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Sergio [31]
3 years ago
7

Compound X has a molar mass of 416.48 g mol

Chemistry
1 answer:
bazaltina [42]3 years ago
4 0

Answer:

P₂Cl₁₀

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound X = 416.48 g/mol

Percentage of phosphorus (P) = 14.87%

Percentage of Chlorine (Cl) = 85.13%

Molecular formula of X =?

Next, we shall determine the empirical formula of compound X. This can be obtained as follow:

P = 14.87%

Cl = 85.13%

Divide by their molar mass

P = 14.87 / 31 = 0.480

Cl = 85.13 / 35.5 = 2.398

Divide by the smallest

P = 0.480 / 0.480 = 1

Cl = 2.398 / 0.480 = 5

Empirical formula of compound X is PCl₅

Finally, we shall determine the molecular formula of compound X. This can be obtained as follow:

Molar mass of compound X = 416.48 g/mol

Empirical formula = PCl₅

Molecular formula =?

Molecular formula= [Empirical formula]ₙ

[PCl₅]ₙ = 416.48

[31 + (35.5 × 5)]ₙ = 416.48

[31 + 177.5]n = 416.48

208.5n = 416.48

Divide both side by 208.5

n = 416.48 / 208.5

n = 2

Molecular formula = [PCl₅]ₙ

Molecular formula = [PCl₅]₂

Molecular formula = P₂Cl₁₀

Therefore, the molecular formula of compound X is P₂Cl₁₀

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Answer:

c. Many of their bonds are C-C and C-H

Explanation:

The majority of bonds in  carbohydrates and lipids( being an organic compound) are C-C and C-H. Like glucose, fructose or galactose ,etc.

These bonds are strong and do require a lot of energy to break. Thus, a lot of energy are required to break carbs and lipids into simpler compounds.Therefore, carbohydrates and lipids have high potential energy.

The correct answer is c.

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Which combination best represent the mass and charge of a neutron?
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The correct answer is 1 amu; neutral
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Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv
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The characteristics of the α and β particles allow to find  the design of an experiment to measure the ²³⁴Th particles is:

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  • The neutrons cannot be detected in this experiment because they have no electrical charge.

In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.

The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.

The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.

Thorium has several isotopes, with different rates and types of emission:

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In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.

Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.

In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:

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Learn more about radioactive emission here: brainly.com/question/15176980

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2 years ago
When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The struc
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Answer:

The structure is shown below.

Explanation:

The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:

FC = V - (L + S/2)

Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.

The possible formulas, from the empiric one, are:

SCl, S₂Cl₂, and S₃Cl₃.

To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:

FC = 7 - (6 + 2/2) = 0

So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)

FC = 6 - (4+ 2/2) = +1

For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so

FC = 6 - (4 + 4/2) = 0

So, it is the correct structure. The lewis structure represents the bonds by lines and the lone pairs of electrons by dots, and it is shown below.

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