Umm yo no se no ablo engles
Answer: a
Explanation:
Industry uses only about 18% while the others use around 70-90% of water.
YW!!! please mark branlest!!!! =^.^=
Answer:
The final mass of sample is 1.3 g.
Explanation:
Given data:
Half life of H-3 = 12.32 years
Amount left for 15.0 years = 3.02 g
Final amount = ?
Solution:
First all we will calculate the decay constant.
t₁/₂ = ln² /k
t₁/₂ =12.32 years
12.32 y = ln² /k
k = ln²/12.32 y
k = 0.05626 y⁻¹
Now we will find the original amount:
ln (A°/A) = Kt
ln (3.02 g/ A) = 0.05626 y⁻¹ × 15.0 y
ln (3.02 g/ A) = 0.8439
3.02 g/ A = e⁰°⁸⁴³⁹
3.02 g/ A = 2.33
A = 3.02 g/ 2.33
A = 1.3 g
The final mass of sample is 1.3 g.
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
I think is 5 because you use density is found by mass divide by volume.
