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Liono4ka [1.6K]
3 years ago
11

One part nitrogen gas combines with one part oxygen gas to form how many part(s) dinitrogen monoxide (nitric oxide)?

Chemistry
1 answer:
____ [38]3 years ago
7 0

Answer : The one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

When nitrogen gas combines with oxygen gas then it react to give dinitrogen monoxide or nitrous oxide.

The balance chemical reaction will be:

2N_2(g)+O_2(g)\rightarrow 2N_2O(g)

By the stoichiometry we can say that, 2 parts of nitrogen gas combines with 1 part of oxygen gas to give 2 parts of dinitrogen monoxide or nitrous oxide.

First we have to determine the limiting reagent.

From the reaction we conclude that,

As, 2 moles of nitrogen gas combine with 1 mole of oxygen gas

So, 1 moles of nitrogen gas combine with 0.5 mole of oxygen gas

It means that, oxygen gas is an excess reagent because the given moles are greater than the required moles and nitrogen gas is a limiting reagent and it limits the formation of product.

Now we have to determine the moles of dinitrogen monoxide.

As, 2 moles of nitrogen gas combine to give 2 mole of dinitrogen monoxide

So, 1 mole of nitrogen gas combine to give 1 mole of dinitrogen monoxide

Thus, the one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

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Many industrial reactions, like the reaction of nitrogen gas (N2) and hydrogen gas (H2) to produce ammonia for fertilizers, have
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How many moles are in 297g of nh3?<br><br> Please show work, will give brainliest.
rusak2 [61]

Answer:

17.5moles

Explanation:

The number of moles in a substance can be calculated by using the formula;

Number of moles (n) = mass (m) ÷ molar mass (MM)

According to this question, mass of ammonia (NH3) = 297g

Molar Mass of NH3 = 14 + 1(3)

= 17g/mol

n = 297/17

n = 17.47

Number of moles of NH3 = 17.5moles

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3 years ago
Which statements true about the total mass of the reactants during a chemical change? (5 points)
Lapatulllka [165]
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5 0
3 years ago
A hypothetical element consists of two isotopes of masses 86.95amu and 88.95amu with abundances of 35.5% and 64.5% respectively.
Art [367]

Answer: The average atomic mass of the element = 88.242amu

Explanation:

The abundance of the first isotope is =35.5%

 Atomic mass of first isotope = 68.9257

The average atomic mass of the first isotope =86.95amu X 35.5%  =86.95amu X 0.355 =30.8725 amu

The abundance of the second isotope =64.5%

Atomic mass of the second isotope =88.95amu

The average atomic mass of second isotope =88.95amu x 64.5% = 88.95amu x 0.645= 57.37275 amu

Now the average atomic mass =30.8725 +57.37275 = 88.242amu

OR using the formulae

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

{(86.95amu X 35.5 )+(88.95amu x 64.5)}/100

8,824/100

=88.24amu

5 0
3 years ago
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