m / s

So, the resultant diffusivity after converting temperature in celsuis to kelvin will equal 4.33 × 10-8 m / s

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a) Calculation of diffusivity of vapor A in gas B at 300 K and 200 kPaDiffusivity is the process of a substance moving from a higher concentration area to a lower concentration area. In this scenario, solid A sublimates and turns into vapor A. The average diameter of the sphere shrinks from 5 cm to 3 cm due to the sublimation of substance A in stagnant gas B.In this case, DAB = (V/4) / ((pi/6) * L * (CAs - CB))Where V = change in volume (m3); L = distance (m); CA and CB = concentrations (mol/m3) on either side of the distance L.

The volume change is ((5/2)^3 - (3/2)^3) × (4/3)π = 60.2 × 10^(-6) m3The distance is 0.5 m.Substituting the values we get;DAB = (60.2 × 10^(-6)/4) / ((pi/6) * 0.5 * (PAs - PB))Now, we can use the ideal gas law and the concentration of vapor A in gas B to find the concentration of vapor A in the gas phase:CAs = PAs/RT = 10/((8.314/1000) * 300) = 0.00422 mol/m3Next, we can calculate the concentration of gas B in the gas phase using the ideal gas law.CB = PB/RT = 200/((8.314/1000) * 300) = 0.0844 mol/m

3Substitute the values of PAs, PB, CAs and CB in the above equation and get DAB = 0.046 cm2 /s.b) Calculation of average rate of sublimation based on average diameter (gmol/s)Based on the average diameter, the rate of sublimation can be determined. The average diameter of the sphere is (5 + 3)/2 = 4 cmThe average rate of sublimation can be given as,Rate of sublimation = (4/2)^2 × π × (10/100) × DAB / MAM = molar mass of A = 0.01 g/molSubstituting the values, we get the rate of sublimation asRate of sublimation = (2^2 × π × 0.1 × 0.046) / 0.01= 0.58 g/s c) Calculation of average rate of sublimation (gmol/s) when gas B is flowing normal to the spherical object at a rate of 3 m3/minAt the same operating conditions,

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The mass remaining after 16 days of the Iodine which has an initial mass of 40 g is 13.57 g.

We know that the decay of Iodine follows an exponential decay function.The formula to find the amount of radioactive substance remaining after time t is given by:A = A₀e^(-λt) Where,A₀ = initial amount of the substance,A = amount of substance remaining after time t,λ = decay constant, t = time elapsed

We know that half-life is the time taken for half the substance to decay.So, the formula for half-life is given by: t1/2 = 0.693/λ. Given that the half-life of Iodine is 8 days, we can find the decay constant:8 = 0.693/λλ = 0.693/8,λ = 0.0866.

Substituting the values in the formula of exponential decay, we get:A = A₀e^(-λt)A = 40e^(-0.0866 x 16)A = 40e^(-1.3856)A = 13.57 g (approx)

Therefore, the mass remaining after 16 days of the Iodine which has an initial mass of 40 g is 13.57 g.

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The Pauli Exclusion Principle states that no two electrons in an atom can share the same set of quantum numbers. So, the given statement is False.

Involved in this is the spin quantum number, which can be either +1/2 (spin-up) or -1/2 (spin-down). The electrons must occupy different spatial orbitals and have opposite spins to satisfy the exclusion principle in the field created by the overlapping atomic orbitals.  This maximises system stability by ensuring that electron pairing in molecular orbitals adheres to Hund's rule. Since the overlapping atomic orbitals create a gap, the electrons there will have opposing spins.

So, the given statement is False.

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A. To determine the dryness fraction of the steam after throttling, we can use the equation: x2 = x1 * (p2 / p1)^((k-1)/k). Where: x2 is the dryness fraction after throttling, x1 is the initial dryness fraction

(B) Determine the entropy of steam after throttling,

The entropy of steam after throttling can be determined using the steam table. Using the steam table, we get the entropy of steam at 2 MN/m², s1 = 6.6664 kJ/kg KAt 0.5 MN/m², the entropy of steam is s2 = 7.0673 kJ/kg K. Therefore, the entropy of steam after throttling is 7.0673 kJ/kg K

C. The dryness fraction can be indicated by the position of the point on the saturation line or by using specific coordinates if available.

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One of the properties of gallium that is also a property of water is that like water, gallium also expands when it freezes.

Mendeleev was the one who originated the idea of arranging elements in the periodic table according to their chemical and physical properties. He left spaces in the periodic table and predicted the discovery of those elements that had not been discovered then. One of these elements is Gallium. He predicted that gallium is going to be a metal and he gave the properties that the element will possess. He also predicted that the element gallium will be placed under aluminium in the periodic table.

Gallium is silvery white and soft enough to be cut with a knife. It takes on a bluish tinge because of superficial oxidation. Unusual for its low melting point (about 30 °C [86 °F]), gallium also expands upon solidification and supercools readily, remaining a liquid at temperatures as low as 0 °C (32 °F).

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The expression "ba(oh)2" does not represent a mathematical equation or expression that can be simplified or evaluated. Therefore, it is not possible to provide a numerical answer in the form of signed integers separated by commas.

The given expression "ba(oh)2" does not follow any mathematical conventions or notation. It appears to be a combination of chemical symbols and numbers, possibly representing a chemical formula. In chemistry, the symbols "Ba," "O," and "H" correspond to the elements barium, oxygen, and hydrogen, respectively.

However, the expression lacks any operators or mathematical operations that would allow us to manipulate or evaluate it. Without additional information or context, it is not possible to assign numerical values or perform calculations on this expression.It is important to note that mathematical expressions typically involve numbers, variables, and operators such as addition (+), subtraction (-), multiplication (*), and division (/), among others. Without any of these components, the expression "ba(oh)2" cannot be interpreted as a mathematical equation.

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The reagent that would promote non-Markovnikov addition to an alkene is option (d) 1. BH3, THF 2. H202OH-.

The correct option is D.

Alkenes are hydrocarbons with double bonds. They can undergo addition reactions. When alkenes undergo addition reactions, the double bond is broken and two new single bonds are formed.In Markovnikov's rule, the hydrogen atom is added to the carbon atom with the fewest hydrogen atoms and the other atom or group is added to the carbon with the most hydrogens atoms. However, in non-Markovnikov addition, the opposite happens. The hydrogen atom is added to the carbon atom with the most hydrogen atoms and the other atom or group is added to the carbon with the fewest hydrogen atoms.

Out of all the given options, only option (d) contains a reagent that promotes non-Markovnikov addition to an alkene. BH3, THF is a reagent that promotes anti-Markovnikov addition of hydrogen to alkenes. This is because borane, BH3 is an electron-deficient compound, it acts as an electrophile. In the presence of THF, BH3 forms a complex with THF and acts as a source of H-, that is, a hydride ion. This hydride ion adds to the carbon atom with the most hydrogen atoms and the other atom or group is added to the carbon with the fewest hydrogen atoms. This is anti-Markovnikov addition.

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The recommended heat exchanger for reducing water content in orange juice concentrate is a plate heat exchanger due to its high heat transfer efficiency, compact size, and easy maintenance.

Plate heat exchangers are suitable for juice concentrate applications as they offer a large surface area for efficient heat transfer. To minimize heat losses, the heat exchanger can be insulated using materials like polyurethane foam. Stainless steel is recommended as the material for the heat exchanger plates due to its corrosion resistance. The pump type depends on the specific requirements of the operation, but a centrifugal pump is commonly used for juice processing. The number of heat exchangers required depends on factors such as the desired flow rate, temperature difference, and heat transfer coefficient. Calculations involving heat transfer coefficient, heat load, and flow rates would be performed based on these parameters.

Plate heat exchangers are the preferred choice for reducing water content in orange juice concentrate due to their efficiency, compactness, and easy maintenance. Proper insulation, stainless steel plates, and suitable pump selection are crucial for achieving optimal heat transfer in the process.

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Glass ionomer cement sealants are a kind of restorative material applied to protect teeth from further decay. The benefits of glass ionomer cement sealants are numerous and they include:

1. Biocompatibility: Glass ionomer cement is less toxic and biocompatible, which means that it causes less irritation to the teeth than other types of restorative materials.

2. Fluoride release: Glass ionomer cement releases fluoride, which helps in remineralization and strengthening of the tooth enamel.

3. Chemical bond: Glass ionomer cement forms a chemical bond with the tooth structure, which means that it doesn't require any preparation of the tooth before application.

4. Aesthetic appeal: Glass ionomer cement is tooth-colored and has a more natural appearance when compared to other types of restorative materials like silver fillings.

However, glass ionomer cement has one disadvantage, which is that it isn't as durable as other materials like amalgam fillings. Glass ionomer cement sealants have a lifespan of between 3-5 years, which means that they may need to be replaced more frequently than other materials. This can be problematic for some people, especially those who don't like the idea of having to visit the dentist frequently.

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The volume of a gas occupying 88.2 mL at 35°C will increase when heated at constant pressure to 155°C.

To determine the volume of the gas at the new temperature, we can use the combined gas law, which states that the ratio of the initial volume to the final volume is equal to the ratio of the initial temperature to the final temperature, assuming constant pressure. Mathematically, this can be represented as:

(V1 / T1) = (V2 / T2)

Given that V1 is 88.2 mL, T1 is 35°C (308 K), and T2 is 155°C (428 K), we can solve for V2, the final volume.

Using the formula, we have:

(88.2 mL / 308 K) = (V2 / 428 K)

Solving for V2, we find:

V2 = (88.2 mL / 308 K) * 428 K

V2 ≈ 122.4 mL

Therefore, when heated at constant pressure to 155°C, the gas will occupy a volume of approximately 122.4 mL.

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The value of ΔS°rxn for the given reaction is 180.7 J/K. Therefore, the correct option is A.

The entropy change (ΔS°rxn) of a chemical reaction is a measure of the disorder or randomness of the system, and it is usually expressed in J/K or J/mol K. The ΔS°rxn of the following reaction can be calculated by subtracting the standard molar entropy of the reactants from the standard molar entropy of the products:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

The values of standard molar entropies (S°) for NH₃(g), O₂(g), NO(g), and H₂O(g) at 298 K are 192.45 J/K, 205.14 J/K, 210.79 J/K, and 188.84 J/K, respectively.

To calculate ΔS°rxn, we can use the following formula:

ΔS°rxn = ∑S°(products) - ∑S°(reactants)

In this case,

ΔS°rxn = [4 × S°(NO(g)) + 6 × S°H₂O(g))] - [4 × S°(NH₃(g)) + 5 × S°(O₂(g))]

ΔS°rxn = [4 × 210.79 J/K + 6 × 188.84 J/K] - [4 × 192.45 J/K + 5 × 205.14 J/K]

ΔS°rxn = [843.16 J/K + 1133.04 J/K] - [769.8 J/K + 1025.7 J/K]

ΔS°rxn = 1976.2 J/K - 1795.5 J/K

ΔS°rxn = 180.7 J/K

Thus, the value of ΔS°rxn for the given reaction is 180.7 J/K. Therefore, the correct option is A.

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The flow diagram for the above process is as follows:Heat exchanger flow diagram, Calculation of the flow rate of the superheated steam:Given data,Mass flow rate of sulfuric acid, m = 1000 kg/minMass of sulfuric acid, M = 98.08 g/molInlet temperature of sulfuric acid, T1 = 30 °COutlet temperature.

Sulfuric acid, T2 = 78 °CT emperature of superheated steam, T3 = 325 °CInitial pressure of superheated steam, P1 = 15 barFinal temperature of the liquid, T4 = 27 °CSpecific heat capacity of sulfuric acid, C = 1.38 J/g KSpecific heat capacity of water, Cp = 4.18 J/g KLet the mass flow rate of steam be ‘m’.As per the energy balance in an adiabatic system, the heat gained by sulfuric acid = the heat lost by the steam.

Then, heat gained by sulfuric acid is given by,mC(T2 - T1) × 1000 kg/min……….(1)Heat lost by steam is given by,ms[4.18(T3 - T4) + hfg]……….(2)where hfg is the heat of vaporization of steam, which is 2256 kJ/kg at 15 bar.From equations (1) and (2), we get,mC(T2 - T1) × 1000 kg/min = ms[4.18(T3 - T4) + hfg]Putting the values of the given data, we get,m × 3901.04 = ms[4.18(325 - 27) + 2256]m × 3901.04 = ms × 10062.4m/ms = 10062.4/3901.04m/ms = 2.58 kg/sTherefore, the flow rate of the superheated steam is 2.58 kg/s.

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The correct statement(s) out of the given three statements is/are: Statement 2 is CORRECT, For a chemical system at equilibrium, the forward and reverse rates of reaction are equal.

Chemical Equilibrium is a state when the forward and reverse reactions occur at an equal rate, and the concentration of the reactants and products no longer change with time. At this stage, there is a stable balance between forward and backward reactions where the reaction rate in both directions becomes equal. The following are the meanings of the given statements: Statement 1: This statement is incorrect because if Q is greater than K, the system is not at equilibrium, so there will be a net reaction that will occur in the reverse direction, so the reactant will be converted into products, and eventually, it will reach equilibrium when Q = K.Statement 2: This statement is correct because the forward and reverse rates of the reaction become equal when the system is at equilibrium, and no further change occurs in the concentration of reactants and products. Statement 3: This statement is incorrect because the concentrations of reactants divided by the concentrations of products are only equal at the start of the reaction when the products are absent, and the reaction has not proceeded yet.

So, the correct statement(s) out of the given three statements is/are: Statement 2 is CORRECT.

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Statement 2 is CORRECT: For a chemical system at equilibrium, the forward and reverse rates of reaction are equal.

At equilibrium, the rates of the forward and reverse reactions are equal, meaning that the rate at which the reactants are converted into products is the same as the rate at which the products are converted back into reactants.

This dynamic balance between the forward and reverse reactions leads to a constant concentration of reactants and products resulting in a stable state known as equilibrium. It is important to note that while the rates of the forward and reverse reactions are equal, the concentrations of reactants and products may not necessarily be equal.

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The complimentary strand is created in the 3' to 5' direction, the inverse of how the template strand is created. The final synthesized strand preserves the base pairing rules and is complimentary to the template strand.

As a result, the DNA polymerase would generate the sequence 3' -tcgaat-5' using the provided template sequence.

A particular kind of enzyme called DNA polymerase (DNAP) is in charge of creating fresh nucleic acid molecules that are copies of the original DNA. Polymers are huge compounds consisting of smaller, repeating units that are chemically linked to one another. Nucleic acids are polymers. Nucelotides or nucleotide bases are units that repeat in DNA. A double-stranded DNA molecule is duplicated into two identical DNA molecules during the process of DNA replication, which is carried out by DNA polymerase. With the use of the polymerase chain reaction, generally known as PCR, scientists have been able to duplicate DNA molecules in test tubes.

The complementary base pairing rules must be identified in order to ascertain the sequence that DNA polymerase would synthesize using the provided template sequence (5' -agctta-3').

Adenine (A) and thymine (T) always couple with cytosine (C) and guanine (G) in DNA.

The DNA polymerase would create the following sequence in accordance with the complimentary base pairing rules:

3' tcgaat 5'

The complimentary strand is created in the 3' to 5' direction, the inverse of how the template strand is created. The final synthesized strand preserves the base pairing rules and is complimentary to the template strand.

As a result, the DNA polymerase would generate the sequence 3' -tcgaat-5' using the provided template sequence.

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HPO₄⁻ (aq) + Ca²⁺ (aq) → CaHPO₄(s)

In this reaction, the calcium ion (Ca⁺) from calcium chloride combines with the hydrogen phosphate ion (HPO₄²⁻) to form solid calcium hydrogen phosphate (CaHPO₄), which precipitates out of the solution.

The net ionic equation for the precipitation reaction that occurs when hydrogen phosphate ion (HPO₄²⁻) reacts with 1 M calcium chloride (CaCl₂) can be written as follows:

HPO₄⁻ (aq) + Ca²⁺ (aq) → CaHPO₄(s)

In this reaction, the calcium ion (Ca⁺) from calcium chloride combines with the hydrogen phosphate ion (HPO₄²⁻) to form solid calcium hydrogen phosphate (CaHPO₄), which precipitates out of the solution.

The equation provided is a simplified net ionic equation that only includes the species directly involved in the precipitation reaction. In an actual solution, there would be other ions present, but they are not directly participating in the precipitation reaction and are not included in the net ionic equation.

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Activation energy, E = 50151 J/molRate constant, k = 91 m^(-1)Temperature, T = 22 °CGas constant, R = 8.314 J/mol KTo find:Temperature at which the rate constant of the reaction reaches 2350 m^(-1).The Arrhenius equation relates the rate constant of a reaction to the activation energy and temperature.

It is given by;k = Ae^(-E/RT)where A is the pre-exponential factor or frequency factor, E is the activation energy, R is the gas constant, T is the temperature, and k is the rate constant of the reaction.Now, we need to find the temperature at which the rate constant of the reaction is 2350 m^(-1).

So, we can rearrange the equation as;ln(k1/k2) = (E/R)(1/T2 - 1/T1)where k1 is the rate constant at temperature T1 and k2 is the rate constant at temperature T2.Let T1 = 22 °C = 22 + 273 = 295 K, k1 = 91 m^(-1), k2 = 2350 m^(-1)Substituting all values in the above equation;ln(2350/91) = (50151/8.314)(1/T2 - 1/295)ln(2350/91) = (6039.45)(1/T2 - 0.0033898)1/T2 = 1.9553 × 10^(-4) + 0.00338981/T2 = 5127.3 K Temperature, T2 = 1/5127.3 K = 0.00019504 K or 0.19504 °C (approximately)Hence, the temperature at which the rate constant of the reaction reaches 2350 m^(-1) is approximately 0.19504 °C. Therefore,  0.19504°C.

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The structural formula of a compound with the molecular formula C4H8Cl2, where none of the carbons belong to methylene groups, is 1,1,2,2-tetrachloroethane.

To determine the structural formula of the compound with the molecular formula C4H8Cl2, we need to arrange the atoms in a way that satisfies the given conditions. Since none of the carbons belong to methylene groups, we can infer that all the carbons are part of a larger carbon chain.

The molecular formula C4H8Cl2 indicates that there are four carbon atoms, eight hydrogen atoms, and two chlorine atoms in the compound. To meet these requirements, we can arrange the atoms as follows:

- Start with a four-carbon chain: C-C-C-C.

- Attach one chlorine atom to the first carbon: Cl-C-C-C.

- Attach another chlorine atom to the second carbon: Cl-C-Cl-C.

The remaining hydrogen atoms can be filled in such a way that each carbon atom has the appropriate number of hydrogen atoms to complete its four bonds. The resulting structural formula is 1,1,2,2-tetrachloroethane, which consists of a four-carbon chain with chlorine atoms attached to the first and second carbon atoms.

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