Because pure silicon is a perfect semiconductor.
For room temperature, it rarely conducts, you can search for the threshold temperature, the characteristic equation is fairly complicated.
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>
Answer : The amount of reactant left in excess is 13.1075 grams.
Explanation : Given,
Mass of 
= 26.5 g
Mass of 
= 10 g
Molar mass of = 53.5 g/mole
Molar mass of = 40 g/mole
First we have to calculate the moles of and .
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 1 mole of
So, 0.25 moles of react with 0.25 moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles
Now we have to calculate the mass of .
Therefore, the amount of reactant left in excess is 13.1075 grams.
Answer:
8.9L is the volume of the gas that must be dissolved.
Explanation:
For a weak base, we can find [(CH₃)₃N] using the equation:
Kb = [OH⁻] [[(CH₃)₃NH⁺] / [(CH₃)₃N]
As [OH⁻] = [[(CH₃)₃NH⁺] and [OH⁻] = 10^-pOH = 3.16x10⁻³M:
6.3x10⁻⁵ = [3.16x10⁻³M][3.16x10⁻³M] / [(CH₃)₃N]
[(CH₃)₃N] = 0.1587M
As the volume is 2.5L, moles are:
2.5L * (0.1587mol / L) = 0.3968moles
Using:
PV = nRT
We can solve for volume of the gas as follows:
P = 1atm at STP; n = 0.3968moles; R = 0.082atmL/molK; T = 273.15K at STP
V = 0.3968mol*0.082atmL/molK*273.15K/1atm
V = 8.9L is the volume of the gas that must be dissolved.
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