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Ka (mol/L ) = <span>0.00002340</span>
V1 = 2.00 L
<span>T1 = 25 + 273 = 298 K </span>
<span>V2 = 6.00 L </span>
<span>T2 = ? </span>
<span>Assuming the pressure is to remain constant, then </span>
<span>V1/T1 = V2/T2 </span>
<span>T2 = T1V2/V1 = (298)(6)/(2) = 894 deg K</span>
Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get
Now we have to calculate the half-life.
Therefore, the half life of 28-Mg in hours is, 6.94
Answer:
Positive: a and b
Negative: c
Explanation:
The entropy (S) is the measure of the randomness of the system, and it intends to increase. The randomness can be determined by the energy of the molecules, their velocity and how distance they are between the other molecules.
When the entropy increases, ΔS is positive, when the entropy decreases, ΔS is negative. So, when gasoline mix with air in a car engine, the process intends to continue, the randomness increases and ΔS is positive. When hot air expands, the distance between the molecules increases, so ΔS is positive.
But, when humidity condenses, the molecules stay closer, so there's a decrease in the randomness, then ΔS is negative.
