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zysi [14]
3 years ago
13

What intermolecular force(s) is/are present in solid SO3?

Chemistry
1 answer:
guapka [62]3 years ago
5 0

Answer:

1. London dispersion

Explanation:

Sulphur trioxide ( SO₃ ) -

The chemical compound SO₃ is planar in structure , the only intermolecular forces shown by SO₃ is the London forces .

dipole - dipole is not observed in this compound , as it is not possible to generate poles between the sulfur and oxygen atom due to very less difference in the electronegativity .

Hydrogen bonding is also not observed , because there is not hydrogen atom .

Hence , only London forces are observed in SO₃ .

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The potential energy of a 25 kg bicycle resting at the top of a 3 m high hill is what?
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Potential energy<span> is the </span>energy<span> that is stored in an object due to its position relative to some zero position.  It is calculated by the expression as follows:

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4 0
3 years ago
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A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)
rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }

<u>Given:</u>

A mysterious white powder could be,

  • powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
  • cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
  • codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
  • norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
  • fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

  • The solute = the powder
  • The solvent = ethanol
  • The freezing point of the solvent = −114.6°C
  • The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

  • Mass of solute = 82 mg = 0.082 g
  • Mass of solvent = density x volume, i.e., \boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg  \ }

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

Now we combine it with the formula of freezing point depression.

\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }

\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }

\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }

We get \boxed{ \ Mr = 153.65 \ }

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>
  1. The molality and mole fraction of water brainly.com/question/10861444
  2. About the mass and density of ethylene glycol as an  antifreeze brainly.com/question/4053884
  3. About the solution as a homogeneous mixture  brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

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3 years ago
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Which quantity is equivalent to the product of the absolute index of refraction of water and the speed of light in water?
givi [52]

The quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.

<h3>What is the speed of light?</h3>

Speed of light is the rate of speed though the light travels. To find the speed of light in any medium, the following formula is used.

v=\dfrac{c}{n}

Here, (n) is the index of reaction and (c) is the speed of light in the vacuum. The speed of light in the vacuum is almost equal to the 3.0×10⁸ m/s.

Now the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water has to be find out.

The above formula can be written as,

v=\dfrac{c}{n}\\vn=c\\c=vn

Here, the product of index of refraction and speed of light is equal to the speed of light in vacuum. This will be true for water as well.

Thus, the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.

Learn more about the speed of light here;

brainly.com/question/104425

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1 year ago
Why is scientific notation useful?
BlackZzzverrR [31]
It condenses very long strings of numbers while retaining the general accuracy of the figure.
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