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anzhelika [568]
3 years ago
11

What is the observation of heating copper carbonate???.?

Chemistry
1 answer:
damaskus [11]3 years ago
7 0

Answer:

1)This type of reaction is called thermal decomposition . Copper carbonate is green and copper oxide is black. You can see a colour change from green to black during the reaction. The carbon dioxide produced can be detected using limewater, which turns milky. 2) The extremely low pressure of the partial vacuum inside the can made it possible for the pressure of the air outside the can to crush it. A can is crushed when the pressure outside is greater than the pressure inside, and the pressure difference is greater than the can is able to withstand.

Explanation:

hope this helps you :)

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What is the difference between a scientist theory and a scientific law
Andru [333]

Answer:

A scientific theory is a widely accepted belief on why something happens in the natural world while a scientific law is proven to a fact that shows what happens.

Explanation:

3 0
3 years ago
How much thermal energy is added to 10.0 g of ice at −20.0°C to convert it to water vapor at 120.0°C?
Sonbull [250]

Answer:

7479 cal.

31262.2 joules

Explanation:

This is a calorimetry problem where water in its three states changes from ice to vapor.

We must use, the calorimetry formula and the formula for latent heat.

Q = m . C . ΔT

Q = Clat . m

First of all, let's determine the heat for ice, before it melts.

10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal

Now, the ice has melted.

Q = Clat heat of fusion . 10 g

Q = 79.7 cal/g . 10 g → 797 cal

We have water  at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.

Q = m . C . ΔT

Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal

Water is ready now, to become vapor so let's determine the heat.

Q = Clat heat of vaporization . m

Q = 539.4 cal/g . 10 g → 5394 cal

Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°

Q = m . C . ΔT

Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal

Now, we have to sum all the heat that was added in all the process.

100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.

We can convert this unit to joules, which is more acceptable for energy terms.

1 cal is 4.18 Joules.

Then, 7479 cal are (7479 . 4.18) = 31262.2 joules

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Answer:

Explanation:

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