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satela [25.4K]
2 years ago
12

38. Identify the most important types of interparticle forces pres

Chemistry
1 answer:
zhannawk [14.2K]2 years ago
4 0

Answer: im thinking its gonna be d.C2H6 and also

the explanation is on the research i had did before i had answered this question so i really hope this help :)

Explanation:

Ar = van de waals forces or london forces

C

H

4

= van de waals forces or london forces

HCl=permanent dipole-dipole interactions

CO = permanent dipole-dipole interactions

HF = hydrogen bonding

N

a

N

O

3

= permanent dipole-dipole interactions

C

a

C

l

2

= van de waals forces or london forces

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How many grams are in 1.6 moles of potassium bromide?​
slava [35]

Answer:

190.4g

Explanation:

1.6mol of KBr (119.002g KBr/1 mol) = 190.4g

since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.

6 0
3 years ago
Describe the overall enthalpy of the chemical reactants compared to the enthalpy of the chemical products in the combustion of p
Veronika [31]

Answer:

A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:

C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol

The value given for ΔH∘rxnΔHrxn∘ means that:

a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.

b. the reaction is endothermic.

c. the enthalpy of formation of propane is 2202 kJ/mol.

d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.

e. None of these.

3 0
1 year ago
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

7 0
3 years ago
Read 2 more answers
A gas exerts a pressure of 350.0 torr at 20.0 Celsius. What would be the new pressure in atmospheres if the temperature is raise
IgorC [24]

Answer:

373.88 torr

Explanation:

P1 = 350 torr

T1 = 20°C = (20 + 273.15)K = 293.15K

P2 = ?

T2 = 40°C = (40 + 273.15)K = 313.15K

From pressure law,

Pressure of a given mass of gas is directly proportional to its temperature.

P = KT

K = P / T

P1 / T1 = P2 / T2

Solve for P2

P2 = (P1 * T2) / T1

P2 = (350 * 313.15) / 293.15

P2 = 109602.5 / 293.15

P2 = 373.878 torr

P2 = 373.88 torr

The new pressure of the gas would be 373.88 torr.

5 0
3 years ago
Read 2 more answers
A certain ionic compound X has a solubility in water of 40.3 g/L at 20. degrees C. Calculate the mass X of required to prepare 5
tino4ka555 [31]

Answer:

20.1 g

Explanation:

The solubility indicates how much of the solute the solvent can dissolve. A solution is saturated when the solvent dissolved the maximum that it can do, so, if more solute is added, it will precipitate. The solubility varies with the temperature. Generally, it increases when the temperature increases.

So, if the solubility is 40.3 g/L, and the volume is 500 mL = 0.5 L, the mass of the solute is:

40.3 g/L = m/V

40.3 g/L = m/0.5L

m = 40.3 g/L * 0.5L

m = 20.1 g

7 0
3 years ago
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