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solniwko [45]
3 years ago
12

1. A light bulb is filled with 3.54 g of argon at 25 C and 725 mmHg. What is

Chemistry
1 answer:
Ugo [173]3 years ago
3 0

Answer: The volume of given gas is 2.27 L.

Explanation:

Given: Mass = 3.54 g

Temperature = 25^{o}C = (25 + 273) K = 298 K

Pressure = 725 mm Hg (1 mm Hg = 0.00131579) = 0.95 atm

As moles is the mass of substance divided by its molar mass.

So, moles of argon (molar mass = 40 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{3.54 g}{40 g/mol}\\= 0.0885 mol

Formula used to calculate the volume of given gas is as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times V = 0.0885 mol \times 0.0821 L atm/mol K \times 298 K\\V = 2.27 L

Thus, we can conclude that the volume of given gas is 2.27 L.

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As a result of losing electrons in the process, a reactant oxidizes. When a reactant obtains electrons during a reaction, reduction takes place. This frequently happens when acid and metals react.

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❃ The following points should be kept in mind to write and balance a chemical equation :

Step 1 : Write the molecular formula of all the reactants and products correctly.

Step 2 : Separate reactants and products by a sign of arrow. If reactants or products are more than one, connect them by a sign of a plus.

Step 3 : Balance the atoms of O and H at last [ The atoms used at many places in an equation should be balanced at last ]. For balancing , the number should be added as coefficient i.e in the front of the molecules.

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\large{ \tt{❁ \: LET'S \: GET \: STARTED}} :

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Step 2 : CS₂ + O₂ ⟶ CO₂ + SO₂

Step 3 : In the reactant side , there is two ' S ' but on the other side , there is one ' S '. So , add 2 as a coefficient before S on the product side. Now , There are two ' O ' in the reactant side but six ' O ' in the product side. So , add 3 as a coefficient before O on the reactant side. Now , there are equal atom of C , S and O on both sides

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Step 2 : Ag + HNO₃ ⟶ Ag ( NO₃ ) + NO₂ + H₂O

Step 3 : In the reactant side , There is one ' H ' but on the other side , there are two ' H '. Now add 2 before H on the reactant side. There are equal atom of ' Ag ' , ' H ' , ' N ' , and ' O '.

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- The last step is a bit more confusing I guess. So , which balancing , count the atoms in following ways :

  • The number written at the right lower corner of an atom is counted for that atom only. For example : In MgSO₄ , there are one ' Mg ' , one ' S ' and four ' O '

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- Hope this helps! :)

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