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Norma-Jean [14]
3 years ago
13

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25°C. How many grams of chlorophyll, C55H72MgN4O5, a nonvolatile,

nonelectrolyte (MW = 893.5 g/mol), must be added to 187.4 grams of diethyl ether to reduce the vapor pressure to 457.45 mm Hg ?
Physics
1 answer:
mixer [17]3 years ago
8 0

Answer:

29.4855 grams of chlorophyll

Explanation:

From Raoult's law

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 457.45 mmHg ÷ 463.57 mmHg = 0.987

Mass of solvent (diethyl ether) = 187.4 g

MW of diethyl ether (C2H5OC2H5) = 74 g/mol

Number of moles of solvent = mass/MW = 187.4/74 = 2.532 mol

Let the moles of solute (chlorophyll) be y

Total moles of solution = moles of solute + moles of solvent = (y + 2.532) mol

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

0.987 = 2.532/(y + 2.532)

y + 2.532 = 2.532/0.987

y + 2.532 = 2.565

y = 2.565 - 2.532 = 0.033

Moles of solute (chlorophyll) = 0.033 mol

Mass of chlorophyll = moles of chlorophyll × MW = 0.033 × 893.5 = 29.4855 grams

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MariettaO [177]
With that information you can only suppose a uniformly accelerated motion.  This is, acceleration is constant.

Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2

Then the equation for velocity, V is

V = Vo + a*t = Vo + 2 (km/h^2)  * t = Vo +  2t

Vo is the initial velocity, which you can find using V = 54km/h and t = -2

Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h

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2 years ago
A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially relea
grandymaker [24]

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

A = 1.803 ft

Period = \frac{\pi}{2} seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = \frac{w}{g}

m= \frac{32}{32}

= 1 slug

Also from Hooke's Law

2 k = 32

k = \frac{32}{2}

k = 16 lb/ft

Using the function:

\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0

x(0) = -1        (because of the initial position being above the equilibrium position)

x(0) = -6          ( as a result of upward velocity)

NOW, we have:

x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)

However;

x(0) = -1 means

-1 =c_1\\c_1 = -1

x(0) =-6 also implies that:

-6 =4(c_2)\\c_2 = - \frac{6}{4}

c_2 = -\frac{3}{2}

Hence, x(t) =-cos4t-\frac{3}{2} sin 4t

A = \sqrt{C_1^2+C_2^2}

A = \sqrt{(-1)^2+(\frac{3}{2})^2 }

A=\sqrt{\frac{13}{4} }

A= \frac{1}{2}\sqrt{13}

A = 1.803 ft

Period can be calculated as follows:

= \frac{2 \pi}{4}

= \frac{\pi}{2} seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

x* \frac{\pi}{2} = 4 \pi

x \pi = 8 \pi

x=8 cycles

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3 years ago
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Gala2k [10]

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3 years ago
Suppose we have a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it trie
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Answer:

2.64 m/s

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Given that a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna

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3 years ago
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V= 4300cm3 + (-8.000cm3)

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3 0
3 years ago
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