<span>1. What is the molar mass of gold?
Molar mass is a unit that expresses the mass of a molecule per one mol. The molar mass can be obtained by adding the neutron with the proton of the atoms. Gold has atomic number 79 so the proton is 79. The number of the neutron is 118. Then the molar mass would be: 79 + 118 = </span>197 g/mol<span>
</span><span>2. Calculate the number of moles of gold (Au) in the sample. Show your work.
</span>In this question, you are given the mass of the gold and asked for how many moles the sample has. To find the number of moles you just need to divide the weight by the molar mass.
For 45.39 grams of gold, the number of moles would be:
45.39 / (197g/mol)= 0.23 moles
3. Calculate the number of atoms of gold (Au) in the sample. Show your work.Moles is unit of a number of molecules but 1 mol doesn't represent 1 molecule. The number of atoms can be obtained by multiplying the number of moles with Avogadro number. The calculation would be:
0.23 moles * (6.023 * 10^23 molecules/mol)= 1.387 * 10^23 molecules
Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka= ![\frac{ [H^{+}] [A^{-} ]}{ [HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BH%5E%7B%2B%7D%5D%20%20%5BA%5E%7B-%7D%20%5D%7D%7B%20%5BHA%5D%7D)
Ka= 
Ka= 
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka=
=
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:
ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.
ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)
k2 = 0.3325 L / mol-s