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siniylev [52]
1 year ago
6

the pressure of a sample of argon gas was increased from 2.32 atm2.32 atm to 7.16 atm7.16 atm at constant temperature. if the fi

nal volume of the argon sample was 18.3 l,18.3 l, what was the initial volume of the argon sample? assume ideal behavior.
Chemistry
1 answer:
SOVA2 [1]1 year ago
6 0

initial volume of the argon sample = 5.93L according to Boyle's law

What is Boyle's law ?

Boyle's law, also known as Mariotte's law, is a relationship describing how a gas will compress and expand at a constant temperature. The pressure (p) of a given quantity of gas changes inversely with its volume (v) at constant temperature, according to this empirical connection, which was established by the physicist Robert Boyle in 1662. In equation form, this means that pv = k, a constant.

According to Boyle's law

P1/V1 = P2/V2

P1 = initial pressure

P2 = final pressure

V1 =initial volume

V2= final volume

V1 = P1*V2/P2

V1 = 2.32*18.3/7.16 = 5.93L

initial volume of the argon sample = 5.93L according to Boyle's law

To know about Boyle's law from the link

brainly.com/question/26040104

#SPJ4

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A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. a 100.0 mL volume of the so
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Answer : The molar mass of unknown substance is, 39.7 g/mol

Explanation : Given,

Mass of unknown substance = 9.56 g

Volume of solution = 100.0 mL

Molarity = 2.41 M

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{Mass of unknown substance}\times 1000}{\text{Molar mass of unknown substance}\times \text{Volume of solution (in mL)}}

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

2.41M=\frac{9.56g\times 1000}{\text{Molar mass of unknown substance}\times 100.0mL}

\text{Molar mass of unknown substance}=39.7g/mol

Therefore, the molar mass of unknown substance is, 39.7 g/mol

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In which location would you expect to find the greatest species diversity?!
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2. Fe(s) + O2(g) → Fe3O4(s) a. When 13.54 g of O2 is mixed with 12.21 g of Fe, which is the limiting reactant? b. What mass in g
julsineya [31]

The percent yield shows the extent to which the reactants are converetd into products . The limiting reactant is used up in the reaction.

<h3>What is a limiting reactant?</h3>

A limiting reactant is the reactant that is in the least amount in the system.  Now;

Number of moles of Fe =  12.21 g/56 g/mol = 0.22 moles

Number of moles of O2 = 13.54 g/32 g/mol = 0.42 moles

Balanced reaction equation;

3Fe(s) + 2 O2(g) = Fe3O4(s)

If 3 moles of Fe reacts with 2 mole of O2

0.22 moles of Fe reacts with  0.22 moles * 2 mole/3 moles = 0.15 moles

Hence, Fe is the limiting reactant

If 3 mole of Fe produces 1 mole of  Fe3O4(s)

 0.22 moles  of O2   produces   0.22 moles * 1 mole/3 moles of  Fe3O4(s) = 0.1073 moles

Mass of  Fe3O4(s) =0.1073 moles  * 232 g/mol =16.9 g

Number of moles of excess reactant = 0.42 moles - 0.15 moles = 0.27 moles

Mass of excess reactant = 0.27 moles * 32 g/mol = 8.64 g

percent yield = 15.88 g/16.9 g * 100/1

= 93.4%

Learn more about percent yiled: brainly.com/question/13463225

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