Answer
solubility product = 3.18x 10^-7
Explanation:
We were given the pressure in torr then we need to convert to atm for consistency, ten we have
21torr/760= 0.0276315789 atm
21 Torr = .0276315789 atm
P = i M S T
M = P / iRT
Where p is osmotic pressure
T= temperature= 25C+ 273= 298K
for XY vanthoff factor i = 2
S = 0.0821 L-atm / mol K
M = .0276315789 atm / (2)(0.0821 L atm / K mole)(298 K)
M = 0.000564698046 mol/liters
solubility= 0.000564698046 mol/liters
Ksp = [X+][Y-]
Ksp = X^2
Ksp = [Sr^+2] * [SO4^-2]
Ksp = X^2
Ksp = (0.000564698046)^2
Ksp = 3.18883883 × 10-7
Ksp = 3.18x 10^-7
solubility product = 3.18x 10^-7
Therefore, the solubility product of this salt at 25 ∘C∘C is 3.18x 10^-7
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Answer:
The Earth, Stars, and Mars I guess...
Answer:
b Different amounts of food samples were used.
Explanation:
The mass of the two samples needs to be the same in order for the test to be accurate.
Mass = molarity x molar mass( NaNO₃) x volume
mass = 1.50 x 85.00 x 4.50
mass = 573.75 g of NaNO₃
hope this helps!
