Pentanone was treated with excess sodium cyanide in HCl (aq) followed by hydrogen gas has over Pd. This produced:________
A. 2-amino-1-hexanol
B. 1-amino-2-methylpentan-2-ol
C. 1-cyano-1-pentanol
D. 2-aminomethylpentan-1-ol

Answer:

a

Explanation:

answer is a on edg

Answer: C. the same amount of energy as

Explanation:

A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.

Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.

[tex]A+B\rightleftharpoons C+D[/tex]

Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.

The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.

Answer:

C.) the same amount of energy as

Explanation:

I got it correct on founders edtell

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the cyclohexanone.

See figure 1

I hope it helps!

Answer:

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

Explanation:

When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.

Answer:

[tex][H^+]=0.000123M[/tex]

[tex]pH=3.91[/tex]

Explanation:

Hello,

In this case, dissociation reaction for acetic acid is:

[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]

For which the equilibrium expression is:

[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

Which in terms of the reaction extent [tex]x[/tex] could be written as:

[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]

Thus, solving by using a solver or quadratic equation we obtain:

[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]

And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:

[tex][H^+]=0.000123M[/tex]

Now, the pH is computed as follows:

[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]

Best regards.

Answer:

461.85 degrees Celsius

Answer:

Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.

Answer: The monomers of bakelite are formaldehyde and phenol

Explanation:

Answer:

Explanation:

Answer in attached file .

Answer:

[tex]Ksp=1.2[/tex]

Explanation:

Hello,

In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):

[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]

Next, as its dissociation reaction is:

[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]

The equilibrium expression is:

[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]

And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):

[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]

Best regards.

The given question is incomplete.

The complete question is:

When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?

Answer: 4 grams of methane were needed for the reaction

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]

Given:  mass of oxygen = 16 g

Mass of carbon dioxide = 11 g

Mass of water = 9 g

Mass of products = Mass of carbon dioxide + mass of water = 11 g  +9 g = 20 g

Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g

As mass of reactants = mass of products

mass of methane + 16 g= 20 g

mass of methane  = 4 g

Thus 4 grams of methane were needed for the reaction

Answer:

804 J

Explanation:

Step 1: Given data

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.0235J·g⁻¹K⁻¹ × (1.50 × 10³g) × [15.0°C-(-7.8°C)]

Q = 804 J

Answer:

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

Explanation:

Hello,

In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).

- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).

- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).

Best regards.

Answer:

All of the above processes have a ΔS < 0.

Explanation:

ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.

The question requests us to identify the process that has a negative change of entropy.

carbon dioxide(g) → carbon dioxide(s)

There is  a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.

water freezes

There is  a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.

propanol (g, at 555 K) → propanol (g, at 400 K)

Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.

This reaction highlights a drop in temperature which means a negative change in entropy.

methyl alcohol condenses

Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.

Answer:

Lone pairs cause more repulsion than bond pairs

Explanation:

A lone pair takes up more space around the central atom than bond pairs of electrons. This is because, a lone pair is attracted to only one nucleus while bond pairs are attracted to two nuclei.

Hence the repulsion between lone pairs is far greater than the repulsion between bond pairs or repulsion between a lone pair and a bond pair. The presence of a lone pair therefore distorts a molecule away from the ideal shape predicted on the basis of the valence shell electron pair repulsion theory.

Lone pairs are found to decrease the observed bond angles in a molecule.

Answer:

Empirical formula is: C₂H₅

Explanation:

The chemical equation of burning of a compound that conatins only Carbon and Hydrogen is:

CₓHₙ + O₂ → XCO₂ + n/2H₂O

That means the moles of CO₂ produced are the moles of Carbon in the compound and moles of hydrogen are twice moles of water. Empirical formula is the simplest ratio between moles of each element in the compound. Thus, finding molse of C and moles of H we can find empirical formula:

Moles C and H:

Moles C = Moles CO₂:

7.851g CO₂ ₓ (1mol / 44g) = 0.1784 moles CO₂ = Moles C

Moles H = 2 Moles H₂O

4.018g H₂O ₓ (1mol / 18.01g) = 0.2231 * 2 = 0.4417 moles H

Ratio C:H

The ratio between moles of hydrogen and moles of Carbon are:

0.4417 moles H / 0.1784 moles C = 2.5

That means there are 2.5 moles of H per mole of Carbon. As empirical formula must be given only in whole numbers,

Density measures how tightly packed particles are.

If particles are tightly packed together, they will be more dense.

If they are loosely together, they will be less dense.

However, a common mistake is thinking that if something

is more dense it means that it's heavier.

However, that's not the case.

It has to do with how particles are packed in an object.

Answer:

[tex]V_2=1.17L[/tex]

Explanation:

Hello,

In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:

[tex]V_1C_1=V_2C_2[/tex]

Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:

[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]

Best regards.

Answer:

Below

Explanation:

Let n be the quantity of matter in the Calcium Bromide

● n = m/ M

M is the atomic weight and m is the mass

M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)

M = 40.1 + 2×79.9

● 0.422 = m/ (40.1+2×79.9)

●0.422 = m/ 199.9

● m = 0.422 × 199.9

● m = 84.35 g wich is 88.4 g approximatively

88.4 g approximatively is  the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.

Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .

To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,

Let n be the quantity of matter in the Calcium Bromide

M is the atomic weight and m is the mass

M of CaBr2 is the sum of the atomic weight of its components

Mass of  Ca = 40.1 , Mass of Br = 79.9

M = 40.1 + 2×79.9

  0.422 = m/ (40.1+2×79.9)

  0.422 = m/ 199.9

  m = 0.422 × 199.9

  m = 84.35 g which is 88.4 g approximatively .

Thus ,88.4 g approximatively is  the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .

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Answer:

  30. 5 planes are shown

  31. 1 plane

  32. CEF

  33. on line AB

  34. E or F

   35. ABCD or BCEF or CDEF or ACEF

Explanation:

30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.

__

31. 3 points define one plane only.

__

32. The only points shown on the same line segment are points E, F, and C.

__

33. If G is to be collinear with A and B, it must lie on line AB.

__

34. The only points shown that are not on plane N are points E and F. Either of those will do.

__

35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...

Plane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)

Answer:

rate = [NO]²[H₂]

Explanation:

2NO + H2 ⟶N2 + H2O2 (slow)

H2O2 + H2 ⟶2H2O (fast)

From the question, we are given two equations.

In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.

This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.

This means our rate law is;

rate = [NO]²[H₂]

Answer:

-195.8º < -191.5º < 100º

Explanation:

Water, or H20, starts boiling at 100ºC.

Nitrogen, or N2, starts boiling at -195.8ºC.

Carbon monoxide, or C0, starts boiling at -191.5ºC.

When we place these in order from decreasing boiling point:

-195.8º goes first, then -191.5º, and 100º goes last.

Answer:

therefore, N2, CO, H20

Decreasing boiling point

Explanation:

the bond existing in H2O is hydrogen bond

bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction

bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction

hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction

the stronger the bond , the higher the boiling point

therefore, N2, CO, H20

-------------------------------------->

Decreasing boiling point

Answer:

The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"

Explanation:

[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]

In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor  and each other suspend

[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]

[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]

The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.

The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.

The cell potential has been changed when there has been a change in the potential of the half cells.

The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.

With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.

Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

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Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

Answer:

(R)-but-3-en-2-ylbenzene

Explanation:

In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.

The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.

See figure 1

I hope it helps!

Answer:

Entropy change of ice changing to water at 0°C is equal to 57.1 J/K

Explanation:

When a substance undergoes a phase change, it occurs at constant temperature.

The entropy change Δs, is given by the formula below;

Δs = q/T

where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur

From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J

Δs = 15600 J / 273.15 K

Δs = 57.111 J/K

Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K

The entropy change of ice changing to water will be "57.1 J/K".

The shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.

According to the question,

Temperature, T = 0°C or,

                          = 273.15 K

Heat, q = 15.6 KJ or,

            = 15600 J

We know the formula,

Entropy change, Δs = [tex]\frac{q}{T}[/tex]

By substituting the values, we get

                                 = [tex]\frac{15600}{273.15}[/tex]

                                 = 57.11 J/K

Thus the above answer is correct.    

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Answer:

Atomic no = 12 = Mg

Explanation:

It is given that,

The atomic number of two elements that are represented by letter Q and R are 9 and 12.

We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.

For R, atomic number = 12

Its electronic configuration is : 2,8,2

It has two valance electrons in its outermost shell. The element is Magnesium (Mg).

Answer:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Explanation:

Hello,

In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Best regards.

from

1moles=iatom

Mole=mass÷avogardos

Where

Avogadro's= 6.02×10²³

So moles = 65.0÷6.02×10²³

Atoms of zinc = 391.6 ×10²³

The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].

Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.

Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately  [tex]6.022 \times10^{23}[/tex]atoms per mole.

First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.

Number of moles = Mass / Molar mass

Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol

Next, multiply the number of moles by Avogadro's number to find the number of atoms.

Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]

Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol

Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.

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