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2 years ago

A) 24 g B) 36 g C) 48 g D) 60 g

Chemistry

1 answer:

Hitman42 [59]2 years ago

5 0

Answer:

I'm not sure if this right because i've only taken one year of Chemistry but the answer I got was

A. 24 g

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What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

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

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

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3 years ago

CAVA Chemistry 302/303B Unit 2 Lab Report

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2 years ago

An investigation into an unknown object in space resulted in the following observations:

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2 years ago

3. A 4.00 gram sample of solid gold was heated from 274K to 314K. How much energy was involved?

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Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).

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2 years ago

Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera

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Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

$Q=m*Ce*ΔT$

Q= heat (unknown)

m= mass (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature (2.75 °C)

Latent heat:

$ΔE=∝mΔHvap$

ΔE= latent heat

m= mass (unknown)

∝= mass fraction (unknown)

ΔHvap= enthalpy of vaporization (539.4 cal/g)

Since Q and E are equal, we can match both equations:

$m*Ce*ΔT=∝*m*ΔHvap$

Mass fraction is:

$∝=\frac{Ce*ΔT}{ΔHvap}$

$∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}$

∝=0,0051

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2 years ago