Answer:
AuCl
Explanation:
Given parameters:
Mass of Gold = 2.6444g
Mass of Chlorine = 0.476g
Unknown:
Empirical formula = ?
Solution:
Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.
Elements Au Cl
Mass 2.6444 0.476
Molar mass 197 35.5
Number of moles 2.6444/197 0.476/35.5
0.013 0.013
Divide by the
smallest 0.013/0.013 0.013/0.013
1 1
The empirical formula of the compound is AuCl
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
<h3>2.75 = pKa</h3>
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
<h3>1.78x10⁻³ = Ka</h3>
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
They are both made up of the same protein called "histones".
Ca=40.07841 ,Nitrogen=14.006722 ,Oxygen15.99943
molar mass= (1xMW of Ca)+(2xMWof N)+(6x MW of oxygen)
Molar mass (molecular weight) of Ca(NO3)2 is 164.0884 g/mol
When it comes to equilibrium reactions in chemistry, there are a lot of equilibrium constants that can be used. In the case of solubility, the appropriate one to use is the equilibrium constant of solubility product denotes as Ksp. This is the concentration of products raised to their coefficients. For example,
cC ⇔ aA + bB
Ksp = {[A^a][B^b]}
Now, for the this problem, the reaction is
BaSO₄ ⇔ Ba²⁺ + SO₄²⁻
The reaction is already balanced. Since we don't know the value of Ba²⁺ and SO₄²⁻, let's denote this at x.
1.1 × 10⁻¹⁰ = [x][x] =[x²]
[x] = [Ba²⁺] = [SO₄²⁻] = [BaSO₄] = 1.049 × 10⁻⁵ M
