Unknown A melts at 113- 114oC. Known compounds 3-Nitroaniline and 4-Nitrophenol both melt at 112-114 oC. If A is mixed with 3-Nitroaniline and the melting point becomes broad and depressed, what must A be __________A) 3-Nitroaniline B) 4-Nitrophenol C) Both

Answers

Answer 1

Answer:

Answer:

C) Both

Explanation:

Whenever we mix any pure form of a compound with some other form of a compound which is not in the other standard pure state, this results in the melting point of mixture to get dispersed and it becomes broad form.

Thus, when a known compound of 3-Nitroaniline mixes with both 3-Nitroaniline and 4-Nitrophenol, the melting point of the compound becomes depressed and board.

Thus the correct option is (C).

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Compound X has the formula C8H14.X reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form a mixture of cis- and trans-1,2-dimethylcyclohexane. Treatment of X with ozone follwed by zinc in aqueous acid gives a ketone plus formaldehyde (CH2=O). What is the structure of X?

Answers

Answer:

Compound X has the formula C8H14.

X reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form a mixture of cis- and trans-1,2-dimethyl cyclohexane. Treatment of X with ozone followed by zinc in aqueous acid gives a ketone plus formaldehyde (CH2=O). What is the structure of X?

Explanation:

The degree of unsaturation in the given molecule C8H14 is:

DU=(Cn+1)-Hn/2-Xn/2+Nn/2

where,

Cn=number of carbon atoms

Hn=number of hydrogen atoms

Xn=number of halogen atoms

Nn=number of nitrogen atoms

C8H14:

DU=(8+1)-14/2

=>DU=9-7 =2

Hence, the given molecule will have either two double bonds or one double bond and one ring or two rings.

X reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form a mixture of cis- and trans-1,2-dimethylcyclohexane.

This indicates that the molecule X has one double bond and one ring that is cyclohexane ring.

Treatment of X with ozone follwed by zinc in aqueous acid gives a ketone plus formaldehyde (CH2=O).

So, the molecule has a ring and double bond CH2.

Based on the given data the structure of compound X is shown below:

The reaction sequence is shown below:

Final answer:

Compound X, with the formula C8H14, reacts with hydrogen to form 1,2-dimethylcyclohexane. When treated with ozone and zinc, it yields a ketone and formaldehyde. Therefore, the structure of X is 1,2-dimethylcyclohexane.

Explanation:

X is a compound with the formula C8H14. The reaction of X with hydrogen in the presence of a palladium catalyst produces a mixture of cis- and trans-1,2-dimethylcyclohexane. When X is treated with ozone followed by zinc in aqueous acid, it forms a ketone and formaldehyde (CH2=O).

This information tells us that compound X is a cycloalkane with two methyl groups. Since it reacts with hydrogen to form 1,2-dimethylcyclohexane, we know that X must have a cyclohexane ring with two methyl groups at the 1 and 2 positions. The cis and trans isomers of 1,2-dimethylcyclohexane have different arrangements of the methyl groups relative to each other.

Thus, the structure of compound X is 1,2-dimethylcyclohexane.

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Scaled Synthesis of Alum. Show your calculations for:a.the experimental scaling factor giving rise to a 15.0 g theoretical yield;b.the corrected volumes of KOH and H2SO4; andc.the theoretical yield of alum based on the actual amount of Al used.Make sure you carefully show each step for these calculations.

Answers

Answer:

Explanation:

You don't give your experimental data, so I shall assume:

Mass of Al = 1.07 g

20 mL of 3 mol·L⁻¹ KOH

20 mL of 9 mol·L⁻¹ H₂SO₄

The overall equation for the reaction is

Mᵣ: 26.98 474.39

2Al + 2KOH +4H₂SO₄ + 22H₂O ⟶ 2K[Al(SO₄)₂]·12H₂O + 3H₂

m/g: 1.07

(c) Theoretical yield of alum

(i) Moles of Al

$\text{Moles of Al} = \text{1.07 g Al} * \frac{\text{1 mol Al}}{\text{26.98 g Al}} = \text{0.039 66 mol Al}$

(ii) Moles of alum

$\text{Moles of alum} = \text{0.039 66 mol Al} * \frac{\text{2 mol alum }}{\text{2 mol Al}} = \text{0.039 66 mol alum \n}$

(iii) Theoretical yield of alum

$\text{Mass of alum} = \text{0.039 66 mol alum} * \frac{\text{474.39 g alum}}{\text{1 mol alum}} = \textbf{18.8 g alum}$

(a) Scaling factor for 15.0 g alum

You want a theoretical yield of 15.0 g, so you must scale down the reaction.

$\text{Scale factor} = (15.0)/(18.8) = \mathbf{0.798}$

(b) Corrected volumes of NaOH and H₂SO₄

V = 0.798 × 20 mL = 16 mL

A solution containing 292 g of Mg(NO3)2 per liter has a density of 1.108 g/mL. The molality of the solution is:A) 2.00 m

B) 1.77 m

C) 6.39 m

D) 2.41 m

E) none of these

Answers

Answer: D) 2.41 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molality=(n)/(W_s)$

where,

n = moles of solute

$W_s$ = weight of solvent in kg

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}=(292g)/(148g/mol)=1.97moles$

volume of solution = 1L = 1000 ml (1L=1000ml)

Mass of solution= ${\text {Density of solution}}* {\text {Volume of solution}}=1.108g/ml* 1000ml=1108g$

mass of solute = 292 g

mass of solvent = mass of solution - mass of solute = (1108- 292) g = 816g = 0.816 kg

Now put all the given values in the formula of molality, we get

$Molality=(1.97moles)/(0.816kg)=2.41mole/kg$

Therefore, the molality of solution will be 2.41 mole/kg

Final answer:

In this problem, we calculate molality by using the given mass of the solute, the mass of the solvent, and the molar mass of the solute. After performing the necessary calculations, we find that the molality is 2.41 m.

Explanation:

The subject of this student's question is molality, which is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. To find the molality (m), we need to know the mass of the solute and the mass of the solvent in the solution.

Given, that the solution contains 292g of Mg(NO3)2 per liter (which is the mass of the solute). The density of the solution is 1.108g/mL. We know that 1L = 1000mL, so the mass of the solution is density x volume = 1.108g/mL x 1000mL = 1108g.

We need to find the mass of the solvent (water). The mass of the solution is the mass of the solute + the mass of the solvent. So, the mass of the solvent is 1108g(mass of the solution) - 292g(mass of solute) = 816g or 0.816gkg.

The molar mass of Mg(NO3)2 is 148.31452 g/mol. So, the number of moles of Mg(NO3)2 in the solution is moles = mass / molar mass = 292g / 148.31452 g/mol = 1.97 moles.

Now we can calculate molality (m) = moles of solute/mass of solvent in kg = 1.97 moles / 0.816 kg = 2.41 m. Therefore, the answer is D) 2.41 m.

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Which of the following statement best defines matter?

Answers

show us the statements

When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. If 100 g of nitrogen react with 6 g of hydrogen, how much ammonia will be produced? 106 g 34 g 128 g 40 g

Answers

Answer:

34 g

Explanation:

Let's consider the following balanced equation.

N₂ + 3 H₂ → 2 NH₃

The theoretical mass ratio of N₂ to H₂ is 28g N₂ : 6g H₂ = 4.6g N₂ : 1g H₂.

The experimental mass ratio of N₂ to H₂ is 100g N₂ : 6g H₂ = 16.6g N₂ : 1g H₂.

As we can see, hydrogen is the limiting reactant.

According to the task, we 6 g of H₂ react completely, 34 g of ammonia are produced.

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 500. mL flask with 3.3 atm of sulfur dioxide gas and 0.79 atm of oxygen gas at 31.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.47 atm Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.

Answers

Answer:

0.051

Explanation:

Let's consider the following reaction.

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

We can compute the pressures using an ICE chart.

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I 3.3 0.79 0

C -2x -x +2x

E 3.3-2x 0.79-x 2x

The partial pressure of sulfur trioxide gas is 0.47 atm. Then,

2x = 0.47

x = 0.24

The pressures at equilibrium are:

pSO₂ = 3.3-2x = 3.3-2(0.24) = 2.82 atm

pO₂ = 0.79-x = 0.79-0.24 = 0.55 atm

pSO₃ = 0.47 atm

The pressure equilibrium constant (Kp) is:

Kp = pSO₃² / pSO₂² × pO₂

Kp = 0.47² / 2.82² × 0.55

Kp = 0.051

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