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Elena-2011 [213]
3 years ago
11

The concentration of ozone in los angeles is 0.67 ppm on a summer day. what is the partial pressure of o3 if the total pressure

is 735 torr?
Physics
1 answer:
Katarina [22]3 years ago
5 0
We use the Raoult's Law to answer this question.

Partial Pressure = Mole Fraction*Total Pressure

When concentration is expressed in ppm, it means 'parts per million'. So, there can be 0.67 mol of ozone per 10⁶ mol of mixture.

Mole fraction = 0.67/10⁶ = 0.67×10⁻⁶

Partial pressure = 0.67×10⁻⁶(735 torr) = <em>4.92×10⁻⁴ torr</em>
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An object experiences a net acceleration to the left. Which of the following statements about this object are true? There may be
dedylja [7]

Answer:

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

Explanation:

When an object experiences acceleration to the left, the net force acting on this object will also be to the left.

From Newton's second law of motion, the acceleration of the object is given as;

a = ∑F / m

a = -F / m

The negative value of "a" indicates acceleration to the left

where;

∑F is the net force on the object

m is the mass of the object

At a constant force, F = ma ⇒ m₁a₁ = m₂a₂

If the mass of the object was doubled, m₂ = 2m₁

a₂ = (m₁a₁) / (m₂)

a₂ = (m₁a₁) / (2m₁)

a₂ = ¹/₂(a₁)

Therefore, the following can be deduced from the acceleration of this object;

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

6 0
3 years ago
If you put a total of 8.05×106×106 electrons on an intially electrically neutral wire of length 1.03 m, what is the magnitude of
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Answer:

The magnitude of the electric field is 0.1108 N/C

Explanation:

Given;

number of electrons, e = 8.05 x 10⁶

length of the wire, L = 1.03 m

distance of the field from the center of the wire, r = 0.201 m

Charge of the electron;

Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)

Q = 1.2896 x 10⁻¹² C

Linear charge density;

λ = Q / L

λ = (1.2896 x 10⁻¹² C) / (1.03 m)

λ = 1.252 x 10⁻¹² C/m

The magnitude of electric field at r = 0.201 m;

E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201}  \\\\E = 0.1108 \ N/C

Therefore, the magnitude of the electric field is 0.1108 N/C

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