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2 years ago

13

Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "

catch" all of the butane (from a lighter) at room conditions, if you released all of the butane from the lighter.

1 answer:

Hatshy [7]2 years ago

8 0

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

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When the reaction equation is:

CaSO3(s) → CaO(s) + SO2(g)

we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

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when P is the pressure = 1.1 atm

and V is the volume = 14.5 L

n is the moles' number (which we need to calculate)

R ideal gas constant = 0.0821

and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

1.1 * 14.5 L = n * 0.0821 * 285.5

∴ n = 1.1 * 14.5 / (0.0821*285.5)

= 0.68 moles SO2

∴ moles CaSO3 = 0.68 moles

so we can easily get the mass of CaSO3:

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<u>Answer:</u> The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

<u>Explanation:</u>

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To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

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Putting values in above equation, we get:

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To calculate the number of moles, we use the equation:

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Putting values in above equation, we get:

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Molarity of methylamine solution = 0.017 M

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As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$

So, $1.7\times 10^{-4}mol$ of methyl amine will produce = $\frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+$

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$pK_b=9.41$

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Putting values in above equation, we get:

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