1) You need to use the atomic mass of copper.
You can find it in a periodic table. It is 63.546 amu.
2) The atomic mass is the weigthed mass of the different isotopes.
This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:
=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 * atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.
3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x
=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278
=> 63.546 = 62.9296x + 64.9278 - 64.9278x
=> 64.9278x - 62.9296 = 64.9278 - 63.546
=> 1.9982x = 1.3818
=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%
=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%
Answer:
Cu-63 69.15%;
Cu-65 : 30.85%
In lower temperatures, the molecules of real gases tend to slow down enough that the attractive forces between the individual molecules are no longer negligible. In high pressures, the molecules are forced closer together- as opposed to the further distances between molecules at lower pressures. This closer the distance between the gas molecules, the more likely that attractive forces will develop between the molecules. As such, the ideal gas behavior occurs best in high temperatures and low pressures. (Answer to your question: C) This is because the attraction between molecules are assumed to be negligible in ideal gases, no interactions and transfer of energy between the molecules occur, and as temperature decreases and pressure increases, the more the gas will act like an real gas.
Answer: the pH of the solution is 4.52
Explanation:
Consider the weak acid as Ha, it is dissociated as expressed below
HA H⁺ + A⁻
the Henderson -Haselbach equation can be expressed as;
pH = pKa + log( [A⁻] / [HA])
the weak acid is dissociated into H⁺ and A⁻ ions in the solution.
now the conjugate base of the weak acid HA is
HA(aq) {weak acid} H⁺(aq) + A⁻(aq) {conjugate base}
so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;
pKₐ = -logKₐ
pKₐ = -log ( 7.4×10⁻⁵ )
pKₐ = 4.13
now thw pH is
pH = pKₐ + log( [A⁻] / [HA])
pH = 4.13 + log( [0.540] / [0.220])
pH = 4.13 + 0.3899
pH = 4.5199 = 4.52
Therefore the pH of the solution is 4.52
Glycolysis--The breakdown of a glucose molecule into two three-carbon pieces called pyruvate. You will notice that very little ATP is produced in this step and no oxygen is required. ... This step is also where other molecules besides glucose may be fed into the cell respiration<span> process, especially lipids.</span>
