Draw structures for (a) a chain isomer, (b) a positional isomer, and (c) a functional isomer of hexan-1-ol
(i.e., 1-hexanol)
a. Chain isomer
b. Positional isomer
c. Functional isomer

Answers

Answer 1

Answer:

See attached picture.

Explanation:

Hello,

In this case, we should define each type of structural formula as shown below:

- Chain isomers: molecules with the same molecular formula, but different arrangements.

- Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups but differ from each other in the location of the functional groups.

- Functional isomers are structural isomers that have the same molecular formula (that is, the same number of atoms of the same elements), but the atoms are connected in different ways so that the groupings are dissimilar.

Regards.

Draw Structures For (a) A Chain Isomer, (b) A Positional Isomer, And (c) A Functional Isomer Of Hexan-1-ol(i.e.,

Related Questions

You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes are additive, what is the molarity of chloride ion in the mixture.

Answers

Answer:

[tex]M=0.727M[/tex]

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

[tex]n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-[/tex]

So the total mole of chloride ions:

[tex]N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-[/tex]

And the total volume by adding the volume of each solution in L:

[tex]V=0.500L+0.425L=0.925L[/tex]

Finally, the molarity turns out:

[tex]M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M[/tex]

Best regards.

State five difference between ionic compound and covalent compound​

Answers

Answer:

Compound are defined as the containing two or more different element .

(1) Ionic compound and (2) Covalent compound.

Explanation:

Covalent compound :- covalent compound are the sharing of electrons two or more atom.

Covalent compound are physical that lower points and compared to ionic .

Covalent compound that contain bond are carbon monoxide (co), and methane .

Covalent compound are share the pair of electrons.

Covalent compound are bonding a hydrogen atoms electron.

Ionic compound a large electrostatic actions between atoms.

Ionic compound are higher melting points and covalent compound.

Ionic compound are bonding a nonmetal electron.

Ionic electron can be donate and received ionic bond.

Ionic compound bonding kl.

If D+2 would react with E-1, what do you predict to be the formula?

Answers

Answer:

DE2

Explanation: for every one D+2 you need two E-1 because +2=-2

A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the percent mass of the tin and percent by mass of oxygen in the sample

Answers

Answer:

Percentage mass of Tin = 96.3%

Percentage mass of oxygen = 6.40%

Explanation:

The product of the reaction is an oxide of tin.

Assuming all of the 0.500 g sample of tin reacted with oxygen to produce the oxide:

Mass of oxide = 0.534 g

Mass of tin present in the oxide = 0.500 g

Mass of oxygen in the oxide = 0.534 g of oxide - 0.500 g Sn = 0.034 g O

Percentage composition = mass of element/mass of compound × 100%

Percentage composition of Sn = 0.500 g/0.534 g × 100 = 93.6% Sn

Percentage composition of oxygen = 0.034 g/0.534 g × 100 = 6.40%

A certain reaction has the following general form. aA → bB At a particular temperature and [A]0 = 2.80 ✕ 10−3 M, concentration versus time data were collected for this reaction, and a plot of 1/[A] versus time resulted in a straight line with a slope value of +3.40 ✕ 10−2 L mol−1 s−1. (a) Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. (Rate expressions take the general form: rate = k . [A]a . [B]b.) rate law:

Answers

Answer:

Rate law: [tex]r=k[A]^2[/tex]

Integrated rate law: [tex]\frac{1}{[A]}=kt+ \frac{1}{[A]_0}[/tex]

Rate constant: [tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]

Explanation:

Hello,

In this case, since the slope is obtained by plotting 1/[A] and it has the units L/(mol*s) or 1/(M*s), we can infer the reaction is second-order, therefore, its rate law is:

[tex]r=k[A]^2[/tex]

The integrated rate law:

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

That is obtained from the integration of:

[tex]\frac{d[A]}{dt}=-k[A]^2[/tex]

And of course, since the slope equals the rate constant, its value is:

[tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]

Regards.

For the product of the reaction below, which proton is removed irreversibly by NaNH2 base, thus preventing any isomerization of the alkyne bond in the product?

Answers

Answer:

The Highly acidic proton joined to one of the carbon in the ALKYNE bond.

(Kindly Check the attachment for the drawing because the solution will need us to draw).

Explanation:

So, let us start by defining some major key terms in this particular Question given above;

(1). ISOMERIZATION: isomerization can simply be defined as the kind is of chemical rearrangement whichay lead to the breaking and the formation of new bonds.

(2). NaNH2 BASE: Sodium amide is a Chemical compound which has a Molar mass of 39.01 g/mol and Heat capacity (C) of 66.15 J/mol K. It is also known as sodamide. It is a good nucleophile.

(3). ALKYNE BOND: it is a C-C joined together by three bonds.

The chemical reaction given in the Question is given in the attachment too.

Therefore, The Highly acidic proton joined to one of the carbon in the ALKYNE bond  is removed irreversibly by NaNH2 base.

1)The average lethal dose of Valium is 1.52 mg/kg of body weight. Estimate how many grams of Valium would be lethal for a 200.-lb woman. Show all your calculations. (1lb = 453.6 g)

2) A patient in hospital is receiving the antibiotic amoxcillin IV at the rate of 50. mL/h. The IV contains 1.5 g of the antibiotic in 1000. mL. (IV stands for intravenous). Calculate the mg/min of the drip. Show all your calculations

Answers

Answer:

1. 0.138g of valium would be lethel in the woman

2. 125mg/min is the drip of the patient

Explanation:

1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.

A person that weighs 200lb requires:

200lb × (453.6g / 1lb) × (1kg / 1000g) = 90.72kg (Weight of the woman in kg)

90.72kg × (1.52mg / kg) =

137.9mg ≡

0.138g of valium would be lethel in the woman

2. The IV contains 1.5g = 1500mg/mL.

If the patient is receiving 5.0mL/h, its rate in mg/h is:

5.0mL/h × (1500mg/mL) = 7500mg/h

Now as 1h = 60min:

7500mg/h × (1h / 60min) =

125mg/min is the drip of the patient

The displacement of a bromine atom by an amine is a substituion reaction. Write out the mechanism of this reaction (2-->3) Why might you expect that the reaction you have performed, using t-BuNH2, to be much slower than the same reaction using methylamine

Answers

Answer:

An alkyl halide can undergo SN2 reaction with an amine

Explanation:

The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.

The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).

This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.

The detailed mechanism of this reaction has been attached to this answer.

If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of B is 0.82 g/ml; the concentration of C is 0.64 M; and the concentration of D is 0.38 M.
A(s) + 3 B(l) _____ 2(aq) + D(aq)
Pick the correct statement about this system.
A. Q < K and reaction shifts left
B. Q > K and reaction shifts left
C. Q > K and reaction shifts right
D. Q = K and reaction does not shift
E. Q < K and reaction shifts right

Answers

Answer:

E. Q < K and reaction shifts right

Explanation:

Step 1: Write the balanced equation

A(s) + 3 B(l) ⇄ 2(aq) + D(aq)

Step 2: Calculate the reaction quotient (Q)

The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.

Q = [C]² × [D]

Q = 0.64² × 0.38

Q = 0.15

Step 3: Compare Q with K and determine in which direction will shift the reaction

Since Q < K, the reaction will shift to the right to attain the equilibrium.

A local barista serves coffee at 85 C. You add ice to the coffee to cool it to 55 C. Assume that an ice cube is 24g and -18.5 degrees Celsius. Hiw many ice cubes would you need to add to your 355mL cup of coffee to bring it to 55 degrees Celsius?.. The specific heat of ice is 2.95J/g degrees Celsius, the specific heat is 4.184 J/g degrees Celsius, and the specific heat of fusion of water is 334 J/g. Remember that an ice cube will need to be warmed to 0 degrees Celsius, will melt, and then the newly melted water will be warmed to 55 degrees Celsius.

A .1
B .3
C .4
D .2

Answers

Answer:

B. 3

Explanation:

To decrease the temperature of your coffee from 85°C to 55°C your system need to absorb energy. This energy will be absorbed from the addition of some ice.

How many energy must be absorbed? You can use:

Q = C×m×ΔT

Where Q is heat (Energy), C is specific heat of your solution (4.184J/g°C), m is its mass (mass of 355mL of coffee = 355g) and ΔT is change in temperature (85°C-55°C = 30°C)

Replacing, your ice needs to absorb:

Q = C×m×ΔT

Q = 4.184J/g°C×355g×30°C

Q = 44559.6J

The energy that is taken from an ice cube to change its temperature from -15°C to 55°C is:

Energy from -15°C to 0°C (C of ice = 2.95J/g°C):

Q = C×m×ΔT

Q = 2.95J/g°C×24g×15°C

Q = 1062J

Now the energy taken to pass the ice from solid to liquid is:

Q = ΔHf×m

Q = 334J/g×24g

Q = 8016J

And the energy to increase the temperature of 0°C to 55°C of 24g of ice:

Q = 4.184J/g°C×24g×55°C

Q = 5522.9J

And the total energy that 1 ice cube needs is:

Q = 1062J + 8016J + 5522.9J

Q = 14600.9J

But you need 44559.6J to decrease the temperature of your coffee, that is:

44600J / 14600.9J = 3.05

≈ 3 ice cubes to decrease the temperature of the coffee.

Right solution:

B. 3

Answer:

3

Explanation:

I’m not positive but I think it’s correct

How many moles of aqueous sodium ions and sulfide ions are formed when 2.05 mol of sodium sulfide dissolved in water

Answers

Explanation:

We have to put out the balanced chemical equation before proceeding. The equation for the dissociation of sodium sulfide in water is given as;

Na₂S  --> 2Na⁺ + S²⁻

From the stoichiometry of the reaction we can tell that;

1 mol of Na₂S produces 2 moles of Na⁺ and 1 mol of S²⁻.

Sodium ion:

1 mol of Na₂S  = 2 mol of Na⁺

2,05 = x

upon solving for x;

x =  2 * 2.05 = 4.10 moles

Sulfide ion:

1 mol of Na₂S  = 1 mol of S²⁻

2.05 = x

upon solving for x;

x = 1 * 2.05 = 2.05 moles

Answer:

HEY ITS RIGHT ABOVE!!!

Explanation:

i did 1.85 instead of 2.05 and it was right... im so sorry

A reaction mixture at 175 K initially contains 522 torr of NO and 421 torr of O2. At equilibrium, the total pressure in the reaction mixture is 748 torr. Calculate Kp at this temperature. Express your answer to three significant figures.

Answers

Answer:

[tex]Kp=0.0386[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]

For which the equilibrium expression is:

[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]

Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:

[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]

And the total pressure:

[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]

Yet it is 748 torr, for which the extent is:

[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]

Therefore, Kp turns out:

[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]

Best regards.

When 3-methylpent-2-ene is treated with mercury(II) acetate in methanol and the resulting product isreacted with NaBH4, what is the primary organic compound which results

Answers

Answer:

3-methylpentan-3-ol

Explanation:

In this case, we have an "Oxymercuration reaction". With this in mind, we will have to add an "OH" to the most substituted carbon of the double bond and we will obtain 3-methylpentan-3-ol. To understand  how this molecule is produced we have to check the mechanism:

The mercury(II) acetate ([tex]Hg(OAC)_2[/tex]) is an ionic substance. So, this substance can be ionized into his ions and we will have the cation [tex]HgOAc^+[/tex] and the anion [tex]AcO^-[/tex]. The cation will attack the double bond and vice-versa to produce a "cyclic intermediate". Then a water molecule will attack the most substituted carbon and the cyclic compound would be broken producing a new bond C-O with a positive charge in the oxygen. Then a deprotonation step takes place and finally, the [tex]NaBH_4[/tex] would reduce the compound to produce the final alcohol.

See figure 1

I hope it helps!

An oxide has a chemical formula with the form X2O3. Which group is element X more likely to be a member of? Select the correct answer below: group 12 group 13 group 14 group 2

Answers

Answer:

Group 13

Explanation:

You know X has 3 valence electrons, as oxygen has a subscript of 3. This means X has an ionic charge of +3. Group 13 consists mainly of metalloids but it also has metals such as aluminum, which has a +3 charge. If you use aluminum as an example, you know that when combined with oxygen, it forms Al2O3. Group 12 has transition metals that don't have +3 ionic charges, group 14 has metalloids, metals that don't have ionic charges of +3, and nonmetals, and group 2 has metals with ionic charges of +2. Group 13 is the answer.

When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4. The balanced equation is:

Answers

Answer:

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

Explanation:

The general schemefor a reaction is given as;

Reactants --> Products

In this question, the reactants are AgNO3 and Na2SO4. The product is Ag2SO4.

The equation is given as;

AgNO3 + Na2SO4 --> Ag2SO4

The other poduct formed in this reaction is NaNO3.

The full reaction is given as;

AgNO3 + Na2SO4 --> Ag2SO4 + NaNO3

The above reaction is not balanced because there are unequal number of atoms of the elements on both sides of the reaction.

The balanced equation is given as;

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

In this equation, there are equal number of moles of the atoms on both sides.

Select the correct answer. A certain reaction has this form: aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1. What is the half-life of this reaction? A. 23.33 seconds B. 0.043 minutes C. 0.0043 seconds D. 23.33 minutes E. 1680 minutes

Answers

Answer:

[tex]\large \boxed{\text{D. 23.34 min}}[/tex]

Explanation:

1. Find the order of reaction

Use information from the graph to find the order.

If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.

2. Find the  half-life

[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]

The half life of the reaction is 23.33 minutes.

We know that for a first order reaction;

ln[A]t = ln[A]o  - kt

A plot of ln[A]t  against time (t) will yield a straight line graph with a slope of -k.

From the question, the slope is -2.97 x 10-2 min-1.

So, -2.97 x 10-2 min-1 = - k

k = 2.97 x 10-2 min-1

The half life of a first order reaction is obtained from;

t1/2 = 0.693/k

t1/2 = 0.693/2.97 x 10-2 min-1

t1/2 = 23.33 minutes

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Given these data in a study on how the rate of a reaction was affected by the concentration of the reactants,
Experiment [A] [B] [C] Rate (mol L‑1 hr‑1 )
1 0.200 0.100 0.600 5.0
2 0.200 0.400 0.400 80.0
3 0.600 0.100 0.200 15.0
4 0.200 0.100 0.200 5.0
5 0.200 0.200 0.400 20.0
From this data, what is the numerical value of the rate constant, (k), for this reaction (value that would be found using the same units used in the data above)?
a. 2083
b. 694
c. 417
d. 2500
e. 83.3

Answers

Answer:

d. 2500

Explanation:

In a kinetic study with 3 different reactants, you change concentrations of the reactants to see how this concentration affects rate of reaction. General law is:

v = k [A]ᵃ [B]ᵇ [C]ⁿ

If you see 1 and 3 experiments, the concentration of C change from 0.600M to 0.200M but reaction rate doesn't change, thus n=0:

v = k [A]ᵃ [B]ᵇ [C]⁰

v = k [A]ᵃ [B]ᵇ×1

Now, reaction 2 and reaction 4 change B from 0.400M to 0.200M having the other reactants constant. When B is duplicated, rate increase 4 times. That means b = 2:

v = k [A]ᵃ [B]ᵇ

v = k [A]ᵃ [B]²

Finally, if you see 3 and 4 reactions, A change from 0.200M to 0.600M and the reaction rate change from 15.0 to 5.0, That means if the concentration of A is triplicated, reaction rate will be triplicated to. Thus a=1:

v = k [A]ᵃ [B]²

v = k [A] [B]²

Relpacing this equation in any experiment (Experiment 5, for example):

20.0 = k [0.200] [0.200]²

2500 = k

That means right answer is:

d. 2500

Which phase change is an example of an exothermic process?
A.
solid to liquid
B.
solid to gas
C.
liquid to solid
D.
liquid to gas
E.
solid to plasma
Reset

Answers

Answer:

C

Explanation:

Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.

C is an exothermic process. To form solid from a liquid, heat energy must be realised to push particles together and form bonds.
An endothermic process is when heat is absorbed to break bonds between particles (liquid-> gas)

Provide the reagents necessary to carry out the following conversion. Group of answer choices KMnO4, NaOH,H2O KMnO4, H3O , 75oC H2SO4, heat 1. mCPBA 2. H3O none of these

Answers

Answer:

KMnO4,H3O^+,75°C

Explanation:

The conversion of cyclohexene to trans-1,2-cylohexanediol is an oxidation reaction. Alkenes are oxidized in the presence of potassium permanganate and acids to yield the corresponding diols.

These diols may also be called glycols. They are molecules that contain two -OH(hydroxyl) groups per molecule. The reaction closely resembles the addition of the two -OH groups of hydrogen peroxide to an alkene.

The bright color of potassium permanganate disappears in this reaction so it can be used as a test for alkenes.

Which of the following functional groups is formed from the condensation of carboxylic acids???

a. acid anhydride
b. acid halide
c. amide
d. ester
e. ether​

Answers

Answer:

a

Explanation:

its made up of carbon and hydrogen

Suppose you have a container filled with air at 212 oF. The volume of the container 1.00 L, the pressure of air is 1.00 atm. The molecular composition of air is 79% N2 and 21% O2 for simplification. Calculate the mass of air and moles of O2 in the container.

Answers

Answer:

[tex]m_{air}=0.947g[/tex]

[tex]n_{O_2} =0.00686molO_2[/tex]

Explanation:

Hello,

In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:

[tex]T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol[/tex]

In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:

[tex]m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g[/tex]

Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:

[tex]n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2[/tex]

Best regards.

The nutrition label on the back of a package of hotdogs (purchased within the US) indicates that one hotdog contains 100 calories. How many calories does a hotdog actually have?
A. 1,000
B. It depends on how many hotdogs you eat
C. 100
D. 10
E. 100,000

Answers

Answer:

C. 100

Explanation:

Biochemical researches and studies have found out that an average health hotdog has a calorie of between 100 and 150 which is usually dependent on the additives.

Since the nutrition label on the back of a package of hotdogs (purchased within the US) indicates that one hotdog contains 100 calories then it truly contains such amount of calories. The standard number of calories present in a hotdog is independent of the amount eaten by individuals.

What is an example of a molecular compound

Answers

Answer:

Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).

Briefly in your own words, describe the face-centered cubic unit cell in such a way that someone reading only your description, would be able to make a reasonable sketch of the structure.

Answers

Answer:

Face-centered Cubic Unit Cell (FCC) is consists of 8 atoms at the corners (one atom at each corner) and 6 atoms are present at the face-center (one atom at each face). Each eight atoms at corners contribute 1/8 and six atoms at faces contribute  1/2 in each unit cell.

What is the name of this molecule?

Answers

Answer:

[tex]\boxed{Butyne}[/tex]

Explanation:

Triple Bonds => So it is an alkyne

The suffix used will be "-yne"

4 Carbons => The prefix used will be "But-"

Combining the prefix and suffix, we get:

=> Butyne

Answer:

[tex]\boxed{\mathrm{Butyne}}[/tex]

Explanation:

Alkynes have triple bonds ≡. The molecule has one triple bond.

Suffix ⇒ yne

The molecule has 4 carbon atoms and 6 hydrogen atoms.

Prefix ⇒ But (4 carbons)

The molecule is Butyne.

[tex]\mathrm{C_4H_6}[/tex]

Write the molecular equation and net ionic equation for the reaction of hydroiodic acid and potassium hydroxide. Include phases (states). Enter the formula for water as H2O .

Answers

The molecular equation is :[tex]HI(aq)+KOH(aq) -- > KI (aq) + H_2O (l)[/tex]

The net ionic equation is: [tex]H^+(aq) + OH^-(aq) -- > H_2O(l)[/tex]

Reaction between hydroiodic acid and potassium hydroxide:

When hydroiodic acid reacts with potassium hydroxide, this will result in the formation of a salt i.e. potassium iodide, and water is obtained as a by-product.

The molecular equation can be represented as:

[tex]HI(aq)+KOH(aq) -- > KI (aq) + H_2O (l)[/tex]

The net-ionic equation can be represented as:

[tex]H^+(aq) + OH^-(aq) -- > H_2O(l)[/tex]

Find more information about molecular equation here:

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A molecular equation is a balanced chemical equation that represents a chemical reaction by showing the complete chemical formulas of all reactants and products involved. The net ionic equation for the reaction is as follows:

[tex]H+(aq) + OH-(aq) = H_2O(l)[/tex]

The molecular equation for the reaction between hydroiodic acid (HI) and potassium hydroxide (KOH) can be written as follows:

[tex]HI(aq) + KOH(aq) = KI(aq) + H_2O(l)[/tex]

In this equation, (aq) represents aqueous solutions, indicating that the substances are dissolved in water, and (l) represents a liquid phase for water.

To write the net ionic equation, we need to remove the spectator ions that do not participate in the actual chemical reaction. In this case, potassium ion (K+) and iodide ion (I-) are spectator ions, meaning they appear on both sides of the equation without undergoing any change.

Therefore, the net ionic equation for the reaction is as follows:

[tex]H+(aq) + OH-(aq) = H_2O(l)[/tex]

In the net ionic equation, H+ represents the hydrogen ion from hydroiodic acid, and OH- represents the hydroxide ion from potassium hydroxide. These ions combine to form water ([tex]H_2O[/tex]) as the only product.

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What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A? Express your answer using two significant figures.

Answers

Answer

mass of aluminum metal= 7 .0497g of Al

Explanation:

current = 21 A

time = 1 hour = 60 X 60 = 3600 s

quantity of electricity passed = current X time = 21X 3600 = 75600 C

Following the electrolysis the below reaction will occur :

Al3+ + 3e- --------> Al

therefore, 3F i.e. 3 X 96500 C = 289500 C gives 1 mole of Al

so 1 C will produce 1/289500 moles of Al

so 108000 C will produce 1/289500 X 75600 = 0.2611 moles of Al

now 1 mole of aluminium weighs = 27 g/mole

so 0.2611 moles of Al = 0.2611 X 27 = 7 .0497 g

mass of aluminum metal= 7 .0497 g of Al

The mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A is 7.05 g

We'll begin by calculating the the quantity of electricity used. This can be obtained as follow:

Current (I) = 21 A

Time(t) = 1 h = 60 × 60 = 3600 s

Quantity of electricity (Q) =?

Q = it

Q = 21 × 3600

Q = 75600 C

Finally, we shall determine the mass of the aluminum metal produced.

Al³⁺ + 3e —> Al

Recall:

1 mole of Al = 27 g

1 electron (e) = 96500 C

Thus,

3 electrons = 3 × 96500 = 289500 C

From the balanced equation above,

289500 C of electricity produced 27 g of Al.

Therefore,

75600 C of electricity will produce = (75600 × 27) / 289500 = 7.05 g of Al

Thus, the mass of the aluminum metal obtained is 7.05 g

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For the carbonate ion, CO3 2− 1- Draw the electron orbital diagram for the valence electrons of the central carbon before and after hybridization. 2- identify which carbon and oxygen electron orbitals overlap to create each single and double C-O bond in the structure

Answers

Answer:

See explanation below

Explanation:

Carbon has four electrons in its outermost shell. The CO3^2- anion is found to be in the trigonal planar geometry. For a carbon atom in the trigonal planar geometry, the carbon is sp2 hybridized. This implies that an s orbital mixes with two p orbitals to yield the hybrid orbitals in the ion.

Carbon forms three double bonds to three oxygen atoms using these hybrid sp2 orbitals. Recall that the actual bonding in each C-O linkage lies between that of a pure C-O single bond and C-O double bonds.

Note that there are two p orbitals and one s orbital participating in this hybridization hence three hybrid orbitals are expected to be formed.

Predict the product of the following Wittig reaction. Be sure your answer accounts for stereochemistry, where appropriate. If multiple stereoisomers form, be sure to draw all products using appropriate wedges and dashes.


1. PPh3
5-iodo-1-phenyl-1-pentanone →
2. n- BuLi

Answers

Answer:

Final product: cyclopent-1-en-1-ylbenzene

Explanation:

In this case, we have a Wittig reaction. The addition of [tex]PPh_3[/tex] and n-Buli will produce the "Ylide compound". First, we will have an Sn2 reaction in which the iodide is replaced by triphenylphosphine. Then the base n-Buli will remove a hydrogen atom to form a double bond (Ylide compound). Then the double bond will be delocalized to produce a carbanion. This carbanion, will attack the carbon in the carbonyl group generating a negative charge in the oxygen. Then the negative charge will attack the phosphorous atom to produce a cyclic structure. Finally, the cyclic structure is broken to produce the alkene (cyclopent-1-en-1-ylbenzene).

See figure 1

I hope it helps!

In a mixture of argon and hydrogen, occupying a volume of 1.66 L at 910.0 mmHg and 54.9oC, it is found that the total mass of the sample is 1.13 g. What is the partial pressure of argon

Answers

Answer:

Partial pressure (Ar) = 316.1mmHg

Explanation:

In the mixture of Ar and H₂ you can find the total moles of both gases using general gas law and with the mass of the sample and molar weight of each gas find the mole fraction of Argon and thus, its partial pressure.

Moles of gases:

PV = nRT

P = 910.0mmHg ₓ (1atm / 760mmHg) = 1.1974atm

V = 1.66L

n = Moles gases

R = 0.082atmL/molK

T = 54.9°C + 273.15K = 328.05K

PV = nRT

1.1974atm*1.66L = n*0.082atmL/molK*328.05K

0.0739 moles = total moles of the sample

Knowing H₂ = 2.016g/mol and Ar = 39.948g/mol you can write:

1.13g = 2.016X + 39.948Y (1)

Where X = moles of hydrogen and Y = moles of Argon.

Also we can write:

0.0739moles = X + Y (2)

Total moles of the sample are moles of hydrogen + moles Argon

Replacing 2 in 1:

1.13g = 2.016(0.0739-Y) + 39.948Y

1.13 = 0.1564 - 2.016Y + 39.948Y

0.9736 = 37.932Y

0.02567 = Y = moles of Argon

As total moles are 0.0739moles, mole fraction of Ar in the sample are:

XAr = 0.02567mol / 0.0739mol

X Ar = 0.347

Last, partial pressure of Ar = X Ar * total pressure.

Partial pressure (Ar) = 0.347*910.0mmHg

Partial pressure (Ar) = 316.1mmHg

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