I don’t really know but hope you figure it out
Answer:
4.1 atm = 3,116 mmHg = 415.4 kPa
Explanation:
According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:
P₁ x V₁= P₂ x V₂
Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.
From the problem, we have:
V₁= 50.0 L
V₂= 68.0 L
P₂= 3.0 atm
Thus, we calculate the initial pressure as follows:
P₁= (P₂ x V₂)/V₁= (3.0 atm x 68.0 L)/(50.0 L)= 4.08 atm ≅ 4.1 atm
To transform to mmHg, we know that 1 atm= 760 mmHg:
4.1 atm x 760 mmHg/1 atm = 3,116 mmHg
To transform to kPa we use: 1 atm= 101.325 kPa
4.1 atm x 101.325 kPa = 415.4 kPa
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
