T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>
<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>
<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>
<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>
<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>
<span>HL = 14.4 s / 2 = 7.2 seconds </span>
Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
This reaction is decomposition. It is the breakdown of a compound into simpler and smaller elements.
Answer:
A
Explanation:
The Charles law states that the volume of an ideal gas increases when temperature is increased under constant pressure. The pressure inside the balloon is always equal to the atmospheric pressure. Therefore answer A demonstrate the Charles law.