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3 years ago

Which substance is an example of a nonpolar compound?(1 point)

1 answer:

8 0

The answer is B - vegetable oil.

Explanation:

The properties of liquids depend on the attractions the molecules of the liquid have for each other and for other substances.

Liquids can dissolve certain other liquids, depending on the attractions between the molecules of both liquids.

Polar liquids, like water, dissolve other liquids which are polar or somewhat polar.

Polar liquids, like water, do not dissolve non-polar liquids like oil.

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Write a balanced equation for the reaction between aqueous strontium chloride and aqueous lithium phosphate to form solid stront

Marat540 [252]

First, let's write the chemical formula for each of the substances mentioned in the problem.

Strontium Chloride: SrCl₂
Lithium Phosphate: Li₃PO₄
Strontium Phosphate: Sr₃(PO₄)₂
Lithium Chloride: LiCl

So, the balanced chemical reaction is:

<em>3 SrCl₂ (aq) + 2 Li₃PO₄ (aq) ---> Sr₃(PO₄)₂ (s) + 6 LiCl (aq)</em>

5 0

3 years ago

Given the reactions,

enot [183]

The enthalpy change of the reaction is <u>-1347.8 kJ.</u>

<h3>What is the enthalpy change, ΔH, of the reaction?</h3>

The enthalpy change, ΔH, of the reaction is calculated from Hess's law of constant heat summation as follows:

Hess's law states that the enthalpy change of a reaction is the sum of the enthalpies of the intermediate reaction.

Given the reactions below and their enthalpy values;

1. X (s) + 12 O₂ (g)⟶ XO (s) ΔH₁ = −850.5 kJ

2. XCO₃ (s) ⟶ XO (s) + CO₂ (g) ΔH₂ = +497.3 kJ

The enthalpy change, ΔH, of the reaction whose equation is given below, will be:

X (s) + 12 O₂ (g) + CO₂ (g) ⟶ XCO₃ (s)

ΔH = ΔH₁ - ΔH₂

ΔH = − 850.5 kJ - (+497.3 kJ)

ΔH = -1347.8 kJ

Learn more about enthalpy change at: brainly.com/question/14047927

#SPJ1

7 0

2 years ago

What are the fields of science??

Alex Ar [27]

Answer:

Life science is alternatively known as biology, and physical science is subdivided into branches: physics, chemistry, astronomy and Earth science. These branches of natural science may be further divided into more specialized branches. But that is more writting to explain

Explanation:

7 0

3 years ago

A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0

N76 [4]

This question is incomplete, the complete question is;

A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0 mL volumetric flask and diluting to the calibration mark. If 10.97 mL of a 1.63 M solution of potassium hydroxide is added to the buffer, what is the final pH? The Ka for nitrous acid = 4.6 × 10⁻⁴.

Answer:

the final pH is 3.187

Explanation:

Given the data in the question;

Initial moles of HNO2 = 32.6/1000 × 4.90 = 0.15974 mol

Initial moles of NO2- = mass/molar mass = 5.86/68.995 = 0.0849336 mol

Moles of KOH added = 10.97/1000 × 1.63 = 0.0178811 mol

HN02 + KOH → NO2- + H2O

moles of HNO2 = 0.15974 - 0.0178811 = 0.1418589 mol

Moles of NO2- = 0.0849336 + 0.0178811 = 0.1028147 mol

Now,

pH = pka + log( [NO2-]/[HNO2])

pH = -log ka + log( moles of NO2- / moles of HNO2 )

we substitute

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.1028147 / 0.1418589 )

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.724767 )

pH = 3.337242 + (-0.1398 )

pH = 3.187

Therefore, the final pH is 3.187

8 0

3 years ago

The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What

pentagon [3]

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

The heat of vaporization at 75° C = 900 J/mol

8 0

4 years ago