Answer:
d) V = 91.3 L
Explanation:
Given data:
Volume of nitrogen = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Number of atoms of nitrogen = 2.454×10²⁴ atoms
Solution:
First of all we will calculate the number of moles of nitrogen by using Avogadro number.
1 mole = 6.022×10²³ atoms
2.454×10²⁴ atoms × 1 mol / 6.022×10²³ atoms
0.407×10¹ mol
4.07 mol
Volume of nitrogen:
PV = nRT
1 atm × V = 4.07 mol ×0.0821 atm.L /mol.K ×273.15 K
V = 91.3 atm.L /1 atm
V = 91.3 L
Answer:
3.43×10¹ mol
Explanation:
Given data:
Initial number of moles = 12.4 mol
Initial volume = 122.8 L
Final number of moles = ?
Final volume = 339.2 L
Solution:
The number of moles and volume are directly proportional to each other at same temperature and pressure.
V₁/n₁ = V₂/n₂
122.8 L/ 12.4 mol = 339.2 L / n₂
n₂ = 339.2 L× 12.4 mol / 122.8 L
n₂ = 4206.08 L.mol /122.8 L
n₂ = 34.3mol
In scientific notation:
3.43×10¹ mol
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
Answer:
pH = 3.02
Explanation:
Acetic Acid is a weak acid (HOAc) that ionizes only ~1.5% as follows:
HOAc ⇄ H⁺ + OAc⁻.
In pure water the hydronium ion concentration [H⁺] equals the acetate ion concentration [OAc⁻] and can be determined* using the formula [H⁺] = [OAc⁻] = SqrRt(Ka·[acid]) = SqrRt(1.8x10⁻⁵ x 0.0500)M = 9.5x10⁻⁴M.
By definition, pH = -log[H⁺] = -log(9.5x10⁻⁴) = 3.02
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*This formula can be used to determine the [H⁺] & [Anion⁻] concentrations for any weak acid in pure water given its Ka-value and the molar concentration of acid in solution.
