1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
stealth61 [152]
3 years ago
5

Calculate the pressure of 3.2 moles of an ideal gas that occupies a volume of 87 m3 at a temperature of 312 K.

Chemistry
2 answers:
Anestetic [448]3 years ago
4 0

Answer:

                  P  =  0.0009417 atm

Or,

                 P  =  9.417 × 10⁻⁴ atm

Or,

                  P  =  0.0954157 kPa

Or,

                  P  =  0.715677 mmHg (Torr)

Explanation:

Data Given:

                 Moles  =  n  =  3.2 mol

                 Temperature  =  T  =  312 K

                  Pressure  =  P  =  ?

                  Volume  =  V  =  87 m³ = 87000 L

Formula Used:

Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,

                  P V  =  n R T

where;  R  =  Universal Gas Constant  =  0.082057 atm.L.mol⁻¹.K⁻¹

Solving Equation for P,

                  P  =  n R T / V

Putting Values,

                  P  =  (3.2 mol × 0.082057 atm.L.mol⁻¹.K⁻¹ × 312 K) ÷ 87000 L

                  P  =  0.0009417 atm

Or,

                 P  =  9.417 × 10⁻⁴ atm

Or,

                  P  =  0.0954157 kPa

Or,

                  P  =  0.715677 mmHg (Torr)

Solnce55 [7]3 years ago
4 0

Answer:

The correct answer is 95.36 Pa.

Explanation:

It is given that the moles of an ideal gas, n = 3.2 moles

Volume of an ideal gas, V = 87 m³

Temperature, T = 312 K

Pressure, P = x

The ideal gas equation, PV = nRT, here R is the gas constant, and at standard temperature and pressure, one mole of ideal gas holds 22.4 L volume, At STP, T = 0 degree C = 273 K

V = 22.4 L

moles, n = 1 mole

Gas constant, R = PV / nT

R = 1 atm × 22.4 L / 1 mole × 273 K

1 atm = 1.013 × 10⁵ Pa and 1L = 10⁻³ m³

R = 8.31 Pa. m³. mol⁻¹. K⁻¹

Now, there is a need to calculate pressure, P:

PV = nRT

P = 3.2 mol × 8.31 Pa. m³. mol⁻¹. K⁻¹ × 312 K / 87 m³

P = 8296.7 / 87

P = 95.36 Pa

You might be interested in
PLZ HELP! Which of the following statements is true?
Sindrei [870]
<span>1. over thrust plate boundary
2. Its the outer layer and coolest
3. building a bridge
4.hurricanes
5.They can move even if there is no earthquake
6.core, asthenosphere, lithosphere
7.transform (?)
8.10x
9.lithosphere
10.deep crack in the earth's crust
11.?
12.high pressure
13.dome
14.lithospheric plates
15. can travel through liquids and solids
16. the earth's surface
I did this test too

</span>
4 0
3 years ago
Research the compositions of Pennies. What was the composition of each of your Pennies prior to treatment
LenKa [72]

Answer:

History of composition

Years Material Weight (grains)

1944–1946 gilding metal (95% copper, 5% zinc) 48 grains

1947–1962 bronze (95% copper, 5% tin and zinc) 48 grains

1962 – September 1982 gilding metal (95% copper, 5% zinc) 48 grains

October 1982 – present copper-plated zinc (97.5% zinc, 2.5% copper) 38.6 grains

7 0
2 years ago
Oxygen gas was produced in a reaction and collected over water. A 136.1 mL mL sample of gas was collected over water at 25C and
Xelga [282]

Answer:

Explanation:

We shall find volume of gas at NTP or at 273 K , 760 mm of Hg .

Pressure of given gas = 1.06 x 760 mm of Hg less vapor pressure of water .

= 805.6 - 23.76 = 781.84 mm of Hg

For it we use gas law formula ,

P₁V₁ / T₁ = P₂V₂ / T₂

781.84 x 136.1 / ( 273 + 25 ) = 760 x V₂ / 273

= 128.26 mL .

= 128.26  x 10⁻³ L .

22.4 L of oxygen will have mass of 32 g

128.26 x 10⁻³ L of oxygen will have mass of 32 x 128.26 x 10⁻³ / 22.4 g

= 183.22 mg .

4 0
2 years ago
The statements in the table summarize what Johannes Kepler stated in the early 1600s about the motion of planets.
SIZIF [17.4K]

Answer:

it's law

Explanation:

i got it wrong because i put the wrong answer but it was law or c

i hope this help's

4 0
2 years ago
Which of the following pairs contains a conjugate acid-base pair?
zzz [600]
Remember that a conjugate acid-base pair will differ only by one proton. 
None of the options you listed are conjugate acid-base pairs as none of them differ only by one proton (or H⁺)
An example of a conjugate acid-base pair would be NH₃ and NH₄⁺NH₃ + H₂O --> NH₄⁺ + OH⁻NH3 is the base, and NH₄⁺ is the conjugate acid
5 0
3 years ago
Other questions:
  • An object has a mass of 4.9g and a volume of 14.ml.what is the density of the object?
    8·1 answer
  • What do you think is meant by the term pseudoscience?
    13·2 answers
  • The industrial production of ammonia (NH3) involves reacting nitrogen gas with hydrogen gas. If 18.2 kg of ammonia is produced f
    5·1 answer
  • maye Iesult TUI di2... Warm-Up Exercises for Chapter 2 - Acid-Base Reactions roblem 2.68 Part C Which of the following reactions
    14·1 answer
  • Urea spontaneous dissolves in water (to a high concentration). When urea dissolves in water, the solution cools before reaching
    5·1 answer
  • How might you make sure you have the correct amount of nitrogen and hydrogen so that you have enough of each with not wasted sur
    10·1 answer
  • Which of the following is an example of a response to stimulus?
    6·2 answers
  • What is the wavelength of electromagnetic radiation with a frequency of 5.00 x 10^12 Hz? S
    6·1 answer
  • Ionic equation for Anode positive electrode
    5·1 answer
  • assume the given amount of rbf fully dissociates in water. calculate the molarity of a solution if 124.86 g of rbf are dissolved
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!