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olga nikolaevna [1]
3 years ago
6

You are on an alien planet where the names for substances and the units of measures are very unfamiliar. Nonetheless, you obtain

40 quibs of a substance called skvarnick. You can trade this skvarnick for gold coins, but the vendors all measure skvarnick in units of sleps; not quibs. 15 quibs is equal to 13 sleps. If you have 40 quibs of skvarnick, how many sleps do you have? Round your answer to the nearest tenth (one decimal place). Type only the number; not the number and unit. Your Answer:
Chemistry
1 answer:
Kaylis [27]3 years ago
3 0
The lightbulb contains 40 hours.
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What is the value of the activation energy of the uncatalyzed reaction? Express your answer to three significant figures and inc
goblinko [34]

Answer:

175 kJ

Explanation:

Activation energy can be defined as the potential energy that is needed to change reactants to products. This is the minimum energy required for the chemical reaction to take place. Thus, using the given figure:

Activation energy = activation complex - reactant energy

In the given figure, activation complex = 400 kJ

reactant energy = 225 kJ

Therefore:

Activation energy = 400 - 225 = 175 kJ

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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

6 0
3 years ago
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