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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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Answer:

Decomposition

Explanation:

A decomposition reaction is a type of reaction in which a compound is broken down into its constituent elements sometimes under the influence of heat.

When  iron (III) hydroxide is heated,new products are formed according to the equation; 2Fe(OH)3 -----------> Fe2O3 + 3H2O.

This is a thermal decomposition reaction.

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What is the molarity of a solution that contains 17 g of nh3 in 0.50 l of solution?
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Answer:

  • 2.0 M

Explanation:

<u>1) Data:</u>

a) M = ?

b) mass of solue = 17 g

c) solute: NH₃

d) V = 0.5o liter

<u>2) Formulae:</u>

a) number of moles, n = mass in grams / molar mass

b) M = n / V (in liters)

<u>3) Solution</u>

a) Molar mass of NH₃ = 17.03 g/mol

b) n = mass in grams / molar mass = 17 g / 17.03 g/mol = 0.998 mol NH₃

c) M = n / V (in liters) = 0.998 mol / 0.50 liter = 1.996 M

d) Round to the appropiate number of significant figures, 2: 2.0 M.

Answer: 2.0 M

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If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s
AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)

Moles of mercury(II) acetate = \frac{45.101 g}{318.70 g/mol}=0.14152 mol

Moles of sodium dichromate = \frac{12.026 g}{261.97 g/mol}=0.045906 mol

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

\frac{1}{1}\times 0.045906 mol=0.045906 mol of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

\frac{1}{1}\times 0.045906 mol=0.045906 mol of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

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