The half-life equation is written as:
An = Aoe^-kt
We use this equation for the solution. We do as follows:
5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE
Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?
G(t) = 176e^-0.02t
How many grams of goo will remain after 50 minutes?
G(t) = 176e^-0.02(50) = 64.75 g
Well u said ignoring not me
Answer:
[MgSO₄] = 890 mM/L
Explanation:
In order to determine molarity we need to determine the moles of solute that are in 1L of solution.
Solute: MgSO₄ (10.7 g)
Solvent: water
Solution: 100 mL as volume. (100 mL . 1L / 1000mL) = 0.1L
We convert the solute's mass to moles → 10.7 g / 120.36 g/mol = 0.089 moles
Molarity (mol/L) → 0.089 mol/0.1L = 0.89 M
In order to calculate M to mM/L, we make this conversion:
0.89 mol . 1000 mmoles/ 1 mol = 890 mmoles
D. Black holes
When a star dies out it explodes and causes a supernova and after a very long period of time the supernovas gasses vanish and the star can be visibly identified as a black hole!
Answer:
give us something else to work with or this is just the best guess
