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2 years ago

what issues might arise in determining the latent heat of something that sublimes in comparison to something that melts

2 answers:

6 0

The heat of sublimation actually has two components, the heat of solidification (actually melting, since this is going in the reverse direction) and the heat of vaporization since you are transforming the solid to a gas. The energy required is the same, even though you do not actually go through a macroscopic melting phase. So the sublimation energy will be much greater than the heat needed for just melting, given the same mass.

Hope this helps!!

vfiekz [6]2 years ago

5 0

Answer:

Sublimation is defined as the direct transition of a substance from solid state to gaseous state. The latent heat of sublimation is the sum of heat required for conversion from solid to liquid and heat required for conversion from liquid to gas. As in sublimation solid substance directly goes to gaseous state therefore, it is very difficult to find the components of latent heat of sublimation. This is the issue which arises in determining latent heat of sublimation.

On the other hand, when substance is melted it physically goes from solid to liquid. One can easily measure the heat required for this conversion.

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A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0$

$\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2$

So by solving above equation we will have

$v_f = 84.3 m/s$

now in order to find the momentum we can use

$P = mv$

$P = 1.25 \times 84.3$

$P = 105.4 kg m/s$

6 0

2 years ago

At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct n

Andreyy89

Below are the choices that can be found elsewhere:

a. 268 kJ
b. 271 kJ 
c. 9 kJ 
d. 6 kJ

So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have:

Ga (s) --> Ga (g) delta H = 277 kJ/mol 

Ga (l) --> Ga (g) delta H = 271 kJ/mol 

From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). 

Ga (s) --> Ga (l) delta H = 6 kJ/mol 

At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). 

*ANSWER* 
9 kJ/mol (C)

7 0

2 years ago

If a rock is thrown upward on the planet mars with a velocity of 14 m/s, its height (in meters) after t seconds is given by h =

crimeas [40]

Answer:

Velocity of rock after 2 seconds = 6.56 m/s

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Here height of rock in meters, h = $14t-1.86t^2$

Comparing both the equations

We will get initial velocity = 14 m/s(already given) and $\frac{1}{2} a = -1.86$

So, Acceleration, a = -3.72 $m/s^2$

Now we have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

When time is 2 seconds we need to find final velocity.

v = 14 - 3.72 * 2 = 6.56 m/s.

So, Velocity of rock after 2 seconds = 6.56 m/s

6 0

2 years ago

An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t

marissa [1.9K]

The energy transferred to the spring is given by:
$U= \frac{1}{2}kx^2$
where
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
$k=52.9 N/cm = 5290 N/m$
$x=40.0 cm=0.4 m$

so now we can use these data inside the equation ,to find the energy transferred to the spring:
$U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J$

4 0

2 years ago

A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?

Fed [463]

I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
$W_x =mg \sin \theta$
where
m is the mass of the car
g is the acceleration of gravity
$\theta$ is the angle of the hill

Substituting numbers into the formula, we find
$W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N$

6 0

2 years ago