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3 years ago

Rank the following in terms of decreasing miscibility in C₈H₁₈ (octane), a major component of gasoline:

Chemistry

1 answer:

Aleksandr-060686 [28]3 years ago

6 0

Answer:

In order of decreasing miscibility

C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)

Explanation:

The solubility of a solid is a measure of its ability to dissolve in a liquid while for liquids, the miscibility is a measure of thhe liquid to mix with anoyjer liquid resulting in a soltion which can hold any amount of either liquids. Immiscible liquids are those that are not soluble or have very limited solibility with each other.

C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)

In the order of decreasing miscibility as like dissolve like, ability to dissociate and polar and organic characteristics are considered

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2 years ago

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3 years ago

HELP!!!!!

faltersainse [42]

Answer:

5.17.

Explanation:

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

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2 years ago

Read 2 more answers

A gamma ray photon has an energy of 4.75 x 10-14 joules. What is the frequency of this radiation?

Rama09 [41]

The frequency of the radiation is equal to $7.17 \times 10^{19}$ Hertz.

Given the following data:

Photon energy = $4.75 \times 10^{-14}$ Joules

To find the frequency of this radiation, we would use the Planck-Einstein equation.

Mathematically, the Planck-Einstein relation is given by the formula:

$E = hf$

Where:

h is Planck constant.

f is photon frequency.

Substituting the given parameters into the formula, we have;

$4.75 \times 10^{-14} = 6.626 \times 10^{-34} \times F\\\\F = \frac{4.75 \times 10^{-14}}{6.626 \times 10^{-34}}$

Frequency, F = $7.17 \times 10^{19}$ Hertz

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2 years ago

How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

Marat540 [252]

Answer:

$\boxed{\text{2.00 mol}}$

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r: 58.32

Mg(OH)₂ + … ⟶ … + 2HOH

m/g: 58.3

(a) Moles of Mg(OH)₂

$\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}$

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

$\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}$

The reaction will form $\boxed{\textbf{2.00 mol}}$ of water.

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3 years ago