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2 years ago

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At 213.1 K a substance has a vapor pressure of 45.77 mmHg. At 243.7 K it has a vapor pressure of 193.1 mm Hg. Calculate its heat

of vaporization in kJ/mol.

1 answer:

vlabodo [156]2 years ago

8 0

Answer:

$20.3125$ kJ/mol

Explanation:

$P_{i}$ = initial vapor pressure = 45.77 mm Hg

$P_{f}$ = final vapor pressure = 193.1 mm Hg

$T_{i}$ = initial temperature = 213.1 K

$T_{f}$ = final temperature = 243.7 K

$H$ = Heat of vaporization

Using the equation

$ln\left ( \frac{P_{f}}{P_{i}} \right ) = \left ( \frac{-H}{R} \right )\left ( \frac{1}{T_{f}} - \frac{1}{T_{i}}\right)$

$ln\left ( \frac{193.1}{45.77} \right ) = \left ( \frac{-H}{8.314} \right )\left ( \frac{1}{243.7} - \frac{1}{213.1}\right)$

$H = 20312.5$ J/mol

$H = 20.3125$ kJ/mol

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When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:

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1 year ago

In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li

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Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

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Now power of each CFL bulb = 25 W

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2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

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Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

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2 years ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

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$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

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divide 1 and 2

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Therefore spring compresses twice the initial value

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1 year ago

A 5⁢kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant.

Umnica [9.8K]

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

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g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

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Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

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-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

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Then, the magnitude of the gravitational force will be:

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F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

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2 years ago