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Veseljchak [2.6K]
3 years ago
5

20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solu

tion is allowed to reach 90.0°C . The vapor pressure of pure water at this temperature is 528.8 mm Hg. The vapor pressure of the solution is 423.0 mm Hg. How many kg of water was present?
Physics
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

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A 0.0780 kg lemming runs off a
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Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0
3 years ago
4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer
Rudik [331]

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

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3 years ago
A physicist is investigating a beam of laser light of wavelength 550 nm. The light strikes a target that is 189 meters away from
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I hope this helps☺️

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3 years ago
Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep
SIZIF [17.4K]

Answer:

1) iii i= 1m, 2)  iii and iv, 3)  i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

              f = r / 2

              f = 2/2

             f = 1 m

The builder's equation

           1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

              1 / f = 0 + 1 / i

              i = f

              i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

              R₂ = infinity

              R₁ > 0

Cas2 Flat-concave (divergent) lens

             R₂ = infinity

              R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

          1 / f₂ = 1 / (L-f₁) + 1 / i

          1 / i = 1 / f₂ - 1 / (L-f₁)

           1 / i = (L-f₁-f₂) / f₂ (L-f₁)

           i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

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