Answer:
i need help with that too.
Explanation:
Answer:
52.54 %
Explanation:
Half life = 29 years
Where, k is rate constant
So,
The rate constant, k = 0.023902 hour⁻¹
From 1964 to 1991:
Time = 27 years
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
So,
![\frac {[A_t]}{[A_0]}=e^{-0.023902\times 27}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-0.023902%5Ctimes%2027%7D)
![\frac {[A_t]}{[A_0]}=0.5245](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3D0.5245)
<u>The strontium-90 remains in the bone = 52.54 %</u>
Answer:
Static Electricity
Explanation:
Most likely, in the room, Jim was building up static electricity by friction with the floor, which he released upon contact with a metal object.
Fe2O3 + 2Al ---> Al2O3 + 2Fe
Mole ratio Fe2O3 : Al = 1:2
No. of moles of Fe2O3 = Mass/RMM = 250 / (55.8 * 2 + 16 * 3) = 1.56641604 moles
No. of moles of Al = 150/27 = 5.555555555 moles.
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react.
So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared.
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe
