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3 years ago

The most stable conformation of 1,2-dibromoethane is:

Chemistry

1 answer:

zhenek [66]3 years ago

4 0

I have attached an image that shows the most stable conformation of 1,2-dibromoethane. The top image shows the line drawing of the structure and the bottom image shows the Newman project of the structure. The Newman project shows up the perspective of looking down the C1-C2 bond of the compound. From this perspective, we can see the relationship between the two bromo-substituents.

Since the bromo-substituents are the largest groups substituted on each carbon atom, these groups will want to be as far away from each other in space as possible. The result is this anti- relationship shown, where the two bromo atoms are as far away from each other as possible within the molecule to reduce the amount of steric interactions. Therefore, the anti-conformation is the most stable conformation of 1,2-dibromoethane.

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A chemist fills a reaction vessel with mercurous chloride solid, mercury (I) aqueous solution, and chloride aqueous solution at

makvit [3.9K]

Answer:

ΔG° = -533.64 kJ

Explanation:

Let's consider the following reaction.

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)

ΔG° = -533.64 kJ

3 0

3 years ago

What is a nonmetal in the same group as Pb

kupik [55]

Answer:

Carbon

Explanation:

The non metal in the same group as Pb is carbon or C .

I hope this helps you.

6 0

3 years ago

Solution of known concentration is called

Vesna [10]

Answer:

standard solution

Explanation:

4 0

2 years ago

How much heat is required to heat 1.6g of ice from -16c to steam at 112c?

oksian1 [2.3K]

Answer:

Total heat ≅ 49.07 kJ

Explanation:

Given that:

mass = 1.6 g = 0.016 kg

Initial temperature = - 16 ° C

final temperature = 112° C

specific heat for ice = 2.06 kJ/kgC

specific heat of water = 4.186 kJ/kgC

heat fusion of ice = 334 kJ/kg

specific heat for steam = 2.1 kJ/kgK

heat of vaporization of water = 2256 kJ/kg

To heat ice from -16 ° C to 0 ° C

Q₁ = 2.06 kJ/kgC × 0.016 kg × 16 ° C

Q₁ = 0.52736 kJ

To melt Ice at 0° C

Q₂= 334 kJ/kg × 0.016 kg = 5.344 kJ

To heat water from 0° C to 100° C

Q₃ = 4.186 kJ/kgC × 0.016 kg × 100° C

Q₃ = 6.6976 kJ

To vaporize water to steam at 100° C

Q₄ = 2256 kJ/kg × 0.016 kg = 36.096 kJ

To heat steam from 100C to 112° C

Q₅ = 2.1 kJ/kgC × 0.016 kg × 12 C

Q₅ = 0.4032 kJ

Total heat = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Total heat = (0.52736 + 5.344 + 6.6976 + 36.096 + 0.4032) kJ

Total heat = 49.06816 kJ

Total heat ≅ 49.07 kJ

6 0

3 years ago

The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio

iren2701 [21]

Answer : The correct option is, (B) $CO_2$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of unknown gas = $11.9\text{ mL }min^{-1}$

$R_2$ = rate of effusion of oxygen gas = $14.0\text{ mL }min^{-1}$

$M_1$ = molar mass of unknown gas = ?

$M_2$ = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}$

$M_1=44.2g/mole$

The unknown gas could be carbon dioxide $(CO_2)$ that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide $(CO_2)$

5 0

3 years ago