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mote1985 [20]
3 years ago
11

The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?

Chemistry
1 answer:
Firlakuza [10]3 years ago
4 0
Answer is: <span>the empirical formula of enargite is Cu</span>₃AsS₄.<span>
If we use 100 grams of enargite:
n(Cu) = 48,41 g </span>÷ 63.55 g/mol.
n(Cu) = 0.761 mol.
n(As) = 19.02 g ÷ 74.92 g/mol.
n(As) = 0.254 mol.
n(S) = 32.57 g ÷ 32.065 g/mol.
n(S) = 1.016 mol.
n(Cu) : n(As) : n(S) = 0.761 mol : 0.254 mol : 1.016 mol / 0.254.
n(Cu) : n(As) : n(S) = 3 mol : 1 mol : 4 mol.
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Murrr4er [49]

The experimental absolute zero value is less when compared to the accepted value of absolute zero.

<h3>What is absolute zero?</h3>

Absolute zero is defined as the temperature in which the lowest energy possible is attained in a thermodynamic system.

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At absolute zero, it is assumed that the volume of an ideal gas becomes zero. However, it has not been possible to cool any gas to absolute zero.

Based on the graph of temperature against volume of gases, the experimental absolute zero extrapolated from the graph where volume of the gases becomes zero is -285 degrees Celsius.

Therefore, the experimental absolute zero value is less when compared to the accepted value.

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11. What is the specific heat of a substance with a mass of 25.5 g that requires 412 J
Romashka-Z-Leto [24]

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of

1 g

of a given substance by

1

∘

C

.

In your case, aluminium is said to have a specific heat of

0.90

J

g

∘

C

.

So, what does that tell you?

In order to increase the temperature of

1 g

of aluminium by

1

∘

C

, you need to provide it with

0.90 J

of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by

1

∘

C

. So if you wanted to increase the temperature of

10.0 g

of aluminium by

1

∘

C

, you'd have to provide it with

1 gram



0.90 J

+

1 gram



0.90 J

+

...

+

1 gram



0.90 J



10 times

=

10

×

0.90 J

However, you don't want to increase the temperature of the sample by

1

∘

C

, you want to increase it by

Δ

T

=

55

∘

C

−

22

∘

C

=

33

∘

C

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

1

∘

C



10

×

0.90 J

+

1

∘

C



10

×

0.90 J

+

...

+

1

∘

C



10

×

0.90 J



33 times

=

33

×

10

×

0.90 J

Therefore, the total amount of heat needed to increase the temperature of

10.0 g

of aluminium by

33

∘

C

will be

q

=

10.0

g

⋅

0.90

J

g

∘

C

⋅

33

∘

C

q

=

297 J

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

q

=

m

⋅

c

⋅

Δ

T

, where

q

- the amount of heat added / removed

m

- the mass of the substance

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

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