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jok3333 [9.3K]
4 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]4 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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Answer:

a)E=2.88*10^{11}N/C

b)E=1.44*10^{11}N/C

c)F=4.61*10^{-8}N

Explanation:

We use the definition of a electric field produced by a point charge:

E=k*q/r^2

<u>a)Electric Field  due to the alpha particle:</u>

E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C

<u>b)Electric Field  due to electron:</u>

E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C

<u>c)Electric Force on the alpha particle, on the electron:</u>

The alpha particle and electron feel the same force but with opposite direction:

F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N

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Which red line shows the same action spectrum corrected for the unequal number of photons emitted across the visible spectrum?.
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The visible spectrum is composed of red, orange,yellow,  green, blue, violet, indigo.

<h3>What is visible spectrum?</h3>

The visible spectrum refers to the portion of the electromagnetic spectrum that can be seen with the eyes. All other portions of the electromagnetic spectrum are invisible.

The question is incomplete as the details are missing. The visible spectrum is composed of red, orange,yellow,  green, blue, violet, indigo.

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jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
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Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

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