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Ilya [14]
2 years ago
6

A sample of carbon dioxide gas that has a volume of 14.6 L at STP contains how many

Chemistry
1 answer:
Nana76 [90]2 years ago
7 0

Answer:

n = 0.651 mol

Explanation:

Given data:

Volume of CO₂ = 14.6 L

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Number of moles of CO₂ = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

by putting values,

1 atm × 14.6 L = n × 0.0821 atm.L/ mol.K   × 273.15 K

14.6 L. atm =  n × 22.43atm.L/ mol

n = 0.651 mol

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b. Extraction with a solution of sodium hydrogen carbonate, separating the layers, followed by drying and evaporating the organic layer.

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Calculate the concentration of so42− ions in a 0.010 m aqueous solution of sulfuric acid. express your answer to four decimal pl
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<span>Answer: 0.00649M


The question is incomplete,
</span>

<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
</span>
<span>
With that you can solve the question following these steps"
</span>

<span>1) First ionization:
</span>
<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)


Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M


2) Second ionization
</span>

<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
</span>

<span>Do the mass balance:
</span>


<span><span>        HSO₄⁻ (aq)        H⁺        SO₄²⁻</span>
</span>
<span /><span /><span>        0.01 M  - x          x            x


</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
<span /><span>
=> Ka₂ = (x²) / (0.01 - x) = 0.012
</span><span />

<span>3) Solve the equation:


</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
<span /><span>
x² + 0.012x - 0.0012 = 0
</span><span />

<span>Using the quadratic formula: x = 0.00649
</span><span />

<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>

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