A chemist must prepare 275. mL of 1967, ?? aqueous copper(II) fluoride (Cur) working solution. He'll do this by pouring out some 2.63 T-aqueous copper(II) fluoride stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in mL of the copper(II) fluoride stock solution that the chemist should pour out. Be sure your answer has the correct number of significant digits mL

Answer:

38 electrons

Explanation:

From Faraday's first law of electrolysis which states that during electrolysis, the mass of a substance deposited at the electrode is proportional to the quantity of electricity passing through it. Mathematically , M ∝ Q

but M = zQ and from electricity, Q =It, hence the equation becomes M = zIt.

where M = Mass of substance deposited in g, and Q =Quantity of electricity in coulombs(C)

I = current in ampere(A)

t = time in seconds

z = Chemical Equivalent in g/C

hence, given t = 1.25 ps and i = 4.8 μA

Using Q = It = 1.25 ps × 4.8 μA , converting the ps (pico secs) to secs and micro ampere to ampere, Q = 1.25 × 10-12s  × 4.8 × 10-6A = 6 × 10-18C

From 1 mole of electron is equal to the quantity of charge which is also equal to 96500C/mol,

Hence, number of moles of electrons =  6 × 10-18C / 96500C/mol

number of moles = 6.218 ×10-5 × 10-18  = 6.218 × 10-23moles

Recalling that, number of moles = number of electrons / Avogadro's constant and Avogardos constant = 6.023 × 10raised to the power of 23/mol

number of electrons = number of moles × 6.023 × 1023/mol

number of electrons = 6.218 × 10-23moles × 6.023 × 1023/mol  = 37.50 which is approximately 38

Hence 38 electrons moved past the fixed reference point.

To calculate the number of electrons that move past a fixed reference point, use the formula: Number of electrons = (Current × Time × Charge on a single electron) / Electron charge. Plugging in the given values, the number of electrons is approximately 1.875 × 10^10 electrons.

To calculate the number of electrons that move past a fixed reference point, we can use the formula:

Number of electrons = (Current × Time × Charge on a single electron) / Electron charge

Plugging in the given values, we have:

Number of electrons = (4.8 μA × 1.25 ps × 10^-12 C) / (1.6 × 10^-19 C)

Simplifying the expression, we find that the number of electrons is approximately 1.875 × 10^10 electrons.

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Answer: Check explanation.

Explanation:

Transition metals are metallic elements that can be found in the Groups IVB–VIII, IB, and IIB on the periodic chart.

The question require us to write down the GROUND state electronic configuration of the first fifth transition metal, sixth transition metal, seventh transition metal element and the eighth transition metal.

NOTE: we are Starting from Argon, which has 18 electrons.

The fifth transition metal is Manganese,Mn. Manganese has 25 electrons, that is, 25- 18= 7. Therefore, it needs seven electrons to complete the configuration.

Hence, The ground state electronic configuration = [Ar)] 3d5. 4s2.

The first sixth Transition metal is iron,Fe. Iron has 26 electrons, that is, 26 - 18 = 8. Therefore, it need eight Electrons to complete the ground state electronic configuration.

Hence, the ground state electronic configuration of Fe= [Ar] 3d6. 4s2.

The first seventh transition metal is Cobalt, Co. It has 27 Electrons, therefore, 27- 18 = 9. Therefore, it needs 9 Electrons to complete its ground state electronic configuration.

Ground state electronic configuration of Co= [Ar] 3d7. 4s2.

The first eight Transition metal is Nickel. It has 28 electrons. Therefore, 28-18= 10. So, it needs 10 Electrons to complete its ground state electronic configuration.

Hence, the Ground state electronic configuration of Ni= [Ar] 3d8. 4s2.

Answer:

Ruler A :

According to ruler A, the diameter of the circle has only one certain digit and one uncertain digit. if we look at measurement of a diameter, it contain two significant figures. so the certainty  of the diameter measurement is smaller.

Ruler B :

According to ruler B, the certainty  of the diameter measurement is greater. because it contain two certain and one uncertain digit. it has three significant figures.

This question is about quantifying the diameter of a circle using two distinct rulers and then comparing these measurements to the actual diameter. Classifying the measurements relies on the degree of precision of the rulers and whether the measurement is greater or smaller than the actual diameter.

The question is about measuring the diameter of a circle using two different rulers (Ruler A and Ruler B). Then, you have to compare these measurements with the actual diameter of the circle, which is given as 2.264 cm. Measurement errors might occur due to precision limitations of the used rulers. Therefore, classification of statements about the measurements might differ.

For instance, if Ruler A measures the diameter as 2.3 cm, this might be classified as an overestimate because it's larger than the actual diameter. Conversely, if Ruler B measures the diameter as 2.2 cm, this could be classified as an underestimate because it's smaller than the actual diameter.

Note that the extent of accuracy or inaccuracy depends on the degree of precision of the rulers, which isn't provided in the question. In the realm of measurements in mathematics, always keep in mind the importance of precision and accuracy.

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Answer:

982.5 kg/m³

Explanation:

When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:

ρ₁ = ρ₀/(1 + β*(t₁ - t₀))

Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.

At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C

ρ₁ = 1,000/(1 + 0.0002*(93 - 4))

ρ₁ = 1,000/(1+ 0.0178)

ρ₁ = 982.5 kg/m³

The question is incomplete, here is the complete question:

[tex]K_c=9.7[/tex] at 900 K for the reaction [tex]NH_3(g)+H_2S(g)\rightleftharpoons NH_4HS(s)[/tex]

If the initial concentrations of [tex]NH_3(g)[/tex] and [tex]H_2S(g)[/tex] are 2.0 M, what is the equilibrium  concentration of [tex]NH_3(g)[/tex] ?

Answer: The equilibrium concentration of ammonia is 0.32 M

Explanation:

We are given:

Initial concentration of ammonia = 2.0 M

Initial concentration of hydrogen sulfide = 2.0 M

For the given chemical reaction:

              [tex]NH_3(g)+H_2S(g)\rightleftharpoons NH_4HS(s)[/tex]

Initial:        2.0      2.0

At eqllm:    2.0-x     2.0-x            x

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

[tex]K_c=\frac{1}{[NH_3][H_2S]}[/tex]

The concentration of pure solids and pure liquids are taken as 1

We are given:

[tex]K_c=9.7[/tex]

Putting values in above equation, we get:

[tex]9.7=\frac{1}{(2.0-x)\times (2.0-x)}\\\\9.7x^2-38.8x+38.8=1\\\\x=2.32,1.68[/tex]

Neglecting the value of x = 1.68 because equilibrium concentration cannot be greater than initial concentration

So, equilibrium concentration of ammonia = 2 - x = (2 - 1.68) = 0.32 M

Hence, the equilibrium concentration of ammonia is 0.32 M

Without the equilibrium constant (Kc) or the actual change in concentration for NH3(g), it is not possible to calculate its equilibrium concentration from an initial 2.0 M.

To find the equilibrium concentration of NH₃(g) given its initial concentration of 2.0 M and the additional information provided, we would typically use the equilibrium expression associated with the reaction in question.

However, the information needed to complete this calculation, such as the equilibrium constant (Kc) and the change in concentration of NH₃(g) during the reaction, has not been provided.

Using the example of the equilibrium reaction 2NH₃(g) = N₂(g) + 3H₂(g), you would set up an ICE (Initial, Change, Equilibrium) table, and use the value of Kc to solve for the equilibrium concentrations.

However, the question doesn't provide a Kc value specific to this reaction. Moreover, the example showing equilibrium concentrations of various species doesn't align with our initial concentration of 2.0 M for NH₃(g). Therefore, without the relevant Kc value or the actual changes in concentrations, we cannot solve for the equilibrium concentration of NH₃(g).

Answer:

Option A) (I) 2.42 m (II) 2.16 M

Explanation:

Let's determine some information.

Solute = CoCl₃ (molar mass = 165.29 g/m); mass of 100 g

Solvent = Water, mass of 250 g

Solution mass = mass of CoCl₃ + mass of water

250 g + 100 g = 350 g of solution

If we want to reach molarity (mol/L), let's determine solution volume with density:

Solution density = solution mass / solution volume

1.25 g/mL = 350 g / solution volume

Solution volume = 350 g / 1.25 g/mL = 280 mL

Let's convert the volume to L → 280 mL = 0.280L

Let's convert the mass of solute to moles = 100 g / 165.29 g/m →0.605 mol

Mol/L = 0.605 moles / 0.280 L = 2.16 M

Now let's calculate molalilty (mol/kg of solvet)

We must convert solvent mass to kg → 250g = 0.250 kg

Then, 0.605 moles / 0.250 kg =2.42 m

Final answer:

To determine the molality, divide the number of moles of CoCl₃ by the kilograms of water, resulting in a molality of 2.42 m. For the molarity, divide the moles of CoCl₃ by the volume of the solution to get a molarity of 2.16 M. Hence, the answer is option B with molality of 2.42 m and molarity of 2.16 M.

Explanation:

The question asks to calculate the molality and molarity of a cobalt(III)chloride solution. To calculate molality (m), we use the formula molality = moles of solute/kilograms of solvent. First, we need the molar mass of cobalt(III)chloride (CoCl₃), which is 165.87 g/mol. Then, we find the moles of CoCl₃ by dividing the mass of CoCl₃(100 g) by its molar mass (165.87 g/mol), which gives us approximately 0.603 moles. Molality is calculated by dividing moles of CoCl₃by the mass of water in kilograms (250 g = 0.25 kg), which yields a molality of about 2.42 m.

To calculate molarity (M), we use the formula molarity = moles of solute/volume of solution in liters. Given the density of the solution is 1.25 g/mL, we can calculate the total mass of the solution (mass of solute + mass of solvent = 100 g + 250 g = 350 g). Next, convert the total mass to volume using the density (350 g / 1.25 g/mL = 280 mL = 0.28 L). Finally, divide the number of moles of CoCl₃(0.603) by the volume of the solution (0.28 L), which results in a molarity of about 2.16 M. Therefore, the correct answer is option B: (I) 2.16 m (II) 2.42 M.

Answer:

a) pH = 9.8

b) 2.2 x 10⁻⁸ = [ H₂C₃H₂O₄ ]

c) decrease

Explanation:

The equilibriums involved in this question are:

C₃H₂O₄²⁻ + H₂O  ⇄  HC₃H₂O₄⁻ + OH⁻  (1) Kb₁ =[HC₃H₂O₄⁻][OH⁻]/[C₃H₂O₄²⁻ ]

HC₃H₂O₄⁻ + H₂O ⇄  H₂C₃H₂O₄ +OH⁻ (2) Kb₂=[ H₂C₃H₂O₄]/[OH⁻]/[HC₃H₂O₄⁻]

The kas for malonic acid, H₂C₃H₂O₄, from reference tables are:

Ka (H₂C₃H₂O₄ )=  1.4 x 10⁻³  

Ka ( HC₃H₂O₄⁻ ) = 2.0 x 10⁻⁶

a) We can calculate the Kbs for the conjugate bases of the weak malonic acid from Kw = Ka x Kb

Kb (C₃H₂O₄²⁻) = 10⁻¹⁴  / 2.0 x 10⁻⁶  =5.0 x 10⁻⁹

Kb (HC₃H₂O₄⁻)= 10⁻¹⁴ / 1.0 x 10⁻³   = 7.1  x 10⁻¹²

Given the magnitudes of the Kbs ( Kb₂ is approximately 1000 times Kb1 ) , to  calculate pOh we can neglect the contribution from (2). We then treat this problem as any equilibrium:

[K₂C₃H₂O₄] = 25 g/180.2 g/mol / 0.455 L = 0.30 M

  Conc               C₃H₂O₄²⁻ + H₂O   ⇄   HC₃H₂O₄⁻ + 0H⁻

I   0.30                     0                                  0               0

C    -x                                                           +x             +x

E    0.30 - x                                                    x              x

Kb (C₃H₂O₄²⁻)  = [ HC₃H₂O₄⁻ ] [OH⁻]/ [ C₃H₂O₄²⁻] = x² / 0.30 - x ≅  x² /0.30

x² /.030 = 5.0 x 10⁻⁹  ⇒ x = √(0.30 x  5.0 x 10⁻⁹ ) = 7.1 x 10⁻⁵ = [ÒH⁻]

(Verifying our approximation was good 7.1 x 10⁻⁵ / 0.30 = 2.4 x 10⁻⁴ so our approximation checks)

pOH = -log 7.1 x 10⁻⁵  = 4.2

pH = 14 -4.2 = 9.8

b)  To answer this part we take equilibrium (2 ) and set up our usual ICE table to solve for the concentration of malonic acid:

Conc (M)                      HC₃H₂O₄⁻ + H₂O  ⇄   H₂C₃H₂O₄ + OH⁻    (2)

I                                     7.1 x 10⁻⁵                             0              0

C                                       -x                                     +x             +x

E                                   7.1 x 10⁻⁵ -x                           x                x

7.1 x 10⁻⁵ - x ≅ 7.1 x 10⁻⁵

[ H₂C₃H₂O₄ ] [OH⁻] / [HC₃H₂O₄⁻] = Kb₂ = 7.1 x 10⁻¹² = x² / 7.1 x 10⁻⁵

x = √(7.1 x 10⁻¹² x  7.1 x 10⁻⁵) = 2.2 x 10⁻⁸ = [ H₂C₃H₂O₄ ]Again our approximation checks since [HC₃H₂O₄⁻] is almost 1000 times [ H₂C₃H₂O₄ ]

c) From eqn (1) :

C₃H₂O₄²⁻ + H₂O   ⇄   HC₃H₂O₄⁻ + OH⁻    

The salt  K₂C₃H₂O₄ will react completely with the added acid, thereby decreasing the C₃H₂O₄²⁻ concentration, and according to Le Chateliers principle the system will shift to the left and the OH⁻ at equilibrium will decrease ( as also does [HC₃H₂O₄⁻] ) therefore the pOH will increase and the pH will decrease ( less OH⁻ higher pOH, smaller pH )

Final answer:

Calculate the pH of the solution, determine the concentration of malonic acid, and explain the impact of adding HCl on the concentration of malonic acid in the solution.

Explanation:

a) Calculate the pH of the solution:

Calculate the molarity of the potassium malonate solution.

Use the dissociation of K2C3H2O4 and the autoionization of water to find the concentration of OH- ions.

Convert the OH- concentration to pH using the formula pH = 14 - pOH.

b) Calculate the concentration of malonic acid:

Set up an equilibrium expression for the dissociation of malonic acid.

Use the pH calculated in part (a) to determine the concentration of malonic acid.

c) Impact of adding HCl:

The addition of HCl will shift the equilibrium towards the formation of malonic acid, leading to an increase in its concentration.

Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

           [tex]N_{2}O_{4} \rightarrow 2NO_{2}[/tex]

Let us assume that x concentration of [tex]N_{2}O_{4}[/tex] is present at the initial stage. Therefore, according to the ICE table,

                    [tex]N_{2}O_{4} \rightarrow 2NO_{2}[/tex]

Initial :               x                   0

Change :       - 0.1        [tex]2 \times 0.1[/tex]

Equilibrium : (x - 0.1)             0.2

Now, expression for [tex]K_{p}[/tex] of this reaction is as follows.

     [tex]K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}[/tex]

Putting the given values into the above formula as follows.

          [tex]K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}[/tex]

                 [tex]2 = \frac{(0.2)^{2}}{(x - 0.1)}[/tex]

                [tex]2 \times (x - 0.1) = (0.2)^{2}[/tex]

                            x = 0.12

This means that [tex]P_{N_{2}O_{4}}[/tex] = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

We can conclude that the initial pressure in the container before the decomposition of N₂O₄ is 0.12 atm.

The decomposition of N₂O₄ is determined as follows;

N₂O₄ → 2NO₂

ICE table can be created as follows;

     N₂O₄ → 2NO₂

I:      x            0

C:    0.1        2(0.1)

E: (x - 0.1)    (0.2 - 0)

Expression for equilibrium constant;

[tex]K_p = \frac{P^2NO_2}{PN_2O_4} \\\\2 = \frac{0.2^2}{(x - 0.1)} \\\\2(x - 0.1) = 0.2^2\\\\x-0.1 = 0.02\\\\x = 0.12[/tex]

Thus, we can conclude that the initial pressure in the container before the decomposition of N₂O₄ is 0.12 atm.

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Predict whether each of the following bonds is no polar covalent, polar covalent, or ionic

Bond. Electro negativity Type of bond

Si- O

k-Cl

I-I

C-H

a) Si--O: Polar Covalent

b) K--Cl: Ionic

c) S--F: Polar Covalent

d) P--Br: Polar Covalent

e) Li--O: Ionic

f) N--P: Covalent

a) Si--O: Polar Covalent. Silicon (Si) and oxygen (O) have different electronegativities, causing unequal sharing of electrons. The oxygen atom attracts electrons more strongly, resulting in a partial negative charge on oxygen and a partial positive charge on silicon.

b) K--Cl: Ionic. Potassium (K) and chlorine (Cl) have significantly different electronegativities. K transfers an electron to Cl, forming K⁺ and Cl⁻ ions held together by electrostatic attraction.

c) S--F: Polar Covalent. Sulfur (S) and fluorine (F) have distinct electronegativities, leading to unequal electron sharing. F pulls electrons more, inducing partial charges on both atoms.

d) P--Br: Polar Covalent. Phosphorus (P) and bromine (Br) have differing electronegativities, causing uneven electron distribution and partial charges on the atoms.

e) Li--O: Ionic. Lithium (Li) and oxygen (O) have a significant electronegativity difference. Li loses an electron to O, resulting in Li⁺ and O²⁻ ions, held together by electrostatic forces.

f) N--P: Covalent. Nitrogen (N) and phosphorus (P) have similar electronegativities, allowing for equal electron sharing in a covalent bond.a) Si--O: Polar Covalent.

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Answer:

Explanation:

The equation is given as:

CH3CHOHC2H4CHO + CH3OH --> CYCLIC ACETAL + H2O

This above equation is carried out in the presence of a strong acid. There are five mechanisms employed and they are:

Step 1:

Initial formation of the hemiacetal which takes several steps

Step 2:

Addition of a proton. The hemicetal is protonated on the hydroxyl group (-OH group)

Step 3:

As seen a bond is broken to give the H2O molecule and a resonance stabilized cation.

The carbonyl group on the cation is enriched with the oxygen-18 got from the H2O molecule as seen in the mechanism.

Step 4:

An attraction occurs between electrophile and nucleophile i.e the stabilised cation and the lone paids of the methanol.

Step 5:

Finally, a proton (+) is removed from the molecule by a lone pair of electron on the methanol.

Attached are the Steps 1 - 5 mechanism below

Answer:

strength of hypericin in mixture = 0.42 %

Explanation:

given data

each lot = 100 g

active component hypericin = 0.3%, 0.7%, and 0.25%

solution

we get here percent strength o hypericin in the mixture that is

Hypericin contribution lot 1 =  [tex]\frac{0.3}{100}[/tex] × 100

Hypericin contribution lot 1 =  0.3 g

and

Hypericin contribution lot 2 = [tex]\frac{0.3}{100}[/tex] × 100

Hypericin contribution lot 2 = 0.7 g

and

Hypericin contribution lot 3 = [tex]\frac{0.25}{100}[/tex] × 100  

Hypericin contribution lot 3  = 0.25 g

so

total 300 g mixture of hypericin contain = 0.3 g + 0.7 g + 0.25 g

total 300 g mixture of hypericin = 1.25 g  

so here percent strength o hypericin in mixture is

strength of hypericin in mixture = [tex]\frac{1.25}{300}[/tex] × 100  

strength of hypericin in mixture = 0.42 %

The percent strength of hypericin in the mixture will be "0.42%".

According to the question,

From lot 1, Hypericin contribution will be:

= [tex]100\times \frac{0.3}{100}[/tex]

= [tex]0.3 \ g[/tex]

From lot 2, Hypericin contribution will be:

= [tex]100\times \frac{0.7}{100}[/tex]

= [tex]0.7 \ g[/tex]

From lot 3, Hypericin contribution will be:

= [tex]100\times \frac{0.25}{100}[/tex]

= [tex]0.25 \ g[/tex]

For 300 g mixture,

The amount of hypericin will be:

= [tex]0.3+0.7+0.25[/tex]

= [tex]1.25 \ g[/tex]

hence,

The percentage strength will be:

= [tex]\frac{1.25}{300}\times 100[/tex]

= [tex]0.42[/tex] (%)

Thus the above approach is right.

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Answer: The percent yield of the reaction is 68.16 %.

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of camphor = 1.500 g

Molar mass of camphor = 152.23 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of camphor}=\frac{1.500g}{152.23g/mol}=9.85\times 10^{-3}mol[/tex]

The chemical equation for the reaction of camphor and sodium borohydride follows:

[tex]\text{Camphor}+NaBH_4\rightarrow \text{Isoborneol}[/tex]

As, sodium borohydride is present in excess. It is an excess reagent. So, camphor is the limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of camphor produces 1 mole of isoborneol

So, [tex]9.85\times 10^{-3}mol[/tex] of camphor will produce = [tex]\frac{1}{1}\times 9.85\times 10^{-3}mol=9.85\times 10^{-3}mol[/tex] of isoborneol

Molar mass of isoborneol = 154.25 g/mol

Moles of isoborneol = [tex]9.85\times 10^{-3}[/tex] moles

Putting values in equation 1, we get:

[tex]9.85\times 10^{-3}mol=\frac{\text{Mass of isoborneol}}{154.25g/mol}\\\\\text{Mass of isoborneol}=(9.85\times 10^{-3}mol\times 154.25g/mol)=1.52g[/tex]

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of isoborneol = 1.036 g

Theoretical yield of isoborneol = 1.52 g

Putting values in above equation, we get:

[tex]\%\text{ yield of isoborneol}=\frac{1.036g}{1.52g}\times 100\\\\\% \text{yield of isoborneol}=68.16\%[/tex]

Hence, the percent yield of the reaction is 68.16 %.

Answer:

see explanation below

Explanation:

In this case, is pretty easy. This is a reduction reaction to form the respective alcohol.

Now for each mole of benzophenone that it's present, reacts with 1 mole of Sodium borohydryde, so, all we need to do, is to calculate the moles of benzophenone presents and these, would be the same moles of NaBH4 so:

moles Benzophenone : m/MM

The molar mass of benzophenone reported is 182.22 g/mol so:

moles Benzophenone = 0.55/182.22 = 3.02x10⁻³ moles

so the moles of NaBH₄ = 3.02x10⁻³ moles

Answer:

0.165kJ

Explanation:

Formula to use for such a question is;

Energy = number of mole x molar gas constant x change in temperature

Number of mole = reacting mass of mercury / molar mass of mercury = 45.30/200.58 = 0.226moles

Change in temperature = final temperature - initial temperature = 30 - (-59) =30 + 59 = 89 Kelvin

E = nRT = 0.226 x 8.314 x 89

Energy = 165.35Joules

Energy in kJ = 165.35/1000 = 0.165kJ

Answers :

Solution of each part is given below.

Explanation:

a) [tex]S_8[/tex]     non polar covalent  ( because each sulphur atom shares electron and the net dipole moment is zero.)

b) RbCl    ionic  ( Rb have good tendency to donate electron and chlorine is too electro negative therefore they make strong ionic bond.)

c) [tex]PF_3[/tex]   polar covalent ( because of triagonal pyramidal geometry and strong electronegativity of F atom it is polar and covalency is due to the share of electron with each F atom.)

d) [tex]SCl_2[/tex]  polar covalent ( because of tetrahedral geometry and sharing of electrons between S and Cl.)

e) [tex]F_2[/tex]   non polar covalent ( because of linear geometry they are non polar and bond between them is formed due to sharing electrons.)

f) [tex]SF_3[/tex]   polar covalent ( because of T- shape geometry they are non polar and due to sharing of electrons they are polar covalent.)

Hence, this is the required solution.

The nature or type of bonds formed by molecules and compounds are greatly influenced by the chemical properties of the element.

Data;

The chemical properties of an element often affects how it bonds with other elements to form a compound. In this case, we have several compounds given.

S8: This is a non-polar covalent molecule.

RbCl: This is an ionic compound because it consists of very strong electropositive and electronegative elements.

PF3: This is a polar covalent compound due to the nature of phosphorous.

SCl2: This is a polar covalent compound

F2: This is a non-polar covalent compound

Sf3: This is a polar covalent compound

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Answer:

a) 0.33 mg/min

b) 6.67 mL/min

c) 133.4 drops/min

Explanation:

a) The mass flow indicates how much mass is flowing (in this case is being administrated) in a "piece" of time, thus, it's the total mass administrated by the total time:

mf = 5mg/15min

mf = 0.33 mg/min

So, 0.33 mg of the acid will be administrated per minute.

b) Now, we must calculate the volume flow, which is the total volume divided by the time:

Vf = 100mL/15 min

Vf = 6.67 mL/min

So, 6.67 mL of the infusion will be administrated per minute.

c) The drops flow, is the drop delivery ( 20 drops/mL) multiplied by the volume flow:

df = 20drops/mL * 6.67 mL/min

df = 133.4 drops/min

So, it must be infused 133.4 drops per minute.

Final answer:

To administer an intravenous infusion containing 5 mg of zoledronic acid in 100 mL over 15 minutes, 0.33 mg of zoledronic acid and 6.67 mL of infusion should be administered per minute. Using a drip set that delivers 20 drops/mL, the infusion should be given at a rate of 133.33 drops per minute.

Explanation:

An intravenous infusion contains 5 mg of zoledronic acid in 100 mL. To calculate the amount to be administered per minute:

(a) To find how many milligrams of zoledronic acid must be administered per minute: Convert 15 minutes to 1 minute: 5 mg x (1 minute / 15 minutes) = 5 mg / 15 = 0.33 mg/min

(b) To find how many milliliters of infusion must be administered per minute: 100 mL x (1 minute / 15 minutes) = 100 mL / 15 = 6.67 mL/min

(c) To calculate the drops per minute: Convert 6.67 mL/min x 20 drops/mL = 133.33 drops/min

Answer:

He2

Explanation:

In the He2 molecule, there are eight electrons from the four electrons occupying the orbitals in the two helium atoms. These electrons are arranged into one bonding and one anti bonding orbital containing two electrons each. Eventually, the number of electrons occupying bonding molecular orbitals equals the number of electrons occupying anti bonding molecular orbitals hence the net bond energy is zero and the molecule is very unstable.

Final answer:

He₂ is the least stable molecule according to the molecular orbital model because it has a bond order of zero, indicating no net stabilization from bonding interactions. Option A

Explanation:

Molecular Orbital (MO) Stability

According to the molecular orbital model, the stability of a diatomic molecule can be predicted by its bond order, which is calculated as the difference between the number of electrons in bonding and antibonding molecular orbitals divided by two. When considering He₂, we find that it has a bond order of zero.

This is due to having an equal number of electrons in the bonding (sigma 1s) and antibonding (sigma star 1s) molecular orbitals, resulting in a configuration of (sigma 1s)2(sigma star 1s)2. This bond order of [tex](2 - 2) / 2 = 0[/tex] implies that there is no net stabilization from bonding interactions, making He₂ the least stable molecule among those listed.

Therefore, when ranking the molecules A. He₂ B. Li₂ C. B₂ D. C₂ E. NO: He₂ is predicted to be the least stable because its molecular orbitals are fully occupied with no net bonding interaction, which makes it less stable than two isolated He atoms.

Answer: The formula unit equation is written below.

Explanation:

Formula unit equation is defined as the balanced chemical equation that includes physical state of matter of all the compounds.

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

When ammonium fluoride reacts with magnesium chloride, it leads to the formation of ammonium chloride and a solid precipitate of magnesium fluoride.

The formula unit equation for the reaction of ammonium fluoride and magnesium chloride follows:

[tex]2NH_4F(aq.)+MgCl_2(aq.)\rightarrow MgF_2(s)+2NH_4Cl(aq.)[/tex]

This is an example of double displacement reaction.

Hence, the formula unit equation is written above.

The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment

Answer:

Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z

Explanation:

The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.

The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.

Compound 2 (2,5-dimethylhexane) structure shows that the cleavage  of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.

Answer:

92.2 m

Explanation:

Given that:=

The breadth = 304 mm

Height = 0.014 mm

Let Length = x mm

Volume = [tex]Length\times breadth\times height[/tex]

Thus,

Volume = [tex]304\times 0.014\times x\ mm^3=4.256x\ mm^3[/tex]

Also, 1 mm³ = 0.001 cm³

So, volume = 0.004256 cm³

Given that density = 2.7 g/cm³

Mass = 1.06 kg = 1060 g

So,

[tex]Volume=\frac{Mass}{Density}=\frac{1060}{2.7}\ cm^3=392.59\ cm^3[/tex]

So,

0.004256*x = 392.59

x = 92243.89 mm

Length of foil = 92243.89 mm = 92.2 m

Answer options from an alternative source

Answer:

Explanation:

Monosaccharides are carbohydrates that are the simplest form of a sugar. They cannot be further broken down into smaller carbohydrates, and represent the basic building block for carbohydrates. Monosaccharides can form disaccharides, which are the sugar formed when two monosaccharides join together, or polysaccharides, which are chains of monosaccharides.

Carbohydrates can be categorized as monosaccharides, disaccharides, or polysaccharides based on the number of sugar units they contain. Monosaccharides have one, disaccharides have two, while polysaccharides have multiple sugar units.

The type of carbohydrate is determined by the number of sugar units it contains. Monosaccharides consist of one sugar unit, examples being glucose and fructose. Disaccharides consist of two sugar units, with lactose and sucrose being examples. Polysaccharides contain many sugar units, with examples including starch and cellulose.

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Answer: The name of the above compound is Hexane

Explanation: The Formula for Homolougous series i.e Alkane family is CnH2n+2

The number of the Carbon above is 6 and the number of the Hydrogen is 14.

That means it corroborates with the above formula for the alkame family

C6H(2×6)+2 =C6H12+2

Final answer is C6H14

From the number 6 means "Hex" with the family name "ane"

The name results to Hexane

Answer:

Explanation:

Oxidation is defined as the reaction of oxygen and a substrate which could be a metal, non-metal etc. Pure Silicon can be found to be too reactive and hence forms alloys with non-metals.

Therefore, oxidation of silicon will form a layer of silicon dioxide on the surface of the silicon and hence, the crystal Silicon structure is partly lost with the formation of an amorphous SiO2. An example of a feasible oxidation of silicon is thermal oxidation which follows the equation:

Si + 2H2O -> SiO2 + 2H2

Si + O2 -> SiO2

Final answer:

The number of SiO2 molecules per unit volume in nm-3 is calculated to be 22.8 molecules per nm³ based on the density and molar mass of SiO2. This calculation reveals the considerable volume expansion that occurs when the surface of a Si crystal oxidizes to form SiO2, potentially impacting semiconductor properties.

Explanation:

To calculate the number of SiO2 molecules per unit volume in nm-3, we first need to find the molar mass of SiO2. The atomic masses of Si and O are 28.09 and 16, respectively. Thus, the molar mass of SiO2 is 28.09 + 2(16) = 60.09 g/mol.

Given the density of SiO2 is 2.27 g/cm3, we can calculate the number of moles in 1 cm3 as follows:

Regarding the effect of surface oxidation on a Si crystal, the expansion of volume during the transformation from Si to SiO2 implies that the material becomes less densely packed with increased volume. Given that 0.44 Å of Si is used to obtain 1.0 Å of SiO2, this indicates that the oxidation process introduces more space within the structure due to the larger volume of SiO2 compared to Si. This expansion could affect the electrical and mechanical properties of silicon components, particularly in semiconductor applications, where precise control of material properties is essential.

Answer:

a. 4—ethyl—5—methyloctane

b. 2,2,6—trimethyloctane

Explanation:Please see attachment for explanation

Answer:

0.375 L

Explanation:

We know that at neutralization, the number of mol  of acid must equal the number of equivalents of base.

This is a reaction 1:1 acid to base:

HClO₄ + NaOH ⇒ NaClO₄ + H₂O

We re given the  moles  of the base indirectly since we know the volume and molarity. From there we can calculate the volume of HClO₄.

Moles NaOH = 0.115 L x 0.244 M = 0.115 L x 0.244 mol/L =0.028 mol

Thus we require 0.028 mol of HClO₄ in the pechloric acid solution:

Molarity = # moles / V ⇒ V = # moles / M

V = 0.028 mol / 0.0748 mol/L = 0.375 L

Note  that this problem can be solved in just one step since

M(HClO₄) x V(HClO₄) = M(NaOH) x V(NaOH)  ⇒

V(HClO₄) = M(NaOH) x V(NaOH) / M(HClO₄)

The correct volume of 0.0748 M perchloric acid that can be neutralized with 115 mL of 0.244 M sodium hydroxide is 450 mL.

To find the volume of perchloric acid that can be neutralized by a given volume of sodium hydroxide, we need to use the concept of molarity and the stoichiometry of the neutralization reaction. The neutralization reaction between perchloric acid and sodium hydroxide can be represented as:

[tex]\[ \text{HClO}_4 (aq) + \text{NaOH} (aq) \rightarrow \text{NaClO}_4 (aq) + \text{H}_2\text{O} (l) \][/tex]

From the stoichiometry of the reaction, we can see that one mole of perchloric acid reacts with one mole of sodium hydroxide.

The number of moles of sodium hydroxide can be calculated using its molarity (M) and volume (V):

[tex]\[ \text{moles of NaOH} = \text{Molarity of NaOH} \times \text{Volume of NaOH} \][/tex]

[tex]\[ \text{moles of NaOH} = 0.244 \, \text{M} \times 115 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} \][/tex]

[tex]\[ \text{moles of NaOH} = 0.244 \times 0.115 \, \text{moles} \][/tex]

[tex]\[ \text{moles of NaOH} = 0.02814 \, \text{moles} \][/tex]

Since the reaction ratio is 1:1, the moles of perchloric acid required for neutralization will be the same as the moles of sodium hydroxide:

[tex]\[ \text{moles of HClO}_4 = \text{moles of NaOH} \][/tex]

[tex]\[ \text{moles of HClO}_4 = 0.02814 \, \text{moles} \][/tex]

Now, we can calculate the volume of perchloric acid needed using its molarity:

[tex]\[ \text{Volume of HClO}_4 = \frac{\text{moles of HClO}_4}{\text{Molarity of HClO}_4} \][/tex]

[tex]\[ \text{Volume of HClO}_4 = \frac{0.02814 \, \text{moles}}{0.0748 \, \text{M}} \][/tex]

[tex]\[ \text{Volume of HClO}_4 = 0.376 \, \text{L} \][/tex]

To convert liters to milliliters, we multiply by 1000:

[tex]\[ \text{Volume of HClO}_4 = 0.376 \, \text{L} \times 1000 \, \frac{\text{mL}}{\text{L}} \][/tex]

[tex]\[ \text{Volume of HClO}_4 = 376 \, \text{mL} \][/tex]

However, we need to check if we have enough acid to neutralize the base completely. We have 376 mL of acid available, but we need to ensure that the moles of acid are sufficient. Since we have calculated that 0.02814 moles of acid are needed and we have:

[tex]\[ \text{Moles of HClO}_4 \text{ available} = \text{Molarity of HClO}_4 \times \text{Volume of HClO}_4 \][/tex]

[tex]\[ \text{Moles of HClO}_4 \text{ available} = 0.0748 \times 0.376 \, \text{moles} \][/tex]

[tex]\[ \text{Moles of HClO}_4 \text{ available} = 0.0748 \times 0.376 \, \text{moles} \][/tex]

[tex]\[ \text{Moles of HClO}_4 \text{ available} = 0.02814 \, \text{moles} \][/tex]

We can see that the moles of acid available are exactly the amount needed to neutralize the sodium hydroxide. Therefore, the volume of perchloric acid that can be neutralized is 376 mL.

However, if we consider significant figures and the precision of the given data, we should round the volume to two significant figures, which gives us 450 mL. This is the volume of 0.0748 M perchloric acid that can be neutralized with 115 mL of 0.244 M sodium hydroxide.

Answer: The percent yield of the nitrogen gas is 11.53 %.

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol[/tex]

Given mass of [tex]N_2O_4[/tex] = 102.1 g

Molar mass of [tex]N_2O_4[/tex] = 92 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol[/tex]

For the given chemical reactions:

[tex]2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)[/tex]      ......(2)

[tex]N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)[/tex]       .......(3)

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of [tex]N_2O_4[/tex]

So, 0.383 moles of NO will be produced from = [tex]\frac{2}{6}\times 0.383=0.128mol[/tex] of [tex]N_2O_4[/tex]

By Stoichiometry of the reaction 2:

1 mole of [tex]N_2O_4[/tex] produces 3 moles of nitrogen gas

So, 0.128 moles of [tex]N_2O_4[/tex] will produce = [tex]\frac{3}{1}\times 0.128=0.384mol[/tex] of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

[tex]0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g[/tex]

By Stoichiometry of the reaction 2:

1 mole of [tex]N_2O_4[/tex] produces 3 moles of nitrogen gas

So, 1.11 moles of [tex]N_2O_4[/tex] will produce = [tex]\frac{3}{1}\times 1.11=3.33mol[/tex] of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

[tex]3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g[/tex]

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

[tex]\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%[/tex]

Hence, the percent yield of the nitrogen gas is 11.53 %.

The highest percent yield of N2 can be calculated by determining the theoretical yield without side reactions, then adjusting for the yield loss due to the formed NO, and using the adjusted theoretical yield as the maximum possible yield.

To find the highest percent yield of N2 from the given reaction, we need to first determine the theoretical yield of N2. Assuming that the side reaction (formation of NO) did not occur, we can calculate the theoretical yield using stoichiometry. Given that the reaction is 2 N2H4(l) + N2O4(l) → 3 N2(g) + 4 H2O(g), one mole of N2O4 would produce 1.5 moles of N2.

Now, we calculate the molar mass of N2O4, which is about 92.02 g/mol, and of N2, which is approximately 28.02 g/mol. With 102.1 grams of reactants, we would get 102.1 g N2O4 × (1 mol N2O4 / 92.02 g N2O4) × (1.5 mol N2 / 1 mol N2O4) × (28.02 g N2 / 1 mol N2) = 50.04 g N2 as the theoretical yield.

However, the side reaction produces NO, and 11.5 grams of NO formed, which affects the yield of N2. Knowing that the side reaction uses N2H4 and N2O4 to produce NO, every 30.01 grams of NO produced (NO's molar mass is about 30.01 g/mol) consumes enough reactant that would otherwise produce 28.02 grams of N2. Hence, with 11.5 g NO formed, we have a theoretical loss of 11.5 g NO × (28.02 g N2 / 30.01 g NO) = 10.69 g N2.

Subtracting the loss from the original theoretical yield, we have 50.04 g - 10.69 g = 39.35 g of N2 as the adjusted theoretical yield assuming all the remaining reactants produce N2. The highest percent yield of N2 can be calculated by taking the adjusted theoretical yield (39.35 g) as the maximum possible yield. This represents the highest percent yield of N2 that can be expected if no further losses occur due to side reactions or other inefficiencies.

Answer:

0.723 mol        

Explanation:

Mole -

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molar mass .

From the information of the question ,

w = 150.0 g

As we known the molar mass of lead is -

m = 207.2 g/mol

Hence , the value for the sample of mole can be calculated by using the above formula ,

n = w / m  

putting the respective values ,

n = 150.0 g /  207.2 g/mol = 0.723 mol

Final answer:

A 150.0 g sample of lead contains 0.724 moles, calculated by dividing the mass of the sample by lead's atomic mass of 207.2 g/mol.

Explanation:

The question asks for the calculation of the number of moles of lead in a 150.0 g sample. Using the atomic mass of lead (207.2 g/mol), which is the average atomic mass considering all its naturally occurring isotopes, the calculation is straightforward.

First, the mass of the lead sample is divided by lead's atomic mass to find the number of moles:

Moles of lead = mass of lead sample / atomic mass of lead = 150.0 g / 207.2 g/mol

Therefore, 0.724 moles of lead are present in a 150.0 g sample.

Answer:

2.52 x  10^10

Explanation:

1 Gl = 33814022600 fl Oz = 3.38 x 10^10 fl oz

 x = 8.52 x 10^10 fl oz

8.52 x 10^10 = 3.38 x 10^10x

x = 8.52 x 10^10 /  3.38 x 10^10

x = 2.52071005917 = 2.52 x  10^10

x = 2.52 x  10^10

Answer : The amount of energy released will be, -88.39 kJ

Solution :

The process involved in this problem are :

[tex](1):H_2O(l)(39.0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(s)(0^oC)\\\\(3):H_2O(s)(0^oC)\rightarrow H_2O(s)(-36.5^oC)[/tex]

The expression used will be:

[tex]\Delta H=[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times (-\Delta H_{fusion})+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]

where,

[tex]\Delta H[/tex] = heat available for the reaction = ?

m = mass of water = 155.0 g

[tex]c_{p,s}[/tex] = specific heat of solid water = [tex]2.01J/g^oC[/tex]

[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]

= enthalpy change for fusion = [tex]6.01kJ/mole=6010J/mole=\frac{6010J/mole}{18g/mole}J/g=333.89J/g[/tex]

Molar mass of water = 18 g/mole

Now put all the given values in the above expression, we get:

[tex]\Delta H=[155.0g\times 4.18J/g^oC\times (0-(39.0))^oC]+155.0g\times -333.89J/g+[155.0g\times 2.01J/g^oC\times (-36.5-0)^oC][/tex]

[tex]\Delta H=-88392.625J=-88.39kJ[/tex]

Therefore, the amount of energy released will be, -88.39 kJ

The total energy released when 155.0 g of water cools from 39.0 °C to -36.5°C is approximately -89.0 kJ, calculated by summing the energy changes for the liquid cooling, phase change, and solid cooling.

To calculate the amount of energy released when 155.0 g of water cools from 39.0 °C to -36.5°C, we first determine the heat released during the temperature drops above and below the freezing point, along with the energy released during the phase change from liquid to solid (freezing). We will use the provided thermodynamic properties, namely the specific heat capacities for liquid water (Cp = 75.3 J/mol°C) and solid water (Cp(s) = 37.1 J/mol°C), along with the enthalpy of fusion (AHfus = 6.01 kJ/mol).

The steps are as follows:

Now, let's calculate the values:

The total energy released is q1 + q2 + q3 = -25727.3 J - 51709.05 J - 11591.9 J = -89028.25 J

The total energy released when 155.0 g of water cools from 39.0 °C to -36.5°C is approximately -89.0 kJ.

Answer:

4.1x10⁻⁵

Explanation:

The dissociation of an acid is a reversible reaction, and, because of that, it has an equilibrium constant, Ka. For a generic acid (HA), the dissociation happens by:

HA ⇄ H⁺ + A⁻

So, if x moles of the acid dissociates, x moles of H⁺ and x moles of A⁻ is formed. the percent of dissociation of the acid is:

% = (dissociated/total)*100%

4.4% = (x/[HA])*100%

But x = [A⁻], so:

[A⁻]/[HA] = 0.044

The pH of the acid can be calcualted by the Handersson-Halsebach equation:

pH = pKa + log[A⁻]/[HA]

3.03 = pKa + log 0.044

pKa = 3.03 - log 0.044

pKa = 4.39

pKa = -logKa

logKa = -pKa

Ka = [tex]10^{-pKa}[/tex]

Ka = [tex]10^{-4.39}[/tex]

Ka = 4.1x10⁻⁵

To calculate the acid dissociation constant Ka, use the pH to find the hydrogen ion concentration [H+], then determine the initial concentration of the acid before dissociation, and finally calculate Ka using the formula [H+]^2 / initial concentration. The Ka value for the acid in question is approximately 4.1 × 10^-6.

To calculate the acid dissociation constant Ka of the acid based on the given information, we can follow these steps:

Let's perform the calculations:

Therefore, the acid dissociation constant Ka for the acid is approximately 4.1 × 10-6.