POH = - log [ OH⁻ ]
pOH = - log [ 1 x 10⁻⁹ ]
pOH = 9
Answer C
hope this helps!
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>
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Option B
dumbbell is the shape of 3p atomic orbital
<u>Explanation:</u>
Atomic orbitals are three-dimensional places inside an atom where there is a tremendous chance of detecting electrons. The p orbital, which develops in complexity and ought 2 spaces encompassing the atom core, or is defined as possessing a dumbbell pattern. The 3p atomic orbital is at energy level 3, as seen by the number 3 filed ere the character.
These orbitals have identical appearances but are arranged asymmetrically in location. p orbitals are wavefunctions with l=1. They ought an angular frequency that is ununiform at each angle. They have an appearance that is much defined as a "dumbbell".
Answer:
19,700 Joules
Explanation:
Quantity of heat (Q) = mc(T2-T1)
m (mass of ethanol) = 183g
c (specific heat capacity of ethanol) = 2.44J/g°C
T2 (boiling temperature of ethanol) = 78°C
T1 (initial temperature of ethanol) = 33.9°C
Q = 183×2.44(78 - 33.9) = 183×2.44×44.1 = 19691.532 = 19,700 Joules (to three significant digits)
