Fuse should only affect voltage
They're elements. Elements are the simplest form of any substance that cannot be broken down by any chemical methods, including electrolysis or even heat. Examples include oxygen, hydrogen etc.
Answer:
15 h
Explanation:
Okay, the first thing that we all have to know before we can answer this question is that this Topic that is, Chemistry of Radioactivity is related to kinetics in a way that Radioactive disintegration follows the first order of Reaction which is under kinetics. So, we will be using the first order kinetics rate law to answer this question. Using the equation (1) below;
k =[ 2.303/ t ]×log ([N°}/ [Nr]) --------(1).
We are given from the question that N° = initial sample = 0.8 mg and Nr= sample remaining = 0.2 and the time taken = t= 30.0 h.
k= (2.303/ 30.0 h ) × log (0.8/0.2).
k=0.076768 h^-1 × log (4).
k= 0.076768 h^-1 × 0.6021.
k= 0.0462 h^-1.
Therefore, using the formula for Calculating half life below for first order kinetics we will be able to find out answer.
k = ln 2/ t(1/2). Where t(1/2) is the half life.
t(1/2) = ln 2/ k.
t(1/2) = ln 2 / 0.0462 h^-1.
t(1/2)= 0.6931/0.0462 h^-1.
t(1/2)=15 h
Answer:
Kb = 1.77x10⁻⁵
Explanation:
When NH₃, a weak base, is in equilibrium with waterm the reaction that occurs is:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
And the dissociation constant, Kb, for this equilibrium is:
Kb = [NH₄⁺] [OH⁻] / [NH₃]
To find Kb you need to find the concentration of each species. The equilibrium concentrations are:
[NH₃] = 0.950M - X
[NH₄⁺] = X
[OH⁻] = X
<em>Where X is reaction coordinate.</em>
You can know [OH⁻] and, therefore, X, with pH of the solution, thus:
pH = -log [H⁺] = 11.612
[H⁺] = 2.4434x10⁻¹²
As 1x10⁻¹⁴ = [H⁺] [OH⁻]
1x10⁻¹⁴ / 2.4434x10⁻¹² = [OH⁻]
4.0926x10⁻³ = [OH⁻] = X
Replacing, concentrations of the species are:
[NH₃] = 0.950M - X
[NH₄⁺] = X
[OH⁻] = X
[NH₃] = 0.9459M
[NH₄⁺] = 4.0926x10⁻³M
[OH⁻] = 4.0926x10⁻³M
Replacing in Kb expression:
Kb = [NH₄⁺] [OH⁻] / [NH₃]
Kb = [4.0926x10⁻³M] [4.0926x10⁻³M] / [0.9459M]
<h3>Kb = 1.77x10⁻⁵</h3>
