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4 years ago

4) What is the IUPAC name of this compound?

Chemistry

1 answer:

zheka24 [161]4 years ago

8 0

Answer:

1,3- dichlorobutane

Explanation:

The IUPAC name of given compound is 1,3- dichlorobutane.

First of all we will select the longest chain.

The given compound is straight chain. So the compound have four carbon atoms means parent name will butane.

Then we will see the position of functional group. Gives the lowest possible number to the functional group attached.

In given compound one chlorine atom is attached on left side and other is on right side. we will start the numbering from left because in this way chlorine will get lowest possible number which 1 and 3.

As we start the numbering from left the one chlorine atom get number one and other is attached on carbon 3 will have number 3.

There are two chlorine atoms so we will give name "dichloro"

The name will be,

⁴CH₃-³CHCl-²CH₂-¹CH₂Cl

1,3- dichlorobutane.

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What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?

Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

$\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})$

Hence, the probability of finding the particle in the one-dimensional box is:

$P = \int_{x_{1}}^{x_{2}} \psi^{2} dx$

$P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx$

$P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx$

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

$P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]$

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})$

Solving for n=4:

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})$

$P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})$

$P = \frac{1}{8}$

I hope it helps you!

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3 years ago

Lewis Dot diagram for NaCi*

Georgia [21]

Answer:

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Explanation:

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3 years ago

What are some of the techniques by which mixtures can be resolved into their components?

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4 years ago

Please help!! A compound with a molar mass of 544.0 g/mol is made up of 26.5 grams Carbon, 2.94 grams

lukranit [14]

Answer:

1. Empirical formula = C₃H₄O₆

2. Molecular formula = C₁₂H₁₆O₂₄

Explanation:

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

Molar mass of compound = 544 g/mol

Mass of Carbon (C) = 26.5 g

Mass of Hydrogen (H) = 2.94 g

Mass of oxygen (O) = 70.6 g

Empirical formula =?

Molecular formula =?

1. Determination of the empirical formula of the compound.

C = 26.5 g

H = 2.94 g

O = 70.6 g

Divide by their molar mass

C = 26.5 / 12 = 2.208

H = 2.94 / 1 = 2.94

O = 70.6 / 16 = 4.4125

Divide by the smallest

C = 2.208 / 2.208 = 1

H = 2.94 / 2.208 = 1.33

O = 4.4125 / 2.208 = 2

Muitiply through by 3 to express in whole number.

C = 1 × 3 = 3

H = 1.33 × 3 = 4

O = 2 × 3 = 6

Empirical formula = C₃H₄O₆

2. Determination of the molecular formula of the compound.

Molar mass of compound = 544 g/mol

Empirical formula = C₃H₄O₆

Molecular formula = [C₃H₄O₆]ₙ

[C₃H₄O₆]ₙ = 544

[(12×3) + (4×1) + (16×6)]n = 544

[36 + 4 + 96]n = 544

136n = 544

Divide both side by 136

n = 544 / 136

n = 4

Molecular formula = [C₃H₄O₆]ₙ

Molecular formula = [C₃H₄O₆]₄

Molecular formula = C₁₂H₁₆O₂₄

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