<h3><u>Answer;</u></h3>
= 3032.15 kPa
<h3><u>Explanation;</u></h3>
Using the equation;
PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.
Volume = 7.5 L, T = 274 +273 = 547 K, N = 5 moles
Therefore;
Pressure = nRT/V
= (5 × 0.08206 × 547)/7.5 L
= 29.925 atm
But; 1 atm = 101325 pascals
Hence; Pressure = 3032150.63 pascals
<u>= 3032.15 kPa</u>
Answer:
E. 6.9 E-7 M
Explanation:
0.10M 0.10M 0.10M
eq: S S 2S + 0.1
∴ Ksp = 6.9 E-9 = [Mg2+][F-]² = (S)(2S+0.1)²
if we compared the concentration ( 0.1 M ) with the Ksp ( 6.9 E-9 ); then we can neglect the solubility as adding:
⇒ Ksp = 6.9 E-9 = (S)(0.1)² = 0.01S
⇒ S = 6.9 E-9 / 0.01 = 6.9 E-7 M
The boiling point increase of a solution is a colligative property, which means that it is related with the solvent and the concentration of the solute, as per this formula:
ΔT = i * kb * m
Where, ΔT is the increase in the boiling point, i is the van't Hoof factor (which accounts for the numberof particles that are dissolved), kb is the boiling point and m the molality of the solution.
Gvien the normal boiling point of 100°C for pure water, ΔT = 101.4 °C - 100.0 °C = 1.4 °C.
Kb = 0.512 °C / m
m = 1.2 m
Therefore, i = ΔT / (kb * m) = 1.4°C / (0.512 °C/m * 1.2m) = 2.28
Answer: 2.28
