1-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield.

Answer: Br-CH₂-CH₂-C≡N (3-bromopropanenitrile)

Explanation:

The question tells us to " Deduce the identity of the following compound from the spectral data given. C3H4BrN: H NMR, δ 2.98 (2H, triplet), 3.53 (2H, triplet) 13C NMR, δ 21.05 (triplet), 23.87 (triplet), 118.08 (singlet) (ppm):JR, 2963, 2254 cm" Express your answer as a chemical formula".

Answer: Another Alternative to this is seen below

The Double Bond Equivalent of C₃H₄BrN = 3+1-4/2-1/2+1/2 = 2

2254 cm-1 Vc ≡N [IR analysis]

2963.4 = sp³Vc-H streek [IR analysis]

where the first CH₂ = b = 2.98 (2H,triplet) and

the second CH₂ = a = 3.53 (2H,triplet)

Note:  where both are 1HNMR.

where C = c = 118.08 (singlet)

CH₂ = b = 21.05 (triplet)

CH₂ = a = 23.87 ( triplet)

Note: where all are 13CNMR

Important:

I have attached a copy of the solution to enhance better understanding in case the typed solution isn't clear enough.

Answer:

Explanation:

We are to write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag.

And to determine which one of the two metals, (copper and silver) is the anode and which is the cathode?

The chemical equation for the reaction is as follows:

[tex]\mathbf{Cu + 2Ag^+ \to Cu^{2+}+2Ag}[/tex]

The atoms of the Copper electrodes undergo oxidation and loses two electrons each to form copper II ions [tex](Cu^{2+})[/tex] which go into the solution. The copper then becomes negatively charged and functions as the negative electrode i.e the anode

The oxidation half reaction is:

[tex]\mathsf{Cu \to Cu^{2+} + 2e^-}[/tex]

The silver ions [tex]Ag^+[/tex] becomes reduced by gaining  two electrons each from the metallic copper which was deposited into the silver electrode. The  silver electrode thus becomes positively charged and functions as the positive electrode. i.e the cathode.

The reduction half reaction is:

[tex]\mathtt{Ag^+ + 1 e^- \to Ag}[/tex]

Answer:

NHCOR

Explanation:

In electrophilic aromatic substitution, certain substituents on the aromatic ring such as

-NHCOR are known to enhance ortho-para substitution. These ortho-para activating groups stabilizes the intermediate cation.

These ortho-para activating groups usually consists of species that donate electron density towards the reaction centre thereby stabilizing the carbocation and making the reaction faster.

The question is incomplete. Here is the complete question.

Acetic acid and water react to form hydronium cation and acetate anion, like this:

[tex]HCH_{3}CO_{2}_{(aq)}+H_{2}O_{(l)} -> H_{3}O^{+}_{(aq)}+CH_{3}CO_{2}^{-}_{(aq)}[/tex]

Imagine 226 mmol of [tex]CH_{3}CO_{2}^{-}[/tex] are added to a flask containing a mixture of [tex]HCH_{3}CO_{2}[/tex], [tex]H_{2}O[/tex],[tex]H_{3}O^{+}[/tex] and [tex]CH_{3}CO_{2}^{-}[/tex] at equilibrium and then answer the following questions:

1) What's the rate of the reverse reaction before any [tex]CH_{3}CO_{2}^{-}[/tex] has been added to the flask?

a) 0

b) Greater than 0, but less than the rate of the forward reaction

c) Greater than 0, but equal to the rate of the forward reaction

d) Greater than 0, but  greater than the rate of the forward reaction

2) What is the rate of the reverse reaction just after the [tex]CH_{3}CO_{2}^{-}[/tex] has been added to the flask?

a) 0

b) Greater than 0, but less than the rate of the forward reaction

c) Greater than 0, but equal to the rate of the forward reaction

d) Greater than 0, but  greater than the rate of the forward reaction

3) What is the rate of the reverse reaction when the system has again reached equilibrium?

a) 0

b) Greater than 0, but less than the rate of the forward reaction

c) Greater than 0, but equal to the rate of the forward reaction

d) Greater than 0, but  greater than the rate of the forward reaction

4) How much more [tex]CH_{3}CO_{2}^{-}[/tex]  is in the flask when the system has again reached equilibrium?

a) None

b) Some, but less than 226 mmol

c) 226 mmol

d) More than 226 mmol.

Answer: 1) c) Greater than 0, but equal to the rate of the forward reaction

2) d) Greater than 0, but  greater than the rate of the forward reaction

3) c) Greater than 0, but equal to the rate of the forward reaction

4) b) Some, but less than 226 mmol

Explanation: A reversible chemical reaction reaches its equilibrium when forward and reverse reaction are at the same rate. At that point, equilibrium has a constant called K.

Equilibrium constant depends on the concentration of its products and reagents.

For the reaction [tex]HCH_{3}CO_{2}_{(aq)}+H_{2}O_{(l)} -> H_{3}O^{+}_{(aq)}+CH_{3}CO_{2}^{-}_{(aq)}[/tex],

the forward is towards production of [tex]CH_{3}CO_{2}^{-}[/tex] and reverse is towards the production of [tex]HCH_{3}CO_{2}[/tex].

1) When equilibrium is reached, forward and reverse are at the same rate and are different from zero.

2) When adding a compound, the equilibrium is broken. So, to go back to the equilibrium, reaction tend to counteract the change. In the case of acetic acid and water above, when adding acetate anion, the reverse reaction will produce more acetic acid to restore equilibrium, so reverse reaction will be at a greater rate and different from 0.

3) After a while when the system is back to the equilibrium, the rate will be equal again.

4) After second equilibrium, acetate anion will have less mmol than when this new equilibrium state started.

Answer:

D

Explanation:

In this time there are made a lot of nuclear bombs so itatters on how what how much damage it makes according to its capacity in my opinion.

Nuclear reaction of which nuclear fusion is a type  which depends on the nucleons that fuse as the total mass of single nucleus is less than that of two original nuclei.

There are two types of nuclear reactions which are nuclear fusion and nuclear fission .They involve the combination and disintegration of the element's nucleus respectively.

In nuclear fission, the nucleus of the atom is bombarded with electrons of low energy which splits the nucleus in to two parts .Large amount of energy is released in the process.It is used in nuclear power reactors as it produces large amount of energy.

In nuclear fusion,on the other hand, is a reaction which occurs when two or more atoms combine to form a heavy nucleus.Large amount of energy is released in the process which is greater than that of the energy which is released in nuclear fission process.

Learn more about nuclear reactions,here:

https://brainly.com/question/12786977

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Answer:

0 kJ/mol.

Explanation:

Hello,

In this case, since the chemical potential can be represented in terms of the Gibbs free energy of formation:

[tex]\mu = \frac{\Delta _fG}{n}[/tex]

Thus, since the Gibbs free energy of formation of an element is zero, the chemical potential is also zero, or just 0 kJ/mol.

Best regards.

Answer:

The value is  [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]

Explanation:

From the question we are told that

  The  no of moles of the sample is  n = 0.0447 mole

  The formula weight is  [tex]M = 114 \ g/mol[/tex]

   The  mass of water is  [tex]m = 6.19 *10^{2}\ g[/tex]

   The amount of the fuel is  [tex]f= 6.13*10^{-1} \ nutritional \ Cal[/tex]

   The temperature rise is  [tex]\Delta T = 5.05^o[/tex]

Generally

      [tex]1 \ nutritional \ Cal => 4.184*10^{3} \ kJ/kg[/tex]

=>  [tex]f= 6.13*10^{-1} \ nutritional \ Cal \to x[/tex]

=>     [tex]x = \frac{6.13 *10^{-1} * 4.184 *10^{3}}{1}[/tex]

=>     [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]

The correct option is A.

Ordinary sea salt is 97 percent sodium chloride whereas Dead Sea salt is a mixture of lots of different chloride and bromide salts.

Dead Sea Salt differs greatly from other sea salts in mineral content, is made up of sodium chloride with a high percentage of magnesium, sulfates, and potassium. You can bring that ancient, therapeutic experience home with our guaranteed authentic Dead Sea bath salts.

Learn more about salt here https://brainly.com/question/2626194

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Answer:

E. In the external circuit, elections flow from the other compartment to the Cu^2+|Cu^+ compartment

Explanation:

In a galvanic cell, electrons flow from anode to cathode. We must remember that oxidation occurs at the anode and reduction occurs at the cathode. Hence the process; M(s) -------> M^+(aq) + e occurs at the anode.

The electrons lost at the anode are conveyed to the cathode where they are accepted by other chemical species and are reduced to the corresponding reduced species according to the reaction; M^+(aq) + e -----> M(s)

Hence the Cu^2+ | Cu^+ half cell, being the cathode accepts electrons from the other half cell for this reduction reaction to take place, hence the answer

Answer:

I don't know the second one but one of them is Aristotle

Answer:

Aristrotle and John Dalton

Explanation:

Answer:

You would use an electrolytic cell.

Answer:

The sun warms up parts of the oceans. Warm waters rise just like warm air rises. So, as the warmer ocean waters begin to rise in a particular area, the cooler ocean waters from a different area will move in to replace the warmer ocean waters, and this creates our ocean currents.

Hope it helps!

Answer:

D) Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.

Explanation:

Step 1: Given data

Molarity (M): 2.00 M

Volume (V): 500 mL = 0.500 L

Molar mass of NaCl: 58.44 g/mol

Step 2: Calculate the required moles of NaCl

We will use the following expression.

n = M × V

n = 2.00 mol/L × 0.500 L

n = 1.00 mol

Step 3: Calculate the mass corresponding to 1.00 moles of NaCl

1.00 mol × 58.44 g/mol = 58.44 g

Step 4: Describe the procedure to prepare the solution

Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.

Answer:

D) Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.

Explanation:

I got it right in class!

Hope  this helps!! :))

Answer:

The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.

Explanation:

The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:  

H2O (s) ⇒ H2O (l)

The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:  

Entropy change (ΔSsys) = ΔSproduct - ΔSreactant

= (69.9 - 47.89) J mol/K

= 22.0 J mol/K

Therefore, the value of entropy change is positive.  

Now the value of entropy for surrounding ΔSsurr will be,  

ΔSsurr = -ΔHfusion/T  

= -6012 j/mol/273

= -22.0 J/molK

Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.  

Answer:

One oxygen atom is tetrahedral while the other is trigonal planar

Explanation:

Acetic acid contains two oxygen atoms. The two oxygen atoms are different and have different geometries.

The first oxygen atom which is part of the carbonyl moiety in the molecule is in a trigonal planar geometry since it is sp2 hybridized.

The second oxygen atom is in sp3 hybridized state hence it is tetrahedral.

Thus one oxygen atom is tetrahedral and the other is trigonal planar in acetic acid.

Answer:

A. POBr will be consumed in order to establish a new equilibrium.

Explanation:

POBr3(g) =POBr(g) + Brz(g)

The question is based on Le Chatelier's principle. This principle states that whenever a system at equilibrium is stressed, the system would undergo changes to annul that stress.

If more POBr  is added to the system, it means that the concentration of the products have been increased and as a result the equilibrium has been disturbed. To restore equilibrium, the system would have to shift in a direction that reduces the product concentration.

This direction is the backward direction and in doing so, POBr will be consumed.

The correct option is option A.

The statement that describes the behavior of the system is:

A. POBr will be consumed in order to establish a new equilibrium.

POBr₃(g)  ⇄ POBr(g) + Br₂(g)

According to Le Ch-atelier principle,  if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to re-establish an equilibrium.

If more POBr is added to the system, it means that the concentration of the products have been increased and as a result the equilibrium has been disturbed. To restore equilibrium, the system would have to shift in a direction that reduces the product concentration.

This direction is the backward direction and in doing so, POBr will be consumed.

Thus, the correct option is option A.

Find more information about Equilibrium here:

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Answer:

In solving this question it is important to note that entropy is referred to as the degree of randomness in a compound or system.

Entropy is highest in gas and lowest in solids

The order is given as:

Gas > Liquid > Solid

Another criteria is the more the elements in a compound, the more the entropy.

For example: KO3 has more entropy than KO2 due to the presence of more elements.

The last criteria is calculating their masses , the one with a greater mass has the higher entropy.

The option with the least entropy will be a solid. Al(s) and NaF(s) are both solids. NaF(s) has more elements in a compound which makes it have a higher entropy than Al(s). This means NaF(s) > Al(s).

The next to consider are the gaseous compounds which are HF(g) , SiF4(g) and SiH4(g). The least entropy will be HF(g) . This is because it has the least number of elements of the 3 compounds.

SiF4(g) and SiH4(g) have equal number of elements in a compound but the one with the highest mass is SiF4(g) because fluorine has a higher mass when compared to hydrogen. This means the compound with the highest entropy is SiF4(g).

The order of increasing entropy is given below:

Al(s) < NaF(s) < HF(g) < SiH4(g) < SiF4(g)

Answer:

See figure 1

Explanation:

For this question, we can analyze each molecule:

-) (E)-5-Methyl-2-hexene

In this case, we have a carbon chain of 6 carbons. We have a double bond between carbons 3 and 2. Also, we have a methyl group is placed on carbon 5. Finally, the "E" configuration makes reference to the double bond,  the groups on each side of the double bond must have an opposite configuration.

-) 2,5-Diethyl-1-methylcyclohexanol

In this molecule, we have a carbon chain of 2 carbons. In carbon 2 we have two chlorine atoms. In carbon 1 we have a fluorine atom and a chlorine atom. The "S" configuration makes reference to carbon 1 (the chiral carbon), so we have to assign the priorities for each group bonded to carbon 1:

1 = Chlorine

2 = Fluorine

3 = Carbon 2

4 = Hydrogen

If we need an "S" configuration the priorities must be placed in such a way that we have a counterclockwise orientation.

-) 2,5-Diethyl-1-methylcyclohexanol

In this molecule, we have a cyclic carbon chain of 6 carbons. In carbon 1 we have an "OH" group and a methyl group. In carbon s 2 and 5, we have ethyl groups.

See figure 1

I hope it helps!

Answer:

1. HEAT OF SOLUTION IS 1149.47 J

2. HEAT OF SOLUTION PER MOLE OF NaOH IS 41989.77 J/MOL OR 41.989 KJ/MOL

3. THE REACTION IS EXOTHERMIC

Explanation:

Mass of water = 150 g

Mass of NaOH = 1.095 g

Change in temperature = 25.32 - 23.50 = 1.82 C

Total mass of the mixture = 150 + 1.095 = 151.095 g

Specific heat capacity of water = 4.18 J/ g C

So to calculate the heat of solution, we have:

1. Heat of solution = mass * specific heat * change in temperature

Heat of solution = 151.095 * 4.18 * 1.82

Heat of solution = 1149.47 J

The heat of solution is therefore 1149.47 J or 1.15 kJ

2. Heat of solution per mole

1.095 g of NaOH produces 1149.47 J of heat

1 mole of NaOH contains 40 g

In other words:

1.095 g = 1149.47 J

40 g = x J

xJ = 40 * 1149.47 / 1.095

xJ = -41989.77 J

The heat of solution per mole of NaOH is -41.989 kJ / mol

3. The reaction is exothermic that is heat is evolved from the reaction. This is indicated by the negative sign.

Answer:

7.5 g

Explanation:

There is some info missing. I think this is the original question.

Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

[tex]4.9 g \times \frac{1mol}{98.00g} = 0.050mol[/tex]

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

[tex]0.050mol \times \frac{149.09 g}{mol} = 7.5 g[/tex]

Answer:

[tex]7.5x10^{-3}M[/tex]

Explanation:

Hello,

In this case, since the dissociation barium fluoride is represented at equilibrium by:

[tex]BaF_2(s)\rightleftharpoons Ba^{2+}(aq)+2F^-(aq)[/tex]

Hence, the equilibrium expression is:

[tex]Ksp=[Ba^{2+}][F^-]^2[/tex]

Whereas the molar solubility is represented as the reaction extent [tex]x[/tex]:

[tex]Ksp=[x][2x]^2[/tex]

In such a way, we can solve for [tex]x[/tex]:

[tex]1.7x10^{-6}=4x^3\\\\x=\sqrt[3]{\frac{1.7x10^{-6}}{4} } \\\\x=7.5x10^{-3}M[/tex]

Which as said before, is the molar solubility.

Best regards.

Answer:

The answer of the scientific STATEMENT is" Appears That Ants Live In Colonies."

Answer:[tex]\rho=7.87g/mL[/tex]

Explanation:

Hello,

In this case, we are analyzing a problem in which the Archimedes principle is considered, thus, as the volume of the metal rod is determined by the difference between the mass of the water and the mass of the water and the rod:

[tex]V_{rod}=15.00mL-10.00mL=5.00mL[/tex]

In such a way, the density turns out:

[tex]\rho =\frac{m_{rod}}{V_{rod}}=\frac{39.35g}{5.00mL} \\ \\\rho=7.87g/mL[/tex]

Regards.

Answer:

33.85°C

Explanation:

From the question,

Heat lost by the hotter water = heat gained by the colder water

cm'(t₂-t₃) = cm(t₃-t₁)................. Equation 1

Where c = specific heat capacity of water, m' = mass of hot water, m = mass of cold water, t₁ = Initial temperature of cold water, t₂ = Initial temperature of hot water, t₃ =  final temperature of the mixture.

But since the density of water is constant, and mass varies directly as volume, We can replace the mass of water with the volume of water. i.e,

cv'(t₂-t₃) = cv(t₃-t₁)................. Equation 2

Where v' and v are the volume of hot water and cold water respectively

make t₃ the subject of the equation

t₃ = (v't₂+vt₁)/(v'+v)............ Equation 3

Given: v' = 5.0 L, v = 60 L, t₁ = 30°C, t₂ = 80°C

Substitute these values into equation 3

t₃ = (5×80+60×30)/(60+5)

t₃ = 2200/68

t₃ = 33.85°C

Answer:

90%

Explanation:

it will be 81/90 ×100/1

81 over 90 × 100 over 1

Answer:

[tex]\text{Rate} = k\, [\mathrm{H_2\, (g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].

[tex]\rm N_2O\, (g)[/tex] is the intermediate.

Explanation:

Balanced overall reaction: [tex]\rm 2\, H_2\, (g) + 2\, NO\, (g) \to N_2\, (g) + 2\, H_2O\, (g)[/tex].

Proposed mechanism:

Start with the slowest, rate-determining step of the proposed mechanism. Here, the rate-determining step is [tex]\rm H_2\, (g) + 2\, NO\, (g) \to N_2O\, (g) + H_2O \, (g)[/tex].

There are two species on the reactant side of this intermediate reaction: [tex]\rm H_2\, (g)[/tex]  and [tex]\rm NO_2\, (g)[/tex]. The concentration of both of them should be in the rate expression of this step.

On the other hand, the coefficient of [tex]\rm H_2\, (g)[/tex] is one while the coefficient of [tex]\rm NO_2\, (g)[/tex] is two. Therefore, in the rate expression of this step, the concentration of [tex]\rm H_2\, (g)\![/tex] should have a power of one, while the concentration of [tex]\rm NO_2\, (g)\![/tex] should have a power of two.

include the rate constant [tex]k[/tex] to obtain the rate expression of the rate-determining slow step:

[tex]\text{Rate} = k\, [\mathrm{H_2\,(g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].

Make sure that all species in this rate expression are on the reactant side of the overall balanced reaction. Otherwise, further steps would be required to obtain the rate law of the overall reaction.

Therefore, by this proposed mechanism, the rate law of the overall reaction would be [tex]\text{Rate} = k\, [\mathrm{H_2\,(g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].

In a proposed reaction mechanism, a species is an intermediate if it appeared in one of the proposed steps, but not in the balanced overall equation.

The two steps of this proposed mechanism mentioned five species:

With the exception of [tex]\rm N_2O\, (g)[/tex], all the other species appeared in the overall balanced equation. Therefore, [tex]\rm N_2O\, (g)\![/tex] is the intermediate.

Answer: [tex]T_{f}[/tex] = -29.96 °C

Explanation: A solution has a lower freezing point compared to the pure solvent, due to the solute lowering the vapor pressure of the solvent.

The "new" freezing point is calculated as:

[tex]\Delta T_{f} = K_{f}.m[/tex]

[tex]K_{f}[/tex] is the molal freezing-point depression constant

m is molality concentration, i.e., moles of solute per kilogram of solvent

Before calculating freezing point, let's find moles of ethylene glycol ([tex]C_{2}H_{6}O_{2}[/tex]) in the solution:

Molar mass [tex]C_{2}H_{6}O_{2}[/tex] = 62.08 g/mol

For 500 g:

[tex]n = \frac{500}{62.08}[/tex]

n = 8.05 moles

The molality concentration for 0.5kg of water:

m = [tex]\frac{8.05}{0.5}[/tex]

m = 16.11

The freezing point will be:

[tex]\Delta T_{f} = K_{f}.m[/tex]

[tex]\Delta T_{f} = -1.86*16.11[/tex]

[tex]\Delta T_{f} = -29.96[/tex]

Freezing point of a solution of Ethylene Glycol and Water is -29.96°C

Answer:

[tex]m=406.26g[/tex]

Explanation:

Hello,

In this case, since the density is defined as the degree of compactness of a substance and it mathematically defined as the ratio of the mass and volume:

[tex]\rho = \frac{m}{V}[/tex]

By knowing that its density is 1.11 g/mL, the mass in 366 mL is:

[tex]m=\rho V=366mL*1.11\frac{g}{mL}\\ \\m=406.26g[/tex]

Regards.