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Nady [450]
3 years ago
15

-Material A has a melting point of 34°C whereas Material B has a melting point of 56°C. Both materials have been placed in an ov

en at 50°C Describe the state of each material after being being in the oven for a period of time.
I'm confused, if material B is on for the same amount of time does that mean that it has liquified and material A has melted slightly?
Chemistry
1 answer:
attashe74 [19]3 years ago
6 0

Answer:

Explanation:

Material A has a melting point of 34°C

Material B has a melting point of 56°C

Both materials, lets say a metal have been subjected to a temperature of 50°C

After a period of time, both of them would have melted to their liquid state.

The atoms of the solids would vibrates and the bonds would begin to break to form melt.

Material A would be the first to melt as it has a lower melting point. The lower the melting point, the faster and quicker it would reach its melting temperature.

Material B would need to accumulate more heat and its temperature would continue to rise for the phase change to occur. As it reaches the 50°C mark, the bonds are set free and a melt forms.

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Liquid A and liquid B form a solution that behaves ideally according to Raoult's law. The vapor pressures of the pure substances
Rama09 [41]

Answer:

Vapor pressure of solution → 151.1 Torr

Option 2.

Explanation:

Raoult's Law is relationed to colligative property about vapor pressure. A determined solute, can make, the vapor pressure of solution decreases.

ΔP = P° . Xm

where Xm is the mole fraction of solute, P° (vapor pressure of pure solvent)

and ΔP = Vapor pressure of pure solvent - Vapor pressure of solution.

In order to determine the vapor pressure of solution, we need to determine, the vapor pressure of B and A in the solution

B's pressure = P° B . Xm

When we add A to B, A works as the solute and B, as the solvent.

Vapor pressure of pure B is 135 torr. (P° B)

In order to determine, the Xm, we use the moles of A and B

Xm = 5.3 mol of B / (1.28 + 5.3) → 0.806

B's pressure = 135 Torr . 0.806 → 108.81 Torr

If mole fraction of B is 0.806, mole fraction for A (solute) will be (1 - 0.806)

A's pressure = 218 Torr . 0.194 → 42.3 Torr

Vapor pressure of solution is sum of vapor pressures of solute + solvent.

Vapor pressure of solution = 42.3 Torr + 108.81 Torr → 151.1 Torr

6 0
3 years ago
2. Based on what you know about waves and light, do you think that light can be
ss7ja [257]

Answer:

Light as a wave: Light can be described (modeled) as an electromagnetic wave. In this model, a changing electric field creates a changing magnetic field. This changing magnetic field then creates a changing electric field and BOOM - you have light. ... So, Maxwell's equations do say that light is a wave.

Explanation:

Hope this helps

6 0
3 years ago
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Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta
Vilka [71]

Answer:

147.2g

Explanation:

The full solution can be found in the image attached. Graham's law was applied to the problem. The rate of diffusion of a gas is inversely proportional to its molar mass or vapour density. Molar mass= 2vapour density

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3 years ago
What is the frequency of a wave with a wavelength of 6.40x10^4 meters?
yKpoI14uk [10]

Answer:

frequency = 0.47×10⁴ Hz

Explanation:

Given data:

Wavelength of wave = 6.4× 10⁴ m

Frequency of wave = ?

Solution:

Formula:

Speed of wave = wavelength × frequency

Speed of wave = 3 × 10⁸ m/s

Now we will put the values in formula.

3 × 10⁸ m/s = 6.4× 10⁴ m × frequency

frequency = 3 × 10⁸ m/s / 6.4× 10⁴ m

frequency = 0.47×10⁴ /s

s⁻¹ = Hz

frequency = 0.47×10⁴ Hz

Thus the wave with wavelength of 6.4× 10⁴ m have 0.47×10⁴ Hz frequency.

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