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jok3333 [9.3K]
3 years ago
5

A 7.75-l flask contains 0.482 g of hydrogen gas and 4.98 g of oxygen gas at 65°c. What is the partial pressure of oxygen in the

flask?
Physics
1 answer:
dimulka [17.4K]3 years ago
3 0

Answer:

0.558 atm

Explanation:

We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT

Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:

P₀ = P₁ + P₂ + ....

P₀= total pressure

P₁=P₂= is the partial pressure of each gass

If we can consider that each gas is an ideal gas, then:

P₀= (nRT/V)₁ + (nRT/V)₂ +..

Considering the molecular mass of O₂:

M O₂= 32 g/mol

And also:

R= ideal gas constant= 0.082 Lt*atm/K*mol

T= 65°C=338 K

4.98 g O₂ = 0.156 moles O₂

V= 7.75 Lt

Then:

P°O₂=partial pressure of oxygen gas=  (0.156x0.082x338)/7.75

P°O₂= 0.558 atm

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An engine flywheel initially rotates counterclockwise at 6.55 rotations/s. Then, during 20.9 s, its rotation rate changes to 2.1
11Alexandr11 [23.1K]

Answer:

The average angular acceleration is -2.628 rad/s²

Explanation:

Counterclockwise = positive

Clockwise = -negative

Given;

initial rotation of the flywheel, θ₁ = 6.55 rotation/s

final rotation of the flywheel, θ₂ = - 2.19 rotation/s

The average angular acceleration is given by;

\alpha = \frac{\delta \theta}{\delta t}\\\\ \alpha =\frac{\theta _2 - \theta_ 1}{t}\\\\ \alpha =\frac{-2.19 -6.55}{20.9} \\\\ \alpha =\frac{-8.74}{20.9}\\\\ \alpha = -0.4182 \ rotation / s^2\\\\ \alpha = \frac{-0.4182 \ rotation}{s^2}*\frac{2\pi \ radian}{rotation}\\\\ \alpha = -2.628 \ rad/s^2

Therefore, the average angular acceleration is -2.628 rad/s²

7 0
3 years ago
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field
lawyer [7]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = \frac{1}{2} mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = \frac{1}{2}  \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             K_e = K_p

              K_p = ½ m v_e²

              K_p = \frac{1}{2}  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = \frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = \frac{7 \ 10^4 }{ 3 \ 10^8}

        v/c= 2.33 10⁻⁴

8 0
3 years ago
Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird
serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr  and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

  1. db = 2dr
  2. L + (L - x) = 2x
  3. 2L - x = 2x
  4. 2L = 3x
  5. x = \frac{2}{3}L

Insert it in x = \frac{2}{3}L

\frac{2}{3}(2.4km) = 1.6km

Now we use formula db = 2dr

  1. db = 2L - x
  2. db = 2(2.4km) - 1.6km
  3. <u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

  1. Vr = L/ Δt ⇒ Δt = \frac{L}{Vr}
  2. \frac{2.4km}{6.8km/hr}
  3. \frac{6}{17}hr

(Km cancel each other)

  1. Vb = db/Δt ⇒ db = VbΔt
  2. 13.6km/hr(\frac{6}{17}hr )
  3. <u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

6 0
3 years ago
How does the sun's energy most directly influence precipitation in an area?
topjm [15]
The sun's energy influences climate in various ways. For example the latitudes at the equator receive more energy from the sun and therefore have warmer temperatures, On the other hand the sun's energy influences precipitation in a climate by driving the water cycle which determines precipitation.The sun is what makes the water cycle take place. That is the sun provides energy or heat to the earth; the heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds ( precipitation), that in turn give us rain
5 0
3 years ago
If two children, with masses of 16 kg and 25 kg, sit in seats opposite one another in a rotating merry-go-round with an arm leng
Jlenok [28]

Answer:

92.25 kgm^2

Explanation:

Assume both children bodies are point particles. The total moment of inertia about the rotation axis of 2 points particles of mass 16 kg and 25 kg at 1.5 m arm length is

\sum I = m_1r_1^2 + m_2r_2^2

where m_1 = 16 kg, m_2 = 25 kg are the masses of 2 children r_1 = R-2 = 1.5m are their distance to the center of rotation

\sum I = 16*1.5^2 + 25*1.5^2 = 1.5^2(16 + 25) = 92.25 kgm^2

4 0
3 years ago
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