Answer:
C) Q < K, reaction will make more products
Explanation:
- 1/8 S8(s) + 3 F2(g) ↔ SF6(g)
∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³
∴ Q = [ SF6 ] / [ F2 ]³
∴ [ SF6 ] = 2 mol/L
∴ [ F2 ] = 2 mol/L
⇒ Q = ( 2 ) / ( 2³)
⇒ Q = 0.25
⇒ Q < K, reaction will make more products
Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
They have a mass for the particles
There are no totally elastic collisions
There are intermolecular forces
They are eaten by Tertiary consumers
