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madreJ [45]
3 years ago
11

I NEED HELP PLEASE, THANKS SO MUCH! :)

Chemistry
1 answer:
Anon25 [30]3 years ago
7 0

Answer:

Here's what I get  

Explanation:

The molecules of a substance attract each other.

They are also in constant motion.

Their kinetic energy increases with the temperature.

1. Solids

If the temperature is low enough, the molecules will have little kinetic energy.

They will not be able to escape the attractions of their neighbours, so they will be fixed in place in a crystalline array.

All they can do is vibrate about their position in the array.

The substance will be a solid.

2. Liquids

If the temperature is higher, the molecules will have enough kinetic energy to slide past each other, but not enough to escape the attractions of nearby molecules.

The substance will be a liquid.

3. Gases

If the temperature is high enough, the molecules will have enough kinetic energy to escape from the surface and fly off into space.

The molecules will on average so far apart that they won't feel the attractions of others.

When they do collide, they will have energy to bounce off each other rather than stick together.

The substance will be a gas.

 

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Explanation:

When a salt is dissolved , it increases  the boiling point . Increase in boiling point depends upon number of ions . So it is a colligative property .

.19 m AgNO₃ . Each molecule  will ionize into two ions . So effective molar concentration is 0.19 x 2 = .38 m

0.17 m CrSO4.Each molecule  will ionize into two ions . So effective molar concentration is 0.17 x 2 = .34 m

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Biacetyl, the flavoring that makes margarine taste "just like butter," is extremely stable at room temperature, but at 200°C it
Sholpan [36]

Answer:

3.91 minutes

Explanation:

Given that:

Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;

As we known that the half-life for first order is:

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k = constant

The formula can be re-written as:

k = \frac{0.693}{t__{1/2}}

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Also, the amount of butter flavor retained at 200°C (N_t)= 74%

The rate constant k = 0.077 min^{-1}

To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:

\frac{N_t}{N_0}= -kt

t = - (\frac{1}{k}*In\frac{N_t}{N_0}  )

Substituting our values; we have:

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