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Debora [2.8K]
3 years ago
6

A common neutralization reaction, that is used in titrations, involves sodium hydroxide, NaOH, reacting with nitric acid, HNO3.

What are the products of this reaction?
Chemistry
2 answers:
NeX [460]3 years ago
5 0

<u>Answer:</u> The products of the reaction are NaNO_3\text{ and }H_2O

<u>Explanation:</u>

Neutralization reaction is defined as the reaction when an acid reacts with a base to produce a salt and water molecule.

The chemical equation for the reaction of sodium hydroxide and nitric acid follows:

NaOH+HNO_3\rightarrow NaNO_3+H_2O

By Stoichiometry of the reaction:

1 mole of sodium hydroxide reacts with 1 mole of nitric acid to produce 1 mole of sodium nitrate and 1 mole of water.

Hence, the products of the reaction are NaNO_3\text{ and }H_2O

sertanlavr [38]3 years ago
4 0
The  common  neutralization  reaction  that  involve  NaOH  reacting   with  HNO3  produces

NaNO3     and  H2O

The  equation  for  reaction  is      as folows
NaOH  + HNO3  =  NaNO3  +  H2O 
that  is  1  mole  of  NaOH  reacted  with  1  mole  of  HNO3  to  form  1  mole  of  NaNO3  and  1  mole  of  H2O
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If an ice cube weighing 25.0 g with an initial
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Answer:

11

∘

C

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at

0

∘

C

to liquid at

0

∘

C

.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

q

1

+

q

2

=

−

q

3

(

1

)

, where

q

1

- the heat absorbed by the solid at

0

∘

C

q

2

- the heat absorbed by the liquid at

0

∘

C

q

3

- the heat lost by the warmer water sample

The two equations that you will use are

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of water, equal to

4.18

J

g

∘

C

Δ

T

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

and

q

=

n

⋅

Δ

H

fus

, where

q

- heat absorbed

n

- the number of moles of water

Δ

H

fus

- the molar heat of fusion of water, equal to

6.01 kJ/mol

Use water's molar mass to find how many moles of water you have in the

100.0-g

sample

100.0

g

⋅

1 mole H

2

O

18.015

g

=

5.551 moles H

2

O

So, how much heat is needed to allow the sample to go from solid at

0

∘

C

to liquid at

0

∘

C

?

q

1

=

5.551

moles

⋅

6.01

kJ

mole

=

33.36 kJ

This means that equation

(

1

)

becomes

33.36 kJ

+

q

2

=

−

q

3

The minus sign for

q

3

is used because heat lost carries a negative sign.

So, if

T

f

is the final temperature of the water, you can say that

33.36 kJ

+

m

sample

⋅

c

⋅

Δ

T

sample

=

−

m

water

⋅

c

⋅

Δ

T

water

More specifically, you have

33.36 kJ

+

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

0

)

∘

C

=

−

650

g

⋅

4.18

J

g

∘

C

⋅

(

T

f

−

25

)

∘

C

33.36 kJ

+

418 J

⋅

(

T

f

−

0

)

=

−

2717 J

⋅

(

T

f

−

25

)

Convert the joules to kilojoules to get

33.36

kJ

+

0.418

kJ

⋅

T

f

=

−

2.717

kJ

⋅

(

T

f

−

25

)

This is equivalent to

0.418

⋅

T

f

+

2.717

⋅

T

f

=

67.925

−

33.36

T

f

=

34.565

0.418

+

2.717

=

11.026

∘

C

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T

f

=

11

∘

C

Explanation:

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I would say diamond, because it consists only of carbon

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