Isotope 1: 89.905 * 51.45 = 4625.61225 / 100 = 46.2561225
Isotope 2: 90.906 * 11.22 = 1019.96532 / 100 = 10.1996532
Isotope 3: 91.905 * 17.15 = 1576.17175 / 100 = 15.7617075
Isotope 4: 93.906 * 17.38 = 1632.08628 / 100 = 16.3208628
Isotope 5: 95.908 * 2.08 = 268.5424 / 100 = 2.685424
46.2561225 + 10.1996532 + 15.7617075 + 16.3208628 + 2.685424 = 91.22377
actual mass Zr = about 91.22
2.083 Liters of 6.0 M solution sulfuric acid is required. This solved using molecular calculations and Titration.
Solution: 
Moles of hydrogen gas = 
Then 12.5 moles of hydrogen will be obtained from Moles of Sulfuric acid = 12.5 mol
Molarity of the sulfuric acid solution = 6.0 M = 6 mol/ l
6M = 
where V is the volume needed

V = 2.083 l
<h3>
What is Titration?</h3>
- Titration, commonly referred to as titrimetry, is a typical quantitative chemical analysis method used in laboratories to ascertain the unidentified quantity of an analyte .
- Titration is frequently referred to as volumetric analysis because it relies heavily on volume measurements. The titrant or titrator is a reagent that is prepared as a standard solution.
- To determine concentration, a solution of the analyte or titrand reacts with a known concentration and volume of the titrant. The titration volume is the amount of titrant that has responded.
- Titrations come in a variety of forms with various protocols and objectives. Redox and acid-base titrations are the two most typical types of qualitative titrations.
To learn more about titration with the given link
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I think the answer is mutualism.
because mutualism means that the two species benefit each other.
The solution is as follows:
K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106
The molar mass of isoborneol/borneol is 154.25 g/mol
Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol
Use the ICE approach
borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x
Total moles = 0.1458
Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P
0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832
Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>