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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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You conduct an experiment that requires the creation of an ammonia solution. You do this by reacting 50.0 L of nitrogen gas with
ozzi

Answer : The molarity of the resulting ammonia solution is, 0.89 M

Explanation :

The balanced chemical reaction is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

First we have to calculate the moles of nitrogen gas.

As we know that at STP, 1 mole of gas occupies 22.4 L volume of gas.

As, 22.4 L volume of nitrogen gas present in 1 moles of nitrogen gas

So, 50.0 L volume of nitrogen gas present in \frac{50.0}{22.4}=2.23 moles of nitrogen gas

Thus, the moles of nitrogen gas is 2.23 moles.

Now we have to calculate the moles of ammonia gas.

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 2.23 moles of N_2 react to give 2.23\times 2=4.46 moles of NH_3

Now we have to calculate the molarity of the resulting ammonia solution.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of ammonia}}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{4.46mole}{5.0L}

\text{Molarity}=0.89M

Therefore, the molarity of the resulting ammonia solution is, 0.89 M

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3 years ago
Please help! I'm struggling!​
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the answer is 1 or A

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How many electrons are in a neutral atom of iodine-131? 16)
eduard
C) 53

Hope this helps! :)
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3 years ago
Thanks for the help :)
Andre45 [30]
Welcome friend '~'        happy to help
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4 years ago
The two factors that are most important in determining the density of air are:
Pie

Answer:

c. temperature and water vapor content.

Explanation:

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3 years ago
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