The time taken for the isotope to decay is 46 million years.
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
- Original amount (N₀) = 50.25 g
- Amount remaining (N) = 16.75
- Number of half-lives (n) =?
2ⁿ = N₀ / N
2ⁿ = N₀ / N
2ⁿ = 50.25 / 16.75
2ⁿ = 3
Take the log of both side
Log 2ⁿ = 3
nLog 2 = Log 3
Divide both side by log 2
n = Log 3 / Log 2
n = 2
Finally, we shall determine the time.
- Half-life (t½) = 23 million years
- Number of half-lives (n) = 2
t = n × t½
t = 2 × 23
t = 46 million years
Learn more about half-life: brainly.com/question/25927447
Answer is: A) The solution turns blue litmus to red.
Sulfuric acid (H₂SO₄) is a strong acid, it means that the solution of sufuric acid is more acidic (pH<7) than water (pH = 7).
Chemical dissociation of sulfuric acid in water:
H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq).
When solution turns phenolphthalein pink, it means it is basic (pH>7).
Sulfuric acid has more hydrogen ions (H⁺) and less hydroxide ions (OH⁻) than water.
Answer:
5.8μg
Explanation:
According to the rate or decay law:
N/N₀ = exp(-λt)------------------------------- (1)
Where N = Current quantity, μg
N₀ = Original quantity, μg
λ= Decay constant day⁻¹
t = time in days
Since the half life is 4.5 days, we can calculate the λ from (1) by substituting N/N₀ = 0.5
0.5 = exp (-4.5λ)
ln 0.5 = -4.5λ
-0.6931 = -4.5λ
λ = -0.6931 /-4.5
=0.1540 day⁻¹
Substituting into (1) we have :
N/N₀ = exp(-0.154t)----------------------------- (2)
To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:
N = 5.0 μg
N₀ = Unknown
t = 1 day
Substituting into (2) we have
[5/N₀] = exp (-0.154 x 1)
5/N₀ = 0.8572
N₀ = 5/0.8572
= 5.8329μg
≈ 5.8μg
The Chemist must order 5.8μg of 47-CaCO3
Answer:
7000 kg*m/s E
Explanation:
Momentum formula: p=mv
m=200kg
v=35 m/s East
p=(200kg)(35m/s E)
m=7000 kg*m/s E
If you want to simplify it further, m=7*10^3 kg*m/s E
