A chemist measures the amount of iodine solid produced during an experiment. He finds that of iodine solid is produced. Calculate the number of moles of iodine solid produced. Round your answer to significant digits.

Answer:

storage solution , deionized water, stabilizes

Explanation:

A pH meter is a scientific device or instrument that is used to measure the pH of a given aqueous solution thereby determining the nature of the solution whether it is acidic or basic or neutral.

While using the pH meter or taking the measurement using the pH meter --

Answer:

19.7 g.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize this problem can be solved by using a molecules-moles-mass relationship, starting with the given molecules, using the Avogadro's number and the molar mass of nitric acid (63.01 g/mol):

[tex]18.8x10^{22}molec*\frac{1mol}{6.022x10^{23}molec}* \frac{63.01g}{1mol} \\\\=19.7g[/tex]

Regards!

Answer:

Theoretical yield is 138.4 g

Explanation:

In the first step we determine the reaction:

3Mg + N₂ → Mg₃N₂

Mass of reactant is 100 g. We assume the nitrogen is in excess, so we work with Mg. We convert mass to moles:

100 g . 1mol/ 24.3g = 4.11 moles of Mg.

Ratio is 3:1. 3 moles of Mg can produce 1 mol of nitride

Our 4.11 moles, may produce (4.11 . 1)/3 = 1.37 moles of Mg₃N₂

We convert mass to moles, to find the theoretical yield:

1.37 mol . 100.9 g/mol = 138.2 g

Answer:

It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed

Explanation:

A reaction may be endothermic or exothermic. In an endothermic reaction, energy is absorbed by the process while in an exothermic process energy is given out by the process.

Recall that the enthalpy change of a reaction = enthalpy of products - enthalpy of reactants

Hence, where the energy required to break bonds in the reactants is less than the energy released when the products are formed, the reaction is endothermic.

For an endothermic reaction, the enthalpy change of the reaction is positive.

In this case, enthalpy of reaction = 920 - (-750) = 1670 kJ/mol

Answer:

Atmosphere to mmHg Conversion Example. Task: Convert 8 atmospheres to mmHg (show work) Formula: atm x 760 = mmHg Calculations: 8 atm x 760 = 6,080 mmHg Result: 8 atm is equal to 6,080 mmHg.

Explanation:

Answer:

all I can say is town near the ocean atmospheric changes will be cooler, warm, sea breeze, and fresh healthy air. Then when it comes to the mountain lot of change firstly there's a dry air

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction:    NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:  

[NH3] = 0.030M    [NH4+] = 0M    [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 =  x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43          

Initially, the pH of the solution is given by the dissociation of NH₃ in water.  

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻     (1)

The constant of the above reaction is:

[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex]   (2)

At the equilibrium, we have:  

   NH₃    +    H₂O   ⇄   NH₄⁺    +    OH⁻     (3)  

0.030 M - x                      x               x

[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]  

Now, we can calculate the pH of the solution as follows:

[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]

Hence, the initial pH is 10.86.

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻  (4)  

We can calculate the pH of the solution from the equilibrium reaction (3).            

[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)  

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

[tex] n_{b} = n_{i} - n_{HCl} [/tex]     (6)

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]          

[tex] n_{a} = n_{HCl} [/tex]   (7)

[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]

The concentrations are given by:

[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex]   (8)

[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex]      (9)

After entering the values of Ca and Cb into equation (5) and solving for x, we have:  

[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]

Hence, the pH is 9.66.

We can find the pH of the solution from the reaction of equilibrium (3).

The concentrations are (eq 8 and 9):

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]    

[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]    

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:  

[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]

So, the pH is 9.15.

We can find the pH of the solution from reaction (3).

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]      

[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]      

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]

x = 5.013x10⁻⁷ = [OH⁻]      

Then, the pH is:  

[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]  

So, the pH is 7.70.

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]    

[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:                

NH₄⁺    +    H₂O   ⇄   NH₃    +    H₃O⁺

Ca - x                             x               x

[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]  

[tex] Ka(Ca - x) - x^{2} = 0 [/tex]   (10)          

We can find the acid constant as follows:

[tex] Kw = Ka*Kb [/tex]

Where Kw is the constant of water = 10⁻¹⁴

[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]  

The concentration of NH₄⁺ is:

[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]      

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:  

[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]

Hence, the pH is 5.56.

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]      

[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]

Therefore, the pH is 3.43.

Find more about pH here:

brainly.com/question/491373

I hope it helps you!  

Answer:

The answer is growth rate

Explanation:

it will help you

Answer:

e. HCOOH and NaCHOO

Explanation:

For a buffer solution, both an acid and its conjugate base are required.

With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.

Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:

The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO

Sodium metal reacts rapidly with water to form a solution of sodium hydroxide (NaOH) and hydrogen gas (H2). This reaction is exothermic.

Equation:

2Na + 2H²0 --------}- 2NaOH + H²

Answer:

100,000

Explanation:

Answer: The statement it is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed, is true.

Explanation:

A chemical reaction in which heat energy  is released is called an exothermic reaction. For exothermic reactions, the value of [tex]\Delta H[/tex] is always negative.

A chemical reaction in which heat energy is absorbed is called an endothermic reaction. For endothermic reaction, the value of [tex]\Delta H[/tex] is always positive.

In endothermic reactions, energy required for breaking the bonds between reactants is less than the energy when products are formed due to which the value of [tex]\Delta H[/tex] remains positive.

Thus, we can conclude that the statement it is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed, is true.

It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed. The correct option is B.

The above reaction is endothermic because more energy is produced when new bonds form in the products (H = 920 kJ/mol) than is required to break bonds in the reactants (H = -750 kJ/mol).

In an endothermic process, more energy than is generated during bond creation is absorbed from the environment to dissolve existing bonds. This causes a net absorption of energy, which cools the system.

The reaction takes more energy than it releases, proving its endothermic nature, as seen by the positive difference between the energy needed to dissolve bonds and the energy released during bond formation.

Thus, the correct option is B.

For more details regarding endothermic process, visit:

https://brainly.com/question/28909381

#SPJ3

Your question seems incomplete, the probable complete question is:

During a reaction, ΔH for reactants is −750 kJ/mol and ΔH for products is 920 kJ/mol. Which statement is correct about the reaction? (5 points)

Group of answer choices

A. It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.

B. It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.

C. It is exothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.

D. It is exothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.

Answer:

co+ o2+ 2co2 is not balanced reaction

Answer:

MoClBr₂

Explanation:

First we calculate the mass of bromine in the compound:

Then we calculate the number of moles of each element, using their respective molar masses:

Now we divide those numbers of moles by the lowest number among them:

Meaning the empirical formula is MoClBr₂.

Answer:

i think 1. law of muliple proportion

Explanation:

please like

Answer: The type of solvent that should be dried with calcium chloride is esters while magnesium sulfate is diethyl ether

Explanation:

Drying agents are mainly hygroscopic substances that has the ability to absorb water on exposure to the atmosphere but not enough to form solutions. They are used in desiccators. Examples of drying agents include:

--> CALCIUM CHLORIDE: This is a compound of calcium that is found in soil water and sea water. It is prepared by the action of dilute hydrochloric acid on calcium trioxocarbonate(IV). The anhydrous salt is used in drying a wide variety of solvent including esters.

--> MAGNESIUM SULFATE: This is a slightly acidic drying agent. It works well in solvents like diethyl ether. It is a fast drying agent because it comes as a fine powder with a large surface area.

Explanation:

Answer:

For a: Lead iodide is a yellow precipitate.

For b: Barium sulfate is a white precipitate.

For c: Ferric hydroxide is a brown precipitate.

For d: Copper (II) hydroxide is a blue precipitate.

Explanation:

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

The chemical reaction between KI and lead (II) nitrate follows:

[tex]2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)[/tex]

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

The chemical reaction between barium chloride and sulfuric acid follows:

[tex]BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)[/tex]

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

The chemical reaction between NaOH and ferric chloride follows:

[tex]3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)[/tex]

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

The chemical reaction between NaOH and copper sulfate follows:

[tex]CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4[/tex]

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

(a) The yellow precipitate formed in the reaction between KI and Pb(NO3)2 would be PbI2 according to the equation:

[tex]Pb(NO_3)_2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO_3(aq)[/tex]

(b) The white precipitate formed in the reaction between BaCl2 and H2SO4 would be  BaSO4 according to the equation:

   [tex]BaCl_2 (aq) + H_2SO_4 (aq) ---> BaSO_4 (s) + 2 HCl (aq)[/tex]

(c) The brown precipitate formed in the reaction between NaOH and FeCl3 would be Fe(OH)3 according to the equation:

[tex]FeCl_3 (aq) + NaOH (aq) ---> Fe(OH)_3 (s) + NaCl (aq)[/tex]

(d) The blue precipitate formed in the reaction between CuSO4 and NaOH would be Cu(OH)2 according to the equation:

[tex]CuSO_4(aq) + 2 NaOH (aq) ---> Cu(OH)_2 (s) + Na_2SO_4 (aq)[/tex]

More on precipitation reaction can be found here: https://brainly.com/question/24846690

Answer:

I hoped it helps you fod blessed:)

Answer:

Cl

Explanation:

Electronegativity is the ability of an electron to attract electrons.

Now, due to the fact that halogens need just one more electron to become stable in their outermost shell, it means all halogens are electronegative.

However, the smaller the atomic number, the bigger the charge density and thus the more electronegative.

Thus, it is the halogen element with the highest atomic number further down the periodic table that will have the least electro negativity and thus require highest amount of energy to attract other electrons.

Thus, since chlorine (Cl) has the least atomic number of 17, then it means that it will be the one that will easily accept the electrons the most from other elements. Therefore the process of transferring electrons from potassium to chlorine will take the least amount of energy.

Answer:

A person doing physical work needs lots of good carbohydrates to keep their energy levels up and proteins to repair a muscle that might get wear and tear from overexertion. Carbohydrate will help the person work for more extended periods.

Answer: The osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.

Explanation:

Given: Mass = 5.0 g

Volume = 1 L

Molar mass of sucrose = 342.3 g/mol

Moles are the mass of a substance divided by its molar mass. So, moles of sucrose are calculated as follows.

[tex]Moles = \frac{mass}{molarmass}\\= \frac{5.0 g}{342.3 g/mol}\\= 0.0146 mol[/tex]

Hence, concentration of sucrose is calculated as follows.

[tex]Concentration = \frac{moles}{Volume (in L)}\\= \frac{0.0146 mol}{1 L}\\= 0.0146 M[/tex]

Formula used to calculate osmotic pressure is as follows.

[tex]\pi = CRT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure

C = concentration

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]\pi = CRT\\= 0.0146 \times 0.0821 L atm/mol K \times 298 K\\= 0.357 atm (1 atm = 760 torr)\\= 271.32 torr[/tex]

Thus, we can conclude that the osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.

The question is incomplete, the complete question is:

What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.

Answer: 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

Explanation:

We first calculate the number of moles of soft drink in a volume of 10 mL

The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(1)

Taking the concentration of soft drink from the example be = 0.375 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

[tex]0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol[/tex]

Calculating volume of sweetened tea:

Moles of sugar = 0.00375 mol

Molarity of sweetened tea = 0.05 M

Putting values in equation 1, we get:

[tex]0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL[/tex]

Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

Answer:

A hydrometer is a special device used to determine the density of liquids.

Explanation:

I hope this helps you. Have a nice day!

The question is incomplete, the complete question is shown in the image attached

Answer:

3-isopropyl-1-hexyne

Explanation:

Organic compounds can be named from the structure of the compound. The name reveals the arrangement of atoms and bonds in the molecule.

If we look at the compound in the question, we will notice that the parent chain contains six carbon atoms, the triple bond is located at position 1 and the isopropyl substituent is attached to carbon 3.

Hence the proper name of the compound becomes, 3-isopropyl-1-hexyne.

The question is incomplete, the complete question is;

What functional group results when cyclopentanone reacts with ethylamine in the presence of trace acid? A) cyanohydrin B) semicarbazone C) imine D) enamine E) oxime

Answer:

imine

Explanation:

An imine is an unsaturated amine. An imine contains the carbon- nitrogen double bond.

Imines are obtained when a carbonyl compound is condensed with NH3 or an amine. The reaction involves several steps in its mechanism.

Since cyclopentanone is a ketone (carbonyl compound) and ethylamine is an amine,in the presence of trace acid, condensation of the two compounds occur to yield an imine

Answer:

0.74 M

Explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 5.90 M

Volume of stock solution (V₁) = 0.250 L

Volume of diluted solution (V₂) = 2 L

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5.90 × 0.250 = M₂ × 2

1.475 = M₂ × 2

Divide both side by 2

M₂ = 1.475 / 2

M₂ = 0.74 M

Thus, the molarity of the diluted solution is 0.74 M

Answer:

(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms

Explanation:

Alcohols are poor leaving groups.

To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.

Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.

After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.