Two nitro (NO_2) groups are chemically bonded to a patch of surface. They can't move to another location on the surface, but they can rotate (see sketch at right). It turns out that the amount of rotational kinetic energy each NO_2 group can have is required to be a multiple of epsilon, where epsilon = 1.0 times^-24 J. In other words, each NO_2 group could have epsilon of rotational kinetic energy, or 2 epsilon, or 3 epsilon, and so forth - but it cannot have just any old amount of rotational kinetic energy. Suppose the total rotational kinetic energy in this system is initially known to be 39 epsilon. Then, some heat is added to the system, and the total rotational kinetic energy rises to 59 epsilon. Calculate the change in entropy. Round your answer to 3 significant digits, and be sure it has the correct unit symbol.

Answer:

b. atomizes liquids into small solid particles

Explanation:

Spray drying -

It refers to the process of getting a dry powder from the liquid or the slurry ,with the help of hot gas , is referred to as spray drying .

The method is used in the field of pharmacy .

It is a type of atomizer or spray nozzle which helps to disperse the liquid into the controlled drop size spray .

Small solid particles are generated by this method .

Hence , from the given scenario of the question ,

The correct answer is b. atomizes liquids into small solid particles  .

Answer:

a lipid molecule that contains at least one carbohydrate unit

Explanation:

A glycolipid -

It refers to the lipid which consists of the carbohyde group attached via a glycosidic bond , which are basically covalent bonds , is referred to as a glycolipid .

They are present on the surface of the eukaryotic cell membranes .

Glycolipids are important to connect from one tissue to another and facilitate cellular recognition .

Hence , from the given information of the question ,

The correct answer is a lipid molecule that contains at least one carbohydrate unit .

Answer:

The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

Explanation:

For the complete recrystalization,

the amount of hot water should be such that, the benzoic acid is completely soluble in it.

As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.

we conclude that,

As, 68.0 grams of benzoic acid soluble in 1 L of water.

So, 0.700 grams of benzoic acid soluble in  of water.

The volume of water needed = 0.01029 L = 10.29 mL

conversion used : (1 L = 1000 mL)

Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

At a constant temperature of 50 °C, a volume of 1.0 L would contain approximately 0.045 moles of gas, whereas a volume of 3.0 L would contain approximately 0.134 moles of gas. These values are based on the standard molar volume of 22.4 L/mol at Standard Temperature and Pressure (STP).

In order to make this comparison, we can use Avogadro's Law which states that equal volumes of gases, at the same temperature and pressure, will contain an equal number of molecules or moles. We would also use the standard molar volume of an ideal gas at Standard Temperature and Pressure (STP), which is about 22.4 L/mol.

Given that our system's temperature is constant at 50 °C and comparing volumes of 1.0 L and 3.0 L, we simply calculate the number of moles in each volume. For 1.0 L, if we stay consistent with molar volume at STP, it equates to about 0.045 moles of gas (1 L / 22.4 L per mole). For 3.0 L, it would equate to about 0.134 moles of gas (3 L / 22.4 L per mole).

Please note that this approximation is based on the standard molar volume at STP, and accuracy may vary at temperatures significantly different from STP (such as 50 °C).

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Explanation:

Below is an attachment containing the solution.

Answer:

The concentration of the KBr solution is 6.27 mol/L

Explanation:

Step 1: Data given

Mass of KBr = 224 grams

Molar mass KBr = 119.0 g/mol

Volume = 300 mL = 0.300 L

Step 2: Calculate moles KBr

Moles KBr = mass KBr / molar mass KBr

Moles KBr = 224 grams / 119.0 g/mol

Moles KBr = 1.88 moles

Step 3: Calculate concentration KBr solution

Concentration solution = moles KBr / volume

Concentration solution = 1.88 moles / 0.300 L

Concentration solution = 6.27 mol / L

The concentration of the KBr solution is 6.27 mol/L

Answer: Option (b) is the correct answer.

Explanation:

A laboratory device which is used to separate fluids, gases or liquid on the basis of their density is known as a centrifuge. When a vessel containing the sample is spins at a high speed then separation takes place as the centrifugal force pushes the heavier material out of the vessel during this process.  

So, if we want the sample to safely run through the centrifuge then it is important that it should be free from impurities.

Thus, we can conclude that in order to safely run a sample in a centrifuge, it is important to make sure the sample is free from impurities.

When the safety run sample should be in a centrifuge so the sample should be free from impurities.

It is used to distinguish gases or liquids that depend upon the density is called a centrifuge. At the time When a vessel comprised of the sample should be spinning at a high speed so the separation took place since the centrifugal force pushes the heavier material that should be out of the vessel at the time of the process.  

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The question is incomplete, complete question is :

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 348 liters per second of dinitrogen are consumed when the reaction is run at 205°C and 0.72 atm. Calculate the rate at which ammonia is being produced.

Answer:

The rate of production of ammonia is 217.08 grams per second.

Explanation:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

Volume of dinitrogen used in a second = 348 L

Temperature of the gas = T = 205°C = 205+273 K = 478 K

Pressure of the gas = P = 0.72 atm

Moles of dinitrogen = n

[tex]n=\frac{PV}{RT}=\frac{0.72 atm\times 348 L}{0.0821 atm L/mol K\times 478 K}=6.385 mol[/tex]

According to reaction, 1 mole of dinitriogen gives 2 mole of ammonia.Then 6.385 moles of dinitrogen will give:

[tex]\frac{2}{1}\times 6.385 mol=12.769 mol[/tex]

Mass of 12.769 moles of ammonia;

12.769 mol 17 g/mol = 217.08 g

217.08 grams of ammonia is produced per second.So, the rate of production of ammonia is 217.08 grams per second.

Answer:

Ca(OH)2(s) + 2HCl(aq) →  CaCl2(aq) + 2H2O(l)

Explanation:

Step 1: Data given

Mass of Ca(OH)2 = 9.49 grams

Volume of a 0.490 M HCl solution = 30.0 mL = 0.030 L

Step 2: The unbalanced equation

Ca(OH)2 + HCl →  CaCl2 + H2O

On the left side we have 2x O (in Ca(OH)2 )  and on the right side we have 1x O (in H2O). To balance the amount of O we have to multiply H2O, on the right side , by 2.

Ca(OH)2 + HCl →  CaCl2 + 2H2O

On the right side we have 4x H (in 2H2O), and on the left side we have 3x H  (2x in Ca(OH)2 and 1x in HCl). To multiply the amount of H, we have to multiply HCl on the left side by 2.

Ca(OH)2(s) + 2HCl(aq) →  CaCl2(aq) + 2H2O(l)

Answer:

The retention factor, k is 2.49

Explanation:

According to the theory of High-Performance Liquid Chromatography (HPLC), the retention factor (or capacity), k, of a column is the ratio of the retention time of a retained analyte (toluene) to that of the  un-retained solute (methane).

This implies that:

k = [tex]\frac{16.35 (mins)}{ 6.39 (mins)}[/tex]  =  [tex]\frac{995 (secs)}{399 (secs)}[/tex]

 retention factor, k  = 2.49.

Note that there is no unit for retention factor, as it is a ratio.

Final answer:

The retention factor for toluene in the given gas chromatography system, calculated using the retention times of toluene and methane, is approximately 1.56.

Explanation:

The question involves calculating the retention factor (k) for toluene using gas chromatography (GC) data. The retention time is the time a compound spends in the column before being detected, which varies by compound owing to differences in interaction with the stationary phase. The retention factor is a measure of this time relative to an unretained solute, providing insight into the compound's affinity for the stationary phase versus the mobile phase.

To calculate the retention factor (k) for toluene, we use the formula: k = (t-R - t-M) / tM, where tR is the retention time of the analyte (toluene) and tM is the retention time of an unretained solute (methane, in this case). For toluene, t-R = 16.35 min, and for methane, t-M = 6.39 min. Substituting these values into the equation, we find k for toluene as follows: k = (16.35 - 6.39) / 6.39 = 9.96 / 6.39 ≈ 1.56.

Answer:

82.53 % protein in the milk.

Explanation:

Titration formula:

Conc (acid) × Vol (acid) = Conc (base) × Vol (base)

Beginning from back-titration:

0.5 M HCl × Vol  (HCl) = 0.3512 (conc of NaOH) × 34.50 mL (vol of NaOH)

=  [tex]\frac{12.1164}{0.5}[/tex] = 24.2328 mL

⇒ Vol of HCl used in initial titration is 24.2328 mL

So from Initial Titration:

Using same titration formula,

0.5 × 24.2328 = Conc. (base) × 100 ml

Conc (base) = 12.1164 ÷ 100

= 0.121164 M.

But concentration = mass ÷ molar mass

0.121164 = [tex]\frac{100 g}{ molar mass}[/tex]

= 825.3277 g/mol of protein

⇒ [tex]\frac{825.32765}{1000}[/tex] × 100

= 82.53 % protein in the milk.

Fusion and Fission both are nuclear reactions where the major difference between them is Fusion is combining process whereas fission is a breaking process.

Explanation:

Similarity:

Both Fission and Fusion are nuclear reactions that releases vast amount of energy.

Difference:

Answer: Cyclohexene

Explanation:

Cyclohexane belongs to the Alkenes family. Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom get attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed. When bromine is added to cyclohexane in the dark room, there won't be any reaction. If the mixture is exposed to light however, free bromine radicals are generated. In this condition, polybrominated products can be produced as well.

The chemical used to quench the residual bromine at the end of the isomerization reaction is usually an organic compound containing a reducing agent, such as sodium bisulfite (NaHSO3) or sodium hydrogen sulfite (NaHSO3).

The chemical used to quench the residual bromine at the end of the isomerization reaction is usually an organic compound containing a reducing agent, such as sodium bisulfite (NaHSO3) or sodium hydrogen sulfite (NaHSO3). These compounds react with bromine to form non-toxic salts that can be easily removed. For example, NaHSO3 reacts with bromine to form sodium bromide (NaBr) and sodium sulfate (Na2SO4). This reaction effectively removes the residual bromine and prevents it from causing further reactions or harming the desired product.

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Explanation:

The dissociation equations for the carbonic acid are as follows.

    [tex]H_{2}CO_{3}(aq) + H_{2}O(l) \rightarrow HCO^{-}_{3}(aq) + H_{3}O^{+}(aq)[/tex]

  [tex]HCO^{-}_{3}(aq) + H_{2}O(l) \rightarrow (CO^{2-}_{3})(aq) + H_{3}O^{+}(aq)[/tex]

The given buffer contains [tex]KHCO_{3}[/tex] and [tex]K_{2}CO_{3}[/tex]. This means that there will be [tex]HCO^{-}_{3}[/tex] ions and [tex]CO^{2-}_{3}[/tex] ions and since, both of them are present in the second step.

Therefore, [tex]K_{a2}[/tex] is more significant with respect to this reaction.

Thus, we can conclude that [tex]K_{a2}[/tex] is more important to this buffer.

A buffer resists changes in pH. When considering carbonic acid, which is a diprotic acid i.e., it can donate two protons, the first ionization is represented by Ka1, and the second by Ka2. In a buffer solution with physiological pH close to 7, Ka1 is more important because the first ionization of carbonic acid happens more readily at this pH.

Your question asks which Ka value is more important for this buffer, which contains 0.38 M KHCO3 and 0.71 M K2CO3. Carbonic acid is a diprotic acid with Ka1 = 4.5 × 10^−7 and Ka2 = 4.7 × 10^−11.

A buffer solution resists changes in pH by neutralizing added acids and bases. The effectiveness of a buffer is determined by the pKa of the acid component of the buffer, which is the negative logarithm of its Ka. Generally, the most effective buffers are those wherein the pKa is close to the desired pH.

In our case, since carbonic acid is diprotic it has two associated Ka values. The first dissociation (giving HCO3^-) has Ka1 and the second dissociation (giving CO3^2-) has Ka2. Typically in physiological circumstances such as in our blood, where pH is close to 7, the first ionization of carbonic acid matters more because it is the one that happens readily at physiological pH. Therefore, Ka1 would be more important in this buffer solution as it represents the first ionization of the acid.

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Explanation:

In the given solvents -

Final answer:

Toluene is the best solvent choice for separating benzophenone, diphenyl methanol, and biphenyl using TLC on silica gel due to its balanced polarity, which will allow for better differentiation between the non-polar compounds.

Explanation:

In choosing a solvent for separating benzophenone, diphenyl methanol, and biphenyl using TLC on silica gel, toluene appears to be the most appropriate choice. Since phenyl groups are fairly non-polar, a non-polar or moderately polar solvent is needed to achieve good separation. Toluene is a good choice because it has a higher polarity than hexanes but is less polar than the other solvents listed, providing a balanced environment to distinguish the three compounds on the basis of their differing polarities. A solvent like methanol or water would be too polar, causing the compounds to stay near the baseline, while hexanes might be too non-polar, failing to separate the compounds efficiently. Therefore, toluene is the recommended solvent for TLC in this scenario.

Answer:

No mass of hexane could be left over by the chemical reaction.

C₆H₁₄ is the limitin reagent

Explanation:

This is a combustion reaction were the oxygen is one of the reactant and it burns a compound in order to generate water and carbon dioxide.

In this case, we have liquid hexane combustion, so the reaction is:

2C₆H₁₄(l) +  19O₂(g)  →   12CO₂(g) +  14H₂O(g)

In this situation we are asked for the mass of a reactant that could be left over, this is the excess reagent.

We convert the masses of reactants to moles:

9.48 g . 1mol  / 86g = 0.110 moles of C₆H₁₄

43 g . 1 mol/32 g =  1.34 moles of O₂

The hexane may be the excess reagent but we confirm like this:

19 moles of O₂ need 2 moles of hexane to react

Then, 1.34 moles of O₂ will react with (1.34 . 2) / 19 = 0.141 moles

We need 0.141 moles, and we only have 0.110. Hexane is the limiting reagent so no mass could be left over by the chemical reaction.

In conclussion, oxygen is excess reactant. We verify:

2 moles of hexane need 19 moles of O₂ to react

Then, 0.110 moles of hexane will react with (0.110 . 19) / 2 = 1.04 moles of O₂

As I have 1.34 moles of oxygen, value is higher so in this case we are having mass that could be left over by the reaction.

Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = [tex]\frac{mass}{molar mass}[/tex]

Number of moles of acetone = [tex]\frac{31.5}{58.08}[/tex]

Number of moles  of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

Answer:

kp  = 1.55

Explanation:

To get to this reaction, we just need to sum the other two reactions, and multiply coefficients if it's needed so:

C(s) + 2H₂O(g) ⇄CO₂(g) + 2H₂(g)   Kp1 = 3.23    (1)

H₂(g) + CO₂(g) ⇄ H₂O(g) + CO(g)    Kp2 = 0.601   (2)

we will multiply eqn 2 by 2 because we need to eliminate H₂

2H2(g) + 2CO2(g) ⇄ 2H2O(g) + 2CO(g)    Kp = (0.601)²

Now, add both equation

C(s) + 2H₂O(g) ⇄CO₂(g) + 2H₂(g)   Kp1 = 3.23  

2H2(g) + 2CO2(g) ⇄ 2H2O(g) + 2CO(g)    Kp = (0.601)²

we have,

C(s) + CO₂(g) ⇄ 2CO(g)  

Now the value of Kp will be:

kp = 3.23 × (0.693)²

kp  = 1.55

Answer: For the given reaction, the value of [tex]K_c[/tex] is greater than 1

Explanation:

For the given chemical equation:

[tex]CaCO3(s)\rightleftharpoons Ca^{2+}(aq.)+CO_3^{2-}(aq.)[/tex]

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}][/tex]

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression

As, the denominator is missing and the numerator is the only part left in the expression. So, the value of [tex]K_c[/tex] will be greater than 1.

Hence, for the given reaction, the value of [tex]K_c[/tex] is greater than 1

Answer:

radius = 156 pm

Explanation:

The relation between radius and edge length of unit cell of BCC is

r=a[tex]\sqrt{3}[/tex]/4

Given

a = 360 pm

Therefore

r = r = radius = 360[tex]\sqrt{3}[/tex]/4= 155.88 pm

Or

156 pm

Explanation:

Reaction equations for the given species is as follows.

   [tex]H_{3}PO_{4} + 2NaOH \righleftharpoons Na_{2}HPO_{4} + 2H_{2}O[/tex]

   [tex]NaH_{2}PO_{4} + NaOH \rightleftharpoons Na_{2}HPO_{4} + H_{2}O[/tex]

At the first equivalence point we need 2 × 10 mmol NaOH.

At the second equivalence point we need 5 mmol of NaOH.

Hence, total moles of NaOH required is as follows.

               (20 + 5) mmol = 25 mmol

We assume that volume of NaOH required is V.

      [tex]25 mmol NaOH \times \frac{10^{-3} mol NaOH}{1 mmol NaOH} \times \frac{1000 ml V}{0.10 \text{mol NaOH}}[/tex]

      = 250 ml V

Thus, we can conclude that 250 ml of 0.10 M NaOH must be added to reach second equivalence point of the titration of the [tex]H_{3}PO_{4}[/tex] and NaOH.

In the given case, 250 ml of 0.10 M NaOH must be added to attain the second equivalence point.

The reactions taking place in the given case are:

Based on the given information, a solution comprises 10 mmol of H₃PO₄ and 5 mmol of NaH₂PO₄.

For the first equivalence point, there is a need of 2 × 10 mmol NaOH.

For the second equivalence point, there is a need of 5 mmol NaOH.

Now the total moles of NaOH needed is,

= 20 + 5 mmol

= 25 mmol

Now let the volume of NaOH required be V. Now putting the values we get,

[tex]= 25 mmol NaOH * \frac{10^{-3} mole NaOH}{1 mmol NaOH} * \frac{1000 mLV}{0.10 mol NaOH} \\= 250 ml[/tex]

Thus, 250 ml of 0.10 M NaOH is need to be added to attain the second equivalence point of the titration.

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Answer: Option (c) is the correct answer.

Explanation:

It is known that like dissolves like. As cyclohexane is a hydrocarbon and we know that hydrocarbons are non-polar in nature. So, a non-polar compound will be soluble in a hydrocarbon.

For example, NaBr is an ionic compound and when it i added to liquid cyclohexane then it will not dissolve. Similarly, [tex]CH_{2}Cl_{2}[/tex] and HI are polar covalent compounds and they will not dissolve in liquid cyclohexane.

But [tex]CH_3CH_2CH_2CH_2CH_3[/tex] is a non-polar compound. So, when it is added to liquid cyclohexane then it will readily dissolve.

Thus, we can conclude that pentane ([tex]CH_3CH_2CH_2CH_2CH_3[/tex]) will most likely dissolve in cyclohexane to form a solution.

Answer:

0.66 mol

Explanation:

Zero Gauge pressure = 14.7 psi

Pressure read = 173 psi

Actual pressure = 173 psi - 14.7 psi = 158.3 psi

P (psi) = 1/14.696  P(atm)

So, Pressure = 10.77 atm

Given that:

Temperature = 20 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (20+ 273.15) K = 298.15 K

V = 1.50 L

Using ideal gas equation as:

[tex]PV=nRT[/tex]

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

10.77 atm × 1.50 L = n ×0.0821 L atm/ K mol  × 298.15 K

⇒n = 0.66 mol

Answer:

The activation energy is lowered by +85.35 KJ/mol

Explanation:

Rate constant is related to the activation energy and the temperature of the reaction through

K = Ae⁻ᴱᵃ/ᴿᵀ

In K = (In A) - (Ea/RT)

In K = (-Ea/RT) + In A

where k = the activity constant

Ea = activation energy

R = molar gas constant

T = absolute temperature in Kelvin

₁₂

At point 1, without the catalyst

In K₁ = (-Ea₁/RT) + In A (eqn 1)

At point 2 with the catalyst

In K₂ = (-Ea₂/RT) + In A (eqn 2)

Note that the molar gas constant, the absolute temperature in Kelvin and the activity constant are all the same for both points.

Subtract (eqn 1) from (eqn 2)

In K₂ - In K₁ = (-Ea₂/RT) - (-Ea₁/RT) + In A - In A

In (K₂/K₁) = (1/RT) (Ea₁ - Ea₂)

We were told that K₂/K₁ = 2.3 × 10¹⁴, then find the difference in Ea

R = 8.314 J/mol.K, T = 37°C = 310.15 K

In (2.3 × 10¹⁴) = (1/(8.314×310.15) (Ea₁ - Ea₂)

33.1 = (1/2578.5871) (Ea₁ - Ea₂)

(Ea₁ - Ea₂) = 33.1 × 2578.5871 = 85351.23 J/mol

Therefore, the activation energy is lowered by +85.35 KJ/mol

The activation energy is lowered by +85.35 KJ/mol

[tex]K = Ae^{\frac{-E_a}{RT} }\\\\In K = (In A) -( \frac{-E_a}{RT})\\\\In K = (\frac{-E_a}{RT}) + In A[/tex]

where

1. At point 1, without the catalyst

[tex]In K_1 = (-Ea_1/RT) + In A[/tex]............(i)

2. At point 2 with the catalyst

[tex]In K_ = (-Ea_2/RT) + In A[/tex]...........(ii)

On subtracting equation (i) from (ii)

[tex]In K_2 - In K_1 = (-Ea_2/RT) - (-Ea_1/RT) + In A - In A\\\\In (K_2/K_2) = (1/RT) (Ea_1 - Ea_2)[/tex]

Given:

On solving:

[tex]In (2.3 * 10^{14}) = (1/(8.314*310.15) (Ea_1 - Ea_2)\\\\33.1 = (1/2578.5871) (Ea_1 - Ea_2)\\\\(Ea_1 - Ea_2) = 33.1 * 2578.5871\\\\ (Ea_1 - Ea_2)= 85351.23 J/mol[/tex]

Thus, the activation energy is lowered by +85.35 KJ/mol.

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Answer:

A) i) 0.002 ppb (ii) 2ppt (iii) 3.7 x 10^(-6) μM

B) i)0.002 ppm. (ii) 2ppb

Explanation:

A) We know that 1 ppb = 1 μgram per liter, and so the concentration of Mirex in ppb would be 0.002 ppb.

1 ppt = 1 nanogram per liter of water, so the concentration of Mirex in ppt would be 2 ppt;

(0.002 μg/L) (100ng/μg) (1ppt/ng/L) = 2ppt.

Now, MW of Mirex = 540 g/mol ≡ μg/μmol

Thus, 1 μmole = 540 μgram,

Hence, the concentration of Mirex in μmoles would be;

(0.002 μg/L)/(540 μg/μmol) = 3.7 x 10^(-6) μM

B) i) 1 ppm = 1 μgram per gram.

Thus, the concentration of Mirex in ppm would be = 0.002 ppm.

ii) Now, 1 ppb = 1 nanogram per gram.

Thus, concentration of Mirex in ppb would be = (0.002 μg/g) (100ng/μg) (1ppb ng/g) = 2ppb

Final answer:

The concentration of mirex in water samples is 0.002 ppb, 2 ppt, and approximately 3.70 x 10⁻³ μM. In fish samples, the concentration of mirex is 0.002 ppm and 2 ppb.

Explanation:

The student is asking about converting the concentration of a pesticide, mirex, in water and fish samples to various units. The concentration in water samples is given as 0.002 μg/L. The conversions are as follows:

For the fish samples:

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

[tex]\frac{1}{[A]} =\frac{1}{[A]_0} +kt[/tex]

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

[tex]t_{1/2}=\frac{1}{k[A]_0}[/tex]

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

Answer:

   ΔrxnH  = -580.5 kJ

Explanation:

To solve this question we are going to help ourselves with Hess´s law.

Basically the strategy here is to work  in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.

Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-

The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.

This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.

Taking  1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.

      Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l)             ΔrxnHº = 259 / 2 kJ  1/2 eq3

+    2NaOH(s)  + SO₃(g) ⇒  Na₂SO₄ (s) + H₂O (l)   ΔrxnHº = -418 kJ     - eq 2

+    2Na (s) + 2 H₂O (l)  ⇒   2 NaOH (s) + H₂ (g)    ΔrxnHº = -146 x 2    2 eq 1

Na₂O (s) + SO₃ (g)  ⇒ Na₂SO₄ (s)    ΔrxnHº =  259/2 + (-418) + (-146) x 2 kJ

                                                          ΔrxnH  = -580.5 kJ

Final answer:

To find the enthalpy change for the reaction Na2O(s) + SO3(g) → Na2SO4(s), we apply Hess's Law and manipulate the given reactions to find that the enthalpy change is -288.5 kJ.

Explanation:

To determine the enthalpy change (\(\Delta H^o_{rxn}\)) for the reaction (4) Na2O(s) + SO3(g) → Na2SO4(s), we use Hess's Law which states that if a reaction is the sum of two or more other reactions, the enthalpy changes of the constituent reactions can be added to determine the enthalpy change of the overall process. We need to find a combination of reactions (1), (2), and (3) that will result in reaction (4).

Here are the given reactions with their enthalpy changes:

Na(s) + H2O(l) → NaOH(s) + 1/2 H2(g), \(\Delta H^o_{rxn} = -146 kJ\)

Na2SO4(s) + H2O(l) → 2NaOH(s) + SO3(g), \(\Delta H^o_{rxn} = +418 kJ\)

2Na2O(s) + 2H2(g) → 4Na(s) + 2H2O(l), \(\Delta H^o_{rxn} = +259 kJ\)

To solve for the enthalpy of reaction (4), we need to reverse reaction (2) and halve reaction (3) to cancel out unwanted substances and leave Na2O(s) + SO3(g) on the left-hand side and Na2SO4(s) on the right-hand side.

Reversed reaction (2):\[\text{Na2SO4(s) + H2O(l) → 2NaOH(s) + SO3(g)}\], \(\Delta H^o_{rxn} = -418 kJ\) (since we reversed it)

Halved reaction (3):\[\text{Na2O(s) + H2(g) → 2Na(s) + H2O(l)}\], \(\Delta H^o_{rxn} = +259 kJ / 2 = +129.5 kJ\) (since we halved it)

Finally, we add the modified reaction (2) and (3):

(4) Na2O(s) + SO3(g) → Na2SO4(s), the enthalpy change is \(\Delta H^o_{rxn} = (-418 kJ) + (+129.5 kJ) = -288.5 kJ\).

Explanation:

It is known that formula for the ionization energy of hydrogen atom is as follows.

               E = [tex]\frac{13.6 eV}{n^{2}}[/tex]

or,          n = [tex]\sqrt{\frac{13.6}{E}}[/tex]

The value of energy is given as 0.544 eV. Therefore, we will calculate the value of n as follows.

                   n = [tex]\sqrt{\frac{13.6}{E}}[/tex]

                      = [tex]\sqrt{\frac{13.6}{0.544 eV}}[/tex]

                      = 5

Thus, we can conclude that n equals to 5 for a hydrogen atom if 0.544 eV of energy can ionize it.

Final answer:

The principal quantum number n of a hydrogen atom that requires 0.850 eV of energy to ionize from an excited state is 4. This is calculated using the formula for the energy levels of hydrogen, En = -13.6 eV/n², and solving for n.

Explanation:

The question relates to the energy levels of a hydrogen atom and how much energy is required to ionize it when the electron is in an excited state. The ionization energy of hydrogen in the ground state (n = 1) is 13.6 eV. To find the principal quantum number n for a particular amount of energy required to ionize the hydrogen atom from an excited state, we use the formula: En = -13.6 eV/n². We can set this equal to the energy needed to ionize the atom, in this case, 0.850 eV, and solve for n.

First, we rearrange the formula so it looks like this:

-13.6 eV/n² = -0.850 eV
Then we solve for n:

n² = 13.6 eV / 0.850 eV
n² = 16
n = √16
n = 4

Thus, the hydrogen atom is in the n = 4 excited state when 0.850 eV of energy can ionize it.

For the given reaction between rubidium and Boron, only one electron will be transferred in the formation [tex]2 RbBr[/tex].

The potential difference between the two half-cells of an electrochemical cell is known as cell potential. The transfer of electrons from one-half cell to another produces the potential difference.

The given chemical reaction is:

[tex]2Rb_{(s)} + Br_{(l)}[/tex] ⇒ [tex]2RbBr_{(s)}[/tex]

The following electrochemical reaction can be divided into the reaction occurring at the cathode and the reaction occurring at the anode:

Oxidation reaction at the anode: [tex]2Rb_{(s)}[/tex] ⇒ [tex]2Rb^+_{(aq)} + 2e^-[/tex]

Reduction reaction at the cathode: [tex]Br_{2(l)} + 2e^-[/tex] ⇒ [tex]2Br^-_{(aq)}[/tex]

Since there are two units of RbBr in the aforementioned equation and two electrons were transported, only one electron is needed to generate one unit of RbBr.

Thus, only one electron is involved in the formation of RbBr.

Learn more about electrons here:

brainly.com/question/12001116

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Final answer:

In the formation of one formula unit of RbBr, one electron is transferred from rubidium (Rb) to bromine (Br), as Rb loses one electron to form Rb+ and Br gains one electron to form Br-.

Explanation:

The question regards the number of electrons transferred in the formation of one formula unit of RbBr from the reaction 2 Rb + Br2 → 2 RbBr. In order to determine the electron transfer, we need to consider the oxidation states of the elements involved. Rubidium (Rb) is oxidized, meaning it loses electrons, and bromine (Br2) is reduced, meaning it gains electrons.

Rubidium starts with an oxidation state of 0 and becomes Rb+ in RbBr. This indicates that each Rb atom loses one electron. Bromine starts with an oxidation state of 0 as diatomic Br2 and each Br atom gains one electron to become Br- in RbBr. Therefore, for each RbBr formula unit produced, one electron is transferred from Rb to Br.

Answer:

using higher concentration of the nucleophile

Explanation:

In SN2 reaction, the attack of the nucleophile on the substrate occurs simultaneously as the leaving group departs. The entering group normally attacks through the back side of the molecule. The reaction is concerted and bimolecular. This implies that the concentration of the nucleophile is important in the rate equation for the reaction. Hence increasing the concentration of the nucleophile will increase the rate of SN2 reaction.

Final answer:

The molar mass of the unknown protein is 5.79 g/mol.

Explanation:

To find the molar mass of the unknown protein, we can use the formula for osmotic pressure:

π = MRT

Where π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the formula, we have:

M = π / (RT)

Substituting the given values:

M = 2.45 torr / (0.0821 L.atm/(mol.K) * 298 K)

M = 0.0996 mol/L

To convert the molarity to grams per liter, we need to multiply it by the molar mass

Molar mass = (0.0996 mol/L) * (10 mL/1 L) * (5.87 mg/1 mL)

Molar mass = 5.79 g/mol