The solubility of carbon dioxide at 400 kPa at room temperature is ;
( B ) 0.61 CO2/L
<u>Given data </u>
pressure of CO₂ = 400 Kpa = 3.95 atm
Kh of CO₂ = 3.3 * 10⁻² mol/L.atm
<h3>Calculate the solubility of carbon dioxide </h3>
Solubility = pressure * Kh value of CO₂
= 3.95 atm * 3.3 * 10⁻² mol / L.atm
= 0.13 mol/l CO₂
= 0.61 CO₂ / L
Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l CO₂.
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From the options the closest answer is ( B ) 0.61 CO₂ / L
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
A teacher uses a bow and arrow to demonstrate accuracy and precision. she shoots several arrows, aiming at the exact center of the target each time. thedrawing below shows where her arrows hit the target.
<span> The statement that best describes her shots is
</span><span>Her shots were neither accurate nor precise
</span>hope it helps
Answer:
Mercury has a density of 13.6g/mL
