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Natali5045456 [20]
3 years ago
15

What is the concentration of molecular oxygen (O2) in mol/L on a June day in Toronto when atmospheric pressure is 1.0 atm and th

e temperature is 25 oC? (Remember O2 is present at a concentration of 21% (or 21pph) in the atmosphere).
Physics
1 answer:
saveliy_v [14]3 years ago
6 0

Answer:

The concentration of mole evil at oxygen on that day is 0.00858 mol/L

Explanation:

Here, we want to calculate the concentration of molecular oxygen

The pressure on that day is 1.0 atm

Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm

Now let’s calculate the concentration;

From Ideal gas law;

PV = nRT

This can be written as;

P/RT = n/V

The term n/V refers to concentration;

Let’s make substitutions now;

P = pressure = 0.21 atm

R = molar gas constant = 0.0821 L•atm/mol•k

T = temperature = 25 = 25 + 273.15 = 298.15 K

Substituting these values, we have;

n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L

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You place an 8 kg ball on the top of your 2 cm^2 finger tip. Calculate the
Rashid [163]

Answer:

the pressure exerted by the object is 392,000 N/m²

Explanation:

Given;

mass of the object, m = 8 kg

area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²

acceleration due to gravity, g = 9.8 m/s²

The pressure exerted by the object is calculated as;

Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2

Therefore, the pressure exerted by the object is 392,000 N/m²

7 0
3 years ago
A projectile is fired at an upward angle of 29.7° from the top of a 108-m-high cliff with a speed of 130-m/s. What will be its
german

Answer:

79.2 m/s

Explanation:

θ = angle at which projectile is launched = 29.7 deg

a = initial speed of launch = 130 m/s

Consider the motion along the vertical direction

v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s

y = vertical displacement = - 108 m

a = acceleration = - 9.8 m/s²

v = final speed as it strikes the ground

Using the kinematics equation

v² = v₀² + 2 a y

v² = 64.4² + 2 (-9.8) (-108)

v = 79.2 m/s

5 0
3 years ago
A ball is thrown horizontally from a platform so that the initial height of the ball is 6.0 m above the level ground below. The
solniwko [45]
We can first obtain time of flight from vertical fall
Initial velocity U=0, d = 6 m, a = 9.8 m/s² 
d = ut + 1/2 at²
6.0 = 0 + (1/2 × 9.80 t²) 
t = √(12/9.8)
 = 1.106 sec
Horizontal velocity = Vh = Dh/t
  = 24.0 /1.106 s
  = 21.69 m/s
The ball was thrown at a speed of 21.69 m/s
5 0
3 years ago
Even through there is equal and opposite reaction,usually the two forces are not seen balanced.Why?
Tomtit [17]

Answer:

This may refer to a situation like:

"one person pushes a box, if there is equal and opposite reaction why the box moves and the person does not?"

Remember the second Newton's law:

F = m*a

suppose that the mass of the person is 3 times the mass of the box.

So, if the box has a mass M, the person will have a mass 3*M

Then the Newton's equation for the box when the person pushes with a force F is:

F = M*a

solving for the acceleration, we get:

F/M = a

While the person is also pushed by the box with a force with the same magnitude, then the equation for the person is:

F = (3*M)*a'

Solving for the acceleration, we get:

F/(3M) = a'

Now we can compare the acceleration of the box (F/M) with the acceleration of the person (F/3M).

Is easy to see that the acceleration of the box is 3 times the acceleration of the person.

So regardless of the fact that both the box and the person experience a force with the same magnitude, the box will move more due to this force.

This is why in situations like this, the forces do not seem balanced.

5 0
3 years ago
A solid must be given​
alexdok [17]
I- what....? Lolllll
8 0
4 years ago
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