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ANTONII [103]
3 years ago
7

NEED HELP PLSSS

Chemistry
1 answer:
likoan [24]3 years ago
4 0
Are their answer choices? All Ik is that compound a and b have the same ratio of u to g
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Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of
Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

  • We need to calculate the vapor pressure of each component after the stopcocks are opened.
  • Volume after the stopcocks are opened = 3.0 L.

<u><em>1) For N₂:</em></u>

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

<u><em>2) For H₂O:</em></u>

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: <em>(1.0 atm = 760.0 mmHg)</em>.

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

<em><u>3) For C₂H₅OH:</u></em>

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

<em>It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.</em>

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

  • <em>So, </em><em>total pressure</em><em> = </em><em>P of N₂ + P of H₂O + P of C₂H₅OH</em><em> = 0.5 atm + 0.0553 atm + 0.055 atm = </em><em>0.6103 atm</em><em>.</em>
3 0
2 years ago
Is energy transfer the process by which any form of energy changes location?
Elis [28]
Yes, transfer of energy is known as an energy transfer
4 0
3 years ago
What is meant by saying that the latent heat of vaporization of water is 22. 5×10^5 j/kg.
andrey2020 [161]

It means that 22.5×10^5 J of heat is required to change 1 kg of water into steam.

Latent heat of vaporization is amount of energy required to change 1 gram (in this example 1 kilogram) of material from the liquid to the gaseous state at its boiling point.

Boiling point of the water is 100°C.

Joule (J) is the standard unit for energy (in this example heat).

Evaporization is phase change process in which the water changes from a liquid to a gas (water vapor). Fore example, solar radiation can be the source of energy for evaporation.

More about heat of vaporization: brainly.com/question/14679329

#SPJ4

5 0
1 year ago
Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estima
Rzqust [24]

Answer:

See explanation below

Explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.

b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.

h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.

Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵

[OH⁻] = 1.92x10⁻¹⁰ M

PArt C:

In this case, we will use expression of part A, and replace the given values:

[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²

[H⁺] = 3.7x10⁻¹³ M

8 0
2 years ago
Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base
Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

\frac{r_{cation}}{r_{anion}}

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, Na^+ = 102 pm

Radius of cation, Br^- = 196 pm

\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0
3 years ago
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