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Shtirlitz [24]
3 years ago
6

A gas mixture contains sulfur trioxide at a pressure of 1.45 atm and sulfur dioxide at a pressure of 0.32 atms. What is the tota

l pressure of the gas mixture?
Chemistry
1 answer:
aliina [53]3 years ago
3 0

Answer:

1.77 atm

Explanation:

Total pressure in a mixture is, the sum of partial pressures from each gas.

Let's analyse with the Universal Gases Law

Total pressure . V = (total moles)  . R . T

Partial pressure of a gas . V = (mol of that gas) . R . T

Mole fraction:

Partial pressure of a gas / Total pressure = moles of the gas / Total moles

Then, total pressure in this mixture is:

1.45 atm + 0.32 atm = 1.77 atm

You might be interested in
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0
3 years ago
A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k
kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

  • Helium: 0,25 moles ×\frac{4 g}{1 mol} = 1 g of Helium
  • Argon: 0,25 moles ×\frac{40 g}{1 mol} = 10 g of Argon
  • Krypton: 0,25 moles ×\frac{83,8 g}{1 mol} = 20,95 g of krypton
  • Xenon: 0,25 moles ×\frac{131,3 g}{1 mol} = 32,825 g of Xenon

Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium: \frac{1 gHelium}{62,675 g} × 100 = <em>1,6%</em>

<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em>  </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

\frac{4+40+83,8+131,3}{4} = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

\frac{P}{R.T} = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145\frac{L.kPa}{K.mol}

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² \frac{mol}{L} × \frac{64,775 g}{mol} × \frac{1L}{1000 mL} =

2,3×10⁻³ g/mL

I hope it helps!

7 0
3 years ago
NaCl has a Hfus = 30.2 kJ/mol. What is the mass of a sample of NaCl that needs 732.6 kJ of heat to melt completely? 24.3 g 82.7
alina1380 [7]
Hello.

The answer is  1417.7

Its 1417.7 grams becasue you have  to convert the moles to grams.


<span /><span><span>A mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.023 X 1023).

</span><span>a coherent, typically large body of matter with no definite shape.

Have a nice day</span></span>
4 0
3 years ago
Read 2 more answers
One of the pieces of evidence supporting energy quantization was the line spectra of elements. Why does this demonstrate energy
bezimeni [28]

Answer:

c) There are sharp emission lines demonstrating discrete energy levels.

Explanation:

When an element emits energy in the form of radiation, it produces a spectrum of colors on a photographic plate. This spectrum can either be continuous or discrete. In continuous spectrum the spectrum continues without any discrimination between two regions. This represents the continuous emission of radiation, and thus the continuous emission of energy without any break.

On the other hand, the line spectrum consists of discrete and sharp lines, which shows the emission of radiation in a certain amount in a certain time, with a break between emission. Hence, the line spectra supports the quantization of energy.

The correct option is:

<u>c) There are sharp emission lines demonstrating discrete energy levels.</u>

8 0
3 years ago
Calculate the work (w) and ΔEo, in kJ, at 298 K and 1 atm pressure, for the combustion of one mole of C4H10 (g). First write and
Paul [167]

Answer:

Explanation:

2C₄H₁₀ + 13O₂ = 8 CO₂ + 10H₂O

Change in number of moles Δn = 18 - 15 = + 3 moles .

ΔHo = -2658.3 kJ/mol.

ΔHo = ΔEo+ Δn RT

Δn = 3

For one mole Δn = 1.5

ΔHo = ΔEo+ W

W = Δn RT

= 1.5 x 8.31 x 298

= 3714.5 J

= 3.7 kJ /mole

ΔHo = ΔEo+ W

ΔEo =  ΔHo -  W

= -2658.3 - 3.7  kJ

= - 2662 kJ .

3 0
3 years ago
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