Answer:
3.1°C
Explanation:
Using freezing point depression expression:
ΔT = Kf×m×i
<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>
Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:
9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m
Replacing in freezing point depression formula:
ΔT = 5.12°cm⁻¹×0.472m×1
ΔT = 2.4°C
As freezing point of benzene is 5.5°C, the new freezing point of the solution is:
5.5°C - 2.4°C =
<h3>3.1°C</h3>
<em />
Answer:
The presion is 0.6 atm
Explanation:
P1V1=P2V2
P2 = P1V1/V2
P2 = (4.00 atm * 0.30 L) / 2.0 L
P2= 0.6 atm
Answer:
D
Explanation:
that should be the answer
Explanation:
Characteristic of matter that is not associated with its change in chemical composition.
Ca(OH)₂: strong base
pOH = a . M
a = valence ( amount of OH⁻)
M = concentration
Ca(OH)₂ ⇒ Ca²⁺ + 2OH⁻ (2 valence)
so:
pOH = 2 x 0.005
pOH = 0.01
pH = 14 - 0.01 = 13.99
