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Deffense [45]
3 years ago
10

Calculate the concentration of hi when the equilibrium constant is 1x10^5

Chemistry
1 answer:
exis [7]3 years ago
5 0
Answer is: concentration of hydrogen iodide is 6 M.

Balanced chemical reaction: H₂(g) + I₂(g) ⇄ 2HI(g).
[H₂] = 0.04 M; equilibrium concentration of hydrogen.
[I₂] = 0.009 M; equilibrium concentration of iodine.
Keq = 1·10⁵.
Keq = [HI]² / [H₂]·[I₂].
[HI]² = [H₂]·[I₂]·Keq.
[HI]² = 0.04 M · 0.009 M · 1·10⁵.
[HI]² = 36 M².
[HI] = √36 M².
[HI] = 6 M.
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E is solution of HCI of unk concentration f contain 4.8g of NaOH in 25cm point of salution​
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Please give me the answer please
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Answer:

A. 30cm³

Explanation:

Based on the chemical reaction:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>

<em />

To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:

<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>

1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles

<em>Moles HCl:</em>

50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles

For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:

2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.

The moles produced of CO₂ are:

2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂

Using PV = nRT

<em>Where P is pressure = 1atm assuming STP</em>

<em>V volume in L</em>

<em>n moles = 1.25x10⁻³ moles CO₂</em>

<em>R gas constant = 0.082atmL/molK</em>

<em>T = 273.15K at STP</em>

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V = nRT / P

1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V

0.028L = V

28cm³ = V

As 28cm³ ≈ 30cm³

Right option is:

<h3>A. 30cm³</h3>

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