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Fed [463]
1 year ago
15

specify whether each of the items listed is hydrophilic or hydrophobic by dragging the labels into the appropriate box.

Chemistry
1 answer:
erik [133]1 year ago
4 0

Hydrophilic;

Polar compounds

water

ions

glucose

Hydrophobic;

Triglycerides

cholesterol

Lipids

Oil

Carbohydrates

<h3>What are hydrophilic substances?</h3>

The term hydrophilic substances is used to describe the substances that could dissolve in water. They are those substances that are water soluble. Somehow, the substances must be polar because in chemistry like dissolves like.

The hydrophobic substances are those substances that do not dissolve in water. They are nonpolar and interact with other nonpolar substances.

The classification of the substances took place below;

Hydrophilic;

Polar compounds

water

ions

glucose

Hydrophobic;

Triglycerides

cholesterol

Lipids

Oil

Carbohydrates

Learn more about hydrophilic substances:brainly.com/question/4692308

#SPJ1

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Answer:

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3 years ago
The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.42×10−4s−1 at a certain temperature. What is t
Keith_Richards [23]

Answer:

a) Half life of the decomposition = 4951.1 s ≈ 4950 s

b) Time it will take for the concentration of SO₂Cl₂ to decrease to 25% of its initial concentration = 9900 s

c) If the initial concentration of SO₂Cl₂ is 1.00 M, time it will take for the concentration to decrease to 0.78 M is 1775s

d) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.146 M

e) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.1398 M

Explanation:

Let C₀ represent the initial concentration of SO₂Cl₂

And C be the concentration of SO₂Cl₂ at anytime.

a) Rate of a first order reaction is represented by

dC/dt = - KC

dC/C = - kdt

Integrating the left hand side from C₀ to C₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)

In [(C₀/2)/C₀] = - k t(1/2)

In (1/2) = - k t(1/2)

- In 2 = - k t(1/2)

t₍₁,₂₎ = (In 2)/k

t₍₁,₂₎ = (In 2)/(1.4 × 10⁻⁴)

t₍₁,₂₎ = 4951.1 s

b) dC/C = - kdt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = - kt

C/C₀ = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C = 25% of C₀ = 0.25C₀

0.25C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

- kt = In 0.25

- kt = - 1.386

t = 1.386/(1.4 × 10⁻⁴) = 9900 s

c) C = C₀ e⁻ᵏᵗ

C = 0.78 M; C₀ = 1.00 M

0.78 = 1 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.78

- kt = In 0.78

- kt = - 0.2485

t = 0.2485/(1.4 × 10⁻⁴) = 1775 s

d) C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 2 × 10² s = 200 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 200) = 0.972

C = 0.15 × 0.972 = 0.146 M

e) C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 5 × 10² s = 500 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 500) = 0.9324

C = 0.15 × 0.9324 = 0.1398 M

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