<em>K</em> = 2.4 × 10^(-72)
<em>Step 1</em>. Determine the <em>value of n
</em>
Zn^(2+) + 2e^(-) → Zn
2Cl^(-) → Cl_2 + 2e^(-)
Zn^(2+) + 2Cl^(-) → Zn + Cl_2
∴ <em>n</em> = 2
<em>Step 2</em>. Calculate <em>K</em>
log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62
<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)
Volume = a x a x a
V = 2 cm x 3 cm x 4 cm => 24 cm³
Density = 19.3 g/cm³
Mass = ?
Therefore:
m = D x V
m = 19.3 x 24
m = 463.2 g
Answer:
The force of gravity acting on the car is <u>9800 N vertically downward.</u>
Explanation:
Given:
Mass of the car given is 1000 kg.
We know that the force of gravity is the force applied by the center of Earth on any body. The force of gravity is also called the weight of the body and always act towards the center of the Earth.
From Newton's second law, we know that the force acting on a body is equal to its mass and acceleration.
Here, the acceleration acting on the car is due to gravity and thus has a constant value of 9.8 m/s² on the surface of Earth.
Therefore, the force of gravity acting on the car is given using the Newton's second law as:
Force of gravity = Mass of car 
Acceleration due to gravity.
Force of gravity = (1000 kg) (9.8 m/s²)
Force of gravity = 9800 N [1 kg.m/s² = 1 N]
Therefore, the force of gravity acting on the car is 9800 N vertically downward.
b) law of conservation of energy
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
