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Lapatulllka [165]
3 years ago
9

A compound is found to have a molar mass of 598 g/mol. if 0.0358 g of the compound is dissolved in enough water to make 175 ml o

f solution at 25°c, what is the osmotic pressure of the resulting solution?
Physics
1 answer:
zalisa [80]3 years ago
3 0

π=iMRT

Where, π is Osmotic pressure,
i=1 for non-electrolytes,
M is molar concentration of dissolved species (units of mol/L)
R is the ideal gas constant = 0.08206 L atm mol⁻¹K⁻¹,
T is the temperature in Kelvin(K),

Here, to calculate M convert into standard units mg tog, ml to L, c to Kelvin
M= (\frac{35.8}{598} *10⁻³ )/ 0.175 =(5.987 *10⁻⁵)mol / 0.175L = 34.21*10⁻⁵ mol/L

π=iMRT=(1)*(34.21*10⁻⁵)*(0.08206)*(298.15)=837×10⁻⁵= 8.37×10⁻³ atm
=6.36 torr
(1 atm=760 torr, 1 Kelvin =273.15 °C, 1L=1000ml, 1g=1000mg)

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What happens when a ray of light is directed at a mirror, a glass block and a prism?
zimovet [89]
When it is directed at a mirror it is reflected
When it is directed at a glass block it refracts
When it is directed through a prism the light splits to form a spectrum ( red orange yellow green blue indigo and violet )

Hope this is ok
5 0
3 years ago
A worker on the roof of a house drops his 0.46 kg hammer, which slides down the roof at constant speed of 9.88 m/s. The roof mak
nekit [7.7K]

Answer:

The horizontal distance travelled in that time lapse is 12.94 m

Explanation:

In order to solve this problem, we'll need:

  1. The horizontal speed
  2. the time the hammer takes to fall from the roof to the ground

At the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get

v_{x}=v*cos(27)=9.88 m/s *cos(27)=8.80 m/s \\v_{y}=v*sen(27)=9.88 m/s *sen(27)=4.49 m/s

Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates

x(t)=8.8 m/s *t

y(t)=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}

Now we calculate the time the hammer takes to get to the floor

17.1m=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}\\t=1.47s or t=-2.38s

Now, we keep the positive time result and calculate the horizontal distance travelled

x(1.47s)=8.8m/s*1.47s=12.94m

3 0
3 years ago
A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises
atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s²  * t²

-2.0 m = -4.9 m/s²  * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t      where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t²           y = 1.5 m       y0 = 0 m   t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

4 0
3 years ago
When a trebuchet counterweight is hoisted by soldiers, what form of energy is
vladimir1956 [14]
10. A safe place to use the trebuchet would be away from other buildings and people. A good example of a place would be a large field with no nearby structures. 
14. Many factors need to be kept consistent throughout the experiment. One example of a variable that would need to be consistent is the weight and size of the projectile. 
15. It is important to do many trials so that you can make sure that the results of each trial are nearly the same. If they are all vastly different, then it means that something has gone wrong. 

Sorry I was only able to answer a few questions, but I hope these few answers help! :) 
4 0
3 years ago
Read 2 more answers
A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds af
disa [49]

Answer:

Explanation:

Given

Cannon is fired with a velocity of u=72.50\ m/s

Using Equation of motion

y=ut+\frac{1}{2}at^2

where

y=displacement

u=initial\ velocity

a=acceleration

t=time

after time t=3.3 s

y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2

y=239.25-53.36

y=185.89\ m

So after 3.3 s cannon ball is at a height of 185.89 m

6 0
3 years ago
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