Answer:
An Educated Guess
Explanation:
After the scientist is done with the experiment he should do an educated guess, they use the data from the experiments to make charts and graphs to communicate the results of the experiment. After the scientists makes the hypothesis, they perform this procedure.
<span>1.15x10^24 molecules of hypothetical substance b
Making the assumption that each molecule in hypothetical substance a reacts to produce a single molecule of hypothetical substance b, then the number of molecules of substance b will be the number of moles of substance a multiplied by avogadro's number. So
Moles hypothetical substance a = 29.9 g / 15.7 g/mol = 1.904458599 moles
This means that we should also have 1.904458599 moles of hypothetical substance b. And to get the number of atoms, multiply by 6.0221409x10^23, so:
1.904458599 * 6.0221409x10^23 = 1.146892x10^24 molecules.
Rounding to 3 significant figures gives 1.15x10^24</span>
I believe it is density but i'm not for certain try looking it up somewhere else
Q1)
As Kemmi pipettes a volume of 25.00 ml of the solution
density of pure propanol is 0.803 g/ml
This means that in 1000 ml of solution - 0.803 g of pure propanol
Therefore in 25.00 ml of solution - 0.803 g x 25.00 ml / 1000 ml
= 0.0201 g
Using molar mass, number of moles can be calculated= 0.0201 g / 60.09 g/mol
= 3.35 x 10⁻⁴ mol
therefore the number of pure propanol moles in exactly 25.00 ml is
3.35 x 10⁻⁴ mol
Q2)
molarity is the concentration of the solution. It can be defined as the number of moles of solute per liter of solution
we know the number of moles in 25.00 ml of solution. When its diluted in a 100.00 ml volumetric flask, number of moles remain constant but now the volume over which the moles of solute are dissolved is increased.
therefore number of moles = 3.35 x 10^(-4) mol
volume over which its dissolved - 100.00 / 10³ dm³
= 1.0000 x10⁻¹ dm³
the molarity = 3.35 x 10⁻⁴ mol / 1.0000 x10⁻¹ dm³
= 3.35 x 10⁻³ mol/dm³
Kingdom Monera, organisms in this Kingdom are prokaryotes, they don't have a nuclear envelope, thus they have a nucleoid.
