Bc they dooooooooooooooooooo
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.
A(t) = A(o) (1/2)^(t/d)
where t is the certain period of time. Substituting the known values,
A(t) = (20 mg)(1/2)^(85.80/14.30)
Solving,
A(t) = 0.3125 mg
Hence, the answer is 0.3125 mg.
The two first ones because it’s always the opposite
Answer:
Explanation:
Given parameters:
pH = 3.50
Unknown:
concentration of [H₃0⁺] = ?
concentration of [OH⁻] = ?
Solution:
In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.
pH = -log₁₀[H₃O⁺]
[H₃O⁺] = inverse log₁₀ (-pH) =
= 
[H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³
For the [OH⁻]:
we use : pOH = -log₁₀ [OH⁻]
Recall: pOH + pH = 14
pOH = 14 - pH = 14 - 3.5 = 10.5
Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]
[OH⁻] = 
[OH⁻] =
= 3.2 x 10⁻¹¹moldm⁻³
The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.
Answer:
sorry don't know the answer but i really need the points sorry
Explanation: