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3 years ago

IF 3.25 mol of argon gas occupies a volume of 100. L at a particular temperature and

Chemistry

1 answer:

Serga [27]3 years ago

3 0

Answer:

435.38 L

Explanation:

From the question given above, the following data were obtained:

Initial mole (n₁) = 3.25 mole

Initial volume (V₁) = 100 L

Final mole (n₂) = 14.15 mole

Final volume (V₂) =?

The final volume occupied by the gas can be obtained as follow:

V₁/n₁ = V₂/n₂

100 / 3.25 = V₂ / 14.15

Cross multiply

3.25 × V₂ = 100 × 14.15

3.25 × V₂ = 1415

Divide both side by 3.25

V₂ = 1415 / 3.25

V₂ = 435.38 L

Thus, the final volume of the gas is 435.38 L

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Substance is matter with a composition that is ?

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Answer:

A substance is a form of matter that will have a constant composition of evenly distributed products.

Explanation:

(Has to be the same everywhere)

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In an exothermic reaction, the average kinetic energy of the particles of matter ib the products _____

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Answer:

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Explanation:

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3 years ago

For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be prod

natima [27]

Answer:

(a)

Moles of ammonium chloride = 0.5 mole

Mass of ammonium chloride formed = 26.7455 g

(b)

Mole of $CS_2$ = 0.125 mole

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

Mole of $H_2S$ = 0.25 mole

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

Explanation:

(a)

For the first reaction:-

$NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}$

The mole ratio of the reactants = 1 : 1

0.5 moles of ammonia react with 0.5 moles of hydrochloric gas to give 0.5 moles of ammonium chloride

So, Moles of ammonium chloride formed = 0.5 moles

Molar mass of ammonium chloride = 53.491 g/mol

Mass = Moles * Molar mass = 0.5 * 53.491 g = 26.7455 g

(b)

For the first reaction:-

$CH_4_{(g)}+4S_{(s)}\rightarrow CS_2_{(l)}+2H_2S_{(g)}$

The mole ratio of the reactants = 1 : 4

It means

0.5 moles of methane react with 2.0 moles of sulfur to give 0.5 moles of Carbon disulfide and 1.0 moles of hydrogen sulfide gas.

But available moles of S = 0.5 moles

Limiting reagent is the one which is present in small amount. Thus, S is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

4 moles of S produces 1 mole of $CS_2$

Thus,

0.5 moles of S produces $\frac{1}{4}\times 0.5$ mole of $CS_2$

Mole of $CS_2$ = 0.125 mole

Molar mass of $CS_2$ = 76.139 g/mol

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

4 moles of S produces 2 moles of $H_2S$

Thus,

0.5 moles of S produces $\frac{2}{4}\times 0.5$ mole of $H_2S$

Mole of $H_2S$ = 0.25 mole

Molar mass of $H_2S$ = 34.1 g/mol

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

7 0

3 years ago

+ c) FeCl3 + NH4OH Fe(OH)3 NHACI

bixtya [17]

The question is incomplete, the complete question is:

Write the net ionic equation for the below chemical reaction:

(c): $FeCl_3+3NH_4OH\rightarrow Fe(OH)_3+3NH_4CI$

Answer: The net ionic equation is $Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)$

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

(c):

The balanced molecular equation is:

$FeCl_3(aq)+3NH_4OH(aq)\rightarrow Fe(OH)_3(s)+3NH_4Cl(aq)$

The complete ionic equation follows:

$Fe^{3+}(aq)+3Cl^-(aq)+3NH_4^+(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq)+3Cl^-(aq)$

As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)$

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3 years ago