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
<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>
Explanation:
A)
We know, each mole contains atoms.
It is given that mass of one oxygen atom is m= .
Therefore, mass of one mole of oxygen, .
Putting value of n and ,
B)
Given,
Mass of water in glass=0.050 kg = 50 gm.
From above part mass of one mole of oxygen atoms = 16.0 gm.
Therefore, number of mole of oxygen equivalent to 50 gm oxygen
LEARN MORE :
Avogadro's number
brainly.com/question/12902286
Answer: high pressure because it's stable
Explanation:
Air mass is volume of air which has stable temperature, humidity and pressure horizontally. Over time, each air mass acquires properties of the region by residing over same part of a surface.
Areas of low pressure and high pressure occur where there is warm air and cold air respectively. An air mass usually forms over an area of high pressure. Warm air rises up and cold air takes its place. Warm air has low density and low pressure where as cold air has high density and pressure and therefore, sinks to the bottom. This is a stable condition. The movement of air mass is responsible for maintenance of temperature conditions on Earth.
Answer: 2.17 x 10^23 molecules
Explanation:
1mole of H2O contains 6.02x10^23 molecules.
Therefore 0.360mole of H2O will contain = 0.36 x 6.02x10^23 = 2.17 x 10^23 molecules
<span>they assign a numerical date to each rock layer studied.</span>