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

Answer:

b. glass and charcoal

Explanation:

Step 1: Given data

Step 2: Determine which material will float in molten lead

Density is an intrinsic property of matter. Less dense materials float in more dense materials. The materials whose density is lower than that of lead and will therefore float on it are glass and charcoal.

Answer:2C4H10+2C12+12O2 4CO2+CC14+H20

Answer:

It has a negative charge and is located in orbitals around the nucleus

Explanation:

The statement describes an electron is " It has a negative charge and is located in orbitals around the nucleus."

The electron would be a subatomic particle with a negatively one elementary charge electric charge.

Protons, that have a positive charge, as well as neutrons, which have no electrical charge, make up the nucleus. Quarks were subatomic particles that make up protons but also neutrons.

Electrons were present surrounding the atom's nucleus, in contrast to protons as well as neutrons, that are contained within the nucleus at its core. Negative electrons were drawn to the positive nucleus since the electric charges of opposite polarity attract one another.

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Answer:

2.07459

Explanation:

this is the correct answer.

Answer:

option a

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Answer:

Calculate the moles of N2 molecules in 3.94 grams of nitrogen.

Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules.

Explanation:

Calculate the moles of N2 molecules in 4.73 liters of nitrogen gas. FALSE. You can't make this conversion using only the conversion factor with units of g/mol. To convert liters to moles are necessaries pressure, temperature and volume of the gas to use PV = nRT

Calculate the grams of N2 in 10.58 liters of nitrogen gas. FALSE. As explained, you need, P,V and T to find the moles of the gas. With the moles you can find the mass using the conversion factor of 28.02g/mol

Calculate the moles of N2 molecules in 3.94 grams of nitrogen. TRUE. You can find the moles of N2 as follows:

3.94g N2 * (1mol/28.02g) = 0.14 moles of N2 molecules

Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules. TRUE. The mass in 5.03x10²⁰ moles of nitrogen molecules is:

5.03x10²⁰ moles * (28.02g/mol) = 1.4x10²²g of nitrogen.

Answer:

At higher temperatures, particles move faster and collide more, increasing solubility rates.

Agitation increases solubility rates as well, by bringing fresh solvent into contact with the undissolved solute

The smaller the particle size, the higher (faster) solubility rate. Vice versa, the bigger the particle size, the lower (slower) solubility rate.

Explanation:

Answer:

9.36 g

Explanation:

The equation of the reaction is;

C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)

Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles

1 mole of octane yields 9 moles of water

0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water

Number of moles of oxygen = 23g/32g/mol = 0.72 moles

12.5 moles of oxygen yields 9 moles of water

0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water

Hence oxygen is the limiting reactant;

Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g

Answer:

64 g

Explanation:

Step 1: Write the balanced equation

2 Na + Cl₂ ⇒ 2 NaCl

Step 2: Calculate the moles corresponding to 25 g of Na

The molar mass of Na is 22.98 g/mol.

25 g × 1 mol/22.98 g = 1.1 mol

Step 3: Calculate the moles of NaCl formed from 1.1 moles of Na

The molar ratio of Na to NaCl is 2:2. The moles of NaCl formed are 2/2 × 1.1 mol = 1.1 mol.

Step 4: Calculate the mass corresponding to 1.1 moles of NaCl

The molar mass of NaCl is 58.44 g/mol.

1.1 mol × 58.44 g/mol = 64 g

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:

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Explanation:

Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.

In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

In a solution it is fulfilled:

Ci* Vi = Cf* Vf

where:

In this case:

Replacing:

1.40 M* 20 mL= 0.088 M* Vf

Solving:

[tex]Vf=\frac{1.40 M* 20 mL}{0.088 M}[/tex]

Vf= 318.18 mL

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Answer:

SN1 mechanism

Explanation:

The mechanism of this reaction is shown in the image attached.

The Di- n- pentyl ether is first protonated. The CH3(CH2)4OH is now a good leaving group as shown.

The attack of the bromide ion on the cation formed completes the mechanism to yield 1- bromopentane as shown in the mechanism.

Answer:

The equivalent weight of calcium hydroxide is 1/2 he mass of a mol of calcium hydroxide. 1 mol Ca(OH)2 = 74 grams Ca(OH)2 ; 1 equivalent Ca(OH)2 = 37 grams Ca(OH)2......

Explanation:

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Answer:

Danger

Explanation:

Answer:

molar heat of combustion = -5156 *10³ kJ/mol

Explanation:

A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

Step 1: Data given

Mass of naphthalene = 1.435 grams

Initial temperature of water = 20.28 °C

Final temperature of water = 25.95 °C

heat capacity of the bomb plus water was 10.17 kJ/°C

Molar mass naphtalene = 128.2 g/mol

Step 2:

Qcal = Ccal * ΔT

⇒with Qcal =the heat of combustion

⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C

⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C

Qcal = 10.17 kJ/°C * 5.67 °C

Qcal = 57.7 kJ

Step 3: Calculate moles

Moles naphthalene = 1.435 grams / 128.2 g/mol

Moles naphthalene = 0.01119 moles

Step 4: Calculate the molar heat of combustion

molar heat of combustion = Qcal/ moles

molar heat of combustion = -57.7 kJ/ 0.01119 moles

molar heat of combustion = -5156 *10³ kJ/mol

Answer:

Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.

[tex]2CO(g) + O2(g) <=> 2CO2[/tex]

Explanation:

When the reactants concentration increases, then the equilibrium will shift towards products and when the concentration of products increases, then equilibrium will shift towards reactants.

So, increases in concentration of carbon monoxide (CO) shifts the equilibrium to favor the formation of carbondioxide.

Similarly increase in concentration of oxygen also favor the formation of product carbon dioxide.

Increase in concentration of CO2 favors the formation of CO and O2.

Decrease in product concentration also favors the formation of product.

Decrease in reactant concentration favors the formation of reactants only.

Answer:

the product and the reactant must be balanced

if u are required to give the mechanism if the reaction it must be written

Answer:

The correct answer is "-268.667°C".

Explanation:

Given:

Temperature,

= 4.483 K (below)

Now,

The formula of temperature conversion will be:

⇒ [tex]T(^{\circ} C)=T(K)-273.15[/tex]

By putting the values, we get

⇒            [tex]=4.483-273.15[/tex]

⇒            [tex]=-268.667^{\circ} C[/tex]

Thus the above is the correct answer.

Answer:

The correct options are a, b and d

Explanation:

A catalyst is a substance that increases the rate of a chemical reaction by reducing the activation energy. Le Catelier's  principle explains how a substance or an "action" can affect a reaction in equilibrium.

The principle states that when a change is made to the conditions of a reacting system at equilibrium, the position of the equilibrium moves to counteract the change made. These changes are change in temperature, pressure, volume and/or concentration. These changes will either cause the equilibrium to shift forward or backward.

However, the presence of a catalyst DOES NOT affect a chemical equilibrium/equilibrium constant nor does it affect the reaction quotient because the same amount of reactants and products are available just as in uncatalyzed reaction except that the reaction proceeds faster (which does not affect equilibrium).

The rate of reaction is given as the time required by the reactant to convert into the product. The addition of catalyst increases the rate of reaction, while the reaction quotient and the equilibrium remain unaffected.

A catalyst is a chemical or compound that adds to the reaction and lowers the activation energy by providing an alternative path to the reaction.

The catalyst takes part in the reaction but did not consume in the chemical reaction.

The equilibrium and the reaction quotient are dependent on the conversion of the reactant to the product. The catalyst is not used in the reaction and thus did not affect the reaction quotient or the equilibrium.

Hence, options A, B, and D are correct for the use of catalysts in the chemical reaction.

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Answer:

88.9%

Explanation:

Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:

Después calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:

Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:

Finalmente calculamos el rendimiento porcentual:

Answer:

1.88 A

Explanation:

Let's consider the reduction of copper in an electrolytic cell.

Cu²⁺ + 2 e⁻ ⇒ Cu

We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

The charge used is:

[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]

We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

The current used is:

I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A

Answer:

Melting-point apparatus

Answer:

 Lewis structure of Hydronium ion is shown below :                          

Explanation:

Lewis structure : It is a representation of valence electrons on the atoms in a molecule

Here , Hydronium ion is given , which contains 1 atom of oxygen and 3 atoms of hydrogen .

Oxygen has a total of 6 valence electrons and hydrogen contains 1 valence electron .

Oxygen share its 3 valence electrons with 3 hydrogen atoms and left with 3 valence electrons. From these three valence  electrons of oxygen atom  two electrons will be shown as a pair of electrons on oxygen atom but a single electron can not be shown . So , to simplify this, one positive charge is shown overall .  

Resonance structure will be same as the hybrid structure because all  three atoms are same , that is hydrogen .

Answer:

Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol

Explanation:

According to IUPAC rules, the name of a compound is:

Prefix+root word+suffix

1) Select the longest carbon chain and it gives the root word.

2) The substituents give the prefix.

3) The functional group gives the secondary suffix and the type of carbon chain gives the primary suffix.

The structure of the given compounds are shown below:

Answer:

The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.

The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.

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

The Urms refers to the root mean square speed of the gas. The order of increasing Urms for the gases shown in the question; NF3 < H2S < H2O < He.

The Urms refers to the root mean square speed of the gas. This is ultimately dependent on the relative molecular mass of the gases when they are maintained at the same temperature.

Now, let us look at the order of increasing Urms for the gases shown in the question; NF3 < H2S < H2O < He.

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Answer:

d. van der waals force

Explanation:

Van der Waals force :

the weakest intermolecular forceand consist of dipole-dipole force and dispersion force.

Answer:

All of the above.

Explanation:

For a scientific hypothesis to be considered a hypothesis, it has to be testable. When conducting a lab experiment, it also allows the tester to predict what might occur during and after the experimentation. They are also explanatory. For example, theories are hypotheses that have been verified and can explain why something in nature takes place.