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2 years ago

Differences between sublimation and condensation

Physics

1 answer:

zaharov [31]2 years ago

5 0

The difference between sublimation and condensation is that

Sublimation is used to in a separation technique to separate sublime substance from a mixture while condensation is the

is the process where water vapor becomes liquid

<h3>What is separation technique?</h3>

This technique is used in chemistry to separate two or more constituents from a mixture

In conclusion, sublimation differs from condensation because sublimation is used to in a separation technique to separate sublime substance from a mixture while condensation is the

is the process where water vapor becomes liquid

Learn more about separation technique:

brainly.com/question/10862519

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When comparing the three atomic models provided, which statement best describes the diagram?

aliina [53]

C. Three isotopes of the same element because the number of electrons and protons remain the same but the number of neutrons in the nucleus are different. This means that the isotopes have different masses to each other but the same chemical properties.

6 0

3 years ago

A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

IrinaK [193]

Answer : The specific heat capacity of the alloy $1.422J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$C_1$ = specific heat of alloy = ?

$C_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of alloy = 21.6 g

$m_2$ = mass of water = 50.0 g

$T_f$ = final temperature of system = $31.10^oC$

$T_1$ = initial temperature of alloy = $93.00^oC$

$T_2$ = initial temperature of water = $22.00^oC$

Now put all the given values in the above formula, we get

$21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC$

$c_1=1.422J/g^oC$

Therefore, the specific heat capacity of the alloy $1.422J/g^oC$

6 0

2 years ago

Match the terms to the correct descriptions. Question 1 options:

faust18 [17]

1) Refraction
2)Reflection
3)Concave
4)Convex

I took the test and got this right so you can believe me :)
Hope this helps

4 0

3 years ago

Read 2 more answers

In order to be accepted, a scientific theory must be

bija089 [108]

Answer:

The answer is the 1st one

6 0

3 years ago

If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat

Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0

3 years ago