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stich3 [128]
3 years ago
11

How many moles of nitric acid are present in 35.0 ml of a 2.20 M solution?

Chemistry
1 answer:
vodomira [7]3 years ago
8 0

Answer:

There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

Explanation:

Molarity of the solution = 2.20 M

Molarity=\frac{number\:of\:moles}{Volume\:of\:Solution\:in\:L}\\\\Number\:of\:moles=Molarity\times(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35\times10^{-3}L\\\\Number\:of\:moles=2.20\times35\times10^{-3}=77\times10^{-3}\:moles\:of\:HNO_{3}

Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

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The temperature of a sample of water changes from 10°C to 20°C when the water absorbs 100 calories of heat. What is the mass of
Vlad1618 [11]

Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be  

10 g

.

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of  

1 g

of that substance by  

1

∘

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.

Water has a specific heat of approximately  

4.18

J

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∘

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. This tells you that in order to increase the temperature of  

1 g

of water by  

1

∘

C

, you need to provide  

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of  

1 g

of water by  

10

∘

C

?

Well, you'd need  

4.18 J

to increase it by  

1

∘

C

, another  

4.18 J

to increase it by another  

1

∘

C

, and so on. This means that you'd need

4.18 J

×

10

=

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

.

Now look at the value given to you. If you need  

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

, what mass of water would require  

10

times as much heat to increase its temperature by  

10

∘

C

?

1 g

×

10

=

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

q

=

m

⋅

c

⋅

Δ

T

 

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

418

J

=

m

⋅

4.18

J

g

∘

C

⋅

(

20

−

10

)

∘

C

m

=

418

4.18

⋅

10

=

10 g

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