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:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
Explanation:
We identify the reactants:
CuBr₂ and Pb(CH₃COO)₂
The products will be: Cu(CH₃COO)₂ and PbBr₂
You may know these information:
Salts from acetate are soluble.
Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺
PbBr₂ is formed, so this will be our precipitate
The equation is:
CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓
The reaction will produce solid copper and aluminium chloride salt.
Explanation:
Copper chloride (CuCl₂) in solution will react with aluminium to form solid cooper and aluminium chloride (AlCl₃).
3 CuCl₂ (aq) + 2 Al (s) → 3 Cu (s) + 2 AlCl₃ (aq)
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numerical problems with copper chloride and aluminium
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Answer:
The pressures will remain at the same value.
Explanation:
A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.
While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.
Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).
The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.
Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.
